Important Questions – Page 8 – Maharashtra Board Solutions

Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy

July 25, 2024July 24, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 8 Plant Tissues and Anatomy Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 8 Plant Tissues and Anatomy

Question 1.
How plant tissues are classified on the basis of their ability to divide?
Answer:
Plant tissues are classified into meristematic tissues and permanent tissues based on their ability to divide.

Question 2.
Identify the labels i, ii and iii in the given figure of meristematic tissue and write its characteristics.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 1
Answer:
1. Cell wall
2. Nucleus
3. Cytoplasm
Characteristics of meristematic tissue:

It is a group of young, immature cells.
These are living cells with ability to divide in the regions where they are present.
These are polyhedral or isodiametric in shape without intercellular spaces.
Cell wall is thin, elastic and mainly composed of cellulose.
Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
Cells show high rate of metabolism.

Question 3.
With the help of neat and labelled diagram explain the classification of meristematic tissue based on its position.
Answer:
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 4.
Complete the given table representing types of meristematic tissue based on its function.
Answer:

Types of meristematic tissue	Function
1. Protoderm	It is found in young growing region of a plant forming a protective covering like epidermis around the various organs.
2. Procambium	It is involved in developing primary vascular tissue.
3. Ground meristem	It forms structures like cortex, endodermis, pericycle, medullary rays, pith.

Question 5.
Which are the simple permanent tissues in plants?
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 6.
Complete the given chart by giving characteristics of following tissues:
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 2

Question 7.
Name the type of tissue in the given figure, identify labels ‘a’ and ‘b’ and write its characteristics.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 3
Answer:
1. The given figure represents simple permanent tissue i.e. Parenchyma.
2. a: Vacuole, b: Intercellular air spaces.

Characteristics of parenchyma: Parenchyma:

It is a type of simple permanent tissue.
Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
Cell wall is composed of cellulose.
Cells are living with prominent nucleus and cytoplasm with large vacuole.
Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
This is less specialized permanent tissue.
Occurrence:
These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
Functions:
These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
secondary growth.

Question 8.
Identify the type of tissue shown in the given figure and write its characteristics.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 4
Answer:
The given figure represents Collenchyma tissue.
Characteristics of Collenchyma:

It is a simple permanent tissue made up of living cells.
The cell wall is cellulosic but shows uneven deposition of cellulose and pectin especially at comers.
The walls may show presence of pits.
Cells are similar like parenchyma, containing cytoplasm, nucleus and vacuoles but small in size and without intercellular spaces. Thus, the cells appear to be compactly packed.
The cells are either circular, oval or angular in transverse section.

Function:
Collenchyma is a living mechanical tissue and serves different functions in plants.
a. It gives mechanical strength to young stem and parts like petiole of leaf.
b. It allows bending and pulling action in plant parts and also prevents tearing of leaf.
c. It also allows growth and elongation of organs.
d. Collenchyma is usually absent in monocots and roots of dicot plant.

Question 9.
With the help of neat and labelled diagrams explain the Sclerenchyma Tissue.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 5
Answer:
Sclerenchyma Tissue:

It is simple permanent tissue made up of compactly arranged thick walled dead cells.
The cells are living at the time of production but at maturity they become dead.
Cells are devoid of cytoplasm.
Their walls are thickened due to uniform deposition of lignin.
Cells remain interconnected through several pits.

Types of Sclerenchyma:
Sclerenchyma cells are categorized into two types on the basis of their size and shape as
fibres and sclereids:
a. Fibres:
Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique. They provide mechanical strength.

b. Sclereids:
Sclereids are usually broad, with blunt end walls.
These occur singly or in loose groups and their pits are deep branched and straight.
These are developed due to secondary thickening of parenchyma cells and provides stiffness only.

Functions:
a. This tissue functions as the main mechanical tissue.
b. It permits bending, shearing and pulling.
c. It gives rigidity to leaves and prevents it from falling.
d. It also gives rigidity to epicarps and seeds.

Question 10.
Give a brief account of water-conducting tissues in higher plants.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 6
1. Xylem is the water-conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

2. Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

3. Vessels:
a. Vessels are longer than tracheids with perforated or dissolved ends and formed by union of several vessels end to end.
b. These are involved in conduction of water and minerals.
c. Their lumen is wider than tracheids and the thickening is due to lignin and similar to tracheids.
d. In monocots, vessels are rounded where as they are angular in dicot angiosperms.
e. The first formed xylem vessels (protoxylem) are small and have either annular or spiral thickenings while latter formed xylem vessels are larger (metaxylem) and have reticulate or pitted thickenings.
f. When protoxylem is arranged towards pith and metaxylem towards periphery it is called as endarch
e. g. in stem and when the position is reversed as in the roots is called as exarch.

4. Xylem parenchyma:
a. Xylem parenchyma cells are small associated with tracheids and vessels.
b. This is the only living tissue among this complex tissue.
c. The function is to store food (starch) and sometimes tannins.
d. Xylem parenchyma are involved in lateral or radial conduction of water or sap.

5. Xylem fibres:
a. Xylem fibres are sclerenchymatous cells and serve mainly mechanical support. These are called wood fibres.
b. These are also elongated, narrow and spindle shaped.
c. Cells are tapering at both the ends and their walls are lignified.

Question 11.
Draw neat and labelled diagram of xylem tissue and vascular bundle.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 7

Question 12.
Match the following.

Column I	Column II
1. Protoxylem	(a) Xylem with larger vessels
2. Endarch Xylem	(b) Protoxylem arranged towards pith
3. Metaxylem	(c) Metaxylem arranged towards pith
4. Exarch xylem	(d) First formed xylem vessels

Answer:

Column I	Column II
1. Protoxylem	(d) First formed xylem vessels
2. Endarch Xylem	(b) Protoxylem arranged towards pith
3. Metaxylem	(a) Xylem with larger vessels
4. Exarch xylem	(c) Metaxylem arranged towards pith

Question 13.
Describe the structure of phloem.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 8
Answer:
Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Question 14.
Draw a diagram of phloem.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 9

Question 15.
Name the types of tissue systems in plants.
Answer:
The types of tissue systems in plants are epidermal tissue system, ground tissue system and vascular tissue system.

Question 16.
Write a short note on Epidermis.
Answer:
Epidermis:

It is the outermost protective cell layer made up of compactly arranged cells without intercellular spaces.
Cells show presence of central large vacuole, thin cytoplasm and a nucleus.
The outer side of the epidermis is often covered with a waxy thick layer called the cuticle which prevents the loss of water.
Root epidermis (Epiblema) has root hairs. These are unicellular, elongated and involved in absorption of sap from the soil.
In stem, epidermal hairs are called trichomes. These are generally multicellular, branched or unbranched, stiff or soft or even secretory. These help in preventing water loss due to transpiration.

Question 17.
Draw a diagram representing epidermal tissue system.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 10

Question 18.
Write a short note on Structure of stomata.
Answer:
Structure of stomata:

Small gateways in the epidermal cells are called as stomata.
Stoma is controlled or guarded by specially modified cells called guard cells.
These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.
Guard cells have chloroplasts to carry out photosynthesis.
Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
Stomata are further covered by subsidiary cells.
Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question 19.
Write the information related to diagrams given below.
Answer:
1. The given diagram represents stoma in dicot leaf.
[Note: We have given additional label of ‘chloroplast ’for better understanding of students]
2. Structure of stomata:

Small gateways in the epidermal cells are called as stomata.
Stoma is controlled or guarded by specially modified cells called guard cells.
These guard cells may be kidney shaped (dicot) or dumbbell shaped (monocot), collectively called as stomata.
Guard cells have chloroplasts to carry out photosynthesis.
Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
Stomata are further covered by subsidiary cells.
Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question 20.
Explain the term ground tissue.
Answer:
Ground tissue:

All the plant tissues excluding epidermal and vascular tissue is ground tissue.
It is made up of simple permanent tissue e.g. parenchyma.
It is present in cortex, pericycle, pith and medullary rays in the primary stem and root.
Collenchyma and sclerenchyma in the hypodermis and chloroplasts containing mesophyll tissue in leaves is also ground tissue.

Question 21.
Describe various types of vascular bundles.
Answer:
Vascular bundles occur in the form of distinct patches of the complex tissue viz. Xylem and Phloem. On the basis of their arrangement in the plant body they are classified as follows:
1. Radial vascular bundles:
When the complex tissues (xylem and phloem) are situated separately on separate radius as separate bundle, vascular bundle is called Radial vascular bundle. This is a common feature of roots.

2. Conjoint vascular bundles:
When the complex tissue (xylem and phloem) is collectively present as neighbours of each other on the same radius, vascular bundle is called Conjoint vascular bundle.
They are of two types:
a. Collateral vascular bundle:
In this type of vascular bundle, xylem lies inwards and the phloem lies outwards.
These bundles may be further of open type (secondary growth takes place) containing cambium in between xylem and phloem and closed type if cambium is not present (secondary growth absent).
b. Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

3. Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 11

Question 22.
Explain how formation of cambial ring occurs in dicot stem.
Answer:

The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
With the onset of favourable season, meristematic cells of intrafascicular cambium become active.
Simultaneously, the ray parenchyma cells, both fusiform initials and ray initials become meristematic. This is known as dedifferentiation.
These form patch of cambial cells (meristematic cells) in between the adjacent bundles and produce interfascicular cambium.
Both intrafascicular and interfascicular cambium join and form a complete ring, known as the cambial ring. This is possible because they lie in one plane.

Question 23.
‘Secondary growth is observed in most of the dicot and gymnospermic root.’ State whether the given statement is true or false and justify your answer.
Answer:

The given statement is true.
Secondary growth is observed in most of the dicot and gymnospermic root by producing secondary vascular tissue and periderm.
Secondary growth is produced by vascular cambium and cork cambium respectively.
Conjunctive parenchyma cells present on the inner edges of primary phloem bundles become meristematic.
These cells add secondary xylem and secondary phloem on the inner and outer side respectively which results in secondary growth.

Question 24.
Differentiate between heartwood and sap wood.
Answer:

Heartwood	Sap wood
1. It is central region of secondary xylem (wood).	It is the peripheral region of secondary xylem (wood).
2. It is darker in colour due to deposition of oils, gums, resins, tannins, etc	It is lighter in colour and without any depositions.
3. It is non- functional part of secondary xylem.	It is functional part of secondary xylem.
4. It is resistant to pathogens.	It is more susceptible to pathogens
5. It is not involved in conduction of sap.	It is involved in conduction of sap.
6. It is also called as duramen.	It is also called as alburnum.

Question 25.
What are tyloses?
Answer:
Tyloses:
1. Tracheary elements of heartwood are plugged by in-growth of adjacent parenchyma cells are known as tyloses.
2. Tyloses are fdled by oils, gums, resins, tannins called as extractives.

Question 26.
Explain how periderm is formed?
Answer:
Formation of periderm:
As the stem increase in diameter due to activity of vascular cambium, the outer cortical and epidermal layer get ruptured. Thus, it becomes necessaiy to replace these cells by new cells.

Phellogen (cork cambium) develops in extrastelar region (cortex region) of the stem.
The outer cortical cells of cortex become meristematic and produce a layer of thin walled, rectangular cells. These cells cut off new cells on both sides.
The cells produced on outer side develop phellem (cork), whereas on the inner side produce phelloderm (secondary cortex).
The cork is impervious in nature and does not allow entry of water due to suberized walls. Secondary cortex is parenchymatous in nature.
Phellogen, phellem and phelloderm constitute periderm.

Question 27.
Explain the given terms:
1. Bark
2. Lenticels
3. Anomalous secondary growth
Answer:
1. Bark:
a. Bark is non-technical term referring to all cell types found external to vascular cambium including secondary phloem.
b. Bark of early season is soft and of the late season is hard.

2. Lenticels:
a. Lenticels are aerating pores present as raised scars on the surface of bark.
b. These are portions of periderm, where phellogen activity is more.
c. Lenticels are meant for gaseous and water vapour exchange.

3. Anomalous secondary growth:
a. Monocot stems lack cambium hence secondary growth does not take place.
b. However, accessory cambium development in plants like, Dracaena, Agave, Palms and root of sweet potato shows presence of secondary growth. This is called as anomalous secondary growth.

Question 28.
With the help of neat and labelled diagram explain the anatomy of dicot root.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 12
The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Question 29.
With the help of neat and labelled diagram explain the anatomy of monocot root.
Answer:
The transverse section of a typical monocotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
Endodermis: It is innermost layer of cortex. The cells of endodermis are thick walled except the passage cells which lie just opposite to the protoxylem.
Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Pericycle is present below the endodermis.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Polyarch condition of xylem is observed, (xylem bundles are more than six).
Pith: Pith is large and well developed.
Secondary growth does not occur due to absence of cambium.

Question 30.
What is polyarch condition of root?
Answer:
Polyarch condition is the one in which roots possess more than six xylem bundles.

Question 31.
Explain in detail anatomy of sunflower stem.
Answer:
A transverse section of sunflower (dicot) stem shows the following structures:
1. Epidermis: It is a single, outermost layer with multicellular outgrowth called trichomes. A layer of cuticle
is usually present towards the outer surface of epidermis.

2. Cortex: Cortex is situated below the epidermis and is usually differentiated into three regions namely, hypodermis, general cortex and endodermis.
a. Hypodermis: It is situated just below the epidermis and is made of 3-5 layers of collenchymatous cells. Intercellular spaces are absent.
b. General cortex: It is made up of several layers of large parenchymatous cells with intercellular spaces.
c. Endodermis: It is an innermost layer of cortex which is made up of barrel shaped cells. It is also called starch sheath, as it is rich in starch grain.

3. Stele: It is differentiated into pericycle, vascular bundles and pith.
a. Pericycle: It is the outermost layer of vascular system situated between the endodermis and vascular bundles. In sunflower, it is multi-layered and also called hard bast.
b. Vascular bundles: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
c. Pith: It is situated in the centre of the young stem and is made up of large-sized parenchymatous cells with conspicuous intercellular spaces.

Question 32.
With the help of neat and labelled diagram explain the anatomy of maize stem.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 13
A transverse section of maize (monocot) stem shows the following structures:

Epidermis: It is single layered and without trichomes.
Hypodermis: It is sclerenchymatous.
Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
Pith: Pith is absent.

Question 33.
With the help of a neat and labelled diagram, describe the internal structure of dorsiventral leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 14
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question 34.
With the help of a neat labelled diagram, describe the anatomy of isobilateral leaf.
Answer:
The parts of isobilateral leaf are as follows:
Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 15
1. Epidermis:
It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 35.
Compare between dorsiventral and isobilateral leaf.
Answer:

Dorsiventral leaf	Isobilateral leaf
1. Dorsiventral Leaf is very common in dicotyledonous plants.	Isobilateral leaf is very common in monocotyledonous plants.
2. In this mesophyll tissue is differentiated into palisade and spongy parenchyma.	In this mesophyll tissue is not differentiated into palisade and spongy parenchyma.
3. The leaves are commonly horizontal in orientation with distinct upper and lower surfaces. The upper surface which faces the sun is darker than the lower surface.	In this leaf both the surfaces are equally illuminated as both the surface can face the sun, and show similar structure. The two surfaces are equally green.
4. Stomata is absent on the upper surface of these leaves.	Stomata is present on both the upper and lower surfaces of these leaves.

Question 36.
Distinguish between anatomy of dicot root and monocot root.
Answer:

Anatomy of dicot root	Anatomy of monocot root
1. Pith is narrow.	Pith is large and well developed.
2. Diarch, triarch or tetrarch condition can be observed. (Xylem bundles vary from two to six number)	Polyarch condition is observed, (xylem bundles are more than six)
3. Cambium is formed in later stage between xylem and phloem which causes secondary growth.	Secondary growth is absent.

Question 37.
Distinguish between anatomy of dicot stem and monocot stem.
Answer:

Anatomy of Dicot stem	Anatomy of Monocot stem
1. Epidermis shows presence of multicellular trichomes.	Epidermis is without trichomes.
2. Hypodermis is made up of collenchymatous cells.	Hypodermis is made up of sclerenchymatous cells.
3. Medullary rays are present between vascular bundles.	Medullary rays are absent.
4. Vascular bundles are arranged in the form of a ring.	Vascular bundles are scattered in the ground tissue.
5. It is conjoint, collateral and open (Cambium present)	They are conjoint, collateral and closed (cambium is absent).
6. Vascular bundle is not surrounded by a sclerenchymatous bundle sheath.	Vascular bundle is surrounded by a sclerenchymatous bundle sheath.
7. Secondary growth takes place due to presence of cambium.	Secondary growth does not occur due to absence of cambium.
8. Pith is present.	Pith is absent.

Question 38.
Apply Your Knowledge

Question 1.
Which plant part would show the following:

Radial vascular bundles.
Large and well-developed pith.
Differentiation of mesophyll into palisade and spongy tissue.
Presence of stomata on both upper and lower epidermis.

Answer:

Root
Monocot root and Dicot stem,
Dicot leaf
Monocot leaf

Question 2.
When a tree is debarked, which tissues are removed?
Answer:
The bark is made up of tissues like cork, cork cambium and secondary cortex, which are removed when a tree is debarked.

Question 3.
While eating fruits like pear or guava, it feels gritty. What gives stiffness to these fruits?
Answer:
Sclereids are found in pulp of fruits like pear and guava which gives them stiffness and thus we feel gritty while eating these fruits.

Question 39.
Quick Review

Maharashtra Board Class 11 Biology Important Questions Chapter 8 Plant Tissues and Anatomy 16

Question 40.
Exercise

Question 1.
Define tissue.
Answer:
A group of cells having essentially a common function and origin is called as tissue.

Question 2.
Classify the meristematic tissue based on its origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Question 3.
Explain in detail classification of meristematic tissue based on its position.
Answer:
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

Question 4.
Give any two examples of secondary meristematic tissue.
Answer:
Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Question 5.
Draw a diagram of meristematic cells.
Answer:
1. Cell wall
2. Nucleus
3. Cytoplasm
Characteristics of meristematic tissue:

It is a group of young, immature cells.
These are living cells with ability to divide in the regions where they are present.
These are polyhedral or isodiametric in shape without intercellular spaces.
Cell wall is thin, elastic and mainly composed of cellulose.
Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
Cells show high rate of metabolism.

Question 6.
Write a short note on tracheids.
Answer:
Tracheids:
a. These are elongated, tubular and dead cells (without protoplasm).
b. The ends are oblique and tapering.
c. The cell walls is unevenly thickened and lignified. This provides mechanical strength.
d. Tracheids contribute 95% of wood in gymnosperms and 5% in angiosperms.
e. The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

Question 7.
Describe parenchyma in detail.
Answer:
Cell is the component that brings about important processes in the living organisms.

Question 8.
Describe the structure of xylem in detail.
Answer:
1. Xylem is the water conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 9.
What are Sclerenchyma fibres?
Answer:
a. Fibres:
Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique. They provide mechanical strength.

Question 10.
Write the functions of parenchyma cells.
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 11.
Write function of collenchyma tissue.
Answer:
Function:
Collenchyma is a living mechanical tissue and serves different functions in plants.
a. It gives mechanical strength to young stem and parts like petiole of leaf.
b. It allows bending and pulling action in plant parts and also prevents tearing of leaf.
c. It also allows growth and elongation of organs.

Question 12.
Which are the different types of tracheids based on the types of thickenings on their walls?
Answer:
The different types of thickening patterns are seen on their walls such as annular (in the form of rings), spiral (in the form of spring/helix), scalariform (ladder like) and pitted (small circular area). Pitted is the most advanced type of thickening which may be simple or bordered.

Question 13.
Death of companion cell causes death of sieve tube cells and vice versa. Justify.
Answer:
Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

Question 14.
Which are the three types of simple permanent tissues?
Answer:
Parenchyma, Collenchyma and Sclerenchyma are the simple permanent tissues in plants.

Question 15.
Write the functions of sclerenchyma tissue.
Answer:
Functions:
a. This tissue functions as the main mechanical tissue.
b. It permits bending, shearing and pulling.
c. It gives rigidity to leaves and prevents it from falling.
d. It also gives rigidity to epicarps and seeds.

Question 16.
What are the components of xylem?
Answer:
1. Xylem is the water conducting tissue in higher plants. It is a dead complex tissue.
It also provides mechanical strength to the plant body.
Components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 17.
Name the living component of xylem.
Answer:
This is the only living tissue among this complex tissue.

Question 18.
Name the dead component of phloem.
Answer:
Phloem fibres are the only dead tissue among this unit.

Question 19.
What is closed vascular bundle?
Answer:
When cambium is not present between xylem and phloem, it is known as closed vascular bundle.

Question 20.
Describe two types of conjoint vascular bundles.
Answer:
Conjoint vascular bundles:
When the complex tissue (xylem and phloem) is collectively present as neighbours of each other on the same radius, vascular bundle is called Conjoint vascular bundle.
They are of two types:
a. Collateral vascular bundle:
In this type of vascular bundle, xylem lies inwards and the phloem lies outwards.
These bundles may be further of open type (secondary growth takes place) containing cambium in between xylem and phloem and closed type if cambium is not present (secondary growth absent).
b. Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 21.
Write the function of trichomes.
Answer:
In stem, epidermal hairs are called trichomes. These are generally multicellular, branched or unbranched, stiff or soft or even secretory. These help in preventing water loss due to transpiration.

Question 22.
What are bicollateral vascular bundle?
Answer:
Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 23.
Name the tissue that are not included in ground tissue.
Answer:
All the plant tissues excluding epidermal and vascular tissue is ground tissue.

Question 24.
Which type of conjoint – vascular bundles are found in members of Cucurbitaceae family?
Answer:
Bicollateral vascular bundle:
When phloem is present in a vascular bundle on both the sides of xylem and intervening cambium tissue, it is called bicollateral vascular bundle. It is a feature of family Cucurbitaceae.

Question 25.
What is concentric vascular bundle?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question 26.
Define intrafascicular cambium.
Answer:
The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.

Question 27.
What is the difference between spring wood and autumn wood?
Answer:
During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

Question 28.
Explain how growth rings are formed in trees?
Answer:
1. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.
2. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.
3. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

Question 29.
Which tissues are together called as periderm?
Answer:
Phellogen, phellem and phelloderm constitute periderm.

Question 30.
What is bark?
Answer:
Bark:
a. Bark is non-technical term referring to all cell types found external to vascular cambium including secondary phloem.
b. Bark of early season is soft and of the late season is hard.

Question 31.
What is the function of lenticels?
Answer:
Lenticels are meant for gaseous and water vapour exchange.

Question 32.
Explain the term anomalous secondary growth.
Answer:
Anomalous secondary growth:
a. Monocot stems lack cambium hence secondary growth does not take place.
b. However, accessory cambium development in plants like, Dracaena, Agave, Palms and root of sweet potato shows presence of secondary growth. This is called as anomalous secondary growth.

Question 33.
Explain in detail anatomical structure of a dicot stem.
Answer:
A transverse section of sunflower (dicot) stem shows the following structures:
1. Epidermis: It is a single, outermost layer with multicellular outgrowth called trichomes. A layer of cuticle
is usually present towards the outer surface of epidermis.

Question 34.
Draw neat and labelled diagrams of dicot and monocot root and differentiate between their anatomical characters.
Answer:
The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

Question 35.
Which type of vascular bundles are observed in isobilateral leaf?
Answer:
Vascular bundle:
These are conjoint, collateral and closed.

Question 36.
Describe the internal structure of a leaf in which mesophyll is differentiated in palisade and spongy parenchyma.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question 37.
Multiple Choice Questions:

Question 1.
Meristematic tissues are found
(A) only in stems of the plants
(B) in both roots and stems
(C) in all growing tips of the plant body
(D) only in roots of the plants
Answer:
(C) in all growing tips of the plant body

Question 2.
The tissue responsible for translocation of food material is _________
(A) xylem
(B) cambium
(C) parenchyma
(D) phloem
Answer:
(D) phloem

Question 3.
________ are used in making ropes and rough clothes.
(A) Phloem parenchyma
(B) Trachieds
(C) Phloem fibres
(D) Sieve tube elements
Answer:
(C) Phloem fibres

Question 4.
_______ are the only dead tissue among the phloem.
(A) Phloem parenchyma
(B) Sieve tubes
(C) Companion cells
(D) Phloem fibres
Answer:
(D) Phloem fibres

Question 5.
Phloem was named as _______ by Haberlandt as similar to xylem.
(A) Bast
(B) Leptome
(C) Wood fibres
(D) Casparian
Answer:
(B) Leptome

Question 6.
The sieve tube cell is connected to companion cell through phloem parenchyma by
(A) Plasmodesmata
(B) Interfascicular cambium
(C) Pericycle
(D) Hypodermis
Answer:
(A) Plasmodesmata

Question 7.
Which of the following tissues is with dead thick-walled cells without intercellular spaces?
(A) parenchyma
(B) collenchyma
(C) sclerenchyma
(D) phloem
Answer:
(C) sclerenchyma

Question 8.
The tissue which is present in between xylem and phloem of stem is called
(A) apical meristem
(B) pericycle
(C) vascular cambium
(D) cork cambium
Answer:
(C) vascular cambium

Question 9.
In stem, epidermal hairs are called as
(A) Cuticles
(B) Casparian strip
(C) Trichomes
(D) Companion cells
Answer:
(C) Trichomes

Question 10.
_______ forms the outer covering of plant body and is derived from protodenn or dermatogen.
(A) Ground tissue system
(B) Interfascicular cambium
(C) Vascular tissue system
(D) Epidermal tissue system
Answer:
(D) Epidermal tissue system

Question 11.
_______ play a vital role in exchange of gases and water vapour.
(A) Vascular bundles
(B) Stomata
(C) Ground tissues
(D) Trichomes
Answer:
(B) Stomata

Question 12.
In which of the following leaf possesses dumbbell shaped guard cell’?
(A) Pisum sativum
(B) Wheat
(C) Datura
(D) Sunflower
Answer:
(B) Wheat

Question 13.
Which of the following is NOT a characteristic of spring wood?
(A) Tracheids with wide lumen
(B) Less number of fibres
(C) Narrow xylem band
(D) Lighter colour
Answer:
(C) Narrow xylem band

Question 14.
Periderm consists of
(A) Phellogen
(B) Phellem
(C) Phelloderm
(D) All of these
Answer:
(D) All of these

Question 15.
Which of the following is essential for secondary growth?
(A) Xylem
(B) Pith
(C) Phloem
(D) Cambium
Answer:
(D) Cambium

Question 16.
Vascular bundles of dicot root are
(A) radial exarch
(B) radial endarch
(C) conjoint exarch
(D) conjoint endarch
Answer:
(A) radial exarch

Question 17.
In which of the following characters, a monocot root differs from dicot root?
(A) Open vascular bundle
(B) Large pith
(C) Radial vascular bundles
(D) Scattered vascular bundles
Answer:
(B) Large pith

Question 18.
Which of the following plant shows isobilateral leaves?
(A) Hibiscus
(B) Maize
(C) Mangifera indica
(D) Sunflower
Answer:
(B) Maize

Question 19.
Secondary growth does not occur in
(A) Maize stem
(B) Mango leaf
(C) Carina root
(D) All of these
Answer:
(D) All of these

Question 20.
Stele of a dicot stem consists of all the given below, EXCEPT
(A) Pericycle
(B) Cortex
(C) Vascular bundles
(D) Pith
Answer:
(B) Cortex

Question 21.
Hypodermis is collenchymatous in
(A) monocot stem
(B) dicot stem
(C) monocot root
(D) both (A) and (B)
Answer:
(B) dicot stem

Question 22.
Lysigenous cavity filled with water is present in
(A) dicot stem
(B) monocot stem
(C) monocot root
(D) dicot root
Answer:
(B) monocot stem

Question 23.
The vascular bundles in a dicot stem are
(A) collateral and open
(B) radial
(C) bicollateral and open
(D) collateral and closed
Answer:
(A) collateral and open

Question 38.
Competitive Corner

Question 1.
Phloem in gymnosperms lacks:
(A) companion cells only
(B) both sieve tubes and companion cells
(C) albuminous cells and sieve cells
(D) sieve tubes only
Answer:
(B) both sieve tubes and companion cells

Question 2.
Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following:
(A) Shrinkage of air spaces in spongy mesophyll
(B) Tyloses in vessels
(C) Closure of stomata
(D) Flaccidity of bulliform cells
Hint: Grass leaves curl inwards to the minimize water loss.
Answer:
(D) Flaccidity of bulliform cells

Question 3.
Which of the statements given below is NOT true about formation of ‘annual rings’ in trees?
(A) Activity of cambium depends upon variation in climate.
(B) Annual rings are not prominent in trees of temperate region.
(C) Annual ring is a combination of spring wood and autumn wood produced in a year.
(D) Differential activity of cambium causes light and dark bands of tissue – early and late wood respectively.
Hint: Annual rings are formed due to activity of cambium. The activity of cambium is under the control of many physiological and environmental factors. In temperate regions, the climatic conditions are not uniform throughout the year due to this, annual rings are formed.
Answer:
(B) Annual rings are not prominent in trees of temperate region.

Question 4.
Regeneration of damaged growing grass following grazing is largely due to:
(A) secondary meristem
(B) lateral meristem
(C) apical meristem
(D) intercalary meristem
Hint: Intercalary meristems occur in grasses at the base of intemode, which regenerates the grass damaged due to grazing.
Answer:
(D) intercalary meristem

Question 5.
In the dicot root, the vascular cambium originates from:
(A) intrafascicular and interfascicular tissue in a ring
(B) tissue located below the phloem bundles and a portion of pericycle tissue above protoxylem
(C) cortical region
(D) parenchyma between endodermis and pericycle
Answer:
(B) tissue located below the phloem bundles and a portion of pericycle tissue above protoxylem

Question 6.
Casparian strips occur in
(A) Cortex
(B) Pericycle
(C) Epidermis
(D) Endodermis
Answer:
(D) Endodermis

Question 7.
Plants having little or no secondary growth are
(A) Conifers
(B) Deciduous angiosperms
(C) Grasses
(D) Cycads
Hint: Secondary growth takes place in stems and roots of dicotyledons and gymnosperms, but does not occur in monocotyledons.
Answer:
(C) Grasses

Question 8.
Secondary xylem and phloem in dicot stem are produced by
(A) Phellogen
(B) Vascular cambium
(C) Apical meristems
(D) Axillary meristems
Answer:
(B) Vascular cambium

Question 9.
The vascular cambium normally gives rise to
(A) Phelloderm
(B) Primary phloem
(C) Secondary xylem
(D) Periderm
Answer:
(C) Secondary xylem

Question 10.
Identify the wrong statement in context of heartwood.
(A) Organic compounds are deposited in it
(B) It is highly durable
(C) It conducts water and minerals efficiently
(D) It comprises dead elements with highly lignified walls
Hint: In old trees, secondary xylem (wood) becomes physiologically non active. It does not conduct water and becomes dark due to organic deposits (tannins, resins, oils, aromatic substances, etc.) It comprises of dead elements and called as heart wood. It is non conductive, hard, durable and resistant to microbes and insects.
Answer:
(C) It conducts water and minerals efficiently

Question 11.
Which of the following is made up of dead cell?
(A) Xylem parenchyma
(B) Collenchyma
(C) Phellem
(D) Phloem
Hint: Cork cambium (phellogen) cuts off cells on both the sides. The outer cells differentiate into cork or phellem. The cork is impervious to water due to suberin deposition in the cell wall.
Answer:
(C) Phellem

Maharashtra Board Class 12 Biology Important Questions Chapter 12 Biotechnology

July 25, 2024July 24, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 12 Biotechnology Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 12 Biotechnology

Multiple choice questions

Question 1.
Recombinant DNA technique was established by ………………..
(a) Herbert Boyer
(b) Stanley Cohen
(c) Peter Lobban
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 2.
Modern biotechnology includes ………………..
(a) r-DNA technology and PCR
(b) microarrays, cell culture and fusion
(c) production of curds, cheese, wine
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 3.
How many of the following statements are correct with reference to electrophoresis.
1. DNA is positively charged and it moves towards the negative electrode.
2. DNA is negatively charged and it moves towards the anode.
3. Longer DNA fragments move faster during electrophoresis.
4. Shorter DNA fragments move slowly during electrophoresis.
(a) One
(b) TWo
(c) Three
(d) Four
Answer:
(a) One

Question 4.
PCR technique was developed by ………………..
(a) Stanley Cohen
(b) Herbert Boyer
(c) W. Arber
(d) K. Mullis
Answer:
(d) K. Mullis

Question 5.
……………….. technique is used for in vitro gene cloning or gene multiplication.
(a) Electrophoresis
(b) SDS-PAGE
(c) Spectroscopy
(d) PCR
Answer:
(d) PCR

Question 6.
During denaturation step of polymerase chain reaction, the reaction mixture is heated at ………………..
(a) 40-60°C
(b) 70-75°C
(c) 90-98°C
(d) 36°C
Answer:
(c) 90-98 °C

Question 7.
Annealing temperature of primer is ………………..
(a) 40-60°C
(b) 72°C
(c) 90-98°C
(d) 36°C
Answer:
(a) 40-60 °C

Question 8.
During PCR new strand of DNA is synthesized by ………………..
(a) thermostable Taq DNA polymerase
(b) ligase
(c) phosphorylase
(d) RE
Answer:
(a) thermostable Taq DNA polymerase

Question 9.
Restriction enzymes were discovered by ………………..
(a) K. Mullis
(b) S. Cohen
(c) H. Boyer
(d) Smith, Nathan and Arber
Answer:
(d) Smith, Nathan and Arber

Question 10.
The molecular scissors of DNA are ………………..
(a) ligases
(b) polymerases
(c) endonucleases
(d) transcriptases
Answer:
(c) endonucleases

Question 11.
……………….. generates DNA fragments with sticky ends.
(a) Eco RI
(b) Bam HI
(c) Hind II
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 12.
……………….. is Type III endonulcease.
(a) Hapal
(b) Eco R I
(c) Bgll
(d) Eco K
Answer:
(a) Hapal

Question 13.
Select the incorrect statement.
(a) Type II restriction endonucleases have separate activities for cleaving and methylation.
(b) Type I restriction endonucleases function simultaneously as endonuclease and methylase.
(c) Type II restriction endonucleases cut DNA at specific non-pallindromic sequences.
(d) Thousands of Type II restriction endonucleases have been discovered.
Answer:
(c) Type II restriction endonucleases cut DNA at specific non-palindromic sequences

Question 14.
Ti plasmids are present in ………………..
(a) Bacillus thuringiensis
(b) Agrobacterium tumejaciens
(c) Haemophilus influenza
(d) Escherichia coli
Answer:
(b) Agrobacterium tumefaciens

Question 15.
Plasmid DNA containing foreign DNA is called ………………..
(a) chimeric DNA
(b) passenger DNA
(c) recombinant DNA
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 16.
Bacterial host cell takes up naked r-DNA by process of ………………..
(a) transduction
(b) transfection
(c) transformation
(d) all of these
Answer:
(c) transformation

Question 17.
Virosomes are ………………..
(a) liposome
(b) inactivated HIV
(c) liposome + inactivated HIV
(d) none of these
Answer:
(c) liposome + inactivated HIV

Question 18.
Recombinant protein used in the treatment of emphysema is ………………..
(a) a 1-Antitrypsin
(b) relaxin
(c) Interleukin-1 receptor
(d) urokinase
Answer:
(a) a 1-Antitrypsin

Question 19.
Tissue plasminogen activator and urokinase are used in the treatment of ………………..
(a) blood clots
(b) atherosclerosis
(c) emphysema
(d) asthama
Answer:
(a) blood clots

Question 20.
Bacillus is involved in the production of ……………….. vaccine that melts in mouth.
(a) polio
(b) flu
(c) chicken pox
(d) none of these
Answer:
(b) flu

Question 21.
First transgenic plant produced is …………………..
(a) Bt cotton
(b) tlavr savr tomato
(c) wheat
(d) tobacco
Answer:
(d) tobacco

Question 22.
Insect resistant GMO plants contain ………………..
(a) cowpea trypsin inhibitor gene
(b) cry gene
(c) chalone isomerase gene
(d) (a) or (b)
Answer:
(d) (a) or (b)

Question 23.
‘Cry’ genes are present in ………………..
(a) Agrobacterium tumifaciens
(b) Bacillus thuringiensis
(c) Rhizobium species
(d) Escherichia coli
Answer:
(b) Bacillus thuringiensis

Question 24.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because ………………..
(a) bacteria are resistant to toxin
(b) toxin is immature
(c) toxin is inactive
(d) bacteria encloses toxin in a special case
Answer:
(c) toxin is inactive

Question 25.
The gene coding for a-amylase inhibitor isolated from ……………….. is transferred to tobacco.
(a) mung bean
(b) adzuki bean
(c) E.coli
(d) daffodils
Answer:
(b) adzuki bean

Question 26.
The enzyme affecting the shelf life of Jlavr savr tomato is ………………..
(a) galactosidase
(b) transacetylase
(c) permease
(d) polygalactouranase
Answer:
(d) polygalactouranase

Question 27.
Ferritin, an iron storage protein, isolated from ……………….. and Phaseolus is transferred to ………………… to increase its iron content.
(a) soybean, rice
(b) rice, wheat
(c) maize, soybean
(d) rice, soybean
Answer:
(a) soybean, rice

Question 28.
To improve oil content and oil quality, ……………….. genes are transferred to soybean, oil palm, rapeseed and sunflower.
(a) Arabidopsis
(b) cry
(c) tobacco
(d) canola
Answer:
(a) Arabidopsis

Question 29.
Artemecin is an ……………….. drug developed from transgenic plants.
(a) anticancer
(b) antimalarial
(c) antibacterial
(d) antifungal
Answer:
(b) antimalarial

Question 30.
The transgenic cow born in Scotland, could produce a human protein in her milk for human therapeutics.
(a) Dolly
(b) Molly
(c) Tracy
(d) none of these
Answer:
(c) Tracy

Question 31.
Indian patent does not allow ………………..
(a) process patent
(b) product patent
(c) biopatent
(d) both (b) and (c)
Answer:
(b) product patent

Question 32.
Biopatent are awarded for ………………..
(a) strains of microbes and cell lines
(b) DNA sequences
(c) GMO
(d) All of these
Answer:
(d) All of these

Match the columns

Question 1.

Column A	Column B
(1) Sir Edward Sharpey Schafer	(a) Chemically synthesized DNA sequence of insulin
(2) Hakura	(b) Discovery of insulin
(3) Gilbert and Villokomaroff	(c) PCR
(4) K. Mullis	(d) Insulin production using r-DNA technology

Answer:

Column A	Column B
(1) Sir Edward Sharpey Schafer	(b) Discovery of insulin
(2) Hakura	(a) Chemically synthesized DNA sequence of insulin
(3) Gilbert and Villokomaroff	(d) Insulin production using r-DNA technology
(4) K. Mullis	(c) PCR

Question 2.

Restriction enzyme	Recognition sequence
(1) Alu I	(a) 5′ G-A-A-T-T-C 3′ 3′ C-T-T-A-A-G 5’
(2) Bam HI	(b) 5′ G-T-C-G-A-C3′ 3′ C-A-G-C-T-G 5′
(3) Eco RI	(c) 5′ A-G-C-T 3′ 3′ T-C-G-A5′
(4) Hind II	(d) 5′ G-G-A-T-T-C 3′ 3′ C-C-T-A-A-G 5′

Answer:

Restriction enzyme	Recognition sequence
(1) Alu I	(c) 5′ A-G-C-T 3′ 3′ T-C-G-A5′
(2) Bam HI	(d) 5′ G-G-A-T-T-C 3′ 3′ C-C-T-A-A-G 5′
(3) Eco RI	(a) 5′ G-A-A-T-T-C 3′ 3′ C-T-T-A-A-G 5’
(4) Hind II	(b) 5′ G-T-C-G-A-C3′ 3′ C-A-G-C-T-G 5′

Classify the following to form Column B as per the category given in Column A

(i) Provide tissues for human transplants
(ii) Cancer research
(iii) E. coli hygromycin resistant gene
(iv) Supply of factor IX.

Column A: (Transgenic animal)	Column B : (Application)
(a) Transgenic mice	——————
(b) Transgenic cattle	——————
(c) Pig clones	——————
(d) Transgenic fish	——————

Answer:

Column A: (Transgenic animal)	Column B : (Application)
(a) Transgenic mice	(ii) Cancer research
(b) Transgenic cattle	(iv) Supply of factor IX
(c) Pig clones	(i) Provide tissues for human transplants
(d) Transgenic fish	(iii) E. coli hygromycin resistant gene

Very short answer questions

Question 1.
Who used the term biotechnology for the first time and for what purpose?
Answer:
Karl Ereky in 1919 first used the term biotechnology to describe a process for large scale production of pigs.

Question 2.
Which is the oldest form of biotechnology ?
Answer:
Making curds or bread is the oldest form of biotechnology. Preparation of wine and other alcoholic beverages using microbial fermentation is also old biotechnology.

Question 3.
Modern biotechnology is based on which two core techniques?
Answer:
Modern biotechnology is based on two core techniques – Genetic engineering and chemical engineering.

Question 4.
What is the use of Chemical engineering?
Answer:
Chemical engineering technology is used to maintain sterile environment for manufacturing products like vaccines, antibodies, enzymes, organic acids, vitamins, therapeutics, etc.

Question 5.
What are the different techniques used to characterize macromolecules on the basis of their molecular weight?
Answer:
The techniques used to characterize macromolecules on the basis of molecular weight are gel permeation, osmotic pressure, ion exchange chromatography, spectroscopy, mass spectrometry, electrophoresis, etc.

Question 6.
What are the different types of electrophoresis ?
Answer:
Different types of electrophoresis are agarose gel electrophoresis, PAGE (Polyacrylamide Gel Electrophoresis), SDS – PAGE (Sodium dodecyl Sulphate-PAGE).

Question 7.
What is the use of polymerase chain reaction?
Answer:
Polymerase chain reaction is used for in vitro gene cloning or gene multiplication to produce a billion copies of the desired segment of DNA and RNA, with high accuracy and specificity, in few hours.

Question 8.
What are the requirements of polymerase chain reaction?
Answer:
Polymerase chain reaction requires thermal cycler, DNA containing the desired segment to be amplified, deoxyribonuclueoside triphosphates (dNTPs), excess of two primer molecules, heat stable DNA polymerase and appropriate quantities of Mg⁺⁺ions.

Question 9.
Enlist various biological tools required for transformation of recombinant DNA?
Answer:
Biological tools used for transformation of recombinant DNA are enzymes, cloning vectors (vehicle DNA) and competent host (cloning organisms).

Question 10.
Enlist various enzymes used in recombinant DNA technology?
Answer:
Various enzymes used in recombinant DNA technology are lysozymes, nucleases (exonucleases, endonucleases, restriction endonucleases), DNA ligases, DNA polymerases, alkaline phosphatases, reverse transcriptases, etc.

Question 11.
How do bacteria protect their own DNA from restriction enzymes?
Answer:
The bacteria protect their own DNA from restriction enzymes by methylating the bases at susceptible sites. This chemical modification blocks the action of the enzyme.

Question 12.
What is restriction?
Answer:
Restriction is the process by which the DNA strand is cut into restriction fragments with the help of restriction endonuclease enzymes or REs.

Question 13.
What are molecular scissors? Why are they so called?
Answer:
The enzyme restriction endonucleases are called molecular scissors because they can cut the DNA molecule at a specific point.

Question 14.
What is palindrome in DNA?
Answer:
Palindrome is a DNA sequence which when read on opposite strands of DNA (3′ to 5′ or 5′ to 3′) it reads same.

Question 15.
What are sticky ends?
Answer:
Sticky ends are short extensions of cleaved DNA molecule which can form hydrogen bonded base pairs with other complementary sticky ends.

Question 16.
Give examples of restriction endo¬nucleases that cut DNA at non- pallindromic sequences.
Answer:
Hpal, MboII

Question 17.
Give example of Type I restriction endonucleases.
Answer:
Eco KI

Question 18.
Give example of Type II restriction endonucleases.
Answer:
Eco RI, Bg II

Question 19.
Give the role of plasmids in bacterial cells.
Answer:
Plamids in bacterial cells carry genes for antibiotic resistance (R plasmids) and F plasmids are involved in conjugation.

Question 20.
What are plasmids?
Answer:
The small, extra chromosomal double stranded, circular forms of DNA which are capable of autonomous replication are called plasmids.

Question 21.
What is the source of gene to be cloned in vector?
Answer:
Gene to be cloned is obtained from gene library or by amplification.

Question 22.
What is chimeric DNA?
Answer:
Chimeric DNA is the combination of vector DNA and foreign DNA.

Question 23.
What is meant by transformed cells?
Answer:
The competent host cells which have taken up r-DNA Eire called transformed cells.

Question 24.
What are the different techniques by which foreign DNA can be transferred to host cell without using a vector?
Answer:
Foreign DNA can be transferred to host cell without using a vector by techniques like electroporation, microinjection, lipofection, shot gun, ultrasonification, biolistic method, etc.

Question 25.
Which marker genes are present in plasmid PBR 322?
Answer:
Markers genes in PBR322 plasmid are ampicillin resistant gene and tatracyclin resistant gene.

Question 26.
What is c-DNA?
Answer:
DNA produced by reverse transcription of m-RNA is known as c-DNA.

Question 27.
What are the objectives of CCMB?
Answer:
The objectives of CCMB are to conduct high quality basic research and training in frontier areas of modern biology and promote centralized national facilities for new and modern techniques in the interdisciplinary areas of biology.

Question 28.
Give examples of health problems which develop due to interaction between genetic and environmental factors?
Answer:
Health problems like high cholesterol and high blood pressure develop due to interaction between genetic and environmental factors.

Question 29.
Give examples of diseases which are caused due to single gene defects.
Answer:
Human genetic diseases like sickle-cell anaemia, thalassemia, Tay-sach’s disease, cystic fibrosis, Huntington’s chorea, haemophilia, alkaptonuria, albinism, etc. . are caused by single gene defects.

Question 30.
What is gene therapy?
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.

Question 31.
What is the use of virosomes?
Answer:
Virosomes are used in gene delivery.

Question 32.
Enlist the diseases where clinical trials of somatic cell gene therapy have been employed?
Answer:
The clinical trials of somatic cell gene therapy have been employed for the treatment of disorders like cancer, rheumatoid arthritis, SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle-cell anaemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

Question 33.
Which gene codes for Bt toxin?
Answer:
‘Cry’ gene codes for Bt toxin.

Question 34.
The a-amylase inhibitor gene transferred to tobacco from azuki bean acts against which pests?
Answer:
The a-amylase inhibitor gene transferred to tobacco from azuki bean acts against Zabrotes subjasciatus and Callosobruchus chinensis.

Question 35.
Which gene has been introduced in j sugarbeet for synthesis of fructants?
Answer:
1-sucroese sucrose fructosyl transferase gene has been introduced in sugarbeet for synthesis of fructants.

Question 36.
What percent of world population is affected by iron deficiency?
Answer:
30% of the population is affected by iron deficiency.

Question 37.
What causes softening of tomatoes during ripening?
Answer:
The enzyme polygalacturonase in tomato, breaks down pectin in the middle lamella of cell wall. This results in softening of fruits J during ripening.

Question 38.
What is superglue?
Answer:
Superglue is a biochemical glue for body repairs during surgery.

Question 39.
How is superglue produced?
Answer:
Superglue is produced by tobacco plants which contain genes encoding for adhesive proteins which allow marine mussels to stick to rocks.

Question 40.
Which oncogenes are being analyzed in transgenic mice to find out their role in development of breast cancer?
Answer:
myc and ras oncogenes are being analyzed in transgenic mice to find out their role in development of breast cancer.

Question 41.
How much milk is provided by Holstein cow on an average?
Answer:
Holstein cow provides about 6000 litres of milk per year.

Question 42.
Which gene is introduced in sheep to increase meat production?
Answer:
Human growth hormone gene is introduced in sheep for promoting growth and meat production.

Question 43.
Enlist desirable traits present in transgenic chicken?
Answer:
Desirable traits in transgenic chicken are low levels of fat and cholesterol, high protein containing eggs, in vivo resistance to viral and coccidial diseases, better feed efficiency and better meat quality.

Question 44.
Clones of which animals can provide tissues and organs for human transplants?
Answer:
The pig clone can provide animal organs and tissues for human transplants (xenotransplantation).

Question 45.
Which genes have been introduced in transgenic fish?
Answer:
Transgenic fish are transfected with E.coli hygromycin resistance gene, growth hormone and chicken crystalline protein.

Question 46.
Why does the Indian Government has set up the Genetic Engineering Approval Committee (GEAC)?
Answer:
The Indian Government has set up the Genetic Engineering Approval Committee (GEAC) to make decisions regarding the validity of research involving GMOs and addresses the safety of GMOs introduced for public use.

Question 47.
What is the duration of a biopatent?
Answer:
Duration of biopatents is five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.

Question 48.
What was the first biopatent awarded for?
Answer:
First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.

Question 49.
What is Texmati?
Answer:
Texmati is a trade name of “Basmati rice line and grains” for which Texas based American company Rice Tec Inc was awarded a patent by the US Patent and Trademark Office (USPTO) in 1997.

Give definitions of the following

Question 1.
Biotechnology
Answer:
Biotechnology is defined as ‘the development and utilization of biological forms, products or processes for obtaining maximum benefits to man and other forms of life’.
OR
According to OECD (Organization for Economic Cooperation and Development, 1981), biotechnology is defined as the application of scientific and engineering principles to the processing of materials by biological agents to provide goods and service to the human welfare’.

Question 2.
Genetic engineering/r-DNA technology
Answer:
Genetic engineering is defined as the manipulation of genetic material towards a desired end and in a directed and predetermined way, using in vitro process.
OR
According to John E. Smith (1996), genetic engineering as ‘the formation of new combination of heritable material by the insertion of nucleic acid molecule produced by whatever means outside the cells, into any virus, bacterial plasmid or other vector system so as to allow their incorporation into a host organism in which they do not occur naturally but in which they are capable of continued propagation’.

Question 3.
Nucleases
Answer:
Enzymes that cut the phosphodiester bonds of polynucleotide chains are called as nucleases.

Question 4.
Vector
Answer:
Vectors are DNA molecules that carry a foreign DNA segment and replicate inside the host cell.

Question 5.
Plasmid
Answer:
Plasmids are small, extra chromosomal, double stranded circular forms of DNA that replicate autonomously.

Question 6.
Transformation
Answer:
Insertion of a vector into the target bacterial cell is called transformation. (Learn this as well)

Question 7.
Transfection
Answer:
Insertion of a vector into the eukaryotic cells is called transfection. (Learn this as well)

Question 8.
Transduction
Answer:
Inserting a viral vector in cloning procedures is called transduction. (Learn this as well)

Question 9.
Gene library
Answer:
Gene libarary is a collection of different DNA sequences from an organism where each sequence has been cloned into a vector for ease of purification, storage and analysis.

Question 10.
Genomic library
Answer:
Genomic library is a collection of clones that represent the complete genome of an organism.

Question 11.
c-DNA library
Answer:
c-DNA library is a collection of clones containing c-DNAs inserted into suitable vector like a phage or plasmid.

Question 12.
Passenger DNA
Answer:
Passanger DNA is the foreign DNA which is inserted into a cloning vector.

Question 13.
Gene therapy
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.

Question 14.
Genetically modified organisms
Answer:
Genetically modified organisms are those whose genetic material has been artificially manipulated in a laboratory through genetic engineering to create combinations of plant, animal, bacterial, and viral genes that do not occur in nature or through traditional crossbreeding methods.

Name the following

Question 1.
Restriction endonucleases which produce fragments with sticky ends.
Answer:
Bam HI and Eco RI.

Question 2.
Restriction endonucleases which produce fragments with blunt ends.
Answer:
Alu I, Hind III.

Question 3.
Most commonly used vectors.
Answer:
Plasmid vectors (pBR 322, pUC, Ti plasmid) and bacteriophages (lambda phage, M13 phage.

Question 4.
Bacteriophages used as vectors.
Answer:
Ml3, lambda virus

Question 5.
Constructed plasmids.
Answer:
pBR 322, pBR 320, pACYC 177

Question 6.
Most commonly used plasmid in r-DNA technology.
Answer:
pBR 322

Question 7.
Plasmid vector for plants.
Answer:
Ti plasmid

Question 8.
Soil bacterium that causes a plant disease called crown gall.
Answer:
Agrobacterlum tumejaciens

Question 9.
Bacteria as competent host.
Answer:
Bacillus Haemophilus, Helicobacter pyroli and E. coli.

Question 10.
Most commonly used host in transformation experiments.
Answer:
E. coli

Question 11.
Cloning organisms used in plant biotechnology.
Answer:
Agrobacterium tumejaciens.

Question 12.
Human protein produced by r-DNA technology to treat anaemia.
Answer:
Erythropoeitin

Question 13.
Human protein produced by r-DNA technology to treat asthma.
Answer:
Interleukin 1 receptor

Question 14.
Human protein produced by r-DNA technology to treat antherosclerosis.
Answer:
Platelet derived growth factor

Question 15.
Human protein produced by r-DNA technology to treat parturition.
Answer:
Relaxin

Question 16.
Human protein produced by r-DNA technology to treat cancer.
Answer:
Interferons, tumour necrosis factor, inter-leukins, macrophage activating factor.

Question 17.
Recombinant protein used in the treatment of haemophiliaA.
Answer:
Factor VIII

Question 18.
Recombinant protein used in the treatment of haemophiliaB.
Answer:
Factor IX

Question 19.
Transgenic food crop used to reduce vitamin A deficiency diseases.
Answer:
Golden rice

Question 20.
Human proteins expressed in cow milk.
Answer:
Human lactoferrin, human alpha lactalbumin, human serum albumin, human bile salt.

Question 21.
Bacterial genes concerned with biosynthesis of cystein.
Answer:
cys E, cys M

Question 22.
Transgenic fish.
Answer:
Atlantic salmon, catfish, goldfish, Tilapia, zebra-fish, common carp, rainbow trout.

Give significance or functions of the following

Question 1.
Transgenic plants.
Answer:

Transgenic plants Eire genetically engineered to carry various desirable traits.
They are resistant to bacterial and viral diseases (e.g. tomato, potato, etc), insect pests (Bt cotton), herbicides (e.g. maize and wheat) and abiotic stresses.
Transgenic plants can be used as bioreactors or factories (molecular farming) for production of novel drugs like interferons, humanized antibodies against infective agents like HIV, amino acids and immunotherapeutic drugs.
Some transgenic plants have improved nutritional qualities, e.g. golden rice and golden mustard are biofortified with vitamin A.
Transgenic plants like Flavr savr tomatoes have been engineered to have more shelf life.
Transgenic plants also produce edible vaccines, e.g. Potato, tomato, etc.

Question 2.
Transgenic animals.
Answer:

Transgenic animals are genetically modified animals that are used in a wide range of fields.
In the field of medical research, transgenic animals like mice are used to identify the functions of specific factors through over- or under-expression of a modified gene (the inserted transgene). They are designed to study how genes contribute to the development of disease. These animals are used to investigate development of diseases like cancer, cystic fibrosis, rheumatoid arthritis, etc.
In toxicology field, they are used for detection of toxicants. They are used as responsive test animals.
Transgenic animals are used to evaluate a specific genetic change in molecular biology studies.
In pharmaceutical industry, transgenic animals are used for targeted production of pharmaceutical proteins, drug production and product efficacy testing.
They are also used in study of mammalian developmental genetics.

Transgenic farm animals like cattle, sheep, v poultry, pigs, etc. exhibit many desirable traits.

Improved quantity and quality of meat
Improved quality and quantity of milk
More egg production
Better quality and quantity of wool
Disease resistance
Production of low-cost pharmaceuticals and biologicals

Distinguish between the following

Question 1.
Old or classical biotechnology and Modem biotechnology.
Answer:

Old or classical biotechnology	Modem biotechnology
1. It is mainly based on fermentation technology.	1. It is based on genetic engineering and chemical engineering.
2. It does not involve alteration or modification of genetic material of organisms to develop specific Product.	2. It involves the alteration or modification of genetic material of organisms to develop specific product.
3. It does not involve the use of r-DNA technology, polymerase chain reaction (PCR), microarrays, cell culture and fusion, and bioprocessing to develop products.	3. It involves the use of r-DNA technology, polymerase chain reaction (PCR), microarrays, cell culture and fusion, and bioprocessing to develop specific products.
4. There is no ownership of old biotechnology.	4. It involves ownership of technology.
5. Examples : Preparation of curd, ghee, soma, vinegar, yogurt, cheese making, wine making, etc.	5. Examples : Transgenic organisms.

Question 2.
Genomic library and c-DNA library.
Answer:

Genomic library	c-DNA library
1. It is a collection of clones that represent the complete genome of an organism.	1. It is a collection of clones containing c-DNAs inserted into suitable vectors like phages or plasmids.
2. DNA fragments to be cloned are obtained by cutting genomic DNA by restriction enzymes.	2. c-DNAs are produced by the process of reverse transcription, using complete m-RNA complement obtained from a tissue or an organism.

Question 3.
Germ line gene therapy and somatic cell gene therapy.
Answer:

Germ line gene therapy	Somatic cell gene therapy
1. Germ line gene therapy involves modification of genome of germ cells like sperms, eggs, early embryos.	1. Somatic cell gene therapy involves modification of genome of somatic cells like bone marrow cells, hepatic cells, fibroblasts, endothelium, pulmonary epithelial cells, central nervous system and smooth muscle cells of wall of blood vessels.
2. It allows transmission of the modified genetic information to the next generation.	2. It does not allow transmission of the modified genetic information to the next generation.
3. Its application in human beings is not encouraged because of technical and ethical reasons.	3. It is a feasible option and clinical trials are carried out for treatment of various diseases.

Give reason

Question 1.
The genetic engineering is alternatively called recombinant DNA technology or gene cloning.
Answer:

Genetic engineering involves the manipulation of genetic material in a directed, predetermined, in vitro way to develop a desired end product.
It manipulates the genes for improvement of living organisms,
It involves repairing of the defective genes, replacing of defective genes by healthy genes or normal genes and artificially synthesizing of a totally new gene.
Genetic engineering also involves transfer of a new gene, transfer of genes to a new location or into a new organism, gene cloning, combining of genes from two organisms.
This results in alteration of the genotype and desired products can be developed.
Therefore, the genetic engineering is alternatively called recombinant DNA technology or gene cloning.

Question 2.
Bacteria have restriction enzymes.
Answer:

Restriction endonucleases or restriction enzymes in bacteria help them to recognize and destroy various viral DNAs that might enter the cell.
They cut the phosphodiester back bone at highly specific sites on both strands of DNA.
Thus, these enzymes restrict the potential growth of the virus and protect bacteria.
Hence, bacteria produce restriction enzymes.

Question 3.
All the fragments of a genome are cloned for storing them in genomic library.
Answer:

Genomic DNA is fragmented at the time of preparing genomic library.
It is not known which fragment has the desired gene.
Therefore all the fragments have to be cloned to store the copies of each separately.
Screening for the desired gene is later done through complementation or using DNA probes.
Therefore, all the fragments of a genome are cloned for storing them in genomic library.

Question 4.
The establishment of genomic library is more meaningful in prokaryotes than in eukaryotes.
Answer:

The prokaryotic genome does not contain repetitive DNA.
Eukaryotic DNA genome contains introns, regulatory genes and repetitive DNA.
Hence, the establishment of genomic library is more meaningful in prokaryotes than in eukaryotes.

Question 5.
Flavr savr tomato has longer shelf life.
Answer:

Flavr savr is genetically modified type of tomato.
It is developed by inserting antisense gene which retards ripening.
Due to the presence of this gene a cell wall degrading enzyme called polygalactouronase is produced in lesser amounts.
Owing to the above reason, Flavr savr tomato has longer shelf life.

Question 6.
Pigs are regarded as the most suitable animals to be bred for heart transplant.
Answer:

A pig’s heart is about the same size as a human heart.
Pig heart valves are used in human heart surgery for over a decade.
Hence, pigs are regarded as the most suitable animals to be bred for heart transplant.

Question 7.
Patent jointly issued by Delta and Pineland Company and U. S. department of agriculture under the title ‘control of plant gene expression’ was not granted by Indian government.
Answer:

Patent jointly issued by Delta and Pineland Company and U. S. department of agriculture under the title ‘control of plant gene expression’ is based on a gene that produces a protein toxic to plant and thus prevents seed germination.
Because of such patents, financially powerful corporations would acquire monopoly over biotechnological process.
Thus it would pose a threat to global food security.
Therefore this patent was considered morally unacceptable and fundamentally unequitable and it was not granted by the Indian government.

Write short notes

Question 1.
Electrophoresis.
Answer:

Electrophoresis is a technique that involves migration and separation of charged particles under the influence of electric field.
It is used for the separation of charged molecules like DNA, RNA and proteins, by application of an electric field.
Movement of charged particles is determined by particle size, shape and charge.
DNA is negatively charged and hence it moves towards the positive anode.
It results in size separation of DNA fragments as small fragments of DNA molecules move faster.
Different types of electrophoresis are agarose gel electrophoresis, PAGE and SDA PAGE.

Question 2.
Recognition sequences or restriction sites.
Answer:

Restriction endonucleases have the ability to recognize specific sequences in DNA and cleave it.
They are 4 to 8 nucleotides long and characterized by a particular type of internal symmetry.
The specific site at which restriction endonuclease cuts the DNA is called recognition site or restriction site.
Each restriction endonuclease recognizes its specific recognition sequence.
Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).

For example, recognition sequence of by the enzyme EcoRI is
3′ —– -CTTAA-G —–5′
5′ —– -G A A T T C —–3′
It is as palindrome, i.e. When read on opposite strand of DNA (3′ to 5′ or 5′ to 3′) it reads same.
When the enzyme EcoRI recognizes this sequence, it breaks each strand at the same site in the sequence i.e. between the A and G residues.

Question 3.
Plasmids as cloning vectors.
Answer:

Plasmids are small, extra-chromosomal, double stranded circular forms of DNA that replicate autonomously.
They are seen in bacterial cells, yeast and animal cell.
Plasmids are considered as replicons as they are capable of autonomous replication in suitable host.
The most commonly used vectors in r-DNA technology are plasmids as they replicate in E. coli.

Plasmid as a cloning vector should have a replication origin, a marker gene for antibiotic resistance, control elements like promoter, operator, ribosome binding site, etc. and a region where foreign DNA can be inserted. Naturally plasmids do not have all these features. Hence, they are constructed by inserting gene for antibiotic resistance. pBR 322, pBR320, paCYC177 are the constructed plasmids.

Ti plasmid (for tumor-inducing) of Agrobacterium tumefaciens is an important vector for carrying new DNA in many plants. It contains a transposon, called T DNA, which inserts copies of itself into the chromosomes of infected plant cells. The transposon, with the new DNA, can be inserted into the host cell’s chromosomes. A plant cell containing this DNA, can then be grown in culture or induced to form a new, transgenic plant.

Question 4.
Bt cotton.
Answer:

Bt cotton is well known example of insect resistant transgenic plant which is engineered with a gene from B. thuringiensis.
‘cry’ gene present in B. thuringiensis produces a protein that forms crystalline inclusions in bacterial spores.
When insect ingests it, because of high pH and the proteinase enzymes present in insect’s midgut, the crystalline inclusions are hydrolyzed to release the core toxic fragments.
This toxin causes midgut paralysis and disruption of midgut cells of insect.
Bt toxin acts against many species of Lepidoptera, Diptera and Coleoptera insects.

Question 5.
Golden rice.
Answer:

Golden rice is a transgenic plant developed by Swiss researchers.
It contain genes from the soil bacterium Erwinia and either maize or daffodil plants.
These plants are biofortified to have high content of vitamin A.
The golden colour is due to vitamin A.
Consumption of golden rice and golden mustard can reduce occurrence of vitamin A deficiency diseases (VAD).

Question 6.
Insulin.
Answer:

Insulin is a peptide hormone produced by β -cells of islets of Langerhans of pancreas.
Insulin is essential for the control of blood sugar levels.
Disease Diabetes mellitus is caused due to inability to make insulin.
Insulin was discovered by Sir Edward Sharpey Schafer (1916) while studying Islets of Langerhans.
Hakura et al (1977) chemically synthesized DNA sequence of insulin for two chains A and B and separately inserted into two PBR322 plasmid vector.

Gilbert and Villokomaroff, 1978 produced insulin production using r-DNA technology.

The recombinant plasmids, containing insulin gene inserted by the side of β-galactosidase, were transferred into E. coli host.
The host produced penicillinase pnd pre-pro insulin.
Insulin is later separated by trypsin treatment.

Question 7.
Transgenic cattle.
Answer:

Transgenic cattle are used for food production and for the production of human therapeutic proteins.
Transgenic cattle engineered with additional copies of bovine beta or kappa casein, show 8 to 20% increase in beta casein and a two-fold increase in kappa casein.
Various human proteins like Human lactoferrin, human alpha lactalbumin, human serum albumin and human bile salt stimulated lipase are expressed in the milk of transgenic cattle.
Transgenic cows produce factor IX (plasma thromboplastin component), used in the treatment of haemophilia.
Tracy, the transgenic cow born in Scotland, could produce a human protein in her milk for human therapeutics.
Human antibody products are made using transgenic cows that are immunized with a vaccine containing the disease agent. Antibodies are currently used for treatment of infectious diseases, cancer, transplanted organ rejection, autoimmune diseases and for use as antitoxins.

Question 8.
Transgenic fish.
Answer:

The commercially important fish like Atlantic salmon, catfish, goldfish, Tilapia, zebra-fish, common carp, rainbow trout, etc. are transfected with growth hormone, chicken crystalline protein and E.coli hygromycin resistance gene.
Transgenic fish showed increased cold tolerance and improved growth.

Short Answer Questions

Question 1.
What are the two phases of the development of biotechnology in terms of its growth?
Answer:
Two phases of the development of biotechnology in terms of its growth are as follows:
Traditional biotechnology (old biotechnology) : It is based on fermentation technology that uses microorganisms in the preparation of curd, ghee, soma, vinegar, yogurt, cheese making, wine making, etc.

Modern biotechnology (new biotechnology) :

During 1970 ‘recombinant DNA technology was developed and then established by Stanley Cohen and Herbert Boyer in 1973.
This technique alters or modifies genetic material to develop of new products.
The combination of biology and production technology based on genetic engineering evolved into modern biotechnology.
Modern biotechnology is based on two core techniques-genetic engineering and chemical engineering.

Question 2.
What are the different techniques and devices used in r-DNA technology?
Answer:
(1) Several techniques are used in r-DNA technology to isolate and characterize the macromolecules like DNA, RNA, proteins.
(2) The techniques used on the basis of molecular weight are gel permeation, osmotic pressure, ion exchange chromatography, spectroscopy, mass spectrometry, electrophoresis, etc.

(3) Electrophoresis:

It is used for the separation of charged molecules like DNA, RNA and proteins, by application of an electric field.
Different types of electrophoresis : Agarose gel electrophoresis, PAGE, SDA PAGE.

(4) Polymerase chain reaction (PCR) : It is used for in vitro gene cloning or gene multiplication to produce a billion copies of the desired segment of DNA or RNA, with high accuracy and specificity, in few hours.

Question 3.
What are the basic requirements of PCR technique?
Answer:
The basic requirements of PCR technique are as follows:

DNA containing the desired segment to be amplified.
Excess of forward and reverse primers which are synthetic oligonucleotides of 17 to 30 nucleotide.
They are complementary to the sequences present in DNA.
dNTPs which are of four types such as dATB dGTB dTTP and dCTR
A thermostable DNA polymerase (e.g. Taq DNA polymerase enzyme) that can withstand a high temperature of 90-98°C.
Appropriate quantities of Mg⁺⁺ ions.
Thermal cycler, a device required to carry out PCR reactions.

Question 4.
What are the two types of nucleases? What is their function?
Answer:

The two types of nucleases are exonucleases and endonucleases.
Exonucleases remove nucleotides from the ends of the DNA.
Endonucleases are those enzymes that have ability to make cuts at specific positions within the DNA molecule.
Of the endonucleases, restriction endonucleases serve as the molecular scissors in genetic engineering experiments.
They are used for cutting DNA molecules at specific predetermined sites. This is needed for gene cloning or recombinant DNA technology.

Question 5.
Explain different types of restriction enzymes?
Answer:
Different types of restrictions enzymes are as follows:

Type I – They function . simultaneously as endonuclease and methylase e.g. EcoK.
Type II – They exhibit separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, Bgll. They cut DNA at specific sites within the pallindrome. Thousands of type II restriction enzymes have been discovered.
Type III – They cut DNA at specific non- palindromic sequences e.g. Hpal, MboII.

Question 6.
With the help of a suitable example, illustrate palindrome.
Answer:

Palindrome is a sequence which when read on opposite strands of DNA (3′ to 5’ or 5’ to 3’), reads same.
When the enzyme EcoRI recognizes this sequence, it breaks each between the A and G residues.
In palindrome, the base sequence of second half in DNA represents the mirror image of the base sequence of the first half.
Palindromes are actually groups of letters which form the same word when read in both forward and backward directions.

For example, recognition sequence of by the enzyme EcoRI is a palindrome.
3′ —— – C T T A A G—–5′
5′ —— – G A A T T C—–3′

Same restriction enzyme must be used to cut vector and donor DNA, because it will produce fragments with the same complementary sticky ends, making it bond formation possible between them.

Question 7.
Explain how Ti plasmid of Agrobacterium tumefaciens acts as a vector for transferring genes to plants?
Answer:

Agrobacterium tumefaciens is a soil bacterium that causes crown gall disease in plants.
This disease involves the formation of tumor in the plant.
Ti plasmid in A. tumefaciens contains a transposon called T DNA.
T DNA inserts copies of itself into the chromosomes of infected plant cells.
The transposons, with the foreign DNA, can be inserted into the host cell’s chromosomes.
A plant cell containing this DNA, can then be grown in culture or induced to form a new, transgenic plant.

construction of Genomic library:

When genomic library is constructed, the entire genome or DNA is isolated from a particular organism.
This DNA is fragmented using suitable restriction endonucleases.
These separated fragments are later inserted into cloning vectors.
This develops recombinant vectors.
Such recombinant vectors are transferred into suitable organisms such as bacteria or yeast. Each host cell then contains one fragment.
These transformed organisms are cultured and their clones are thus produced. These clones are stored in the genomic library.

Question 8.
Explain with example how transformed host cells are selected from non- transformed?
Answer:

Transformed recombinant cells are selected using marker genes.
For example, markers genes in pBR 322 plasmid are ampicillin resistant gene and tetracyclin resistant gene.
When pstl restriction enzyme is used, ampicillin resistant gene gets knocked off from the plasmid and recombinant cells become sensitive to ampicillin.

Question 9.
Give the types of human proteins and hormones produced by recombinant DNA techniques.
Answer:

Blood proteins produced by recombinant DNA technique are Erythropoeitin, Tissue plasminogen activator, urokinase, Factor VIII, Factor IX, etc.
Hormones : Insulin, Epidermal growth factor.

Question 10.
What are edible vaccines? How are they produced?
Answer:

Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
Immunogenic protein of certain pathogens are active when administered orally.
The gene encoding for immunogenic protein is isolated and inserted in a suitable vector.
Recombinant vector is then transferred to plant genome.
Expression of this gene in specific parts of the plant results in the synthesis of immunogenic proteins.
When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.

Question 11.
Give an example of ‘melt in mouth’ vaccine and state the advantages of such vaccines.
Answer:

Example of ‘melt in the mouth’ vaccine that can be administered by placing it under tongue, is the production of flu vaccine by Bacillus which melts in the mouth and get delivered into the blood stream.
Advantages of edible oral vaccines are that they can be easily administered, can be easily stored and they are of low cost.

Question 12.
What is meant by recalcitrant seeds? How such plants can be conserved?
Answer:

Recalcitrant seeds are those whose survival is educed upon drying (reduction in moisture below a certain level) and freezing and thus are difficult to store.
It involves subcellular damage of seeds which results in loss of viability, when dried.
Plants which produce recalcitrant sees could be conserved using tissue culture technique.

Question 13.
What are the different ways in which gene therapy is used?
Answer:
Gene therapy is being used as follows:

Replacement of missing or defective genes.
Delivery of genes that speed the destruction of cancer cells.
Supply of genes that cause cancer cells to revert back to normal cells.
Delivery of bacterial or viral genes as a form of vaccination.
Delivery of DNA to antigen expression and generation of immune response.
Supply of gene for impairing viral replication.
Provide genes that promote or impede the growth of new tissue.
Deliver genes that stimulate the healing of damaged tissue.

Question 14.
What are the different types in which genes could be delivered during gene therapy?
Answer:
Genes can be delivered by three ways:

Ex vivo delivery : In this type of gene delivery, viral or non-viral vectors are used to introduce the desired gene in the cells isolated from patient, e.g. Parkinson’s disease, a neurological disorder.
In vivo delivery : In this method, therapeutic genes are directly delivered at the target sites of the cells of diseased tissue. Intravenous infusion genes are injected directly into tumor in the treatment of cancer.
Use of virosomes (Liposome + inactivated HIV), bionic chips.

Question 15.
What are transgenic plants? Explain with any two examples.
Answer:
The genetically engineered crop plants carrying desirable traits are called transgenic plants.
Examples of transgenic plants:

Bt Cotton : Bt cotton is a transgenic plant. Bt toxin gene has been cloned and introduced in many plants to provide resistance to insects without the need of insecticides.
Golden rice : It is a genetically engineered rice with higher beta carotene (provitamin A) content.
Flavr savr tomato : It is developed by inhibiting synthesis of polygalactournonase by inserting antisense gene. This type of tomato has a longer shelf life.

Question 16.
Give any two examples of insect resistant transgenic crops.
Answer:
Examples of insect resistant transgenic crops are as follows:
(1) BT crops:

Insect resistant transgenic plants contain either a gene from B. thuringiensis or the cowpea trypsin inhibitor gene.
‘cry’ gene present in B. thuringiensis produces a protein that forms crystalline inclusions in bacterial spores. When insect ingests it, because of high pH and the proteinase enzymes present in insect’s midgut, they are hydrolyzed to release the core toxic fragments.
This toxin causes midgut paralysis and disruption of midgut cells of insect.
Bt toxin activity has been against many species of insects within the orders of Lepidoptera, Diptera and Coleoptera.

(2) Transgenic tobacco:

The gene of a-amylase inhibitor (aAl-Pv), isolated from adzuki bean (Phaseolus vulgaris) is transferred to tobacco.
This gene works against pests like Zabrotes subfasciatus and Callosobruchus chinensis.

Question 17.
Give any two examples of biofortified transgenic crops.
Answer:
Examples of biofortified transgenic crops are as follows:

Golden rice and Golden mustard These are transgenics rich in vitamin A.
Arabidopsis genes are transferred to soybean, oil palm, rapeseed and sunflower for improvement in oil content and oil quality.
Ferritin, an iron storage protein, isolated from soybean and Phaseolus is transferred to rice to increase its iron content.
Plants deficient in amino acids like methionine, lysine and tryptophan have been engineered to improve protein content.

Question 18.
What factors are responsible for losses during storage and transport of crops? Explain, with example, how genetic engineering can reduce these losses?
Answer:

Diseases and pests, bruising on soft fruits and vegetables, heat and cold storage, over-ripeness, loss of flavours and odours, etc. lead to great deal of losses during storage and transport of crops.
Most of these changes are caused due to endogenous enzyme activities which could be slowed down using genetic engineering.

For example, shelf life of Flavr savr tomatoes has been increased using genetic engineering techniques.
(a) The enzyme polygalacturonase breaks down pectin in the cell wall, leading to softening of fruit during ripening of tomatoes.
(b) In genetically modified Flavr savr tomatoes, polygalactouronase enzyme is inhibited by antisense genes. These tomatoes can remain on the vine until mature and be transported in a firm solid state.

Question 19.
How transgenic animals are produced.?
Answer:

Transgenic animals are produced using recombinant DNA technology.
Foreign DNA is introduced in transgenic animals using r-DNA technology.
It is then transmitted through the germ line so that every cell if the animal contains the same modified genetic material.
This involves cloning of desired gene and introduction of cloned gene into fertilized eggs, successful implantation of modified eggs into receptive female and obtaining progeny carrying cloned genes.

Question 20.
Write any two scientific and commercial values of transgenic animals in favour of human beings.
Answer:
(1) Scientific value of transgenic animals:

Transgenic mice are used medical research to identify the functions of specific factors through over – or under-expression of a modified gene (the inserted transgene). They are designed to study how genes contribute to the development of disease. These animals are used to investigate development of diseases like cancer, cystic fibrosis, rheumatoid arthritis, etc.
In toxicology field, they are used for detection of toxicants. They are used as responsive test animals.

(2) Commercial value of transgenic animals:

Transgenic cattle are used for production of human therapeutic proteins such as human lactoferin, human serum albumin, etc.
Better quality and quantity of wool by transgenic sheep.

Question 21.
How sheep are genetically altered to produce wool of better quality?
Answer:

Bacterial genes, cys E and cys M, are identified, cloned and introduced in sheep.
These genes are involved in biosynthesis of cysteine.
Cysteine is involved in formation of keratin protein found in wool.
Thus, transgenic sheep produce more quantity and better quality of wool.

Question 22.
What is meant by ethics?
Answer:

Ethics is a discipline concerned with moral values or principles.
It deals with certain sets of standards which regulate behaviour of community.
It is concerned with socially accepted norms of moral duty, conduct and judgment.
Ethical concepts differ according to culture and traditions.
They also change with time and get influenced by progress in science and technology.

Question 23.
What are the adverse effects of Biotechnology on the Environment?
Answer:
The adverse effects of biotechnology on the environment are as follows:

Unintended hybrid strains of weeds and other plants can develop resistance to herbicides through cross-pollination. E.g. Crops of Round Up-ready soybeans which are used in agriculture, possibly confer Round Up resistance to neighbouring plants.
Bt corn has adversely affected non target species – Monarch butterfly. It may also prove harmful to neutral or even beneficial species.

Question 24.
Discuss various health concerns regarding the use of GMO crops.
Answer:
Various concerned related to health regarding the use of GMO crops are as follows:

GMO crops may develop some allergies, e.g. A gene from the Brazil nut was transferred to soybean in order to increase methionine content. But this transgenic soybean has caused allergies in some people which are known to suffer from nut allergies (“Biotech Soybeans”).
GMO technology is a recent development and its the long-term effects on health cannot be anticipated at this point.
Potential effects of transgenic proteins, which were never been ingested earlier, on the human body are yet unknown.
The use of GMOs may lead to the development of antibiotic and vaccine- resistant strains of diseases.

Question 25.
What is a patent?
Answer:

Patent is a special right granted to the inventor by the government.
A patent consists of three parts – grant (an agreement with the inventor), specification (subject matter of invention) and claims (scope of invention to be protected).
Patent is a personal property of inventor and it can be sold like any other property.

Question 26.
What is meant by traditional knowledge? What is its importance?
Answer:

Traditional knowledge is a deep understanding of ecological processes and the ability to obtain useful products from the local habitat in a sustainable way.
Most traditional knowledge is handed down through generations. This helps in the development of modern, commercial applications. This saves the makers time, money and effort.

Traditional knowledge includes:

Knowledge about food, crop varieties and agricultural/farming practice.
Sustainable management of natural resources and conservation of biological diversity.
Biologically important medicines.

Chart or Table based Questions

Question 1.

RE	Source	End products
Alu I	————-	Blunt ends
————	Bacillus amyloliquefaciens H	Sticky ends
Eco R I	————–	Sticky ends
Hind II	H. influenza Rd	————–

Answer:

RE	Source	End products
Alu I	Arthobacter luteus	Blunt ends
Bam H I	Bacillus amyloliquefaciens H	Sticky ends
Eco R I	E. coli Ry 13	Sticky ends
Hind II	H. influenza Rd	Blunt ends

Question 2.

Substance	Potential benefit	Crop	Transgene
Provitamin A	Anti-oxidant	————–	Phytoene synthase, Lycopene cyclase
Fructans	————–	Sugarbeet	—————-
Vitamin E	—————-	Canola	γ -tocopherol methyl transferase
Flavonoids	Anti-oxident	Tomato	————–
————–	Iron fortification	Rice	Ferritin, metallothioein, phytase

Answer:

Substance	Potential benefit	Crop	Transgene
Provitamin A	Anti-oxidant	Rice	Phytoene synthase, Lycopene cyclase
Fructans	Low calories	Sugarbeet	I – sucrose : sucrose fructosyl transferase
Vitamin E	Anti-oxidant	Canola	γ -tocopherol methyl transferase
Flavonoids	Anti-oxident	Tomato	Chalone isomerase
Iron	Iron fortification	Rice	Ferritin, metallothioein, phytase

Question 3.

Organization	Expand
OECD	——————
GEAC	——————
USDA	——————
USPTO	——————
CSIR	—————–

Answer:

Organization	Expand
OECD	Organization for Economic Cooperation and Development
GEAC	Genetic Engineering Approval Committee
USDA	US Department of Agriculture
USPTO	US Patent and Trademark Office
CSIR	Council of Scientific and Industrial Research

Diagram Based Questions

Question 1.
(a) Name the reaction shown in the given diagram.
(b) What are the three steps in this reaction?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 12 Biotechnology 1
(a) Polymerase chain reaction
(b) Three steps of polymerase chain reaction are denaturation of DNA, annealing of primer and extension of primer.

Question 2.
(a) Which enzyme has recognition sequence shown in the diagram given below?
(b) What is meant by palindrome?
(c) Enzyme in the given diagram cuts DNA to produce ———- ends.
Maharashtra Board Class 12 Biology Important Questions Chapter 12 Biotechnology 2
Answer:
(a) Enzyme is EcoRI.
(b) Palindrome is a DNA sequence which when read in opposite direction (3’ to 5’ or 5’ to 3’) it reads same.
(c) Enzyme in the given diagram cuts DNA to produce sticky ends.

Question 3.
Draw a labelled diagram of a plasmid showing ori, ampr and a region into which foreign DNA can be inserted.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 12 Biotechnology 3

Question 4.
Draw a diagram showing steps in r-DNA technology.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 12 Biotechnology 4

Long Answer Questions

Question 1.
Explain with examples how transgenic plants can be used as factories or bioreactors?
Answer:

Transgenic plants are potential factories or bioreactors for biochemicals and biopharmaceuticals like starch, sugar, lipids, proteins, hormones, antibodies, vaccines or enzymes.
Various fine chemicals, perfumes, adhesive compounds industrial lubricants, etc. can be isolated from plants.
Plants can be the source of biodegradable plastic and ‘renewable’ energy to replace fossil fuels.
Transgenic plants useful for production of novel drugs like interferons, edible vaccines, antibodies, amino acids, immunotherapeutic drugs, etc. Thus, they are like bioreactors for molecular farming.

Examples:

The gene for Human growth hormone has been inserted into the chloroplast DNA of tobacco plants.
Humanized antibodies against HIV, Respiratory syncytial virus (RSV), Herpes simplex virus (HSV), the cause of “cold sores” are developed using transgenic plants.
Superglue’ is a biochemical glue for body repairs during surgery. It is produced by tobacco plants which contain genes encoding for adhesive proteins which allow marine mussels to stick to rocks.
Protein antigens to be used in vaccines : e.g. Patient-specific antilymphoma vaccines. B-cell lymphomas are clones of malignant B cells expressing a unique antibody molecule on their surface.
Transgenic plants cam be used as factories for producing oil having nutritional value like cod-liver oil. These plants are engineered with a oil encoding gene from marine algae.
Transgenic plants produce the antimalarial drug, Artemisinin.
Genetically engineered opium poppy can be used to produce powerful painkillers.
Transgenic plants like potatoes, tomatoes, bananas, soybeans, alfalfa and cereals can be used as edible vaccine.

Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production

July 25, 2024July 24, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 11 Enhancement of Food Production Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 11 Enhancement of Food Production

Multiple choice questions

Question 1.
Means of in situ germplasm conservation are ………………….
(a) forests
(b) Natural Reserves
(c) botanical gardens, seed banks
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 2.
Means of ex situ germplasm conservation are ………………….
(a) forests and seed banks
(b) natural Reserves
(c) botanical gardens, seed banks
(d) botanical garden and forests
Answer:
(c) botanical gardens, seed banks

Question 3.
Germplasm includes ………………….
(a) only improved varieties of crop
(b) all cultivated varieties and wild relatives of a particular crop
(c) all hybridized varieties only
(d) only mutant varieties of a crop
Answer:
(b) all cultivated varieties and wild relatives of a particular crop

Question 4.
Taichung Native-1 is a variety of rice from ………………….
(a) China
(b) Korea
(c) Malaysia
(d) Taiwan
Answer:
(d) Taiwan

Question 5.
Which of the following is not a fungal disease ?
(a) Late blight of potato
(b) Brown rust of wheat
(c) Red rot of sugar cane
(d) Black rot of crucifers
Answer:
(d) Black rot of crucifer

Question 6.
…………………. variety of wheat is resistant to Hill bunt disease.
(a) Himgiri
(b) Pusa swarnim
(c) Kalyan sona
(d) Pusa A-4
Answer:
(a) Himgiri

Question 7.
Regina-II variety of …………………. is resistant to bacterial rot.
(a) wheat
(b) cabbage
(c) cauliflower
(d) Brassica
Answer:
(b) cabbage

Question 8.
The nectar-less cotton having smooth leaves has resistance against ………………….
(a) bollworms
(b) jassids
(c) aphids
(d) stem borers
Answer:
(a) bollworms

Question 9.
………………… gave concept of in vitro cell culture.
(a) Haberlandt
(b) Frank
(c) Yabuta and Sumiki
(d) None of these
Answer:
(a) Haberlandt

Question 10.
Tissue culture requirements are ………………….
(a) pH of nutrient medium 5 to 5.8
(b) Sterilized glassware, nutrient medium, explants, inoculation chamber
(c) Temperature 18 °C to 20 °C
(d) All of these
Answer:
(d) All of these

Question 11.
Hybrid maize with double the quantity of amino acids …………………. have been developed.
(a) lysine and tryptophan
(b) alanine and aspartic acid
(c) glutamic and proline
(d) histidine and cystine
Answer:
(a) lysine and tryptophan

Question 12.
Inbreeding increases ………………….
(a) homozygosity
(b) heterozygosity
(c) heterosis
(d) hemizygosity
Answer:
(a) homozygosity

Question 13.
Which hormone is used for MOET method?
(a) GH
(b) LH
(c) FSH
(d) ICSH
Answer:
(c) FSH

Question 14.
Find the odd one out.
(a) Sahiwal
(b) Sindhi
(c) Gir
(d) Holstein
Answer:
(d) Holstein

Question 15.
Pullorum is a …………………. disease.
(a) viral
(b) bacterial
(c) fungal
(d) parasitic
Answer:
(b) bacterial

Question 16.
Propolis is ………………….
(a) bee glu
(b) royal jelly
(c) bee venom
(d) none of these
Answer:
(a) bee glu

Question 17.
Select the incorrect pair.
(a) Apis dorsata – Rock bee
(b) Apis indica -Indian bee
(c) Apis florae – Little bee
(d) Apis mellifera – Wild bee
Answer:
(d) Apis mellifera – Wild bee

Question 18.
fishery takes place in Sundarban area of ………………….
(a) Estuarine, West Bengal
(b) Marine, Odisha
(c) Fresh water, West Bengal
(d) None of these
Answer:
(a) Estuarine, West Bengal

Question 19.
Which stage in the life cycle of silk moth secretes silk?
(a) Caterpillar
(b) Egg
(c) Pupa
(d) Adult
Answer:
(a) Caterpillar

Question 20.
Lac insect is a native of ………………….
(a) China
(b) India
(c) Africa
(d) Europe
Answer:
(b) India

Question 21.
Alcoholic fermentation is brought about by ………………….
(a) Lactobacillus
(b) Saccharomyces
(c) Trichoderma
(d) Streptomyces
Answer:
(b) Saccharomyces

Question 22.
…………………. and …………………. are alcoholic beverages produced without distillation.
(a) Wine, rum
(b) Wine, beer
(c) Whisky, brandy
(d) Brandy, beer
Answer:
(b) Wine, beer

Question 23.
Organic acid can be produced directly from glucose or formed as end products from ………………….
(a) pyruvate
(b) ethanol
(c) gluconic acid
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 24.
The microbial source of vinegar is ………………….
(a) Aspergillus niger
(b) Rhizopus arrhizus
(c) Acetobacter aceti
(d) Streptomyces venezuelae
Answer:
(c) Acetobacter aceti

Question 25.
Statins are produced by ………………….
(a) Monascus purpureus
(b) Streptococcus
(c) Aspergillus niger
(d) None of these
Answer:
(a) Monascus purpureus

Question 26.
…………………. is used as a ‘clot buster’.
(a) Pectinase
(b) Statin
(c) Invertase
(d) Streptokinase
Answer:
(d) Streptokinase

Question 27.
Aspergillus niger is used to manufacture ………………….
(a) pectinase, gluconic acid and vitamin C
(b) pectinase, gluconic acid and vitamin B₁₂
(c) invertase, acetic acid and vitamin C
(d) pectinase, citric acid and invertase
Answer:
(a) pectinase, gluconic acid and vitamin C

Question 28.
The first Gibberellin was isolated by ………………….
(a) Frank
(b) Skoog
(c) Yabuta and Sumiki
(d) None of these
Answer:
(c) Yabuta and Sumiki

Question 29.
Once the BOD of waste water is reduced, it is passed into a ………………….
(a) settling tank
(b) primary sedimentation tank
(c) anaerobic sludge digesters
(d) aeration tank
Answer:
(a) settling tank

Question 30.
During biogas production species used to bring about hydrolysis or solubilization is ………………….
(a) Pseudomonas
(b) Rhizopus
(c) Methanococcus
(d) Methanobacillus
Answer:
(a) Pseudomonas

Question 31.
…………………. bacteria are used as herbicides.
(a) Pseudomonas spp., Xanthomonas spp., Agrobacterium spp.
(b) Bacillus thuringiensis, B. papilliae, B. lentimorbus
(c) Pseudomonas spp., Bacillus thuringiensis, B. papilliae
(d) Xanthomonas spp., Agrobacterium spp., Bacillus thuringiensis
Answer:
(a) Pseudomonas spp., Xanthomonas spp., Agrobacterium spp.

Question 32.
The weed Senecio jacobeac is controlled by ………………….
(a) Cactoblastis cactorum
(b) Xanthomonas spp
(c) Bacillus thuringiensis
(d) tyrea moth
Answer:
(d) tyrea moth

Question 33.
Nosema locustae is …………………. pathogen.
(a) bacteria
(b) fungal
(c) protozoan
(d) viral
Answer:
(c) protozoan

Question 34.
Which of the following bacterial pathogen is not used as herbicide ?
(a) Pseudomonas
(b) Xanthomonas
(c) Agrobacterium
(d) Azotobacter
Answer:
(d) Azotobacter

Question 35.
Identity free living bacterial biofertlizer ………………….
(a) Rhizobium
(b) Azotobacter
(c) Nostoc
(d) Bacillus thuringiensis
Answer:
(b) Azotobacter

Question 36.
The ectomycorrhizae form …………………. on the root surface.
(a) root tuber
(b) mantle
(c) root hair
(d) arbuscles
Answer:
(b) mantle

Match the columns

Question 1.

Column I (Biofortified crop)	Column II (Nutrient Enrichment)
(1) Maize	(a) Five times more iron
(2) Rice	(b) Twice the amount of lysine and tryptophan
(3) Wheat Atlas-66	(c) Enriched in vitamin A and minerals
(4) Carrots, spinach	(d) High protein content

Answer:

Column I (Biofortified crop)	Column II (Nutrient Enrichment)
(1) Maize	(b) Twice the amount of lysine and tryptophan
(2) Rice	(a) Five times more iron
(3) Wheat Atlas-66	(d) High protein content
(4) Carrots, spinach	(c) Enriched in vitamin A and minerals

Question 2.

Column I (Organic Acids]	Column II (Microbial source)
(1) Citric acid	(a) Rhizopus arrhizus
(2) Fumaric acid	(b) Acetobacter aceti
(3) Gluconic acid	(c) Aspergillus niger
(4) Acetic acid	(d) Aspergillus niger

Answer:

Column I (Organic Acids]	Column II (Microbial source)
(1) Citric acid	(c) Aspergillus niger
(2) Fumaric acid	(a) Rhizopus arrhizus
(3) Gluconic acid	(d) Aspergillus niger
(4) Acetic acid	(b) Acetobacter aceti

Classify the following to form Column B as per the category given in Column A.

Question 1.
i. Alternaria crassa
ii. Agrobacterium spp.
iii. Cactoblastis cactorum
iv. Beavueria bassiana

Column A (Biocontrol agents)	Column B (Host)
Microbial pesticide	————
Mycoherbicide	————
Insect as herbicide	————–
Bacterial herbicide	—————

Answer:

Column A (Biocontrol agents)	Column B (Host)
Microbial pesticide	Beavueria bassiana
Mycoherbicide	Alternaria crassa
Insect as herbicide	Cactoblastis cactorum
Bacterial herbicide	Agrobacterium spp.

Question 2.
i. Hairy leaves in wheat
ii. Nectar-less cotton having smooth leaves
iii. Hairy leaves in cotton
iv. Solid stem in wheat

Resistance to insects	Morphological characters
Jassids	————
Cereal leaf beetle	————
Stem borers	————–
Bollworms	—————

Answer:

Resistance to insects	Morphological characters
Jassids	Hairy leaves in cotton
Cereal leaf beetle	Hairy leaves in wheat
Stem borers	Solid stem in wheat
Bollworms	Nectar-less cotton having smooth leaves

Very Short Answer Questions

Question 1.
What are the different methods of plant breeding?
Answer:
Different methods of plant breeding are introduction, selection, hybridization, mutation breeding, polyploidy breeding, tissue culture, r-DNA technology and SCP (Single cell protein).

Question 2.
Which variety of sugar cane having high sugar content and better yield is cultivated in South India?
Answer:
Saccharum ojficinarum variety of sugar cane has high sugar content and better yield. It is cultivated in South India.

Question 3.
What are the desirable characteristics in hybrid varieties millets developed in India?
Answer:
Hybrid varieties of millets developed in India are high yielding and resistant to water stress.

Question 4.
Give examples of natural physical mutagens.
Answer:
Natural physical mutagens are high temperature, high concentration of CO₂, X-rays, UV rays.

Question 5.
Give examples of chemical mutagens.
Answer:
Chemical mutagens are nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.

Question 6.
How are seedlings or seeds mutated?
Answer:
Seedlings or seeds are mutated by irradiating them by CO-60 or exposing them to UV bulbs, X-ray machines, etc.

Question 7.
What are the effects of mutagens?
Answer:
Effects of mutagens are gene mutations and chromosomal aberrations.

Question 8.
Which biochemical characters are responsible for resistance against stem borers in maize?
Answer:
High aspartic acid, low nitrogen and sugar content are responsible for resistance against maize stem borers.

Question 9.
What does the plant tissue culture medium consists of?
Answer:
The plant tissue culture medium consists of water, all essential minerals, sources for carbohydrates, proteins and fats, growth hormones like auxins and cytokinins, vitamins. Agar is added to solidify nutrient medium for callus culture.

Question 10.
What are the types of tissue culture based on nature of explants?
Answer:
Cell culture, organ culture, pollen or anther culture, meristem culture and embryo culture are the types of tissue culture based on nature of explants.

Question 11.
What are the types of tissue culture based on the type of in vitro growth?
Answer:
Callus culture and suspension culture are the type of tissue culture based on the type of in vitro growth.

Question 12.
What is the necessity of subculturing?
Answer:
Both the callus and suspension cultures die in due course of time when nutrients get consumed. Therefore, a part of callus or suspension of cells is transferred to the flask containing new nutrient medium. This is subculturing, which is necessary to ensure continuous nutrient supply essential for continuous growth.

Question 13.
Enlist substances that are used as substrate for the production of SCP.
Answer:
Agricultural waste like corn cobs, sugar cane molasses, wood shavings, sawdust, paraffin, N-alkanes, human and animal wastes are the substrates used for the production of SCR.

Question 14.
How biofortified crops are produced?
Answer:
Biofortified crops are produced by conventional selective breeding practices and r-DNA technology.

Question 15.
What per cent of world livestock population is present in India and China and what is the productivity?
Answer:
India and China have 70% of world livestock populations, but the productivity is only 25% of the world farm produce.

Question 16.
What is silage made up of?
Answer:
Silage is fermented fodder made up of legumes, grasses, maize and jowar.

Question 17.
What is the supplementary food to silage?
Answer:
Silage is supplemented with oilcakes, minerals, vitamins and salts.

Question 18.
What is layer?
Answer:
The hen which is reared to obtain eggs is called a layer.

Question 19.
What is broiler?
Answer:
The hen which is reared to obtain meat is called a broiler.

Question 20.
What are the allied professions to poultry?
Answer:
The allied professions to poultry are processing of eggs and meat, marketing of poultry products, compounding and sale of poultry feed, poultry equipment, pharmaceuticals, feed additives, etc.

Question 21.
Which areas are suitable for bee keeping?
Answer:
The areas having sufficient wild shrubs, cultivated crops of sunflower, mustard, safflower, chilly, cabbage, cucumber, legumes, etc. and fruit orchards of apple, mangoes, citrus, etc. are suitable for bee keeping.

Question 22.
What is indicated by yellow spots?
Answer:
Yellow spots indicate shrinking of female lac insect.

Question 23.
What is indicated by orange spots on the eggs of lac insect?
Answer:
Orange spots indicate that eggs are about to hatch.

Question 24.
What is the main function of a fermenter?
Answer:
The main function of a fermenter is to provide a controlled environment for growth of specific microorganisms or a defined mixture of microorganisms, to obtain the desired product.

Question 25.
What is known as Brewer’s yeast?
Answer:
Saccharomyces cerevisiae var. ellipsoids is commonly known as Brewer’s yeast.

Question 26.
Which fermenter is used for large scale preparation of alcohol?
Answer:
Tubular tower fermenter is used for large scale preparation of alcohol.

Question 27.
What is the use of gluconic acid?
Answer:
Gluconic acid is used in medicine for solubility of Ca⁺⁺

Question 28.
What is the use of citric acid?
Answer:
Citric acid is used in confectionary.

Question 29.
What is the use of fumaric acid?
Answer:
Fumaric acid is used in resins as wetting agents.

Question 30.
Give examples of diseases which are treated using antibiotics.
Answer:
Diseases like plague, whooping cough, diphtheria and leprosy are treated using antibiotics.

Question 31.
What is the mechanism of actions of statins?
Answer:
Statins produced by yeast Monascus purpureus are blood cholesterol lowering agents. They are competitive inhibitors of the enzyme that catalyzes synthesis of cholesterol.

Question 32.
Enlist the various enzymes produced using microorganisms.
Answer:
Amylase, cellulase, protease, lipase, pectinase, streptokinase, invertase enzymes are produced using microorganisms.

Question 33.
Gibberellin was first isolated from which plant?
Answer:
Gibberellin was first isolated from rice plant infected by fungus Gibberella Jujikuroi.

Question 34.
How many different types of Gibberellins have been isolated?
Answer:
About 15 different types of Gibberellins have been isolated.

Question 35.
Give the chemical composition of biogas.
Answer:
The biogas consists of methane (50-60%), CO₂ (30 to 40%), H₂S (0-3%) and other gases like CO, N₂, H₂ in traces.

Question 36.
Which substrates are used for biogas production?
Answer:
Substrates like cattle dung (most commonly used substrate, a rich source of cellulose from plants), plant wastes, animal wastes, domestic wastes, agriculture waste, municipal wastes, forestry wastes, etc. are used for biogas production.

Question 37.
Which are the most commonly used models of biogas plants ?
Answer:
Models of biogas plants developed by KVTC (Khadi and Village Industries Commission) and IARI (Indian Agricultural Research Institute) are the most commonly used in India.

Question 38.
Which bacteria transform acetic acid into biogas?
Answer:
The acetic acid is transformed into biogas by methanogenic bacteria like Methanococcus, Methanobacterium and Methanobacillus.

Question 39.
What are the four groups of biocontrol agents?
Answer:
The four groups of biocontrol agents are bacteria, fungi, viruses and protozoans.

Question 40.
What is mycoherbicide ?
Answer:
The pathogenic fungus which kills or inhibits the growth of a weed is called mycoherbicide.

Question 41.
What are the three types of bacterial biofertilizers on the basis of function?
Answer:
On the basis of function, bacterial biofertilizers are of three types – nitrogen fixing, phosphate solubilizing and compost making biofertilizers.

Question 42.
What are the eight different types of mycorrhiza as per recent classification?
Answer:
Nowadays, mycorrhiza are classified into 8 different types – ectomycorrhizae, endomycorrhizae, ectendomycorrhizae, orchidaceous mycorrhizae, ericoid mycorrhizae, arbutoid mycorrhizae, monotrapoid mycorrhizae and ophioglossoid mycorrhizae.

Question 43.
What is heterocyst?
Answer:
Cynobacteria possess specialized colourless cells called heterocysts which are the sites of nitrogen fixation.

Question 44.
Give the role of heterocyst.
OR
Give the importance of heterocyst in cyanobacteria.
Answer:
The heterocysts are the sites of nitrogen fixation in cyanobacteria.

Question 45.
Who discovered mycorrhizae?
Answer:
Mycorrhizae were discovered by Frank (1885).

Give definitions of the following

Question 1.
Food
Answer:
Food is defined as solid or liquid substance, which is swallowed, digested and assimilated in the body, keeping us well.

Question 2.
Plant breeding
Answer:
Plant breeding is the improvement or purposeful manipulation in the heredity of crops and the production of new superior varieties of existing crop plants.

Question 3.
Germplasm collection
Answer:
Germplasm collection is the entire collection having all the diverse alleles for all genes in a given crop.

Question 4.
Mutation
Answer:
Mutation is defined as the sudden heritable change in the genotype, which is caused naturally.

Question 5.
Tissue culture
Answer:
Tissue culture is growing isolated cells, tissues, organs ‘in vitro’ on a solid or liquid nutrient medium, under aseptic and controlled conditions of light, humidity and temperature, for achieving various objectives.

Question 6.
Explant
Answer:
The part of plant used in tissue culture is known as explant.

Question 7.
Totipotency
Answer:
An inherent ability of living plant cell to grow, divide, redivide and give rise to a whole plant is known as totipotency.

Question 8.
Callus
Answer:
Callus is defined as a mass of undifferentiated cells, formed by division and redivision of the cells of explant.

Question 9.
Single cell protein
Answer:
Single cell protein is defined as a crude or a refined edible protein, extracted from pure microbial cultures or from dead or dried cell biomass.

Question 10.
Biofortiflcation
Answer:
Biofortification is a method of developing crops having higher quantity and quality of vitamins, minerals and fats, to overcome problem of malnutrition.

Question 11.
Animal husbandry
Answer:
Animal husbandry is an agricultural practice of breeding and raising livestock.

Question 12.
Inbreeding
Answer:
Breeding of closely related individuals for 4 to 6 generations is known as inbreeding.

Question 13.
Outbreeding
Answer:
Breeding of unrelated animals either of the same breed but having no common ancestors for 4 to 6 generations (outcrossing) or of different breeds (crossbreeding) or even of different species (interspecific hybridization), is known as outbreeding.

Question 14.
Outcrossing
Answer:
Breeding of animals of the same breed but having no common ancestors for 4 to 6 generations is known as outcrossing.

Question 15.
Crossbreeding
Answer:
Breeding of superior male of one breed with superior female of another breed is known as crossbreeding.

Question 16.
Interspecific hybridization
Answer:
Breeding of animals of two different but related species is known as interspecific hybridization.

Question 17.
Apiculture
Answer:
Apiculture is an artificial rearing of honey bees to obtain bee products like honey, wax, pollens, bee venom, propolis (bee glue) and royal jelly and to use honey bees as pollinating agents for crop plants.

Question 18.
Antibiotics
Answer:
Antibiotics are organic substances produced in small amounts by certain microbes to kill or inhibit the growth of other microbes.

Question 19.
Biotechnology
Answer:
Biotechnology is defined as the applications of scientific and engineering principles for the processing of materials by biological agents to provide goods and service to humans or for human welfare.

Question 20.
Biocontrol or biological control:
Answer:
Biocontrol is the natural method of eliminating and controlling insects, pests and other disease-causing agents by using their natural, biological enemies.

Question 21.
Biocontrol agents
Answer:
Biocontrol agents are the organisms like insects, bacteria, fungi, viruses and protozoans which are employed for biocontrol.

Question 22.
Fertilizers
Answer:
Fertilizers Eire the nutrients necessary for plant growth and which increase the productivity of cultivated plants.

Question 23.
Biofertilizers
Answer:
Biofertilizers are commercial preparation of ready-to-use live bacterial, cyanobacterial (mostly N₂ fixing) or fungal formulations which enhance the nutrient quality of soil.

Name the Following

Question 1.
Hybrid wheat varieties in India.
Answer:
Sonalika and Kalyan Sona

Question 2.
Semi-dwarf rice varieties in India.
Answer:
Jaya, Padma and Ratna

Question 3.
Sugar cane varieties developed at Coimbatore, Tamil Nadu.
Answer:
CO-419, 421, 453

Question 4.
Hybrid varieties of millets developed in India.
Answer:
Ganga-3 (maize), CO-12 (Jowar), Niphad (Bajra)

Question 5.
Fungal disease of plants.
Answer:
Brown rust of wheat, Red rot of sugar cane, Late blight of potato

Question 6.
Bacterial disease of plants.
Answer:
Black rot of crucifers

Question 7.
Viral disease of plants.
Answer:
Tobacco mosaic disease

Question 8.
Mutant variety of rice.
Answer:
Jagannath

Question 9.
Mutant variety of wheat.
Answer:
NP 836 (rust resistant)

Question 10.
Mutant variety of cotton.
Answer:
Indore-2 (resistant to bollworm)

Question 11.
Mutant variety of cabbage.
Answer:
Regina-II

Question 12.
The most preferred tissue culture medium.
Answer:
MS (Murashige and Skoog) medium.

Question 13.
High yielding varieties of banana used in Maharashtra.
Answer:
Shrimati, Basarai, G-9

Question 14.
The fungi used for the production of SCP.
Answer:
Aspergillus niger, Trichoderma viride, Saccharomyces cerevisiae, Candida utilis.

Question 15.
Algae used for the production of SCR
Answer:
Spirulina spp, Chlorella pyrenoidosa

Question 16.
Bacteria used for the production of SCR
Answer:
Methylophilus methylotrophus, Bacillus megasterium

Question 17.
A new breed of sheep developed from crossing of Bikaneri ewe and Marino rams in Punjab.
Answer:
Hisardale

Question 18.
Indian breeds of cows.
Answer:
Sahiwal, Sindhi, Gir

Question 19.
Exotic breeds of cows.
Answer:
Jersey, Brown Swiss, Holstein

Question 20.
Breeds of bufaaloes in India.
Answer:
Jaffarabadi, Mehsana, Murrah, Nagpuri, Nili, Surati.

Question 21.
Common breeds of cattle and poultry in the farms, found in your area.
Answer:
Khillari, Gir (breeds of cows), Aseel, Kadaknath (poultry)

Question 22.
American poultry breeds.
Answer:
Plymouth Rock, New Hampshire, Rhode Island Red

Question 23.
Asiatic poultry breeds.
Answer:
Brahma, Cochin and Langshan

Question 24.
Mediterranean poultry breeds.
Answer:
Leghorn, Minorca

Question 25.
English poultry breed.
Answer:
Australorp

Question 26.
Indian poultry breeds.
Answer:
Chittagong, Aseel, Brahma and Kadaknath.

Question 27.
Best layer.
Answer:
Leghorn

Question 28.
Best broilers.
Answer:
Plymouth rock, Rhode Island Red, Aseel, Brahma and Kadaknath

Question 29.
Viral diseases of poultry.
Answer:
Ranikhet, Bronchitis, Avian influenza (bird flu), Bird flu

Question 30.
Bacterial diseases of poultry.
Answer:
Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis.

Question 31.
Fungal diseases of poultry.
Answer:
Aspergillosis, Favus and Thrush.

Question 32.
Parasitic diseases of poultry.
Answer:
Lice infection, roundworm, caecal worm infections, etc.

Question 33.
Protozoan diseases of poultry.
Answer:
Coccidiosis

Question 34.
Domesticated species of bees.
Answer:
Apis mellifera, Apis indica

Question 35.
Lac insect.
Answer:
Trachardia lacca

Question 36.
Silk moth.
Answer:
Bombyx mori

Question 37.
The common fresh water fish.
Answer:
Labeo rohita (rohu), Catla (catla), Cirrihanus mrigala (mrigala) and other carps.

Question 38.
The common marine fish.
Answer:
Harpadon (Bombay duck), Sardinella (sardine), Rastrelliger (mackerel) and Stromateus (pomphret).

Question 39.
Estuaries found in Maharashtra and where these estuaries are located.
Answer:
Thane creek, Manori creek, Rajapuri creek, Kalbadevi Estuary in Ratnagiri, Damanganga estuary and Narmada estuary

Question 40.
Different fish found at an estuary.
Answer:
Clupeids, mullets, catfishes, perches, Mugil cephalus gar fishes, halfbeaks, eels, flatfishes, sharks, rays, oysters and migratory fishes include Hilsa ilisha, Polynemus spp., Pampana, Tachysurus spp, Pangasius spp., etc.

Question 41.
Best Silk.
Answer:
Mulberry silk

Question 42.
Silk of inferior quality.
Answer:
Tussar silk, Eri silk

Question 43.
Distilled alcoholic beverages.
Answer:
Whisky, brandy, rum

Question 44.
Traditional drink made by fermenting the sugar sap extracted from palm plants and coconut palm.
Answer:
Toddy

Question 45.
The wine of Goa, made by fermenting fleshy pedicels of cashew fruits.
Answer:
Fenny

Question 46.
The microbes used in fermentation of dhokla.
Answer:
Lactobacilli

Question 47.
Microbes used in the production of vitamin B₂.
Answer:
Neurospora gossypii, Eremothecium ashbyi

Question 48.
Microbes used in the production of vitamin B₁₂.
Answer:
Pseudomonas denitrtficans

Question 49.
Microbes used in the production of vitamin C.
Answer:
Aspergillus niger

Question 50.
Symbiotic N₂ fixing microorganisms.
Answer:
Rhizobium, Anabaena, Frankia.

Question 51.
Free-living or Non-Symbiotic N₂ fixing microorganisms.
Answer:
Azotobacter, Nostoc, Clostridium, Beijerinkia, Klebsiella, etc.

Question 52.
Phosphate solubilizing biofertilizers.
Answer:
Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Aspergillus spp., etc.

Question 53.
Nitrogen fixing cyanobacteria.
Answer:
Anabaena, Nostoc, Plectonema, Oscillatoria.

Question 54.
Cyanobacteria associated with lichens.
Answer:
Anabaena, Nostoc and Tolypothrix

Question 55.
Cyanobacteria associated with plants like Azolla and Cycas.
Answer:
Anabaena

Question 56.
The aquatic fern commonly used in paddy field as a biofertilizer.
Answer:
Azolla.
Give functions or significance

Question 1.
Hybridization
Answer:

Hybrdization is an effective means of combining together the desirable characters of two or more varieties.
New genetic combinations of already existing characters and new genetic variations can be created by hybridization.
Hybridization exploits and utilizes hybrid- vigour.

Question 2.
Food
Answer:

Food is organic, energy rich, non-poisonous, edible and nourishing substance.
It provides nutrients for growth and development of body.
It provides energy for metabolic reactions.
It keeps us alive, strong and healthy.

Question 3.
Antibiotics
Answer:

Antibiotics are secondary metabolites of therapeutic importance, produced in small amounts by certain microbes like bacteria, fungi and a few algae.
They inhibit growth of other microbial pathogens like fungi and bacteria.
Thus, they are antifungal and antibacterial in nature.
Antibiotics are used in treatment of deadly diseases like plague, whooping cough, diphtheria, leprosy, etc.

Distinguish between the following

Question 1.
Callus culture and Suspension culture.
Answer:

Callus Culture	Suspension culture
1. Solid nutrient medium is used in callus culture.	1. Liquid nutrient medium is used in suspension culture.
2. No shaker or agitator is needed.	2. Shaker or agitator is needed.
3. Cells of explants divide and redivide to form callus.	3. Callus is not formed.
4. Callus a mass of undifferentiated cells.	4. Suspension culture consists of single isolated cells or small groups of cells.
5. It shows slow growth.	5. It shows faster growth.

Question 2.
Outcrossing and Crossbreeding.
Answer:

Outcrossing	Crossbreeding
1. Breeding of animals of the same breed but having no common ancestors for 4 to 6 generations is known as outcrossing.	1. Breeding of superior male of one breed with superior female of another breed is known as crossbreeding.
2. Progeny is known as outeross.	2. Progeny is known as hybrid.
3. New breeds are not developed by outcrossing.	3. New breeds or hybrids are formed by crossbreeding.
4. An outcross helps to overcome inbreeding depression.	4. Hybrids are subjected to inbreeding and new stable breeds are developed by crossbreeding.

Question 3.
Inorganic fertilizers and Organic fertilizers / biofertilizers.
Answer:

Inorganic fertilizers	Organic fertilizers / biofertilizers
1. They are non-renewable nutritional resources.	1. They are renewable nutritional resources.
2. Inorganic fertilizers are synthetic and are in the form of chemicals.	2. They are biological in origin.
3. They are mixtures of mineral salts of NPK in definite proportions.	3. Organic fertilizers are farmyard manure, green manure and compost. Whereas, bio fertilizers are live bacterial, cyanobacterial (mostly N<sub>2</sub> fixing) or fungal formulations which enhance the nutrient quality of soil.
4. Excessive use of inorganic fertilizers results in pollution of soil, groundwater and air.	4. They do not cause pollution.
5. They are not part of sustainable agriculture.	5. They are part of organic farming and sustainable agriculture.

Question 4.
Ectomycorrhizae and Endomycorrhizae.
Answer:

Ectomycorrhizae	Endomycorrhizae
1. Mycelium of ectomycorrhizal fungi form a sheath called mantle on the surface of roots.	1. Most of the endomycorrhizal hyphae penetrate in the root cortex.
2. Few hyphae form hartig-net in the intercellular spaces of root cortex.	2. Fungal hyphae do not produce hartig-net.
3. Arbuscles and vesicles are not formed.	3. Arbuscles and vesicles are formed.

Give reasons

Question 1.
Why are honey bees called as best pollinators?
Answer:

About 80% of insect pollination is carried by honey bees.
They pollinate various crops like sunflower, mustard, safflower, chilly, cabbage, cucumber, legumes, fruits like apple, mango, citrus, etc.
They increase the productivity of crops.
Hence, honey bees are important pollinators.

Question 2.
Regular visit of veterinary doctor to dairy farm is mandatory.
Answer:
Daily visit of veterinary doctor to dairy farm is mandatory to diagnose health problems, diseases and for their rectification.

Question 3.
Apis mellifera and Apis indica are known as domesticated species.
Answer:

Apis mellifera and Apis indica are suitable for apiculture.
Hence, they are known as domesticated species.

Question 4.
Cages of silkworm larvae must be well managed and protected.
Answer:

Silkworm larvae may get infected by protozoans, viruses and fungi.
Ants, crows, other birds and predators may attack these larvae.
Hence, cages of silkworm larvae must be well managed and protected.

Question 5.
Buttermilk is used in the dough of dhokla.
Answer:

Buttermilk contains the lactobacilli.
These lactobacilli bring about the fermentation of gram flour.
CO₂ produced during fermentation increases the volume of the dough.
It escapes during the steaming of dough, making dhokla porous and spongy.
Hence, buttermilk is used in the dough of dhokla.

Question 6.
We include mushrooms in our diet.
Answer:

Mushrooms are directly used as food.
They produce large, fleshy fruiting bodies which are edible.
They are low calorie, sugar free, fat free, but rich in proteins, vitamins, minerals and amino acids.
Hence, we include mushrooms in our diet.

Question 7.
Vitamins are to be consumed through food or tablets or capsules.
Answer:

Vitamins B and K are produced in the body.
But some vitamins like C, D and E are not produced in the body.
If our body does not get these vitamins in the required quantity, then the deficiency of these vitamins may result in various diseases.
Therefore, vitamins are to be consumed through food or tablets or capsules.

Question 8.
Enzymes are essential for survival of living organisms.
Answer:

Enzymes are proteins which act as biocatalysts.
They speed up the chemical reactions without undergoing any change themselves.
They catalyze reactions more quickly and efficiently at body temperature.
Thus they play key role in metabolic reactions.
Hence, enzymes are essential for survival of living organisms.

Question 9.
Biogas plants are more often built in rural areas.
Answer:

Biogas is a non-conventional and renewable source of energy obtained by microbial fermentation.
Cattle dung (the main substrate), domestic wastes, agricultural waste, agro industrial wastes, forestry wastes, etc. are utilized as substrates for production of biogas.
Biogas is eco-friendly and does not cause pollution, can be used as domestic fuel.
As the raw material for its production is readily available, the biogas plants are more often built in rural areas.

Question 10.
Why are healthy root nodules pink in colour?
Answer:

Rhizobium has symbiotic relationship with roots of leguminous plants.
It infects root cortex and form root nodules.
Root nodules are the site of nitrogen fixation.
Enzyme nitrogenase which catalyzes nitrogen fixation, gets inhibited by oxygen.
But root nodule contain a pigment called leghaemoglobin which acts as oxygen scavanger and protects nitrogenase from getting inhibited.
Leghaemoglobin is pink in colour.
Hence, healthy root nodules are pink in colour.

Give Short Notes

Question 1.
Callus culture.
Answer:

In callus culture, nutrient medium is solidified using agar-agar is used.
Shaker or agitator is not required.
Sterilized explant is placed on solid nutritive medium.
The cells of explants absorb nutrients and start multiplying.
This results in the formation of callus.
Callus is a mass of undifferentiated cells, formed by division of the cells of explants.
Growth hormones, auxins and cytokinins are provided to callus in specific proportion to induce formation of organs.
If auxins are in more quantity, roots are formed (rhizogenesis) and if the cytokinins are in more quantity, shoot formation takes place (caulogenesis).
Thus new plantlets are formed.
Callus culture required subculturing to ensure its continuous growth.

Question 2.
Suspension culture.
Answer:

In suspension culture, small groups of cells or a single cell are transferred to liquid nutritive medium as explants.
The liquid medium is constantly agitated by using shakers (agitators).
The agitation serves the purpose of aeration, mixing of medium and prevents the aggregation of cells.
Generally the suspension culture shows a high proportion of single isolated cells and small clumps of cells.
Suspension culture grows much faster than callus culture.
Suspension culture is used for cell biomass production which can be utilized for biochemical isolation, regeneration of new plants, etc.

Question 3.
Applications of micropropagation.
Answer:

Micropropagation involves in rapid multiplication of genetically similar plants (clones).
A large number of plantlets are obtained within a short period and in a small space.
Plants are obtained throughout the year, under controlled conditions, independent of seasons.
As micropropagation results in the formation of clones, desirable characters (genotype and sex) of superior variety can be maintained for many generations.
The rare plant and endangered species are multiplied and conserved using this technique.
With the help of somatic hybrids (cybrids), new variety can be obtained in short time span.
Micropropagation is involved in commercial production of ornamental plants like v orchids, Chrysanthemum, Eucalyptus, etc. and fruit plants like banana, grapes, Citrus, etc.

Question 4.
Applications of tissue culture.
Answer:
Applications of tissue culture are as follows:

Production of healthy plants from diseased plants using apical meristems as explants.
Production of stress resistant plants.
Production of haploid plantlets by pollen culture.
Production of secondary metabolites such as alkaloids, enzymes, hormones, etc.
Multiplication of rare and endangered plants.
Production of somaclonal variants.
Use of micropropagation techniques to produce large number of genetically identical plants.
Protoplast culture
Tissue culture has applications in forestry, agriculture, horticulture, genetic engineering and physiology.

Question 5.
Single cell protein (SCP).
Answer:

Single-cell protein is a crude or a refined edible protein, extracted from pure microbial cultures or from dead or dried cell biomass.
Microorganisms like algae, fungi, yeast and bacteria with high protein content in their biomass, are grown using waste and inexpensive substrates.
Substrates used for growing microbial biomass are wood shavings, sawdust, corn cobs, paraffin, N-alkanes, sugar cane molasses, even human and animal wastes.
SCP is rich in proteins, vitamins, vitamin B complex, minerals and fats.
It can be used as fodder for achieving fattening of calves, pigs, in breeding fish and even in poultry and cattle farmimg.
Fungi like Aspergillus niger, Trichoderma viride, Saccharomyces cerevisiae, Candida utilis, algae like Spirulina spp, Chlorella pyrenoidosa, bacteria like Methylophilus, methylotrophus and Bacillus megasterium are used for the production of SCR

Question 6.
Advantages of Single Cell Protein.
Answer:

Microbes that are used as SCP have very high protein contents in their biomass – 43% to 85% (W/W basis).
SCP is a good source of vitamins, vitamin B complex, fats, amino acids, minerals, crude fibres, etc.
As microorganisms multiply fast, large quantity of biomass can be produced in a short duration.
The microbes can be grown using waste materials and inexpensive substrates. This decreases pollution.
The microbes can be genetically modified to vary the amino acid composition.
SCPs can be used as fodder for achieving fattening of calves, pigs, in breeding fish, poultry and cattle farming.

Question 7.
Inbreeding.
Answer:

Inbreeding is the mating of two closely related individuals of the same breed for 4 to 6 generations.
During inbreeding superior males and superior females of the same breed are identified. The superior males and superior females from this progeny are identified and used for further mating.
Due to inbreeding, homozygosity is increased and harmful recessive genes are exposed. Inbreeding is done when a pure line of an animal is expected.
Inbreeding helps in accumulation of superior genes and elimination of harmful or less desirable genes.
Continued inbreeding usually reduces fertility and productivity. This is called inbreeding depression.

Question 8.
Fish farming.
Answer:
Fish farming is the practice of culturing the edible and commercially important fish species in the ponds, lakes or reservoirs. Fish farming helps in boosting the productivity and the economy of the nation.

For maintaining a fish farm, following aspects are taken into consideration:

selection of the site
excavation of the pond
managing hatchery
nursery
looking after rearing ponds and
stocking ponds besides managing the water source, supplying fertilizer and supplementary feed, etc.

Question 9.
Microbes in industrial vitamin production.
Answer:
(1) Vitamins are nitrogenous organic compounds, required in minute quantities for normal growth and development of the body.

(2) The microbes are involved in the industrial production of vitamins like thiamine (vitamin B₁, riboflavin (vitamin B₂), pyridoxine, folic acid, pantothenic acid, biotin, vitamin B₁₂, ascorbic acid (Vitamin C), beta-carotene (provitamin A) and ergosterol (provitamin D).

(3) Examples of some vitamins produced by fermetaion using different microbial sources are-

Vitamin B₂ – Neurospora gossypii, Eremothecium ashbyi
Vitamin B₁₂ – Pseudomonas denitrijicans
Vitamin C – Aspergillus niger

Question 10.
Industrial uses of enzymes.
Answer:

Enzymes are used to improve the quality of the fabrics in the textile industry.
In the pulp and paper industry, they are involved in biomechanical pulping and bleaching.
In food industry, enzymes are used in the fermentation processes to produce bread and alcoholic beverages such as wine and beer.
They are used in the extraction of carotenoids and olive oil.
Enzymes are also used in cosmetics, animal feed and agricultural industries.
Lipases are used to remove oil stains and to increase the brightness in detergent industry. They have superior cleaning properties.
Streptokinase is used as a ‘clot buster’ for clearing blood clots in the blood vessels in heart patients.

Question 11.
Microorganisms in sewage.
Answer:

Sewage contains pathogenic bacteria, viruses, fungi, protozoa which cause dysentery, cholera, typhoid, polio and infectious hepatitis, etc. It also contains nematodes and algae.
Their number and type of microorganisms in sewage depends on the composition and source of sewage.
Millions of bacteria per ml may be present in raw untreated sewage.
Bacteria in sewage include coliforms, fecal Streptococci, anaerobic spore forming bacilli and other bacteria found in the intestinal tract of humans.
The sewage also contains soil bacteria.
During sewage decomposition, initially aerobic and facultative anaerobic organisms predominate which are followed by strict anaerobic especially methogenic bacteria that produce methane (CH₄) and CO₂.

Question 12.
Rhizobia as a biofertilizer
Answer:

Rhizobia [Singular – Rhizobium) are rod¬shaped, motile, aerobic, gram negative, non- spore forming, nitrogen-fixing bacteria.
They contain Nod genes and Nif genes.
They live in symbiotic association with leguminous roots.
They form nodule on the roots of leguminous plants and multiply inside the nodule.
Nodules are the site of nitrogen fixation. They are pink in colour and contain enzyme nitrogenase and oxygen scavenger leghaemoglobin.
Rhizobia fix atmospheric nitrogen into organic forms, which can be used by plants as nutrients. The host plant in return provides food and water to the bacterium (Rhizobia).
Rhizobia are host specific e.g. leguminosarum is specific to pea R. phaseoli is specific to beans.
In a laboratory, pure cultures of specific Rhizobium species are raised which are used for the preparation of biofertilizer.

Question 13.
Azotobacter as a biofertilizer
Answer:

Azotobacter is a free living, nitrogen fixing, aerobic, non-photosynthetic, non-nodule forming bacterium.
It is associated with roots of grasses and certain plants.
It is used as a biofertilizer for non-leguminous plants like rice, cotton, vegetables, etc.

Question 14.
Azospirillumas as a biofertilizer
Answer:

Azospirillum acts as a biofertilizer for non- leguminous plants like cereals, millets, cotton, oilseed, etc.
It is a free living, aerobic nitrogen fixing bacterium associated with roots of corn, wheat and jo war.
It fixes the nitrogen (20-40 kg N/ha).

Question 15.
Anabaena as a biofertilizer
Answer:

Anabaena is a cyanobacterial biofertilizer.
It is multicellular, filamentous nitrogen fixing organism that exits as plankton.
It can fix nitrogen both in free living conditions as well as by forming symbiotic associations.
Anabaena forms symbiotic relationship with Azolla, Anthoceros and Cycas.
It is found in dorsal leaf lobe in Azolla, thallus of Anthoceros and in coralloid roots of Cycas.
It has specialized colourless cells known as heterocysts.
Heterocysts are the sites for nitrogen fixation.

Question 16.
Benefits of Biofertilizers.
Answer:

Biofertilizers increase soil fertility.
They are low cost and can be used by marginal farmers.
They do not cause pollution.
BGA secret growth promoting substances, organic acids, proteins and vitamins.
Azotobacter supplies nitrogen and antibiotics in the soil.
Use of biofertilizers improves physico-chemical properties of soil-like texture, structure, pH, water holding capacity of soil by providing nutrients and organic matter.
The use of chemical fertilizers gets reduced and the pollution also becomes less.

Short Answer Questions

Question 1.
What are the objectives of plant breeding?
Answer:
Objectives of plant breeding are as follows:

To increase crop yield,
To improve quality of produce.
To increase tolerance to environmental stresses.
To develop varieties of plants resistant to pathogens and insect pest.
To alter the life span.

Question 2.
What are the different types of hybridization in plants?
Answer:
The different types of hybridization in plants are as follows:

Intravarietal : It is the hybridization between plants of same variety.
Intervarietal : It is the hybridization between two varieties of the same species.
Interspecific : It is the hybridization between two species of the same genus.
Intergeneric : It is the hybridization between two genera of the same family.
Wide/distant crosses : These are the crosses between distantly related parental plants.

Question 3.
How aseptic conditions are maintained in tissue culture?
Answer:

Glassware is sterilized by using detergents and hot air oven.
Nutrient medium is autoclaved under constant pressure of 15 lb/sq inch, continuously for 20 minutes to sterilize it.
Explant is treated with 20% ethyl alcohol and 0.1% HgCl₂.
Sterilization of inoculation chamber (Laminar air flow) is done using UV ray tube for 1 hour before actual inoculation of explant on the sterilized nutrient medium.

Question 4.
What are the objectives of biofortification ?
Answer:
Objectives of biofortification are as follows:

Improvement in protein content and quality.
Improvement in oil content and quality.
Improvement in vitamin content.
Improvement in micronutrient content and quality.
To overcome the problem of malnutrition.

Question 5.
What are the objectives of animal breeding?
Answer:
Objectives of animal breeding are as follows:

To increase the yield of animals.
To improve the production of milk, quality of milk product, quality of meat or maximum yield of eggs per year, etc.
To develop breeds with desirable characters.

Question 6.
What is artificial insemination? What are its advantages?
Answer:

Artificial insemination is the technique used for controlled breeding experiments.
Superior males of a particular commercial breed are selected.
Semen from such superior males is collected and injected into the genital tract of female.
This insemination is either done immediately or semen is frozen and used later on.
In frozen semen, sperms can remain alive for long duration. They are also convenient for transport.
Artificial insemination is preferred as it is easy and helpful to overcome several problems of normal mating.

Question 7.
What is Multiple Ovulation Embryo Transfer (MOET) technology? Where is it used?
OR
Explain the technique of multiple ovulation embryo transfer (MOET) in animal breeding.
Answer:

Multiple Ovulation Embryo Transfer (MOET) technology is used to increase chances of successful production of hybrids.
In this technique, cow is administered with Follicle Stimulating Hormone (FSH) which induces follicular maturation and then the super ovulation is brought about.
By such technique in each cycle, 6 to 8 eggs mature simultaneously.
The cow is then either mated with a superior bull or she is artificially inseminated.
The blastocysts at 8 to 32 cell stage are recovered non-surgically and transferred to surrogate mothers.
The genetic mother who gave the egg is then again subjected to another round of super ovulation.
This technology is used for cattle, sheep, rabbits, buffaloes, etc.
The MOET is used to produce high milk yielding breeds of female and high quality meat yielding bulls with lean meat containing less lipids. It helps in increasing favourable herd size in a short period.

Question 8.
As a dairy owner what measures will you adopt to improve the quality of milk?
Answer:

In order to improve the quality of milk, following measures should be taken at every stage of dairy farming:
Good breeds having high yielding potential should be selected.
The breeds selected should be suitable for the local climatic conditions.
The breeds selected should have proper resistance to diseases.
Cattles should be well looked after with proper care.
The feed should be of suitable quality and quantity. Feed includes silage made from legumes, grasses, maize and jowar. Silage should be supplemented with oilcakes, minerals, vitamins and salts.
Utmost care should be taken about cleanliness and hygiene of the cattle as well as the handlers who handle the cattle.
This is especially important during milking, storage and transport of milk and its products. Mechanized processes should be adopted as far as possible as they reduce chance of direct contact of produce with the handlers.
The shed must be clean and well maintained. Similarly the dairy should be spacious with adequate facilities for feeding, watering and light.
Help of veterinary doctor should be sought from time to time for the identification of health problems, diseases and rectification.
Transportation of milk, processing, marketing and distribution play a vital role in dairy industry. If all the above care is taken then the quality of milk will surely improve.

Question 9.
Explain in short the poultry management.
Answer:
For the management of poultry, following aspects are to be taken care of:

Selection of proper and disease free breed, suitable and safe farm conditions.
Proper feeding practice and the quality of feed and water.
Hygiene and health care of the birds.
Management of layers is done by selecting high yielding chicken. Their farms are kept clean, dry and well ventilated. They are given proper feed at proper times. Other aspects such as debeaking, etc. are also taken care of.
In the farm, importance is given to infrastructure such as proper and adequate lighting, placing waterer at places, looking after sanitation, culling and vaccination.
Management of broiler similarly includes selection of breed, housing, temperature, ventilation, lighting, observing the floor space and broiler feed.

Question 10.
What is apiculture? What is its importance?
Answer:

Apiculture is artificial rearing of honey bees to obtain various bee products.
Various products such as honey, wax, pollen, bee venom, propolis (bee glue) royal jelly, etc. are obtained from this cottage industry. Honey is an important item as ayurvedic medicine and in food due to its nutritional value.
Bees also help in the cross pollination of various crops. Hence in pastures, wild shrubs, fruit orchards and cultivated crops, honeybees play an important role as pollinators.
Bee keeping in crop fields increases the productivity of both honey and the crop.
Apiculture itself is a means for employment for rural youth. It is an age old cottage industry which can be done along with agriculture.

Question 11.
What are the requirements of bee keeping?
Answer:

Bee keeping requires bee hive boxes which consist of comb foundation sheets.
In addition, the bee veil, smoker, bee brush, gloves, gumshoes, uncapping knives, swarm net, queen excluder, overall hive tool, etc. are required.
Bee keeping requires familiarity with the habits of bees, selection of suitable location, catching and hiving of swarms, management of hives during different seasons, handling and collection of honey, bee wax and other products.
Successful bee keeping also requires periodic inspection for cleanliness of hive boxes, activity of bees and queen, condition of brood, provision of water.

Question 12.
What are the main divisions of fishery?
Answer:

Fishery can be capture fishery and culture fishery.
Three main divisions of capture fishery are : Inland fishery, estuarine fishery and marine fishery.
Inland fishery : It is culturing and capturing of from fresh water bodies. It is carried out on about 40 to 50 lakh acres of fresh water bodies such as rivers, ponds, lakes and dams.
Marine fishery : It includes capturing fish from sea water. India has a coastline of about 7500 km.
Estuarine fishery : It includes capture of fish from estuaries.
Culture fishery is either of polyculture or of monoculture type. In polyculture, different species are cultured simultaneously at the same time in the same pond. In monoculture, only a single species is cultured.

Question 13.
Can “fishery” be a sustainable job option?
Answer:

Fishery can never be a job. It is a livelihood or can be an occupation.
Fish is a renewable resource. If managed properly, it can be a sustainable resource.
The sustainability is dependent upon the availability of fishes and other edible organisms. But due to climate change, pollution and overexploitation of fish resources, these are depleting rapidly, it is estimated by the scientists that by 2050, no fish will be left in the seas.
However, if aquaculture is done to raise fishes, partly it can be sustaining. But there . are many environmental regulations that can hamper the business of aquaculture.

Question 14.
Describe sericulture in brief.
Answer:

Sericulture is the practice of rearing silkworms for the production of silk.
The silkworm (Bombyx mort) is reared for obtaining best quality of silk called mulberry silk. Tussar silk and Eri silk are other varieties of silk which are inferior to the mulberry silk.
Larvae of silkworm are fed on the mulberry leaves. Quality and quantity of silk depends on the quality of mulberry leaves.
These larvae are reared, developed and well looked after by the skilful labour keeping a constant watch.
Silkworm larvae may be infected by protozoans, viruses and fungi. Ants, crows, birds, and other predators are ready to attack these insects, hence the cages of these larvae must be managed to prevent predators attack.
Silk is obtained from the cocoon of the silkworm.
Sericulture is an age old practice and can be started with low investment and small space. It requires scientific knowledge and skill. Disabled, older and handicapped people also can practise it.

Question 15.
What are the different stages found in the life cycle of silkworm?
Answer:

Stages of development in the life cycle of silkworm are egg, larva, pupa and adult.
The larva is the silkworm caterpillar.
The adult (imago) stage is the silkworm moth.

Question 16.
Describe the process of cocoon formation.
Answer:

The eggs of silkworm hatch into larvae.
The larvae develop into a caterpillar.
Caterpillar feeds on fresh mulberry leaves.
After its growth and moulting, the silkworm secretes a silk fibre to form cocoon.
The silk is a continuous filament comprising fibroin protein, secreted from salivary glands of silkworm and a gum called sericin, which cements the filaments.
The silk solidifies when it contacts the air.
The silkworm spins approximately one mile of filament and completely encloses itself in a cocoon in about two or three days.

Question 17.
Which process is involved in silk production from cocoon?
Answer:

The silk is a continuous filament comprising fibroin protein, secreted from salivary glands of silkworm and a gum ailed sericin.
To remove the sericin, which cements,the filaments, cocoons are placed in hot water.
It frees the silk filaments and readies them for reeling. This is known as the degumming process.
The sillworm pupa gets killed in hot water.
Single filaments are combined to form thread.
The threads are plied to form yarn.
After drying, the raw silk is packed according to quality.

Question 18.
Give the economic importance of lac.
OR
State the economic importance of ‘lac culture’.
Answer:
Lac is used for the following purposes:

For making bangles.
For making different types of toys.
It is used in wood works.
Polish is made from lac.
Inks can be prepared from lac.
Lac is largely used for silvering mirrors.

Question 19.
Tribal people from India have a great contribution in production of lac. But it needs certain trees. Name at least two such trees which give food yield of lac. How is lac purified?
Answer:

Like her, peepal, palas, kusum, babul, etc. are used for feeding lac insects during the practice of lac culture.
Natural lac is contaminated and hence it is washed and purified. This helps to obtain shellac in pure form.

Question 20.
Give an account of alcoholic beverages.
Answer:

Alcoholic beverages are the products of alcoholic fermentation of particular substrates.
Tubular tower fermenters are used to produce alcoholic beverages on a large scale.
Beer is produced from barley by fermentation. For the production of beer, strains of Saccharomyces cerevisiae are used.
Wine is prepared from grapes.
Whisky is prepared by fermenting mixed grains of wheat, barley and corn followed by the distillation of the products of fermentation.
Liquors like beer, wine are produced without distillation.
Whisky, rum and brandy are distilled alcoholic beverages.
Toddy is prepared by fermenting the sugar sap extracted from palms and coconut palms.
Fenny is fermented by fleshy pedicels of cashew fruits.

Question 21.
Industrial production of which substances involves fermentation by microbes? What is the nature of these substances and which factors determine the synthesis of specific products?
Answer:

During fermentation of substrates, various useful products like alcoholic beverages, organic acids, vitamins, growth hormones, enzymes, antibiotics and other molecules of medical significance are produced.
They are secondary metabolites produced during idio phase and are not required for their growth.
A specific secondary metabolite is produced depending on the type of microorganism and the type of substrate.

Question 22.
Can antibiotics kill viruses?
Answer:

Antibiotics work by targeting cell wall or other metabolic pathways in bacteria.
But viruses do not have cell walls and they do not carry out any metabolic reaction when outside the host.
Hence, antibiotics cannot kill viruses.

Question 23.
What are gibberellins? Give the applications of gibberellins.
Answer:
Gibberellins are growth hormones produced by higher plants and fungi.

Applications of gibberellins are as follows:

Gibberellins induce parthenocarpy in fruits like pear and apple.
Gibberellins promote growth by stem elongation.
They break the dormancy of seeds.
They induce flowering in long day plants in short day conditions.
They are used to enlarge the size of grapes.

Question 24.
What is sewage? Describe its composition.
Answer:

Sewage is the waste matter carried off on drainage.
Composition of sewage varies depending upon its industrial source, e.g. textile, chemicals, pharmaceuticals, dairy, canning, brewing, meat packing, tannery, oil refineries and meat industries, etc.
Sewage consists of human excreta, animal dung, household waste, slaughter house waste, dissolved organic matter, algae, nematodes, pathogenic bacteria, viruses and protozoa, discharged waste water from hospitals, industries (contains toxic dissolved organic and inorganic chemicals), tannery and pharmaceutical waste, etc.
Bacteria in sewage include coliforms, fecal Streptococci, anaerobic spore forming bacilli and other bacteria found in human intestinal tract.
Sewage consists of about water (99.5% to 99.9%) and inorganic and organic matter (0.1 to 0.5%) in suspended and soluble form.

Question 25.
State whether BOD will be high or low
(a) in water after primary treatment
(b) in water after secondary treatment.
Answer:
(a) After primary treatment, the primary effluent present in the primary sedimentation tank, still contains large amount of dissolved organic matter. Hence, the BOD is high.

(b) Secondary treatment is a biological treatment. Primary effluent is passed into aeration tank, where aerobic bacteria and fungi form flocks and consume the major part of organic matter in the effluents. Hence, the BOD is reduced.

Question 26.
Enlist the advantages of biogas.
Answer:
Advantages of biogas are as follows:

Biogas is a cheap, safe, non-conventional and renewable source of energy.
It can be easily generated, stored and transported.
Biogas burns with a blue flame without producing smoke.
Biogas is of great help in improving ‘ the sanitation of the surrounding.
Biogas is an eco-friendly gas. It does not cause pollution and imbalance of the environment.
Leftover sludge can be used as fertilizer.
It is used as domestic and industrial fuel. Biogas can be used for domestic lighting, street lighting, cooking, small scale industries, etc.

Question 27.
Explain how Bacillus thuringiensis acts as a bio-control agent.
Answer:

Bacillus thuringiensis (Bt) is an effective biocontrol agent against butterfly, caterpillars.
Dried spores of Bacillus thuringiensis are mixed with water and sprayed onto vulnerable plants such as Brassicas and fruit trees.
When insect larvae eat the leaves, they get killed by the toxin (cry protein) released in their gut by bacteria.

Question 28.
Explain how Trichoderma acts as a bio-control agent.
Answer:

Trichoderma is an effective biocontrol agent against soil borne fungal plant pathogens.
It is a free-living fungus found in the root ecosystem (rhizosphere).
It produces substances like viridin, gliotoxin, gliovirin, etc. that inhibit the soil borne pathogens which infect root and rhizomes to cause rot disease.

Question 29.
What are bioherbicides? Give any two examples.
Answer:

Bacteria, fungi and insects which kill the dicot herbs which acts as weeds in the fields of monocot cereal crops, are known as bioherbicides.
Pseudomonas spp. and Xanthomonas spp. kill several weeds.
Fungus Alternaria crassa controls water hyacinth.

Question 30.
What is composting? Which microorganisms are found in active compost?
Answer:

Composting is a natural process in which organic matter is converted into a dark rich compost or humus.
During composting, microorganisms break down organic matter into compost.
Microorganisms found in active compost are bacteria, fungi, actinobacteria, protozoa and rotifers.

Question 31.
What are cyanobacteria? Give any two examples cyanobacteria as biofertilizers.
Answer:

Cyanobacteria are aerobic, photosynthetic, N₂ fixing microorganisms.
They are aquatic or terrestrial.
They may be free-living or symbiotic.
They may be heterocystous or non- heterocystous. Heterocyst is the site of nitrogen fixation.
e.g. Free living cyanobacteria are Anabaena, Nostoc, Tolypothrix, Plectonema, Oscillatoria.
Anabaena and Nostoc have symbiotic relationship with lichens.
Anabaena is also symbiotically associated with Azolla and Cycas.

Chart based or table based questions

Question 1.
Complete the table given below.

Crop	Variety	Resistant to disease
Wheat	————–	Leaf and stripe rust, Hill bunt
————-	Pusa swarnim	White rust
Cauliflower	————–	Black rot and Curl blight black rot
Chilli	Pusa sadabahar	—————

Answer:

Crop	Variety	Resistant to disease
Wheat	Himgiri	Leaf and stripe rust, Hill bunt
Brassica	Pusa swarnim	White rust
Cauliflower	Pusashubhra	Black rot and Curl blight black rot
Chilli	Pusa sadabahar	Chilli mosaic virus, Tobacco mosaic virus, leaf curl

Question 2.
Complete the table given below.

Crop	Variety	Insect pest
Brassica	Pusa Gaurav	————-
————-	Pusa sem 2 m Pusa sem 3	Jassids, aphids, fruit borer
Okra	————–	Shoot and fruit borer

Answer:

Crop	Variety	Insect pest
Brassica	Pusa Gaurav	Aphids
Flat bean	Pusa sem 2 m Pusa sem 3	Jassids, aphids, fruit borer
Okra	Pusa Sawani, Pusa A-4	Shoot and fruit borer

Diagram based questions

Question 1.
a. Identify A and B in the following diagram.
b. What is organogenesis?
c. What is meant by hardening?
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 1
a. A : Explant, B : Callus

b. When auxins are provided to callus in more quantity compared to cytokinins, it gives rise to roots (rhizogenesis) and when cytokinins are provided in more quantity, there is development of shoot (caulogenesis). This induction of organ formation in callus by providing growth hormones in proper proportion is known as organogenesis.

c. Plantlets produced in tissue culture laboratory are transferred to polythene bags containing sterilized soil and are kept on low light and high humidity conditions for suitable period of time. This is known as hardening.

Question 2.
a. Identify honey bees A, B and C in the given diagram.
b. They belong to which species?
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 2
a. A : Worker honey bee, B : Queen honey bee, C : Drone of honey bee
b. They belong to species Apis mellifera.

Question 3.
Identify A, B, C and D in the given diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 3
Answer:
A : Honey super
B : Queen excluder
C : Hive bodies
D : Entrance reducer

Question 4.
Identify fish A, B, C and D in the given diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 4
Answer:
A : Rohu fish,
B : Mrigal fish,
C : Grass carp and
D : Silver carp

Question 5.
a. Identify A, B and C in the given diagram.
b. The given diagram represent life cycle of _____
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 5
a. A : Mature caterpillar, B : Cocoon and C : Adult female
b. Silk moth (Bombyx mori)

Question 6.
a. Identify stages A and B in the given diagram.
b. Larvae are also called …………..
c. The diagram represent life cycle of ……………
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 6
Answer:
a. A : Hatching, B : Pupa
b. Crawlers
c. Lac insect

Question 7.
Draw a labelled diagram of Tubular tower fermenter.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 7

Question 8.
Draw a labelled diagram of biogas plant.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 8

Question 9.
a. Identify A in the given diagram.
b. Name the bacteria which form ‘A’ in roots of leguminous plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 9
Answer:
a. A : Root nodules
b. Rhizobia form root nodules in leguminous plants.

Question 10.
Draw a labelled diagram of T.S. of root nodule.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 10

Question 11.
Identify and describe the plant in the given diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 11
Answer:
The plant in the given diagram is Azolla.
Azolla is an aquatic fern that has symbiotic association with nitrogen fixing cyanobacterium Anabaena.
It propagates vegetatively and spreads in rice fields very rapidly.
It has a floating rhizome with small overlapping bilobed leaves and roots.
Azolla provides habitat to Anabaena.
The leaf shows dorsal and ventral lobe.
7. Anabaena filaments are present in the aerenchyma of dorsal lobe.

Question 12.
a. Identify A, B and C in the given diagram.
b. Name the plant which has symbiotic relationship with ‘A’.
Maharashtra Board Class 12 Biology Important Questions Chapter 11 Enhancement of Food Production 12
Answer:
a. A : Anabaena, B : Photosynthetic zone. C : Dorsal lobe.
b. Azolla m

Long Answer Questions

Question 1.
Explain concept of outbreeding and its types.
Answer:
(1) Outbreeding involves breeding of two unrelated animals.
(2) It is of three types, viz. outcrossing, cross-breeding and interspecific hybridization.

(3) Outcrossing:

Outcrossing involves mating of animals of same breed, which do not have a common ancestors on either side of mating partners up to 4 to 6 generations.
The progeny obtained from such mating is called an outcross.
Outcrossing is done to overcome inbreeding depression.

(4) Crossbreeding:

In crossbreeding superior males of one breed are mated with superior females of another breed.
New animal breeds of desirable characters are developed by this method.
Example : Hisardale breed of sheep is developed in Punjab by crossing Bikaneri ewes and Marino rams.

(5) Interspecific hybridization:

It involves breeding of animals of two different but related species.
It is used to produce animals with desirable characters from both the parents.
e.g. Mule is a breed obtained from horse and donkey.
It may not be always successful.

Question 2.
Which dairy products are prepared using microorganisms? How?
Answer:

Dairy products prepared using microorganisms are curds, yogurt, butter milk and cheese.
The starter or inoculum used in preparation of dairy products contains millions of lactic acid bacteria (LAB).
Curd is prepared by inoculating milk with Lactobacillus acidophilus. It ferments lactose sugar of milk into lactic acid. Lactic acid causes coagulation and partial digestion of milk protein casein. Thus, milk is changed into curd. It also checks growth of disease causing microbes.
Yogurt is produced by curdling milk with the help of Streptococcus thermophilus and Lactobacillus bulgaricus.
Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.
During the preparation of cheese, the milk is coagulated with LAB. The curd formed is filtered and whey is separated. The solid mass is then ripened with growth of mould that develops flavour in it. Characteristic texture, flavour and taste of cheese are developed by different specific microbes.

The ‘Roquefort cheese is ripened by blue green mold Penicillium roquejortii. Camembert cheese is ripened by blue-green mold P. camembertii. The large holes in Swiss cheese are developed due to production of a large amount of CO₂ by a bacterium known as Propionibacterium shermanii.

Question 3.
What are biofertilizers? Explain what are the different types of biofertilizers with suitable examples.
Answer:

Biofertilizers are commercial ready to use live bacterial, cyanobacterial or fungal formulations.
When they are applied to plants, in soil or in composting pits, soil fertility increases. Biofertilizers are cost effective and eco-friendly.

Types of biofertilizers as follows:
1. Bacterial biofertilizers:

Nitrogen fixing bacterial biofertilizers : They convert atmospheric nitrogen into compounds of nitrogen like ammonia, nitrites and nitrates. The nitrogen fixing bacteria Rhizobium forms symbiotic association with roots of leguminous plants. Free living nitrogen fixing bacteria are Azotobacter and Azospirillum.
Phosphate solubilizing biofertilizers : They are the bacteria which solubilize the insoluble inorganic phosphate compound, e. g. Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Aspergillus spp., etc.
Bacteria are also involved in composting.

2. Cyanobacterial biofertilizers:

They are nitrogen fixing biofertilizers.
Heterocysts are the site of nitrogen fixation.
Free living cyanobacteria are Anabaena, Nostoc, Tolypothrix, Plectonema, Oscillatoria.
Anabaena and Nostoc have symbiotic relationship with lichens
Anabaena is also symbiotically associated with Azolla and Cycas.

3. Fungal biofertilizers:

Fungal biofertilizers are mycorrhizae which form symbiotic association with roots of higher plants. There may be ectomycorrhiza and endomycorrhiza (VAM).
Ectomycorrhizae increase the absorptive surface area of rots and increase uptake of water and nutrients.
Plants with endomycorrhizae grow well even in less irrigated lands.

4. Microbes involved in composting:
During composting, microorganisms break down organic matter into compost. E.g. bacteria, fungi, actinobacteria, protozoa and rotifers.

Question 4.
Describe in details different types of mycorrhizae.
Answer:
Mycorrhizae are fungi that form symbiotic association with the roots of higher plants in humid forests.
There are two types as follows:
(1) Ectomycorrhizae:

Mycelium of these fungi form mantle on the surface of the roots.
Due to this absorptive surface area of roots increases and uptake of water and nutrients (N, P Ca and K) improves.
The plant vigour, growth and yield increase.
Some hyphae may penetrate into the root and form hartig-net in ‘the intercellular spaces of root cortex.

(2) Endomycorrhizae:

Fugal hypahe of endomycorrhizae penetrate the root cortex and form branched arbuscules intracellularly. They also form vesicles mostly in the intercellular spaces.
Hence, they are called Vesiculo Arbuscular Mycorrhizae (VAM). Nowadays they are described as AM fungi.
The plants associated with VAM grow luxuriantly in less irrigated lands.
VAM increase the productivity of field.

Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter

July 25, 2024July 24, 2024 by Prasanna

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 7 Thermal Properties of Matter Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 7 Thermal Properties of Matter

Question 1.
State various units of heat and relate them to SI unit of heat.
Answer:

CGS unit of heat: erg and it is related to SI unit as, 1 J = 10⁷ erg
Thermodynamic unit of heat: calorie (cal) and it is related as 1 cal = 4.184 J

Question 2.
What is thermal equilibrium?
Answer:

When two bodies at different temperatures come into the contact with each other, they exchange heat.
After some time, temperature of two bodies become equal and heat transfer between them stops.
The two bodies are then said to be in thermal equilibrium with each other.

Question 3.
State true of false. Correct the statement and rewrite if false.
i. Heat transfer takes place between the body and the surrounding medium until the body and the surrounding medium are at the same temperature.
ii. Whenever two bodies are in contact, there is a transfer of heat.
Answer:

True.
False.

Whenever two bodies at different temperature are in contact, there is a transfer of heat.

Question 4.
Give reason: Temperature is said to be a measure of average kinetic energy of the atoms/molecules of the body.
Answer:

Matter consists of particles which are in continuous vibrational motion and thereby possess kinetic energy.
When external energy is provided to these particles, internal energy of particles increases.
This increase in internal energy is in the form of increased kinetic energy of atoms/molecules and raises temperature of body (except at melting or boiling point of the body).
Greater the kinetic energy, faster the atoms/ molecules move and temperature of body becomes higher.
Thus, temperature of body is directly proportional to its kinetic energy.
Hence, temperature is said to be a measure of average kinetic energy of the atoms and molecules of the body.

Question 5.
Why do solid particles possess potential energy?
Answer:
The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.

Question 6.
What happens when heat is supplied to a solid at its melting point?
Answer:

When heat is supplied to a solid at its melting point, average kinetic energy of constituent particles does not change.
As a result, temperature of body remains constant.
Supplied energy is used to weaken the bonds between constituent particles.
While order of magnitude of average distance between the molecules remains almost same as solid, substance melts, i.e., changes into liquid state, at melting point.

Question 7.
Why do solids have definite shape and volume?
Answer:
The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.
Hence, solids have definite shape and volume.

Question 8.
Why is density of liquid at melting point nearly same as density of solids at melting point?
Answer:

During change of state from solid to liquid, mass of substance does not change.
Also, mean distance between particles during change of state does not alter at melting point.
Density depends upon mass and volume, in turn, on mean distance between particles.
Hence, density of liquid at melting point is nearly same as density of solids at melting point.

Question 9.
State true or false. If false correct the statement and rewrite.
Due to weakened interatomic bonds liquid do not possess definite volume but have definite shape.
Answer:
False.
Due to weakened bonds liquids do not possess definite shape but have definite volume.

Question 10.
What happens when heat is supplied to liquid its freezing point (melting point)?
Answer:

On heating, the atoms/molecules in liquid gain kinetic energy and temperature of the liquid increases.
If liquid is continued to heat further, at the boiling point, the constituents overcome the interatomic/molecular forces.
The mean distance between the constituents increases so that the particles are farther apart.
At boiling point, the liquid gets converted into gaseous state.

Question 11.
Why, according to kinetic theory of gases, gases have neither definite volume nor shape?
Answer:
As per kinetic theory of gases, for an ideal gas, there are no forces between the molecules of a gas. Hence, gases neither have a definite volume nor shape.

Question 12.
Match the pairs.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 1
Answer:
i – b,
ii – c,
iii – a

Question 13.
Distinguish between an adiabatic wall and diathermic wall.
Answer:

	Adiabatic wall	Diathermic wall
i.	An ideal wall or partition separating two systems such that no heat exchange can take place between the systems is called adiabatic wall.	A wall that allows exchange of heat energy between two systems is said to be diathermic wall.
ii.	It is a perfect thermal insulator.	It is not a perfect thermal insulator.
iii.	It does not exist in reality.	Partition like thin sheet of copper acts as diathermic wall.
iv.	It is generally represented as thick cross-shaded (slanting lines region).	It is represented as a thin dark region.

Question 14.
State and explain zeroth law of thermodynamics.
Answer:
Statement: If two bodies A and B are in thermal equilibrium and also A and C are in thermal equilibrium then B and C are also in thermal equilibrium.
Explanation:

Consider two sections of a container separated by an adiabatic wall containing two different gases as system A and system B.
Systems A and B are independently brought in thermal equilibrium with a system C.
When the adiabatic wall separating systems A and B is removed, there will be no transfer of heat from system A to system B or vice versa.
This indicates that systems A and B are also in thermal equilibrium.
This means, if systems A and B are separately in thermal equilibrium with a system C, then A and B are also mutually in thermal equilibrium.

Question 15.
What is thermometry? What is thermometer?
Answer:
Thermometry is the science of temperature and its measurement. The device used to measure temperature is a thermometer.

Question 16.
State the principle used to measure the temperature of a system using a thermometer.
Answer:
When two or more systems/bodies are in thermal equilibrium, their temperatures are same. This principle is used to measure the temperature of a system by using a thermometer.

Question 17.
Explain how thermal equilibrium is attained between thermometer and the patient holding thermometer, in mouth.
Answer:

Thermometer indicating lower temperature is held in mouth by patient.
As body of patient is at higher temperature, heat energy is transferred from patient to thermometer.
When temperature of thermometer becomes same as temperature of patient, heat exchange stops and thermal equilibrium is attained between thermometer and body of patient.

Question 18.
Define the following terms.
i) Ice point
ii) Steam point
Answer:

Ice point: The temperature at which pure water freezes at one standard atmospheric pressure is called as ice point or freezing point.
Steam point: The temperature at which pure water boils into steam or steam changes to liquid water at one standard atmospheric pressure is called as steam point or boiling point.

Question 19.
Explain Celsius and fahrenheit scale of temperature. Give relation between the two scales with the help of the graph.
Answer:

Celsius scale:
- The ice point (melting point of pure ice) is marked as 0 °C (lower point) and steam point (boiling point of water) is marked as 100 °C (higher point).
- Both are taken at one atmospheric pressure.
- The interval between these points is divided into loo equal pans. Each of these parts is called as one degree celsius and it is written as 1 oc.
Fahrenheit scale:
- The ice point (melting point of pure ice) is marked as 32 °F and steam point (boiling point of water) is marked as 212 °F.
- The interval between these two reference points is divided into 180 equal parts. Each part is called as degree fahrenheit and is written as 1 °F.
Relation between fahrenheit temperature and celsius temperature:
\(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}\) = \(\frac{T_{C}-0}{100}\)
Where, T_F = temperature in fahrenheit scales.
T_C = temperature in celsius scale.
The graph of T_F versus T_C is as shown

Question 20.
What is thermometer? Explain with examples, the thermometric property used in a thermometer.
Answer:

An instrument designed to measure temperature is called as thermometer.
Any property of a substance which changes sufficiently with temperature can be used as a basis of constructing a thermometer and is known as the thermometric property.
There are different types of thermometers.
- In a constant volume gas thermometer, the pressure of a fixed volume of gas (measured by the difference in height) is used as the thermometric property.
- The liquid-in-glass thermometer depends on the change in volume of the liquid with temperature. Small change in temperature changes the volume of liquid considerably. Two such liquids are mercury and alcohol. Mercury thermometers are used for measurement of temperature range -39 °C to 357 °C while alcohol thermometers are used only to measure temperatures near ice point (melting point of pure ice).
- The resistance thermometer uses the change of electrical resistance of a metal wire with temperature.
- Normally in research laboratories, a thermocouple is used to measure the temperature. A thermocouple is a junction of two different metals or alloys eg.: copper and iron joined together.
- When two such junctions at the two ends of two dissimilar metal rods are kept at two different temperatures, an electromotive force is generated that can be calibrated to measure the temperature.
Thermometers are calibrated so that a numerical value may be assigned to a given temperature. The standard fixed points are melting point of ice and boiling point of water.

Question 21.
State characteristics of thermometer.
Answer:

Thermometer must be sensitive, i.e., a noticeable change in the thermometric property should be observed for a very small change in temperature.
It has to be accurate.
It should be easily reproducible.
It is important that the system attains thermal equilibrium with the thermometer quickly.

Question 22.
Explain relation between unknown temperature T and thermodynamic property P_T at temperature T.
Answer:
If the values of a thermometric property are P₁ and P₂ at the ice point (0 °C) and steam point (100 °C) respectively and the value of this property is P_T at unknown temperature T, then T is given by the following equation.
T = \(\frac{100\left(P_{T}-P_{1}\right)}{P_{2}-P_{1}}\)

Question 23.
State true or false. If false correct the statement and rewrite.
Ideally, there should be no difference in temperatures recorded on two different thermometers.
Answer:
True.

Question 24.
List an advantage and a disadvantage of constant volume gas thermometer.
Answer:
Advantage: There is no difference in temperatures recorded on two different constant volume gas thermometers. Hence, it is very accurate.
Disadvantage: Constant volume gas thermometer is bulky instrument. Hence, it is not easily portable.

Question 25.
Give short note on liquid-in-glass thermometer.
Answer:

Liquid-in-glass thermometer depends on the change in volume of the liquid with temperature.
When the bulb is heated, the liquid in a glass bulb expands upward in a capillary tube.
The liquid is such that it is easily seen and expands (or contracts) rapidly and by a large amount over a wide range of temperature.
Most commonly used liquids are mercury and alcohol as they remain in liquid state over a wide range. Mercury freezes at -39 °C and boils at 357 °C; alcohol freezes at -115 °C and boils at 78 °C.

Question 26.
What are thermochromic liquids? Give two examples.
Answer:

Thermochromic liquids are ones which change colour with temperature.
These liquids are very sensitive to temperature, especially in range of room temperature.
Hence, only specific liquids display distinct colour variations at normal temperature.
Examples: Titanium dioxide and zinc oxide are white at room temperature but when heated change to yellow.

Question 27.
Write a note on resistance thermometer.
Answer:

Resistance thermometer uses the change of electrical resistance of a metal wire with temperature.
It measures temperature accurately in the range -2000 °C to 1200 °C is best for steady temperatures.
It is bulky and hence not easily portable.

Solved Examples

Question 28.
If the temperature in the room is 29 °C, what is its temperature in degree fahrenheit?
Solution:
Given: T_C = 29 °C
To find: Temperature in degree fahrenheit (T_F)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 3
Answer:
Temperature in the room in degree fahrenheit is 84.2 °F.

Question 29.
Average room temperature on a normal day is 27 °C. What is the room temperature in °F?
Solution:
T_C = 27 °C
Room temperature in °F
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 4
Answer:
Room temperature in °F is 80.6 °F.

Question 30.
Normal human body temperature in fahrenheit is 98.4 °F. What is the body temperature in °C?
Solution:
T_F = 98.4 °F
Formula: Body temperature in °C (T_C)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 5
Body temperature in °C is 36.89 °C.

Question 31.
The length of a mercury column in a mercury-in-glass thermometer is 25 mm at the ice point and 180 mm at the steam point. What is the temperature when the length is 60 mm?
Solution:
Here the thermometric property P is the length of the mercury column.
Using equation,
T = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}{\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right)}\)
For P₁ = 25 mm,
P₂ = 180 mm,
P_T = 60 mm
T = \(\frac{100(60-25)}{(180-25)}\)
= 22.58 °C
The temperature corresponding to the length of 60 mm is 22.58 °C.

Question 32.
A resistance thermometer has resistance 95.2 Ω at the ice point and 138.6 Ω at the steam point. What resistance would be obtained if the actual temperature is 27 °C?
Solution:
Here the thermometric property P is the resistance. If R is the resistance at 27 °C,
Using equation,
T = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}{\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}\)
For T = 27 °C, P₁ = 95.2 Ω, P₂ = 138.6 Ω .
∴ 27 = \(\frac{100(\mathrm{R}-95.2)}{(138.6-95.2)}\)
∴ R = \(\frac{27 \times(138.6-95.2)}{100}\) + 95.2
= 11.72 + 95.2
= 106.92Ω
The resistance obtained would be 106.92 Ω.

Question 33.
Explain the need for thermodynamic (absolute) scale.
Answer:

The two fixed point scale, Celsius scale and Fahrenheit scale had a practical shortcoming for calibrating the scale.
It was difficult to precisely control the pressure and identify the fixed points, especially for the boiling point as the boiling temperature is very sensitive to changes in pressure.
Hence, a one fixed point scale was adopted to define a temperature scale.
This scale is called the absolute scale or thermodynamic scale.

Question 34.
What is triple point of water? State its physical significance.
Answer:

The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure.
To know the triple point, one has to see that three phases coexist in equilibrium and no one phase in dominating. This occurs for each substance at a single unique combination of temperature and pressure.
Thus, if three phases of water solid ice, liquid water and water vapour coexist, the pressure and temperature are automatically fixed.
Internationally, triple point of water has been assigned as 273.16 K at pressure equal to 6.11 × 10² Pa or 6.11 × 10^-3 atmosphere, as the standard fixed point for calibration of thermometers.
The physical significance of triple point of water is that, it represents unique condition and it is used to define the absolute temperature.

Question 35.
Write a short note on absolute scale of temperature.
Answer:

The absolute scale of temperature, is so termed since ills based on the properties of an ideal gas and does not depend on the property of any particular substance.
The zero of this scale is ideally the lowest temperature possible although it has not been achieved in practice.
It is termed as Kelvin scale after Lord Kelvin with its zero at -273.15 °C and temperature intervals same as that on the Celsius scale. It is written as K (without °).

Question 36.
Draw a neat and well labelled diagram to show comparison of kelvin, Celsius and fahrenheit temperature scales.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 6

Question 37.
Answer the following:
i) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
ii) The absolute temperature (Kelvin scale) T is related to temperature t_c on the Celsius scale by t_c = T – 273.15. Why do we have 273.15 in this relation and not 273.16?
Answer:
i) The triple point of water has been assigned a fixed value of 273.15 K. This number represents a unique value associated with a unique condition of temperature and pressure in which all the three phases of water co-exist. On the other hand, melting point of ice and boiling point of water do not have a unique set of values as they are subject to changes in pressure and volume. For this reason, triple point of water is a standard fixed point in modem thermometry.

ii) On Celsius scale, the melting point of ice at normal pressure has a value 0 °C. The value corresponding to this value on the absolute scale is 273.15 K. The value 273.16 K denotes the triple point of water which has a value,
273.16 – 273.15 = 0.01 °C on Celsius scale as per the given relation.

Question 38.
State Charles’ law and give its formula.
Answer:
Charles’ law:
At constant pressure, volume of a fixed mass of gas is directly proportional to its absolute temperature.
Mathematically,
V ∝ T … ( at constant pressure)
∴ V = kT
where k is constant of proportionality
∴ \(\frac{\mathrm{V}}{\mathrm{T}}\) = k = constant
For two gases, \(\frac{V_{1}}{T_{1}}\) = \(\frac{V_{2}}{T_{2}}\) = constant.

Question 39.
State Pressure law and give its formula.
Answer:
Pressure (Gay Lussac’s) law:
At constant volume, pressure of a fixed mass of gas is directly proportional to its absolute temperature.
Mathematically,
P ∝ T .. .(at constant volume)
∴ P = kT
Where, k is constant of proportionality P
∴ \(\frac{\mathrm{P}}{\mathrm{T}}\) = k = constant
For two gases, \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\) = constant

Question 40.
State Boyle’s law and give its formula.
Answer:
Boyle’s law:
At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.
Mathematically,
V ∝ \(\frac{1}{\mathrm{P}}\).. ..(at constant temperature)
V = k × \(\frac{1}{\mathrm{P}}\)
where, k is constant of proportionality.
PV = k = constant For two gases,
P₁V₁ = P₂V₂ = constant

Question 41.
Derive ideal gas equation PV = nRT.
Answer:

The relation between three variables of a gas i.e., pressure, volume and absolute temperature is called as ideal gas equation.
From Boyle’s law,
V ∝ \(\frac{\mathrm{T}}{\mathrm{P}}\), at constant temperature ….(1)
From Charles’ law,
V ∝ T, at constant pressure … .(2)
Combining equations (1) and (2) we get,
∴ V ∝ \(\frac{\mathrm{T}}{\mathrm{P}}\)
∴ \(\frac{\mathrm{PV}}{\mathrm{T}}\)= constant
For one mole of a gas,
\(\frac{\mathrm{PV}}{\mathrm{T}}\) = R or PV = RT … (3)
where R is the constant of proportionality.
Equation (3) is called ideal gas equation. The value of constant R is same for all gases. Therefore, R is called as universal gas constant. R = 8.31 JK^-1mol^-1.
For ‘n’ moles of gas, i.e. if the gas contains ‘n’ moles, equation (3) can be written as,
PV = nRT

Solved Examples

Question 42.
Express T = 24.57 K in Celsius and fahrenheit.
Solution:
Given: T_K = 24.57 K
To find: Temperature in Celsius (T_C),
Temperature in fahrenheit (T_F)

Calculation:
From formula,
∴ T_C = T_K – 273.15
= 24.57 – 273.15
= -248.58 °C
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 8
Temperature in celsius (T_C) is – 248.58 °C
Temperature in fahrenheit (T_F) is -415.44 °F

Question 43.
Calculate the temperature which has the same value on fahrenheit scale and kelvin scale.
Solution:
Given: T_K = T_F = x
To find: Temperature at which Kelvin and Fahrenheit scales coincide (x)
Formula: \(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}\) = \(\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
Calculation: From formula.
∴ \(\frac{x-32}{180}\) = \(\frac{x-273.15}{100}\)
∴ 5(x – 32) = 9(x – 273.15)
∴ 5x – 160 = 9x – 2458.35
∴ 4x = 2298.35
∴ x = \(\frac{2298.35}{4}\) = 574.6
∴ x = 574.6 °F
The temperature at which kelvin and fahrenheit scales coincide is 574.6 °F.

Question 44.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K, respectively. Express these temperatures on
the celsius and fahrenheit scales. (NCERT)
Solution:
Given: For neon, T_K = 24.57 K
For carbon dioxide, T_K = 216.55 K
To find:
i) Triple point of neon on celsius (T_C_N) and fahrenheit scale (T_F_N)
ii) Triple point of carbon dioxide on Celsius (T_C_C) and fahrenheit scale (T_F_C)

Formulae:
i) T_K – 273.15 = T_C
ii) \(\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\) = \(\frac{\mathrm{T}_{\mathrm{c}}-32}{180}\)
Calculation: From formula (i),
T_C = T_K – 273.15
For neon, T_C_N = 24.57 – 273.15
∴ T_C_N = -248.58 °c
For carbon dioxide.
T_C_C = 216.55 – 273.15
∴ T_C_C = -56.60 °c
From formula (ii),
T_F = \(\frac{9}{5}\)(T_k – 273.15) + 32
For neon, T_k = 24.57 K
∴ T_FN = \(\frac{9}{5}\)[24.57 – 273.15] + 32
∴ T_FN = -415.44 °F
For CO₂, T_K = 216.55 K
∴ T_FC = \(\frac{9}{5}\)(216.55 – 273.15) + 32
∴ T_FC = -69.88 °F
i) Triple point of neon on celsius scale is -248.58 °c and on a fahrenheit scale is -415.44 °F.
ii) Triple point of carbon-dioxide on celsius scale is -56.60 °C and on fahrenheit scale is -69.88 °F.

Question 45.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?
Solution:
Triple point of water is, T = 273.16 K
Since the absolute scales measure the triple point as 200 A and 350 B.
∴ 200A = 350B = 273.16 K
∴ 1A = \(\frac{273.16}{200}\)K and 1B = \(\frac{273.16}{350}\)K
If T_A and T_B are the temperatures on the two scales, then
\(\frac{273.16}{200}\)T_A = \(\frac{273.16}{350}\)T_B
∴ T_A = \(\frac{200}{350}\) T_B = \(\frac{4}{7}\)T_B
\(\frac{\mathbf{T}_{\mathbf{A}}}{\mathbf{T}_{\mathbf{B}}}\) = \(\frac{4}{7}\)
The relation between T_A and T_B is T_A : T_B = 4 : 7

Question 46.
The pressure reading in a thermometer at steam point is 1.367 × 10³ Pa. What is pressure reading at triple point knowing the linear relationship between temperature and pressure?
Solution:
Given: P = 1.367 × 10³ Pa at steam point (T) i.e., at 273.15 + 100 = 373.15 K.
Linear relationship between temperature and pressure means that.
P ∝ T ⇒ P₁T₁ = P₂T₂
To find: Pressure reading (P_triple)
Formula: P_triple = 273.16 × \(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)\)
where P_triple and P are the pressures at temperature of triple point (273.16 K) and T (375.15 K) respectively.
Calculation: From formula,
∴ P_triple = 273.16 × \(\left(\frac{1.367 \times 10^{3}}{373.15}\right)\)
= 1000 × 10³ Pa
Pressure reading is 1.000 × 10³ Pa.

Question 47.
When the pressure of 0.75 litre of a gas at 27 °C is doubled, its temperature rises to 111°C. Calculate the final volume of a gas.
Solution:
Given: V₁ = 0.75 litre = 750 cm³,
T₁ = 27 + 273.15 = 300.15 K,
T₂ = 111 + 273.15 = 384.15 K,
P₂ = 2P₁
To find: Final volume (V₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 9
∴ V₂ = 480 cm³
The final volume of the gas is 480 cm³.

Question 48.
A certain mass of a gas at 20 °C is heated until both its pressure and volume are doubled. Calculate the final temperature.
Solution:
T₁ = 20 + 273.15 = 293.15 K,
P₂ = 2P₁, and V₂ = 2V₁
To find: Final temperature (T₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula,
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\left(2 \mathrm{P}_{1}\right)\left(2 \mathrm{~V}_{1}\right)}{\mathrm{T}_{2}}\)
∴ \(\frac{1}{\mathrm{~T}_{1}}\) = \(\frac{4}{\mathrm{~T}_{2}}\)
∴ T₂ = 4 × 293.15 = 1172.6 K
= 1172.6 – 273.15 = 899.45 °C
∴ T₂ = 1172.6 K or 899.45 °C
The final temperature of the gas is 1172.6 K or 899.45 °C.

Question 49.
Explain how substances expand on the basis of vibrational motion of atoms.
Answer:

The atoms in a solid vibrate about their mean positions.
When heated, they vibrate faster and force each other to move a little farther apart. This results into expansion.
The molecules in a liquid or gas move with certain speed.
When heated, they move faster and force each other to move a little farther apart. This results
in expansion of liquids and gases on heating.
The expansion is more in liquids than in solids; gases expand even more.

Question 50.
What is thermal expansion? List types of thermal expansion.
Answer:

A change in the temperature of a body causes change in its dimensions.
The increase in the dimensions of a body due to an increase in its temperature is called thermal expansion.
There are three types of thermal expansion:
- Linear expansion
- Areal expansion
- Volume expansion

Question 51.
Define linear expansion of solids.
Answer:
The expansion in length of a solid due to thermal energy is called linear expansion.

Question 52.
Define coefficient of linear expansion of solid. State its unit and dimensions.
Answer:

The coefficient of linear expansion of a solid is defined as increase in the length per unit original length at 0 °C per degree rise in temperature.
It is denoted by α and given by,
α = \(\frac{l_{2}-l_{1}}{l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
where l₁ = initial length at temperature T₁ °C
l₂ = final length at temperature T₂ °C
Unit: °C^-1 or K^-1
Dimensions: [L⁰M⁰T⁰K^-1]

Question 53.
Derive an expression for the coefficient of linear expansion in solid.
Answer:
i) If the substance is in the form of a long rod of length l, then for small change ∆T, in temperature, the fractional change ∆l/l, in length is directly proportional to ∆T.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 10
[Note: Linear expansion ∆l is exaggerated for explanation.]
\(\frac{\Delta l}{l}\) ∝ ∆T
∴ \(\frac{\Delta l}{l}\) = α∆T … (1)
where, α is called the coefficient of linear expansion of solid.

ii) Rearranging terms in equation (1),
α = \(\frac{\Delta l}{l \Delta \mathrm{T}}\)
= \(\frac{l_{\mathrm{T}}-l_{0}}{l_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
where,
l₀ = length of rod at 0 °C,
l_T = length of rod when heated to T °C,
T₀ = 0 °C (initial temperature)
T = final temperature,
∆l = l_T – T₀ = change in length,
∆T = T – T₀ = rise in temperature.

iii) If l₀ = 1 m and T – T₀ = 1 °C, then α = l_T – l₀ (numerically).

iv) As magnitude of α varies negligibly with temperature, it is assumed to be constant for a particular material.
Hence, it is not essential to take initial temperature as 0 °C.
This modifies equation (2) into,
α = \(\frac{l_{2}-l_{1}}{l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
where l₁ = initial length at temperature T₁ °C
l₂ = final length at temperature T₂ °C

Question 54.
State true or false. If false correct the statement and rewrite.
i) Coefficient of linear expansion is same for all substances.
ii) Metals have high values for the coefficient of linear expansion, than non-metals.
Answer:

False.
Coefficient of linear expansion is different for different substances.
True.

Question 55.
Define areal expansion of solids.
Answer:
The increase in the surface area of a solid, on heating is called areal expansion or superficial expansion of solids.

Question 56.
Define coefficient of superficial (areal) expansion of a solid, state its unit and dimensions.
Answer:

Coefficient of superficial (areal) expansion of a solid is defined as the increase in area per unit original area at 0 °C per degree rise in temperature.
It is denoted by β and given by,
β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)
where, A₁ = Area of solid sheet at T₁ °C,
A₂ = Area of solid sheet at T₂ °C
Unit: 0°C^-1 or K^-1
Dimensions: [L⁰M⁰T⁰K^-1

Question 57.
Derive expression for coefficient of areal expansion.
Answer:
i) If a substance is in the form of a plate of area A, then for small change ∆T in temperature, the fractional change in area, ∆A/A in figure given below, is directly proportional to ∆T.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 11
[Note: Areal expansion ∆A is exaggerated for explanation]
\(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) ∝ ∆T
∴ \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) = β∆T … (1)
where β is called the coefficient of areal expansion of solid.

ii) Rearranging terms in equation (1),
β = \(\frac{\Delta \mathrm{A}}{\mathrm{A} \Delta \mathrm{T}}\)
= \(\frac{A_{T}-A_{0}}{A_{0}\left(T-T_{0}\right)}\)
where,
A₀ = volume at 0 °C,
A_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature,
∆A = A_T – A₀ = change in area,
∆T = T – T₀ = rise in temperature.
iii) If A₀ = 1 m², T – T_o = 1 °C, then β = A_T – A₀ (numerically).
iv. As, β does not vary significantly with temperature. Hence, if A₁ is the area of a metal plate at T₁ °C and A₂ is the area at higher temperature at T₂ °C, then
β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)

Question 58.
Define volume expansion of solids.
Answer:
The increase in volume due to heating is called volume expansion or cubical expansion.

Question 59.
Define coefficient of cubical (volume) expansion of a solid. State its unit and dimensions.
Answer:
i) Coefficient of cubical (volume) expansion:
Coefficient of cubical (volume) expansion of a solid is defined as increase in volume per unit original volume at O °C per degree rise in temperature.
It is denoted by γ and is given by,
γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
where,
V₁ = Volume of solid at T₁ °C,
V₂ = Volume of solid at T₂ °C
ii) Dimensions: [L⁰M⁰T⁰K^-1]

[Note: Units and dimension of areal expansion in solid (β) and cubical expansion in solid (γ) are same as that of linear expansion in solid (α).

Question 60.
Derive an expression for coefficient of cubical expansion.
Answer:
i) If the substance is in the form of a cube of volume V, then for small change ∆T in temperature, the fractional change, ∆/V in volume is directly proportional to ∆T.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 12
[Note: Volume expansion ∆V is exaggerated for explanation.)
ii) γ = \(\frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}\) = \(\frac{\mathrm{V}_{\mathrm{T}}-\mathrm{V}_{0}}{\mathrm{~V}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
where, V₀ = volume at 0 °C,
V_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature.
∆V = V_T – V₀ = change in volume,
γ = V_T – T₀ = rise in temperature.

iii) If V₀ = 1 m³, T – T₀ = 1 °C, then γ = V_T – V₀ (numerically).

iv) If V₁ is the volume of a body at T₁ °C and V₂ is the volume at higher temperature T₂ °C, then
γ₁ = \(\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
γ₁, is the coefficient of volume expansion at temperature T₁ °C.

Question 61.
How does coefficient of volume expansion depend upon temperature?
Answer:

As compared to coefficient of linear expansion (α) and coefficient of areal expansion (β), γ changes more with temperature.
It is constant only at high temperatures.

Question 62.
Do coefficient of areal expansion and coefficient of volume expansion depend upon nature of material?
Answer:
Yes, coefficient of areal expansion and coefficient of volume expansion depend upon nature of material.

Question 63.
Explain expansion in fluids.
Answer:

Since fluids possess definite volume and take the shape of the container, they exhibit only change in volume significantly.
Equations valid for cubical or volume expansion of fluids are:
γ = \(\frac{\Delta V}{V \Delta T}=\frac{V_{T}-V_{0}}{V_{0}\left(T-T_{0}\right)}\)
where, V₀ = volume at 0 °C,
V_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature,
∆V = V_T – V₀ = change in volume,
∆T = T – T₀ = rise in temperature.
and γ₁ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
where, V₁ is volume of body at T₁ °C, V₂ is volume of body at higher temperature T₂ °C and γ₁ is coefficient volume expansion at T₁ °C.
As fluids are kept in containers, while dealing with the volume expansion of fluids, expansion of the container also needs to be considered.
If expansion of fluid results in a volume greater than the volume of the container, the fluid overflows if the container is open.
If the container is closed, volume expansion of fluid causes additional pressure on the walls of the container.

Question 64.
Use the data given in the table below:

Materials	γ(K^-1)
Invar	2 × 10 ^-6
Steel	(3.3 – 3.9) × 10^-5
Aluminium	6.9 × 10“^-5
Mercury	18.2 × 10^-5
Water	20.7 × 10^-5
Paraffin	58.8 × 10^-5
Gasoline	95.0 × 10^-5
Alcohol (ethyl)	110 × 10^-5

What conclusions can be drawn using the data?
Answer:

Coefficient of volume expansion (γ) is a characteristic of the substance.
It (γ) has higher order of magnitude for liquids than that of solids.

Question 65.
Explain how behaviour of water is different than solids and liquids when heat is supplied.
Answer:

Normally solids and liquids expand on heating. Hence their volume increases on heating.
Since the mass is constant, it results in a decrease in the density on heating.
Water expand on cooling from 4 °C to 0 °C.
Hence its density decreases on cooling in this temperature range.

Question 66.
Derive relation between coefficient of linear expansion (α) and coefficient of areal expansion (β).
Answer:
Consider a square plate of side l₀ at 0 °C and h at T °C.

l_T = l_o (1 + αT) .
If area of plate at 0 °C is A_o, A_o = \(l_{0}^{2}\)
If area of plate at T °C is A_T,
A_T = \(l_{\mathrm{T}}^{2}\) = \(l_{0}^{2}\)(1 + αT)²
or A_T = A₀(1 + αT)² …. (1)
Also,
A_T = A₀(1 + βT) … (2)
Using Equations (1) and (2), A₀(1 + αT)² = A₀(1 + βT)
∴ 1 + 2αT + α²T² = 1 + βT
Since the values of a are very small, the term α²T² is very small and may be neglected, ∴ β = 2α
The result is general because any solid can be regarded as a collection of small squares.

Question 67.
Derive relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).
Answer:

Consider a cube of side l_o at 0 °C and l_T at T°C.
∴ l_T = l_o(l + αT)
If volume of the cube at 0 °C is V₀, V₀ = \(l_{0}^{3}\)
If volume of the cube at T °C is
V_T, V_T = \(l_{\mathrm{T}}^{3}\) = \(l_{0}^{3}\)(1 + αT)³
V_T = V₀ (1 + αT)³ ….(1)
Also,
V_T = V₀(1 + γT) ….(2)
Using Equations (1) and (2),
V_o(1 + αT)³ = V₀(1 + γT)
∴ 1 + 3αT + 3α²T² + α³T³ = 1 + γT
Since the values of a are very small, the terms with higher powers of a may be neglected.
∴ γ = 3α
The result is general because any solid can be regarded as a collection of small cubes.

Question 68.
State the relation between α, β and γ and write their meaning.
Answer:
Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}\) = \(\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion.
γ = coefficient of cubical expansion.

Solved Examples

Question 69.
The length of a rail on a railway line is 25 m at 10 °C. During summer, maximum temperature attained in the region is 50 °C. Find the minimum gap between the rails, (a = 1.2 × 10^-5/°C)
Solution:
Given: L₁ = 25 m, T₁ = 10 °C, T₂ = 50 °C,
α = 1.2 × 10^-5/°C
To find: Minimum gap between rails (L₂ – L₁)
Formula: L₂ – L₁ = L₁α(T₂ – T₁)
Calculation: From formula,
L₂ – L₁ = 25 × 1.2 × 10^-5 × (50 – 10)
= 25 × 1.2 × 10^-5 × 40
= 1.2 × 10^-2 m
∴ L₂ – L₁ = 1.2 cm
The minimum gap between the rails is 1.2 cm.

Question 70.
The length of a metal rod at 27 °C is 4 cm. The length increases to 4.02 cm when the metal rod is heated upto 387 °C. Determine the coefficient of linear expansion of the metal rod.
Solution:
Given: T₁ = 27 °C, T₂ = 387 °C
L₁ = 4 cm = 4 × 10^-2 m
L₁ = 4.02 cm = 4.02 × 10^-2 m
To find: Coefficient of linear expansion
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 15
Coefficient of linear expansion is 1.389 × 10^-5/°C

Question 71.
The length of a metal rod is 150 cm at 25 °C. Find its length when it is heated to 150 °C. (α_steel = 2.2 × 10⁵ /°C)
Solution:
Given. L₁ = 150 cm, T₁ = 25 °C, T₂ = 150 °C,
α_steel = 2.2 × 10⁵ /°C
To find: Length of rod (L₂)
Formula: α = \(\frac{L_{2}-L_{1}}{L_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
L₂ – L₁ = L₁ α(T₂ – T₁)
∴ L₂ = L₁[1 + α(T₂ – T₁)]
= 150[1 + 2.2 × 10⁵ × (150 – 25)]
= 150(1 + 2.2 × 10⁵ × 125)
= 150(1 + 0.00275)
= 150× 1.00275 = 150.4125
∴ L₂ = 150.4cm
Length of the rod at 150 °C) is 150.4 cm.

Question 72.
Length of a metal rod at temperature 27 °C is 4.256 m. Find the temperature at which the length of the same rod increases to 4.268 m. (α for iron = 1.2 × 10⁵ K^-1)
Solution:
Given: T₁ = 27°C, L₁ = 4.256 m, L₂ = 4.268 m,
α = 1.2 × 10⁵ K^-1
To find: Temperature (T₂)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 47
= antilog [log 1114.912 – log 4.256]
= antilog [3.0468 – 0.6290]
= antilog [2.4 1781
= 2.617 × 10²
= 261.7°C
Required temperature is 261.7 °C.

Question 73.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10^-5 °C^-1.
Solution:
Given: d₁ = 4.24 cm, ∆T = 227 – 27 = 200 °C,
α = 1.70 × 10^-5 °C^-1
To find: Change in diameter (∆d)
Formula: α = \(\frac{d_{2}-d_{1}}{d_{1} \Delta T}=\frac{\Delta d}{d_{1} \Delta T}\)
Calculation: From formula.
∆d = α x d₁ x αT
= 1.70 × 10 × 4.24 × 200
∴ ∆d = 1.44 × 10^-2 cm
The change in diameter of the hole is 1.44 × 10^-2 cm.

Question 74.
A thin aluminium plate has an area 286 cm2 at 20 °C. Find its area when it is heated to 180 °C.
(β for aluminium = 4.9 × 10^-5 °C)
Solution:
Given: T₁ = 20°C, T₂ = 180°C, A₁ = 286 cm²
β = 4.9 × 10^-5 °C
To find: Final area (A₂)
Formula: β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
A₂ = A₁ [1 + β(T₂ – T₁)]
= 286[1 + 4.9 × 10^-5 (180 – 20)]
= 286[1 + 4.9 × 10^-5 × 160]
= 286 [1 + 784.0 × 10]
= 286 [1 + 0.00784]
= 286 [1.00784]
∴ A₂ = 288.24 cm²
Its area when its heated is 288.24 cm².

Question 75.
The surface area of the metal plate is 2.4 × 10^-2 m² at 20°C. When the plate is heated to 185 °C, its area increases by 0.8 cm². Find the coefficient of areal expansion of metal.
Solution:
Given: T₁ = 20 °C, A₁ = 2.4 × 10^-2 m²,
T₂ = 185°C,
∆A = 0.8cm² = 0.8 × 10^-4 m²
To find: Coefficient of areal expansion (β)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 48
The coefficient of areal expansion of the metal is 2.02 × 10^-5/ °C.

Question 76.
A liquid occupies a volume of 2 × 10^-4 m³ at 0 °C. Calculate the increase in its volume If It is heated to 80 °C. [The coefficient of cubical expansion of the liquid is 4 × 10^-4 K^-1]
Solution:
Given: V₀ = 2 × 10^-4 m³, T₀ = 0°C,T = 80°C,
γ_r = 4 × 10^-4 K^-1,
T – T₀ = 80 – 0 = 80°C
To find: Increase in volume (∆V)
Formula. ∆V = V₀γ_r(T – T₀)
Calculation: From formula.
∆V = (2 × 10^-4) (4 × 10^-4) × 80
∴ ∆V = 6.4 × 10^-6 m³
Increase m volume of the liquid is 6.4 × 10^-6 m³.

Question 77.
A liquid at O °C is poured in a glass beaker of volume 600 cm3 to fill it completely. The beaker is then heated to 90 °C. How much liquid will overflow?
(γ_liquid = 1.75 × 10^-4/ °C, γ_glass = 2.75 × 10^-5/ °C)
Solution:
Given: V₁ = 600 cm³, T₁ = 0 °C, T₂ = 90 °C
γ_liquid = 1.75 × 10^-4/°C,
γ_glass = 2.75 × 10^-4/°C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
Increase is volume = V₂ – V₁
= γ V₁(T₂ – T₁)
Increase in volume of beaker
= γ_glass × V₁ (T₂ – T₁)
= 2.75 × 10^-5 × 600 × (90 – 0)
= 2.75 × 10^-5 × 600 × 90
= 148500 × 10^-5 cm³
∴ Increase in volume of beaker = 1.485 cm³
Increase in volume of liquid
= γ_liquid × V₁ (T₂ – T₁)
= 1.75 × 10^-4 × 600 × (90 – 0)
= 1.75 × 10^-4 × 600 × 90
= 94500 × 10^-4 cm³
∴ Increase in volume of liquid = 9.45 cm³
∴ Volume of liquid which overflows
= (9.45 – 1.485) cm³
= 7.965 cm³
Volume of liquid that overflows is 7.965 cm³.

Question 78.
The surface area of an iron plate is 80 cm² at 20 °C. Find its surface area at 120 °C. (α_iron = 1.25 × 10^-5 / °C)
Solution:
Given: A₁ = 80 cm², T₁ = 20 °C, T₂ = 120 °C,
α_iron = 1.25 × 10^-5 / °C
To find: Surface area (A₂)
Formula: A₂ = A₁ [1 + β (T₂ – T₁)]
Calculation: β_iron = 2 × α_iron = 2.5 × 10^-5 / °C
From formula,
A₂ = 80[1 + 2.5 × 10^-5 (120 – 20)]
∴ A₂ = 80.2 cm²
Surface area of the iron plate at 120 °C is 80.2 cm².

Question 79.
A sheet of brass is 50 cm long and 8 cm broad at 0 °C. If the surface area at 100 °C is 401.57 cm² find the coefficient of linear expansion of brass.
Solution:
Given: l = 50 cm, b = 8cm.
∴ A₁ = l × b = 50 x 8 = 400 cm²,
T₁ = 0°C, T₂ = 100°C,
A₂ = 401.57 cm²
To find: coefficient of linear expansion (α)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 49
Coefficient of linear expansion of brass is 1.962 × 10^-5/°C.

Question 80.
On heating a glass block of 10.000 cm³ from 25 °C to 40 °C, its volume increases by 4 cm3. Calculate coefficient of linear elipansion of glass.
Solution:
Given: V = 10,000 cm³, ∆V = 4 cm³,
∆T = 40 – 25 = 15°C,
To find: Coefficient of linear expansion (α)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 50
Coefficient of linear expansion of the glass block is 8.89 × 10^-6 /°C.

Question 81.
Explain in detail what is specific heat or specific heat capacity of a substance.
Answer:

Specific heat capacity is defined as the amount of heat per unit mass absorbed or given out by the substance to change its temperature by one uni! (one degree) Le., 1 °C or 1 K.
The amount of heat (∆Q) required to change the temperature of a substance is directly proportional to:
- the mass of the substance (m).
- change in temperature of the substance (∆T).
  ∴ ∆Q ∝ m and ∆Q ∝∆T
  ∴ ∆Q ∝ m∆T
  ∴ ∆Q = sm∆T …………….. (1)
  where ‘s’ is specific heat or specific heat capacity of a substance.
  From equation (1).
  s = \(\frac{\Delta Q}{m \Delta T}\)
  If m = 1 kg and ∆T = 1 °C,then s = ∆Q.
Unit: S.I. unit of specific heat is J kg^-1 °C^-1 or J kg^-1 K^-1 and C.G.S. unit is erg g^-1 K^-1 or erg g^-1 °C^-1.
Example: The specific heat of water is 4.2 J kg^-1 °C^-1
It means that 4.2 J of energy must be added to 1 kg of water to rise its temperature by 1 °C.
The specific heat capacity is a property of the substance.
Specific heat capacity weakly depends on temperature of object. Except for very low temperatures. the specific heat capacity is almost constant for all practical purposes.

Question 82.
State the heat equation.
Answer:
Heat received or given out (Q)
= mass (m) temperature change (∆T) × specific heat capacity (s).
or Q = m × ∆T × s

Question 83.
Write a note on: Molar specific heat.
Answer:

If the amount of substance is specified in terms of moles (μ) instead of mass (m) in kg. then the specific heat is called molar specific heat (C).
It is given by, C = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta \mathrm{T}}\)
The SI unit of molar specific heat capacity is J/mol °C or J/mol K.
Like specific heat, molar specific heat also depends on the nature of the substance and its temperature.

Question 84.
Give reason: Water is used as a coolant in automobile radiators.
Answer:

Water has the highest specific heat capacity compared to other substances.
As a result, water requires higher amount of energy to get heated.
This allows water to absorb heat readily while increasing its temperature minimally.
Hence, water is used as a coolant in automobile radiators.

Question 85.
Give reason: Water is preferred as heater in hot water hag than other liquids.
Answer:

Water has the highest specific heat capacity compared to other substances.
This means certain mass of water heated to certain temperature contains more heat than the same mass of any other liquid heated to same temperature.
As a result, water takes longer time to cool than any other liquid heated to same temperature.
Hence, water is preferred as heater in hot water bag than other liquids.

Question 86.
Explain why specific heat capacity of a gas at constant pressure is greater than that at constant volume.
Answer:

When the gas is heated at constant volume. there is no work done against external pressure.
Hence, all the supplied heat is used in raising the temperature of the gas.
But when the gas is heated at constant pressure, volume of the gas changes.
Due to this, part of the supplied heat is used by the gas to expand against external pressure and remaining part of heat supplied is used to raise the temperature.
Because of this, for the same rise in temperature, the heat to be supplied at constant pressure is greater than that for heating at constant volume.
Hence, specific heat capacity of a gas at constant pressure is greater than specific heat capacity at constant volume.

Question 87.
Define principal and molar specific heat of a gas at constant volunie and constant pressure.
Answer:

Principal specific heat:
- Principal specific heat of a gas at constant volume (s_v):
  Principal specific heat of a gas at constant volume is defined as the quantity of heat absorbed or released for rise or fall temperature of unit mass of a gas through 1 K (or 1 °C), when its volume is kept constant.
- Principal specific heat of a gas at constant pressure (s_p): Principal specfic heat of a gas at constant pressure is defined as the quantity of heat absorbed or released for rise or fall the temperature of unit mass of a gas through 1 K (or 1 °C), when its pressure is kept constant.
Molar specific heat:
- Molar specific heat of a gas at constant volume (C_V): Molar specific heat of a gas at Constant volume is defined as the quantity of absorbed or released for rise or fall the temperature of one mole of the gas through 1 K (or 1 °C), when its volume is kept constant.
- Molar specific heat of a gas at constant pressure (C_P): Molar specific heat of a gas at constant pressure is defined as the quantity of heat absorbed or released for rise or fall the temperature of one mole of the gas through 1 K (or 1 °C), when its pressure is kept constant.

Question 88.
State the relation between principal specific heat capacity and molar specific heat capacity for a gas.
Answer:
Molar specific heat capacity = Molecular weight × principal specific heat capacity.
i.e. C_V = μ × S_P and C_V = μ × S_V
where, μ is the molecular weight of the gas.
[Note: Symbols (S_V) and (S_P) are used as per standard convention.]

Question 89.
What is heat capacity?
Answer:

Heat capacity or thermal capacity of a body is the quantity of heat needed to raise or lower the temperature of the whole body by 1 o°C (or 1 K).
Heat capacity can be written as Heat received or given out
= mass × 1 × specific heat capacity
Heat capacity = Q = m × s
Heat capacity (thermal capacity) is measured in J/°C.

Solved Examples

Question 90.
If the temperature of 4 kg mass of a material of specific heat capacity 300 J/ kg °C rises from 20 °C to 30 °C. Find the heat received.
Solution:
m = 4 kg, s = 300 J/kg °C
∆T = 30 – 20 = 10 °C
To find: Heat received (Q)
Formula: Q = ms∆T
Calculation: From formula,
Q = 4 × 300 × 10
∴ Q = 12000 J
Heat received is 12000 J.

Question 91.
How much heat is required to raise temperature of 750 g of copper pot from 20 to 50 °C?
(The specific heat of copper is 0.094 kcal/kg °C)
Solution:
Given: s = 0.094 kcal/kg°C,
m = 750 g = 0.750kg,
T₁ = 20°C and T₂ = 50°C
Rise in temperature,
∆T = T₂ – T₁ = 50 – 20 = 30°C
To find: Heat required (Q)
Formula: Q = m × s × ∆T
Calculation: From formula,
Q = 0.750 × 0.094 × 30
∴ Q = 2.115 kcal
Heat required to raise the temperature of copper pot is 2.115 kcal.

Question 92.
Calculate the difference in the temperatures between the water at the top and bottom of a water fall 200 m high. Specific heat of water is 4200 J kg^-1 °C^-1.
Solution:
Given: s = 4200 J kg^-1 °C^-1, h = 200 m
To find: Difference in temperatures (∆T)
Formulae:
i) Q = ms∆T
ii) P.E. = mgh
Calculation: From formulae (j) and (ii)
When water falls from top to bottom. assuming no loss in energy, potential energy is converted into heat energy.
∴ Q = P.E.
∴ ms∆T = mgh
∴ s∆T = gh
∴ ∆T = \(\frac{\mathrm{gh}}{\mathrm{s}}=\frac{9.8 \times 200}{4200}\)
∴ ∆T = 0.467 °C
The difference in temperatures between the water at top and bottom is 0.467 °C.

Question 93.
Find thermal capacity for a copper block of mass 0.2 kg, if specific heat capacity of copper is 290 J/kg °C.
Solution:
Given: m = 0.2 kg, s = 290 J/kg °C
To find: Thermal capacity
Formula: Thermal capacity = m × s
Calculation: From formula,
Thermal capacity = 0.2 × 290
= 58 J/ °C
Thermal capacity is 58 J/ °C.

Question 94.
What is calorimetry?
Answer:
Calorimetry is an experimental technique for quantitative measurement of heat exchange.

Question 95.
Explain construction of calorimeter with the help of a labelled diagram.
Answer:

A device in which heat measurement can be made is called calorimeter.
It consists of a cylindrical vessel and stirrer as well as lid of the same material like copper or aluminium.
The vessel is kept inside a wooden jacket which contains heat insulating materials like glass. wool etc. to prohibit any transfer of heat into or out of the calorimeter.

Question 96.
State the principle behind working of calorimeter.
Answer:
Calorimeter being isolated system works on the principle of conservation of energy, where heat gained equals heat lost.

Question 97.
Explain the technique “method of mixtures’.
Answer:

Method of mixtures is a technique used to determine specific heat capacity of a material using calorimeter.
In this technique a sample ‘A’ of the substance is heated to a high temperature which is accurately measured.
The sample ‘A’ is then placed quickly in the calorimeter containing water.
The contents are stirred constantly until the mixture attains a final common temperature.
The heat lost by the sample ‘A’ will be gained by the water and the calorimeter.
The specific heat of the sample ‘A’ of the substance can be calculated as follows:
- Let,
  m₁ = mass of the sample ‘A’
  m2₂ = mass of the calorimeter and the stirrer
  m₃ = mass of the water in calorimeter
  s₁ = specific heat capacity of the substance of sample ‘A’
  s₂ = specific heat capacity of the material of calorimeter (and stirrer)
  s₃ = specific heat capacity of water
  T₁ = initial temperature of the sample ’A’
  T₂ = initial temperature of the calorimeter stirrer and water
  T = final temperature of the combined system
- Using heat equation,
  Heat lost by the sample ‘A’ = m₁s₁ (T₁ – T)
  Heat gained by the calorimeter and the stirrer = m₂s₂ (T – T₂)
  Heat gained by the water = m₃s₃ (T – T₂)
- c. Assuming no loss of heat to the surroundings, the heat lost by the sample goes into the calorimeter, stirrer and water,
  ∴ m₁s₁(T₁ – T) = m₂s₂(T – T₂) + m₃s₃(T – T₂) ………….. (1)
- Knowing the specific heat capacity of water and copper material of the calorimeter and the stirrer, specific heat capacity (si) of material of sample ‘A’ can be calculated.
  e. Rearranging terms of equation (1),
  s₁ = \(\frac{\left(\mathrm{m}_{2} \mathrm{~s}_{2}+\mathrm{m}_{3} \mathrm{~s}_{3}\right)\left(\mathrm{T}-\mathrm{T}_{2}\right)}{\mathrm{m}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}\right)}\)
One can find specific heat capacity of water or any liquid using the following expression, if the specific heat capacity of the material of calorimeter and sample is known
s₃ = \(\frac{\mathrm{m}_{1} \mathrm{~s}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}\right)}{\mathrm{m}_{3}\left(\mathrm{~T}-\mathrm{T}_{2}\right)}-\frac{\mathrm{m}_{2} \mathrm{~s}_{2}}{\mathrm{~m}_{3}}\)

Question 98.
In method of mixtures, why is it essential that density of solid sample be greater than the liquid in calorimeter?
Answer:

In method of mixtures, solid sample gives away heat to liquid, and heat exchange between, solid, liquid and calorimeter is considered as isolated.
If density of solid sample is lesser than liquid in calorimeter, sample will float on liquid.
This will cause partial heat loss to air inside the calorimeter and standard heat equations of calorimeter will no longer be applicable.
Hence, in method of mixtures it is essential that density of solid sample be greater than liquid in calorimeter.

Solved Examples

Question 99.
A sphere of aluminium of 0.06 kg is placed for sufficient time in a vessel containing boiling water so that the sphere is at 100 °C. It is then immediately transferred to 0.12 kg copper calorimeter containing 0.30 kg of water at 25 °C. The temperature of water rises and attains a steady state at 28 °C. Calculate the specific heat capacity of aluminium.
(Specific heat capacity of water, s_w = 4.18 × 10³ J kg^-1 K^-1, specific heat capacity of copper, s_Cu = 0.387 × 10³ J kg^-1 K^-1)
Solution:
Given: Mass of aluminium sphere = m₁ = 0.06 kg
Mass of copper calorimeter = m₂ = 0.12 kg
Mass of water in calorimeter = m₃ = 0.30 kg
Specific heat capacity of copper
= S_Cu = s₂ = 0.387 × 10³ J/kg K = 387 J/kg K
Specific heat capacity of water
= S_w = s₃ = 4.18 × 10³ J/kg K = 4180 J/kg K
Initial temperature of aluminium sphere
= T₁ = 100 °C
Initial temperature of calorimeter and water
= T₂ = 25 °C
Final temperature of the mixture = T = 28 °C
To find: Specific heat capacity of aluminium (s_al)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 27
= 903.08 J/kg K
Specific heat capacity of aluminium is 903.08 J/kg K.

Question 100.
A copper sphere of 100 g mass is heated to raise its temperature to 100 °C and is released in water of mass 195 g and temperature 20 °C in a copper calorimeter. If the mass of calorimeter is 50 g, what will be the maximum temperature of water? (Given: specific heat of copper = 0.1 cal/g °C and specific heat of calorimeter = 0.1 cal/g °C)
Solution:
Let copper sphere, water and calorimeter attain final temperature T °C.
We have,
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 28
Heat lost by copper sphere
Q = m_sphere × S_sphere × ∆T
= 100 × 0.1 × (100 – T)
Heat gained by water in calorimeter
Q₁ = m_water × S_water × ∆T
= 195 × 1 × (T – 20)
Heat gained by calorimeter
Q₂ = m_calorimeter × m_calorimeter × ∆T
= 50 × 0.1 × (T – 20)
According to principle of heat exchange,
Q = Q₁ + Q₂
∴ 10 × (100 – T) = 195 × (T – 20) + 5 × (T – 20)
∴ 1000 – 10T = 200(T – 20)
∴ 210 T = 5000
∴ T ≈ 23.8 °C
Maximum temperature of water will be 23.8 °C.

Question 101.
What is a change of state? When does it occur?
Answer:

Matter normally exists in three states: solid. liquid and gas. A transition from one of these states to another is called a change of state.
This change can occur when exchange of heat takes place between the substance and its surroundings.

Question 102.
Explain the following temperature vs time graph obtained during process of boiling water.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 29
Variation of temperature with time
Answer:

The given temperature v/s time graph demonstrates the behaviour of water when heated continuously and uniformly.
Line segment AB indicates temperature of ice remaining constant at 0 °C for certain period of time.
- This means, amount of heat (latent heat of fusion) supplied to ice is entirely used for changing its state from solid to liquid.
- Thus, line segment AB denotes conversion of ice at 0 °C into water at 0 °C.
Line segment BC indicates continuous rise in temperature of water from 0 °C to 100 °C.
- At point C, boiling point of water is
  reached and heat energy (latent heat of vaporisation) supplied further is used to convert water into steam.
- During this transformation, temperature remains unchanged as represented by line segment CD.
- Thus, line segment CD denotes conversion of water at 100 °C into steam at 100 °C.
Beyond point D, thermometer again shows rise in temperature.

Question 103.
Write a note on latent heat of substance.
Answer:

Latent heat of a substance is the quantity of heat required to change the state of unit mass of the substance without changing its temperature.
Mathematically, if mass m of a substance undergoes a change from one state to the other then the quantity of heat absorbed or released is given by, Q = mL
where. L is known as latent heat.
It is characteristic of the substance.
Its SI unit is J/kg.
The value of L depends on the pressure and is usually quoted at one standard atmospheric pressure.

Question 104.
Explain the following terms.
i) Latent heat of fusion
ii) Latent heat of vaporisation
Answer:

Latent heat of fusion:
- The quantity of heat required to convert unit mass of a substance from its solid state to the liquid state, at its melting point, without any change in its temperature is called its latent heat of fusion.
- The S.I. unit of latent heat of fusion is J/kg and its C.G.S. unit is cal/’g.
Latent heat of vaporisation:
- The quantity of heat required to convert unit mass of a substance from its liquid state to vapour state, at its boiling point, without any change in its temperature is called its latent heat of vaporisation.
- The S.I. unit of latent heat of vaporization is J/kg and its C.G.S unit is cal/g.

Question 105.
Explain why latent heat of vaporisation is much larger than latent heat of fusion.
Answer:

The energy required to completely separate the molecules or atoms in liquids is greater than the energy needed to break the rigidity (rigid bonds between the molecules or atoms) in solids.
Also, when the liquid is converted into vapour, it expands. Work has to be done against the surrounding atmosphere to allow this expansion.
Hence, latent heat of vaporisation is larger than latent heat of fusion.

Question 106.
A plot of temperature versus heat energy for a given quantity of water is shown below. What can be inferred studying it?
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 30
Temperature versus heat for water at one standard atmospheric pressure (not to scale)
Answer:
Inferences:

When heat is added (or removed) during a change of state, the temperature remains constant.
Also the slopes of the phase lines are not all the same, which indicates that specific heats of the various states are not equal.
For water, the latent heat of fusion and vaporisation are L_f = 3.33 × 10⁵ J kg^-1 and L_v = 22.6 × 10⁵ J kg^-1 respectively, i.e., 3.33 × 10⁵ J of heat is needed to melt 1 kg of ice at 0 °C and 22.6 × 10⁵ J of heat is needed to convert 1 kg of water to steam at 100 °C.
This means, steam at 100 °C carries 22.6 × 10⁵ J kg^-1 more heat than water at 100 °C.

Question 107.
Compare change of state from solid to liquid and from liquid to vapour.
Answer:

	Solid to liquid	Liquid to vapour
i.	The change of state from solid to liquid is called melting and from liquid to solid is called solidification.	The change of state from liquid to vapour is called vaporisation while that from vapour to liquid is called condensation.
ii.	Both the solid and liquid states of the substance co-exist in thermal equilibrium during the change of states from solid to liquid or vice versa.	Both the liquid and vapour states of the substance coexists in thermal equilibrium during the change of state from liquid to vapour.
iii.	The temperature at which the solid and the liquid states of the substance are in thermal equilibrium with each other is called the melting point of solid or freezing point of liquid. The freezing point describes the liquid to solid transition while melting point describes solid to liquid transition.	The temperature at which the liquid and the vapour states of the substance coexist is called the boiling point of liquid. This is also the temperature at which water vapour condenses to form liquid.
iv.	It is the characteristic of the substance and also depends on pressure.	It is characteristic of substance and depends on pressure.

Question 108.
What is normal melting point?
Answer:
The melting point of a substance at one standard atmospheric pressure is called its normal melting point.
Example: Normal melting point of water is 0°C.

Question 109.
State true or false. If false correct the statement and rewrite.
Normal freezing point ice is 32 °C.
Answer:
False.
Normal freezing point ice is 0 °C or 32 °F.

Question 110.
What is normal boiling point?
Answer:
The boiling point of a substance at one standard atmospheric pressure is called its normal boiling point.
Example: Normal boiling point of water is 99.97 °C,

Question 111.
Distinguish between boiling and evaporation of liquid.
Answer:

	Boiling of liquid	Evaporation of liquid
i.	Boiling of liquid takes place at boiling point which is fixed for a given pressure and unique for a given liquid.	Evaporation of liquid can take place at any temperature.
ii.	It occurs throughout the liquid.	It occurs only at surface of liquid.
iii.	The process does not depend on area of liquid surface exposed.	The process depends upon area of liquid surface exposed. Higher the exposed surface area, higher the rate of evaporation.
iv.	Source of energy is needed.	Energy is taken from surrounding.
V.	Boiling does not reduce temperature of liquid.	When evaporation takes place, temperature of liquid decreases.
vi.	During boiling process, bubbles are formed in liquid.	During evaporation, no bubbles are formed in liquid.

Question 112.
Explain evaporation in terms of kinetic energy of liquid molecules.
Answer:

Molecules in a liquid are moving about randomly.
The average kinetic energy of the molecules decides the temperature of the liquid.
However, all molecules do not move with the same speed.
Some with higher kinetic energy may escape from the surface region by overcoming the interatomic forces.
This process can take place at any temperature and is termed as evaporation.

Question 113.
Explain the dependence of evaporation on temperature of liquid.
Answer:

If the temperature of the liquid is higher, more is the average kinetic energy.
This implies that the number of fast moving molecules is more.
Hence the rate of losing such molecules to atmosphere will be higher.
Thus, higher is the temperature of the liquid, greater is the rate of evaporation.

Question 114.
Why does evaporation gives a cooling effect to the remaining liquid?
Answer:

In the process of evaporation, faster moving molecules escape from surface of liquid overcoming the interatomic forces.
Since faster molecules are lost, the average kinetic energy of the liquid is reduced.
As a result, the temperature of the liquid is lowered.
Hence, evaporation gives a cooling effect to the remaining liquid.

Question 115.
Explain two applications of evaporation in details.
Answer:

Drying of clothes:
- Clothes dry faster when hanged exposing more surface area than when kept folded.
- Due to more surface area, water in clothes gets evaporated faster, drying clothes quickly.
Using a spirit swab on skin before injecting gives cooling effect:
- Before giving an injection to a patient, normally a spirit swab is used to disinfect the region.
- A cooling effect is experienced by skin of patient due to evaporation of the spirit as explained before.

Question 116.
Write a note on sublimation.
Answer:

All substances do not pass through the three states: solid-liquid-gas.
There are certain substances which normally pass from the solid to the vapour state directly and vice versa.
The change from solid state to vapour state without passing through the liquid state is called sublimation and the substance is said to sublime.
Examples: Dry ice (solid CO₂) and iodine.
During the sublimation process, both the solid and vapour states of a substance coexist in thermal equilibrium.
Most substances sublime at very low pressures.

Question 117.
What is a phase? Give an example.
Answer:
A phase is a homogeneous composition of a material.
Example: Graphite and diamond are two phases of carbon.

Question 118.
What is a phase diagram?
Answer:
A pressure-temperature (P-T) diagram particularly convenient for comparing different phases of a substance is called as a phase diagram.

Question 119.
Study phase diagrams given below and answer the following questions.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 31
i)Explain vaporisation curve (l – v).
ii) Explain fusion curve (l – s).
iii) Explain sublimation curve (s – v).
iv) Explain triple point.
Answer:

- The curve labelled l – v represents those points where the liquid and vapour phases are in equilibrium.
- It is a graph of boiling point versus pressure.
- The l – v curve of water correctly shows that at a pressure of 1 atmosphere, the boiling point of water is 100 °C and the boiling point gets lowered for a decreased pressure.
- The l – v curve for CO₂ yields that CO₂ cannot exist as a liquid under normal atmospheric pressure conditions.
- The curve l – s represents the points where the solid and liquid phases coexist in equilibrium.
- It is a graph of the freezing point versus pressure.
- At one standard atmosphere pressure, the freezing point of water is 0 °C which can be depicted using l – s curve of water.
- At a pressure of one standard atmosphere water is in the liquid phase if the temperature is between 0 °C and 100 °C but is in the solid or vapour phase if the temperature is below 0 °C or above 100 °C.
- Also, l – s curve for water slopes upward to the left i.e., fusion curve of water has a slightly negative slope.
- This is true only of substances that expand upon freezing.
- However, for most materials like CO2, the l – s curve slopes upwards to the right i.e., fusion curve has a positive slope. The melting point of C02 is -56 °C at higher pressure of 5.11 atm.
- The curve labelled s – v is the sublimation point versus pressure curve.
- Water sublimates at pressure less than 0.0060 atmosphere, while carbon dioxide, which in the solid state is called dry ice, sublimates even at atmospheric pressure at temperature as low as -78 °C.
- The temperature and pressure at which the fusion curve, the vaporisation curve and the sublimation curve meet and all the three phases of a substance coexist is called the triple point of the substance.
- The triple point of water is that point where water in solid, liquid and gaseous states coexist in equilibrium and this occurs only at a unique temperature and pressure.
- The triple point of water is 273.16 K and 6.11 × 10^-3 Pa and that of CO₂ is -56.6 °C and 5.1 × 10 ^-5 Pa.

Question 120.
What is critical temperature of a gas?
Answer:
In order to liquefy a gas, it must be cooled to a certain temperature. This temperature is called critical temperature.

Question 121.
Compare gas and vapour.
Answer:

	Gas	Vapour
i.	A substance that is in gaseous phase above its critical temperature is called a gas.	A substance that is in gaseous phase below its critical temperature is call vapour.
ii.	Gas cannot be liquified only by pressure alone.	Vapour can be liquified simply by increasing pressure.
iii.	Gas exerts pressure.	Vapour exerts pressure

Solved Examples

Question 122.
Calculate the amount of heat energy to be supplied to convert 2 kg of ice at 0 °C completely into water at 0 °C if latent heat of fusion for ice is 80 cal/g.
Solution:
Given:
Mass (m) = 2 kg = 2 × 10³ g. latent heat of fusion for ice (L) = 80 cal/g.
To find: Heat energy (Q)
Formula: Q = mL
From formula,
Q = 2 x× 10³ × 80 = 160000 cal
= 160 kcal
Heat energy to be supplied is 160 kcal.

Question 123.
When 0.1 kg of ice at 0 °C is mixed with 0.32 kg of water at 35 °C in a container. The resulting temperature of the mixture is 7.8 °C. calculate the heat of fusion of ice (s_water = 4186 J kg^-1 K^-1).
Solution:
Given:
m_ice = 0.1 kg, m_water = 0.32 kg,
T_ice = 0 °C, T_water = 35 °C, T_F = 7.8 °C
s_water = 4186 J kg^-1 K^-1
To find: Heat of fusion (L_f)

Formula: i) Heat lost by water
Q₁ = m_water × s_water × (T_water – T_F)
ii) Heat required to melt ice
Q₂ = m_ice L_F
iii) Heat required to raise temperature of molten ice (water now) to find temperature
Q₃ = m_ices_water (T – T_ice)
Calculation: From formula (j),
Q₁ = 0.32 × 4186 × (35 – 7.8)
= 36434.944 J
From formula (ii),
Q₂ = 0.1 × L_f
From formula (iii),
Q₃ = 0.1 × 4186 × (7.8 – 0) = 3265.08J
According to principle of heat conservation,
heat lost = heat gained
Q₁ = Q₂ + Q₃
∴ 36434.944 = 0.1 L_f + 3265.08
∴ L_f = \(\frac{36434.944-3265.08}{0.1}\)
= 331698.64 J kg^-1
Rounding off to correct significant figure,
L_f = 3.31699 × 10⁵ J kg^-1
Heat of fusion of ice 3.3 1699 × 10 ⁵ J kg^-1.

Question 124.
If 80 g steam of temperature 97 °C is released on an ¡ce slab of temperature 0 °C, how much ice will melt? How much energy will be transferred to the ice when the steam will be transformed to water?
(Given: Latent heat of melting the ice = L_melt =80 cal/g ; Latent heat of vaporisation of water L_vap = 540 cal/g)
Solution:
Mass of steam (m_s) = 80 g,
Change in temperature (∆T)
= 97 – 0 = 97 °C
We know that: Latent heat of melting of ice = L_melt = 80 cal/g
Latent heat of vaporisation of water = L_vap = 540 cal/g
Specific heat of water c_w = 1 cal /g °C
To find: i) Energy transferred (Q)
ii) Mass of ice that melts (m_i)

Formula: i) Heat released during conversion of steam into water at 97 °C (Q₁)= m_s × L_vap
ii) Heat released during decrease of temperature of water from 97°C to 0°C (Q₂) = m_s × c_w × ∆T
iii) Heat gained by ice (Q)= m_i × L_melt

From formula (i),
Q₁ = 80 × 540 cal
From formula (ii),
Q₂ = 80 × 1 × (97 – 0) = 80 × 97 cal
According to principle of heat conservation,
Total heat gained by ice
Q = Q₁ + Q₂
= 80 × 540 + 80 × 97
= 80 × (540 + 97)
= 80 × 637
= 50960 cal
This energy would cause m; mass of ice to melt,
From formula (iii),
∴ m_i × L_melt = 50960
∴ m_i = \(\frac{50960}{80}=\frac{80 \times 637}{80}\) = 637 g
Energy transferred to ice is 50960 cal and it will melt 637 g of ice.

Question 125.
Name three modes of heat transfer.
Answer:
Three modes of heat transfer are conduction, convection and radiation.

Question 126.
Define conduction. State conditions for conduction of heat.
Answer:
Conduction is the process by which heat flows from the hot end to the cold end of a solid body without any net bodily movement of the particles of the body.
Conditions for conduction:

The two points should be at different temperatures.
There should be a medium between the two points.

Question 127.
Explain process of conduction in solid.
Answer:

Heat passes through solids by conduction only.
When one end of a rod is heated, the molecules near the hot end receive the thermal energy and start oscillating with larger amplitudes.
In doing so, they collide with the neighbouring molecules and transfer a part of their energy to these molecules.
The molecules which receive the energy vibrate with increased amplitudes and collide with the neighbouring molecules. Thus, energy of thermal motion is transferred by molecular collisions down the rod.
As the distance of a molecule from the hot end increases, its amplitude of oscillation decreases and hence there is continuous decrease in temperature.
This transfer of heat continues till two ends of the object are at the same temperature.
In metals, mainly free electrons conduct the heat energy.

Question 128.
Define good conductors and bad conductors.
Answer:
Good conductors:
The substances which conduct heat easily are called good conductors of heat.
All metals are good conductor.
eg: Steel, silver, Aluminium etc.

Insulators:
The substances which do nor conduct heal easily are called insulators or bad condiciors of heal.
eg.: Glass. wood, air, paper. etc.
[Note: In general, good conductors of heat are also good conductors of electricity, while bad conductors of hear are had conductors of electricity.]

Question 129.
Explain why metals are good conductors of heat and electricity.
Answer:

Metals like iron, copper. aluminium etc, contain free electrons in their atoms.
These free electrons assist the atoms in transfer of thermal energy as well as electrical energy.
Therefore, metal are good conductors of heat and electricity.

Question 130.
What is thermal conductivity?
Answer:

Thermal conductivity of a solid is a measure of the ability of the solid ¡o conduct heat through it.
Thus good conductors of heat have higher thermal conductivity than bad conductors.

Question 131.
Give reason: Hot water when poured in glass beaker, it cracks.
Answer:

When hot water is poured in a glass beaker the inner surface of the glass expends on heating.
Since glass is a bad conductor of heat, the heat from inside does not reach the outside surface so quickly.
Hence the outer surface does not expand thereby causing a crack in the glass.

Question 132.
Explain mechanism of thermal conduction and temperature gradient.
Answer:

When one end of a metal rod is heated, the heat flows by conduction from hot end to the cold end.
As a result, the temperature of every section of the rod starts increasing.
Under this condition, the rod is said to be in a variable temperature state.
After some time, the temperature at each section of the rod becomes steady i.e., does not change.
Temperature of each cross-section of the rod now becomes constant though not the same. This is called steady state condition.
Under steady state condition, the temperature at points within the rod decreases uniformly with distance from the hot end to the cold end.
The fall of temperature with distance between the ends of the rod in the direction of flow of heat, is called temperature gradient.
∴ Temperature gradient = \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{x}}\)
Where, T₁ = temperature of hot end
T₂ = temperature of cold end
x = length of the rod

Question 133.
State SI unit and dimensions of temperature gradient.
Answer:
S.I unit: = °C /m or K/m
Dimensions: [L^-1M⁰T⁰K¹]

Question 134.
State S.I. unit and dimensions of coefficient of thermal conductivity.
Answer:
SI unit of coefficient of thermal conductivity is J s^-1 m^-1 °C^-1 or J s^-1 m^-1 K^-1 and its dimensions are [L¹M¹T³K^-1].

Question 135.
Explain how SI unit of coefficient of thermal conductivity be obtained as W/m °C or W/m K.
Answer:

Consider equation, \(\frac{Q}{t}=\frac{k A\left(T_{1}-T_{2}\right)}{x}\)
The quantity Q/t, denoted by P_cond, is the time rate of heat flow (i.e. heat flow per second) from the hotter face to the colder face, at right angles to the faces.
Its SI unit is watt (W).
SI unit of k can therefore be written as W m^-1°C^-1 or W m^-1 K^-1.
[Note: Above equation, using calculus can be written as, \(\frac{d Q}{d t}\) = – kA \(\frac{d T}{d x}\), where \(\frac{d T}{d x}\) is the temperature gradient. The negative sign indicates that heat flow is in the direction of decreasing temperature. If A = 1 m² and \(\frac{d T}{d x}\) = 1, then \(\frac{d Q}{d t}\) = k.]

Question 136.
Define coefficient of thermal conductivity in terms of temperature gradient.
Answer:
Coefficient of thermal conductivity of a material is defined as the rate of flow of heat per unit area per unit temperature gradient when the heat flow is at right angles to the faces of a thin parallel-sided slab of material.

Question 137.
Define conduction rate.
Answer:
Conduction rate (P_cond) is the amount of energy transferred per unit time through a slab of area A and thickness x, the two sides of the slab being at temperatures T₁, and T₂ (T₁ > T₂),
and is given P_cond = \(\frac{Q}{t}\) = kA \(\frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{x}}\)

Question 138.
Explain the analogy between electrical resistance and thermal resistance.
Answer:

Electrical resistance is ratio of \(\frac{V}{I}\) where, V is electrical potential difference between two ends of conductor and I is current or rate flow of charge.
Consider expression for conduction rate,
P_cond = kA \(\frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{x}}\)
⇒ \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\mathrm{cond}}}=\frac{\mathrm{x}}{\mathrm{kA}}\) ……………… (1)
Comparing equation (1) with \(\frac{V}{I}\), (T₁ – T₂) is temperature difference between two ends and P_cond is rate of flow of heat.
Ratio \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\text {cond }}}\) is called as thermal resistance (R_T) of material.
Using (1), thermal resistance R_T = \(\frac{\mathrm{x}}{\mathrm{kA}}\)
Thermal resistance depends on the material and dimensions (length / breadth) of object.

Question 139.
What is thermal resistivity? What does it depend upon?
Answer:
i. Thermal resistivity (ρ_T) is the reciprocal of thermal conductivity (k).
ii. It is characteristic of a material.

Question 140.
Complete the table.
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 33
Answer:

Question 141.
Define convection.
Answer:
The process by which heat is transmitted through a substance from one point to another due to actual bodily movement of the heated particles of the substance is called convection.

Question 142.
Describe the mechanism of heat transfer by convection in liquids and gases.
Answer:

Consider liquid being heated in a vessel from below.
The liquid at the bottom of the vessel is heated
first and consequently its density decreases i.e., liquid molecules at the bottom are separated farther apart.
These hot molecules have high kinetic energy and rise upward to cold region while the molecules from cold region come down to take their place.
Thus, each molecule at the bottom gets heated and rises then cool and descends.
This action sets up the flow of liquid molecules called convection currents.
The convection currents transfer heat to the entire mass of liquid via actual physical movement of the liquid molecules.
Similar process takes place in case of a gas.

Question 143.
Give two applications of convection.
Answer:

Heating and cooling of rooms:
- The mechanism of heating a room by a heater is entirely based on convection.
- The air molecules in immediate contact with the heater are heated up.
- These air molecules acquire sufficient energy and rise upward.
- The cool air at the top being denser moves down to take their place. This cool air in turn gets heated and moves upward.
- In this way, convection currents are set up in the room which transfer heat to different parts of the room.
- The same principle but in opposite direction is used to cool a room by an air-conditioner.
Cooling of transformers:
- Due to current flowing in the windings of the transformer, enormous heat is produced.
- Therefore, transformer is always kept in a tank containing oil.
- The oil in contact with transformer body heats up, creating convection currents.
- The warm oil comes in contact with the cooler tank, gives heat to it and descends to the bottom. It again warms up to rise upward.
- This process is repeated again and again. The heat of the transformer is thus carried away by convection to the cooler tank.
- The cooler tank, in turn loses its heat by convection to the surrounding air.

Question 144.
Distinguish between free convection and forced convection.
Answer:

	Free convection	Forced convection
i.	When a hot body is in contact with air under ordinary conditions, like air around a firewood, the air removes heat from the body by aprocess called free or natural convection.	The convection process can be accelerated by employing a fan to create a rapid circulation of fresh air. This is called forced convection.
ii.	Land and sea breezes are formed as a result of free convection currents in air.	Heat convector, air conditioner, heat radiators in IC engine etc. operate using forced convection.

Question 145.
Define radiation.
Answer:
The transfer of heat energy from one place to another via emission of electromagnetic (EM) energy (in a straight line with the speed of light) without heating the intervening medium is called radiation.

Question 146.
Compare conduction, convection and radiation.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 35

Solved Examples

Question 147.
The temperature difference between two sides of an iron plate. 2 cm thick, is 10 °C. Heat is transmitted through the plate at the rate of 600 kcal per minute per square metre at steady state. Find the thermal conductivity of iron.
Solution:
Given: \(\frac{\mathrm{Q}}{\mathrm{At}}\) = 600 kcal/min m² = \(\frac{600}{60}\) kcal/s m²
= 10 kcal/s m²
x = 2cm = 2 × 10^-2 m
T₁ – T₂ = 10°C
To Find.- Thermal conductivity (k)
Formula: Q = \(\frac{\mathrm{kA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{t}}{\mathrm{x}}\)
Calculation: From formula.
∴ k = \(\frac{\mathrm{Q}}{\mathrm{At} \mathrm{t}} \frac{\mathrm{x}}{\mathrm{T}_{1}-\mathrm{T}_{2}}=\frac{10 \times 2 \times 10^{-2}}{10}\)
= 0.02 kcal / m s
Thermal conductivity is 0.02 kcal / m s °C.

Question 148.
Calculate the rate of loss of heat through a glass window of area 1000 cm² and thickness of 4 mm. when temperature inside is 27 °C and outside is – 5 °C. Coefficient of thermal conductivity of glass is 0.022 cal /s cm °C.
Solution:
Given: A = 1000 cm² 1000 × 10 m²
k = 0.22 cal / s cm °C
= 0.22 × 10² cal/ m °C
x = 4mm = 0.4 × 10^-2 m
T₁ = 27°C, T₂ = -5°C
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 37
= 1.76 × 10³ cal/s = 1.76kcal / s
Rate of loss of heat is 1.76 kcal / s.

Question 149.
Heat is conducted through a copper plate at the rate of 460 cal/s-cm². Calculate the temperature gradient when the steady state is reached. (k_copper = 92 cal/m-s °C)
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 38
Temperature gradient of the copper plate is 5°C/m.

Question 150.
Two parallel slabs of metals A and B of thickness 5 cm and 3 cm respectively are joined together. The outer face of the metal A is maintained at 100 °C and that of metal B is maintained at 40 °C. If the thermal conductivities of metal A and B are 0.045 kcal/m-s K and 0.015 kcal/m-s K respectively, find the temperature of the interface of two plates.
Solution:
For metal A:
T₁ = 100 °C, T₂ = θ
dx₁= 5 cm = 5 × 10^-2 m,
k₁ = 0.045 kcal/m-s K
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 39
∴ 9(100 – T) = 5(T – 40)
∴ 900 – 9T = 5T – 200
∴ 14T = 1100
∴ T = 78.57°C
Temperature of the interlace of two plates is 78.57 °C.

Question 151.
What is the rate of energy loss in watt per square metre through a glass window 5 mm thick if outside temperature in -20 °C and inside temperature is 25 °C? (k_glass = 1 W/m K)
Solution:
Given:
k_glass = 1W/m K, T₁ = 25 °C, T₂ = -20 °C
T₁ – T₂ = 25- (-20)°C = 45 °C
X = 5 mm = 5 × 10^-3 m
As one degree celsius equates to one kelvin, temperature difference of 450 C equals 45 K.
To find: Rate of energy loss per square metre \(\left(\frac{\mathrm{P}_{\text {cond }}}{\mathrm{A}}\right)\)
Formula: P_cond = \(\frac{Q}{t}=k A \frac{T_{1}-T_{2}}{x}\)
Calcula lion: From formula,
∴ The energy loss per square metre,
\(\frac{\mathrm{P}_{\text {oond }}}{\mathrm{A}}=\mathrm{k} \frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{x}}=\frac{1 \times 45}{5 \times 10^{-3}}\)
= 9 × 10³ W/m²
Rate of energy loss per square metre is 9 × 10³ W/m².

Question 152.
A metal sphere cools at the rate of 1.6 °C/min when its temperature is 70 °C. At what rate will it cool when its temperature is 40 °C? The temperature of surroundings is 30 °C.
Solution:
Given: T₁ = 70°C, T₂ = 40 °C, T₃ = 30 °C
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 40
= 0.4 (40 – 30)
= 0.4 °C/mm
Rate of cooling is 0.4 °C/mm.

Question 153.
A body cools at the rate of 0.5 °C/s when it is at 50 °C above the surrounding temperature. What is its rate of cooling when ¡t is at 30 °C above the surrounding temperature?
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 41
Divide equation (2) by (1),
\(\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{2}}{0.5}=\frac{\mathrm{C}(30)}{\mathrm{C}(50)}\)
∴ \(\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{2}\) = 0.5 × \(\frac{30}{50}\) = 0.3 °C/s
The rate of cooling of the body at 30 °C above the surrounding temperature is 0.3 °C/s.

Apply Your Knowledge

Question 154.
A metre scale made up of aluminium (α_a = 24 × 10^-6 /°C) measures length of a steel rod (α_s = 12 × 10^-6 /°C) at room temperature as 50.00 cm. Now the temperature of the room is increased by 100 °C. What can be said about the measured length of the rod at new temperature?
Answer:
At new temperature T °C,
change in length of rod is,
∆L_s = L₀(α_s∆T) = 50.00(12 × 10^-6 × 100)
= 0.06 cm.
Hence, the actual length of rod at T °C,
L_s = 50.06 cm
Due to change in temperature, along with rod, the scale will also increase in length.
For aluminium, at T °C,
∆L_a = L₀(α_a∆t) = 100.00 × 24 × 10^-6 × 100
= 0.24 cm
∴ Length of the scale will be,
L_a = 100.24 cm
As, the expansion in scale is more than that in rod, the reading recorded by the scale at t°C will be less than 50 cm.

Question 155.
A mercury thermometer calibrated to measure temperature in Fahrenheit scale is kept in liquid phosphorus at 150 °F. The liquid phosphorus is then heated continuously until it reaches its boiling point measured by the thermometer to be 500 °F. Find the percentage fractional change in the density of mercury during the whole process. (γ_Hg = 10^-4/°F)
(Assume that no heat is lost to the surrounding during the process.)
Answer:
Given:
T₂ = 500°F, T₁ = 150°F
γ_Hg = 10^-4 /°F
As the liquid phosphorus is heated, the mercury in the thermometer also gets heated.
Due to thermal expansion,
V₂ = V₁(1 + γ∆t)
= V₁ [1 + 10^-4 × (500 – 150)]
= 1.035 V₁
Now, initial density of mercury is,
ρ₁ = \(\frac{\mathrm{m}}{\mathrm{V}_{1}}\)
After heating,
ρ₂ = \(\frac{\mathrm{m}}{\mathrm{V}_{2}}=\frac{\mathrm{m}}{1.035 \mathrm{~V}_{1}}=\frac{\rho_{1}}{1.035}\)
⇒ ρ₂ < ρ₁
∴ change in density of mercury is,
\(\frac{\rho_{1}-\rho_{2}}{\rho_{1}}=\frac{\rho_{1}(1-0.9662)}{\rho_{1}}\) = 0.0338
∴ Percentage fractional change = 3.38%

Question 155.
When ‘m’ g of ice is added to ‘M’g of water at 20 °C, state the conditions for m and M for which
i) temperature of the mixture remains 0° C.
ii) temperature of the mixture exceeds 0 °C. (Specific heat of water = s_w = 4.2 × 10³ J kg^-1 °C^-1, latent heat of fusion = L = 3.36 × 10⁵ J kg^-1 )
Answer:
The heat lost by water in going from 20 °C to 0°C,
Q₁ = Ms_w ∆T = \(\frac{\mathrm{M}}{1000}\) × 4.2 × 10³ × (20) = 84M J
Now, heat required to convert m g of ice into water at 0 °C,
Q₂ = mL = \(\frac{\mathrm{M}}{1000}\) × 3.36 × 10⁵ = 336m J

For temperature of mixture to be 0 °C,
Q₂ > Q₁
⇒ 336m > 84M
⇒ m > \(\frac{\mathrm{M}}{4}\)
For temperature of mixture to exceed 0 °C,
Q₂ < Q₁ ⇒ m < \(\frac{\mathrm{M}}{4}\)

Quick Review

Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 42

Multiple Choice Questions

Question 1.
Heat is transferred between two (or more) systems or a system and its surrounding by virtue of
(A) temperature difference.
(B) material difference.
(C) amount of heat difference.
(D) mass difference.
Answer:
(A) temperature difference.

Question 2.
On celsius scale, the two fixed points are marked as
(A) 0°C and 232°C
(B) 32°C and 100°C
(C) 0°C and 100°C
(D) 100°C and 180°C
Answer:
(C) 0°C and 100°C

Question 3.
If the temperature in a room is 30 °C, temperature in degree fahrenheit is
(A) 22 °F
(B) 62 °F
(C) 86 °F
(D) 96 °F
Answer:
(C) 86 °F

Question 4.
If the temperature on Fahrenheit scale is 140 °F, then the same temperature on kelvin scale will be
(A) 60.15 K
(B) 213.15 K
(C) 333.15 K
(D) 413.15 K
Answer:
(C) 333.15 K

Question 5.
In the gas equation, PV = RT, V stands for volume of
(A) any amount of gas .
(B) one gram mole of gas.
(C) one gram of a gas.
(D) one litre of a gas.
Answer:
(B) one gram mole of gas.

Question 6.
1 litre of an ideal gas at 27 °C is heated at constant pressure so as to attain temperature 297 °C. The final volume is approximately
(A) 1.2 litre
(B) 1.9 litre
(C) 19 litre
(D) 2.4 litre
Answer:
(B) 1.9 litre

Question 7.
How much should the pressure be increased in order to decrease the volume of a gas by 10% at a constant temperature?
(A) 7%
(B) 8%
(C) 10%
(D) 11.11%
Answer:
(D) 11.11%

Question 8.
Two rods of same material are equal in length, but one has cross-sectional area double the other. If they are heated through the same temperature then
(A) thick rod expands more.
(B) thin rod expands more.
(C) both rods will expand equally.
(D) none of these.
Answer:
(C) both rods will expand equally.

Question 9.
Two iron bars of same length with unequal radii are heated for the same rise in temperature. The linear expansion will be
(A) more in thin bar.
(B) more in thick bar.
(C) same for both.
(D) less in thick bar.
Answer:
(A) more in thin bar.

Question 10.
Which of the following has minimum coefficient of linear expansion?
(A) Gold
(B) Copper
(C) Platinum
(D) Invar steel
Answer:
(D) Invar steel

Question 11.
The coefficient of cubical expansion of a solid is the increase in volume per unit original volume at 0 °C per degree rise in _________.
(A) pressure
(B) volume
(C) temperature
(D) area
Answer:
(C) temperature

Question 12.
A disc has an area of 0.32 m2 at 20 °C, what will be its area at 100 °C? (α = 2 × 10^-6 / °C)
(A) 0.12 m²
(B) 0.32 m²
(C) 0.51 m²
(D) 0.71 m²
Answer:
(B) 0.32 m²

Question 13.
The coefficient of linear expansion of iron is 1.1 × 10^-5 per K. An iron is 10 m long at 27 °C. Length of the rod will be decreased by 1.1 mm when the temperature of the rod changes to
(A) 0°C
(B) 10 °C
(C) 17 °C
(D) 20 °C
Answer:
(C) 17 °C

Question 14.
The length of an aluminium rod is 120 cm at 20 °C. What is its length at 80 °C, if coefficient of linear expansion of aluminium is 2.5 × 10^-5/°C?
(A) 130.18 cm
(B) 120.18 cm
(C) 110.18 cm
(D) 100.18 cm
Answer:
(B) 120.18 cm

Question 15.
A metal rod having a coefficient of linear expansion of 2 × 10^-5 /°C has a length of 100 cm at 20 °C. The temperature at which it is shortened by 1 mm is
(A) -40 °C
(B) -30 °C
(C) -20 °C
(D) -10 °C
Answer:
(B) -30 °C

Question 16.
Iron sheet 50 cm × 20 cm is heated through 100 °C. If a = 12 × 10^-6 / °C, the change in area is
(A) 2.4 cm²
(B) 3.4 cm²
(C) 4.2 cm²
(D) 5.3 cm²
Answer:
(A) 2.4 cm²

Question 17.
A liquid with coefficient of volume expansion γ is filled in a container of a material having the coefficient of linear expansion α. If the liquid over flows on heating then
(A) γ = 3α
(B) γ < 3α (C) γ > 3α
(D) γ = 3α²
Answer:
(B) γ < 3α

Question 18.
The volume of a metal block changes by 0.18% when it is heated through 20 °C. Its coefficient at cubical expansion will be
(A) 9 × 10^-5 / °C
(B) 3 × 10^-5 / °C
(C) 18 × 10^-5 / °C
(D) 36 × 10^-5 / °C
Answer:
(A) 9 × 10^-5 / °C

Question 19.
The volume of liquid is 830 m³ at 30 °C and 850 m³ at 90 °C. The coefficient of volume expansion of liquid is
(A) 2 × 10^-4 per °C
(B) 8 × 10^-4 per °C
(C) 4 × 10^-4 per °C
(D) 2.5 × 10^-4 per °C
Answer:
(C) 4 × 10^-4 per °C

Question 20.
The superficial expansivity is 1 / x times the cubic expansivity. The value of x is
(A) 2/3
(B) 3/2
(C) 2
(D) 3
Answer:
(B) 3/2

Question 21.
The unit of molar specific heat is
(A) JK^-1 mole^-1
(B) JK mole^-1
(C) J^-1 K^-1mole^-1
(D) JK^-1 mole
Answer:
(A) JK^-1 mole^-1

Question 22.
The S.I. unit of latent heat is
(A) J^-1 kg
(B) J kg^-1
(C) J k^-1 °C
(D) J^-1 kg °C
Answer:
(B) J kg^-1

Question 23.
The slowest mode of transfer of heat is
(A) conduction
(B) convection
(C) radiation
(D) specific heat
Answer:
(A) conduction

Question 24.
The quantity of heat which crosses unit area of a metal plate during conduction depends on
(A) the density of the metal.
(B) the temperature gradient perpendicular to the area.
(C) the temperature to which the metal is heated.
(D) the area of the metal plate.
Answer:
(B) the temperature gradient perpendicular to the area.

Question 25.
Which of the following is not the unit of thermal conductivity?
(A) J/m s °C
(B) K cal/m s K
(C) Watt/m °C
(D) J/m² s °C
Answer:
(D) J/m² s °C

Question 26.
The most desirable combination for the material of a cooking pot is
(A) high specific heat and high conductivity.
(B) low specific heat and high conductivity.
(C) high specific heat and low conductivity.
(D) low specific heat and low conductivity.
Answer:
(B) low specific heat and high conductivity.

Question 27.
While measuring thermal conductivity of a liquid, we keep the upper part hot and lower part cold, so that
(A) radiation may start.
(B) radiation may stop.
(C) convection may start.
(D) convection may be stopped.
Answer:
(D) convection may be stopped.

Question 28.
Convention currents in air in day time is from
(A) land to sea
(B) sea to land
(C) sea to sky
(D) land to land
Answer:
(B) sea to land

Question 29.
One end of a metal rod one metre long is kept in ice and the other end is at 100 °C. What is temperature gradient throughout the rod?
(A) 10 °C/m
(B) 100 °C/m
(C) 50 °C/m
(D) 1 °C/m
Answer:
(B) 100 °C/m

Question 30.
__________ amount of heat is required to raise the temperature of 100 g of kerosene from 10 °C to 30 °C (Given: specific heat of kerosene is 0.51 kcal/kg °C)
(A) 0.102 kcal
(B) 1.02 kcal
(C) 10.2 kcal
(D) 102 kcal
Answer:
(B) 1.02 kcal

Question 31.
One end of copper rod is in contact with water at 100 °C and the other end in contact with ice at 0 °C. The length of the rod is 100 cm. At a point which is at a distance of 35 cm from the cold end, temperature is (assuming steady state heat flow)
(A) 35 °C
(B) 65 °C
(C) 56 °C
(D) 53 °C
Answer:
(A) 35 °C

Question 32.
In Newton’s law of cooling, the rate of fall of temperature
(A) is constant
(B) increases
(C) decreases
(D) doubles
Answer:
(C) decreases

Question 33.
The metal sphere cools at 1 °C/min, when its temperature is 50 °C. If the temperature of environment is 30 °C, its rate of cooling at 35 °C is
(A) 0.25 °C/min
(B) 0.5 °C/min
(C) 0.75 °C/min
(D) 0.4 °C/min
Answer:
(A) 0.25 °C/min

Competitive Corner

Question 1.
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminum rod is:
(α_Cu = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5 K^-1)
(A) 88 cm
(B) 68 cm
(C) 6.8 cm
(D) 1 13.9 cm
Answer:
(B) 68 cm
Hint:
L_Cu α_Cu ∆T = L_Al α_Al ∆T
∴ 88 × (1.7 × 10^-5) = L_Al(2.2 × 10^-5)
∴ L_Al = \(\frac{88 \times 1.7}{2.2}\)
∴ L_Al = 68 cm

Question 2.
The unit of thermal conductivity is:
(A) W m K^-1
(B) W m^-1 K^-1
(C) JmK ^-1
(D) Jm^-1K^-1
Answer:
(B) W m^-1 K^-1
Hint:
Q = kA \(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) t
Q = quantity of heat conducted.
A = area of cross section
t = time for which heat is passed
\(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) = temperature gradient
∴ K = \(\frac{Q}{A t\left(\frac{\Delta T}{\Delta x}\right)}\)
∴ Unit of k = W m^-1 K^-1

Question 3.
A deep rectangular pond of surface area A, containing water (density = p, specific heat capacity = s), is located in a region where the outside air temperature is at a steady value of -26°C. The thickness of the frozen ice layer in this pond, at a certain instant is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant, would be given by
(A) 26K/ρx(L + 4s)
(B) 26K/ρx(L – 4s)
(C) 26K/(ρx² L)
(D) 26K/(ρxL)
Answer:
(D) 26K/(ρxL)
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 43

Question 4.
An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80°C to 70°C. The time taken to cool for the same object from 70°C to 60°C would be nearly.
(A) 15 min
(B) 10mm
(C) 12mm
(D) 20mm
Answer:
(A) 15 min
Hint:
By Newton’s law of cooling,
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 44

Question 5.
A thermally insulated vessel contains 150 g of water at 0 °C. Then the air from the vessel ispumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0 °C itself. The mass of evaporated water will be closest to :
(Latent heat of vaporization of water = 2.10 × 10⁶ J kg^-1 and Latent heat of Fusion of water = 3.36 × 10⁵ J kg^-1)
(A) 150g
(B) 20g
(C) 130g
(D) 35g
Answer:
(B) 20g
Hint:
m= 150 g = 0.15 kg
The heat required to evaporate ‘m’ grams of water,
∆Q_required = mL_v ………. ( 1)
(015 – m) is the amount of mass that converts into ice
∴ ∆Q_released = (0.15 – m) L_f …………. (2)
Now, amount of heat required = amount of heat released
∴ From (1) and (2),
mL_v = (0.15 – m)L_f
∴ m(L_f + L_v) = 0.15 L_f
∴ m = \(\frac{0.15 \mathrm{~L}_{\mathrm{f}}}{\mathrm{L}_{\mathrm{f}}+\mathrm{L}_{\mathrm{v}}}\)
∴ m = \(\frac{0.15 \times 3.36 \times 10^{5}}{2.10 \times 10^{6}+3.36 \times 10^{5}}\)
∴ m=0.0206kg ≈ 20g

Question 6.
A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, thetemperature of the system is found to be 75 °C. T is given by: (Given: room temperature = 30°C, specific heat of copper 0.1 cal/g °C)
(A) 1250°C
(B) 825°C
(C) 800°C
(D) 885°C
Answer:
(D) 885°C
Hint:
Heat lost by copper ball = Heat gained by calorimeter and water
∴m_bs_c∆T₁ = m_cc_c∆T₂ + m_ws_w∆T₂
∴ (100)(0.1)(T – 75) = (100)(0.1)(75 – 30) + (170)(1)(75 – 30)
10(T – 75) = 450 + 7650 = 8100
T – 75 = 810
T = 885 °C

Question 7.
Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures, which one of the following relations holds good?
(A) α₁²l₂ = α₂²l₁
(B) α₁l₁ = α₂l₂
(C) α₁l₂ = α₂l₁
(D) α₁l₂² = α₂l₁²
Answer:
(B) α₁l₁ = α₂l₂
Hint:
∆L = L(1 + α∆t)
i.e. ∆l₂ = l₂(1 + α₂∆t)
and ∆l₁ =1₁(1 + α₁∆t)
It is given,
l₂– l₁ = ∆tl₂ – ∆ll₁
∴ l₂ – l₁ = l₂ (1 + α₂∆t) – l₁ (1+ α₁∆t)
= l₂ + l₂α₂∆t – l₁ – l₁α₁∆t
∴ – l₂ + l₂ + l₂α₂∆t = -l₁ + l₁ + l₁α₁∆t
∴ l₂α₂∆t = l₁α₁∆t
i.e., l₂α₂ = l₁α₁

Question 8.
A pendulum clock loses 12 s a day if the temperature is 40 °C and gains 4 s a day if the temperature is 20 °C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (a) of the metal of the pendulum shaft are respectively:
(A) 60 πC, a = 1.85 × 10^-4/πC
(B) 30 πC, a = 1.85 × 10^-3/πC
(C) 55 πC, a = 1.85 × 10^-2/πC
(D) 25 πC, a = 1.85 × 10^-5/πC
Answer:
(D) 25 πC, a = 1.85 × 10^-5/πC
Hint:
Period of pendulum, T = 2π\(\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\)
∴ T ∝ \(\sqrt{\mathrm{L}}\)
But, L = L₀ (1 + α∆t)
∴ T ∝ \(\sqrt{\mathrm{L}_{0}(1+\alpha \Delta \mathrm{t})}\)
As L₀ is constant,
⇒ T ∝ (1 + α ∆t)
Calculating fractional change in time period of pendulum, \(\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2}(\alpha \Delta \mathrm{t})\)
For the given pendulum,
T = 24 × 60 × 60 = 864000 s
When t1 = 40 °C, ∆T = 12 s,
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 45

Question 9.
A piece of ice falls from a height h so that it melts completely. Only one — quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4 × 10⁵ J/kg and g = 10 N/kg]
(A) 136 km
(B) 68km
(C) 34km
(D) 544km
Answer:
(A) 136 km
Hint:
When the piece of ice falls from the height h, it possesses potential energy, mgh.
This P.E. is converted to heat energy.
∴ Q = mgh
But only \(\frac{1^{\text {th }}}{4}\) of it is absorbed by ice which is used to change the state.
∴ \(\frac{\mathrm{mgh}}{4}\) = mL
∴ \(\frac{10 \times \mathrm{h}}{4}\) = 3.4 × 10⁵
∴ h = 13.6 × 10⁴ m = 136km

Question 10.
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be
(A) T
(B) \(\frac{7}{4}\) T
(C) \(\frac{3}{2}\) T
(D) \(\frac{4}{3}\) T
Answer:
(C) \(\frac{3}{2}\) T
Hint:
By Newton’s law of cooling,
Maharashtra Board Class 11 Physics Important Questions Chapter 7 Thermal Properties of Matter 46

Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids

July 24, 2024July 23, 2024 by Prasanna

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 6 Mechanical Properties of Solids Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 6 Mechanical Properties of Solids

Question 1.
Explain the equilibrium state of solids at a given temperature.
Answer:

Solids are made up of atoms or a group of atoms placed in a definite geometric arrangement.
This arrangement is decided by nature so that the resultant force acting on each constituent due to others is zero. This is the equilibrium state of a solid at room temperature.
This equilibrium arrangement does not change with time but it can only change when an external stimulus, like compressive force is applied to a solid from all sides.
The constituents vibrate about their equilibrium positions even at very low temperature but cannot leave their fixed positions.
As a result, the solids possess a definite shape and size.

Question 2.
Explain the effects of applied force on a rigid body.
Answer:

When an external force is applied to a solid, the constituents are slightly displaced and restoring forces are developed in it.
These restoring forces try to bring the constituents back to their equilibrium positions so that the solid can regain its shape.
When the deforming forces are removed, the inter-atomic forces tend to restore the original positions of the molecules and thus the body regains its original shape and size.

Question 3.
Explain the concept of deforming force with the help of examples.
Answer:

When a force (within specific limit) is applied to a solid (which is not free to move), the size or shape or both change due to changes in the relative positions of molecules. Such a force is called deforming force.
The larger the deforming force on a body, the larger is its deformation.
Deformation could be in the form of change in length of a wire, change in volume of an object or change in shape of a body.
Examples:
- When a deforming force such as stretching, is applied to a rubber band, it gets deformed (elongated) but when the force is removed, it regains its original length.
- When a similar force is applied to a dough or a clay, it also gets deformed but it does not regain its original shape and size after removal of the deforming force.

Question 4.
Define plasticity.
Answer:
If a body does not regain its original shape and size and retains its altered shape or size upon removal of the deforming force, it is called a plastic body and the property is called plasticity.

Question 5.
When is a body said to be perfectly elastic? Give an example for perfectly elastic body and perfectly plastic body.
Answer:

If a body regains its original shape and size completely and instantaneously upon removal of the deforming force, then it is said to be perfectly elastic.

- There is no solid which is perfectly elastic or perfectly plastic,
- The best example of a near perfectly elastic body is quartz fibre and that of a plastic body is putty.

Question 6.
State SI unit and dimensions of strain.
Answer:
Strain is the ratio of two similar quantities. Hence strain is a dimensionless physical quantity and it has no units.

Question 7.
State and explain longitudinal stress (Tensile stress and Compressive stress).
Answer:
i. Stress produced by a deforming force acting along the length of a body or a rod is called longitudinal stress.

ii. Tensile stress: Consider a force F is applied along a length of a wire, or perpendicular to its cross-section A of a wire (or along its length).

Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 1

[Note: Elongation of wire ∆l is exaggerated for explanation.]
This produces an elongation in the wire and the length of the wire increases, then
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 2

iii. Compressive stress: When a rod is pushed from two ends with equal and opposite forces.
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 3
[Note: Compression of wire ∆l is exaggerated for explanation.]
This restoring force per unit area is called compressive stress.
Compressive stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\)

Question 8.
State and explain longitudinal strain (Tensile strain or Linear strain).
Answer:
The strain produced by a tensile deforming force is called longitudinal strain (tensile strain or linear strain,).
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 4
where, L = Original length,
∆l = Change in length.

Question 9.
State and explain volume stress / hydraulic stress.
Answer:
When a deforming force acting on a body produces change in its volume, the stress is called volume stress.
Volume stress/hydraulic stress:
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 5
[Note: Change in size is exaggerated for explanation.]

Let \(\vec{F}\) be a force acting perpendicular to the entire surface of the body.
It acts normally and uniformly all over the surface area A of the body.
Such a stress which produces change in size but no change in shape is called volume stress.
Volume stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\)
Volume stress produces change in size without change in shape of body, hence it is also called as hydraulic or hydrostatic volume stress.

Question 10.
Explain volume strain.
Answer:

A deforming force acting perpendicular to the entire surface of a body produces a volume strain.
where ∆V = change in volume,
V = original volume.

Question 11.
Define and explain shearing stress.
Answer:
The restoring force per unit area developed due to the applied tangential force is called shearing stress or tangential stress.
Shearing stress:

When a deforming force acting on a body produces change in the shape of a body shearing stress is produced.
Consider ABCD as a front face of a cube, a tangential force is applied to the cube so that the bottom of the cube is fixed and only the top surface is slightly displaced.

Question 12.
Explain shearing strain.
Answer:

The relative displacement of the bottom face and the top face of the cube is called shearing strain.
Shearing strain = \(\frac{\Delta l}{\mathrm{~L}}\) = tan θ
where, ∆l = displaced length,
L = Original length.
When ∆l is very small,
tan θ ≈ θ and shearing strain = θ

Question 13.
State and explain Hooke’s law.
Answer:
Statement: Within elastic limit, stress is directly proportional to strain.
Explanation:

According to Hooke’s law,
Stress-Strain

This constant of proportionality is called modulus of elasticity.
Modulus of elasticity of a material is the slope of stress-strain curve in elastic deformation region and depends on the nature of the material.
The graph of strain (on X-axis) and stress (on Y-axis) within elastic limit is shown in the figure.

Question 14.
Define elastic limit.
Answer:
The maximum value of stress upto which stress is directly proportional to strain is called the elastic limit.

Question 15.
Define modulus of elasticity.
Answer:
The modulus of elasticity of a material is the ratio of stress to the corresponding strain.

Question 16.
State different types of modulus of elasticity.
Answer:

Young’s modulus (Y): It is the modulus of elasticity related to change in length of an object like a metal wire, rod, beam, etc., due to the applied deforming force.
Bulk modulus (K): It is the modulus of elasticity related to change in volume of an object due to applied deforming force.
Shear modulus or Modulus of rigidity (η): The modulus of elasticity related to change in shape of an object is called modulus of rigidity.

Question 17.
Explain the usefulness of Young’s modulus.
Answer:

Young’s modulus indicates the resistance of an elastic solid to elongation or compression.
Young’s modulus of a material is useful for characterization of an object subjected to compression or tension.

Question 18.
Within elastic limit, prove that Young’s modulus of material of wire is the stress required to double the length of wire.
Answer:

(i) Let, L = Initial length of wire
2 L = Final length of wire
∴ Increase in length = ∆l = 2L – L = L

(ii) Longitudinal strain of wire = \(\frac{\Delta l}{\mathrm{~L}}=\frac{\mathrm{L}}{\mathrm{L}}=1\)

(iii)
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 11
∴ Y = Longitudinal stress

(iv) Hence, Young’s modulus of material of wire is the stress required to double the length of wire.

Question 19.
What is bulk modulus? Derive an expression for bulk modulus.
Answer:
Definition:
Bulk modulus is defined as the ratio of volume stress to volume strain.
It is denoted by ‘K’.
Unit: N/m² or Pa in SI system
Dimensions: [L^-1M¹T^-2]
Expression for bulk modulus:

(i) If a sphere made from rubber is completely immersed in a liquid, it will be uniformly compressed from all sides.
Let, F = Compressive force,
dP = Change in pressure,
dV = Change in volume,
V = Original volume.

(ii) Volume stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\) = dP

Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 12

(iii) The negative sign indicates that there is a decrease in volume.
The magnitude of the volume strain is \(\frac{\mathrm{dV}}{\mathrm{V}}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 13

Question 20.
Explain the usefulness of bulk modulus.
Answer:

Bulk modulus indicates the resistance of gases, liquids or solids to change their volume.
Materials with small bulk modulus and large compressibility are easier to compress.

Question 21.
State few applications of bulk modulus.
Answer:

When a balloon is filled with air at high pressure, its walls experience a force from within. It tries to expand the balloon and change its size without changing shape. When the volume stress exceeds the limit of bulk elasticity, the balloon explodes.
A gas cylinder explodes when the pressure inside it exceeds the limit of bulk elasticity of its material.
A submarine when submerged under water is under volume stress. The depth it can reach within water depends upon its limit of bulk elasticity.

Question 22.
Define compressibility. State its unit and dimensions.
Answer:

The reciprocal of bulk modulus of elasticity is called compressibility of the material.
Compressibility is the fractional decrease in volume, (-dV/V) per unit increase in pressure.
Compressibility = \(\frac{-\mathrm{dV}}{\mathrm{V} \mathrm{dP}}\)
Unit: m²/N or Pa^-1 in SI system.
Dimensions: [L¹M^-1T²]

Question 23.
What is modulus of rigidity? Derive an expression for it.
Answer:
Definition: It is defined (is the ratio of shear stress to shear strain within elastic limit.
It is denoted by ‘η’.
Unit: N/m² or Pa in SI system.
Dimension: [L^-1M¹T^-2]

Expression for modulus of rigidity:

Consider a solid cube as shown in the figure.

Let, F = Tangential force
A = Cross sectional area
∆l = Relative displaced length
l = Original length
θ = Shear strain
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 15
[Note: Displacement of upper surface is exaggerated for explanation.]
The forces applied on the block is subjected to a shear stress,
Shearing stress = F/A
The comer angle which changes by a small amount θ (expressed in radian) is given by,
Shearing strain = \(\frac{\Delta l}{l}\) ≈ θ
From definition,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 16

Question 24.
What does modulus of rigidity indicate?
Answer:
Modulus of rigidity indicates resistance offered by solid to change in its shape.

Question 25.
Explain the change of diameter of a wire when it is stretched and compressed.
Answer:
i. When a wire is fixed at one end and a force is applied at its free end so that the wire gets stretched, length of the wire increases and at the same time, its diameter decreases, i.e., the wire becomes longer and thinner as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 17
[Note: Linear expansion ∆l is exaggerated far explanation.]

ii. If equal and opposite forces are applied to an object along its length inwards, the object gets compressed. There is a decrease in dimensions along its length and at the same time there is an increase in its dimensions perpendicular to its length. When length of the wire decreases, its diameter increases.
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 18

[Note: Compression of wire ∆l is exaggerated for explanation.]

Question 26.
Derive expression for Poisson’s ratio.
Answer:
i. Let,
L = original length
l = increase/decrease in length
D = original diameter
d = change in diameter

ii. The ratio of change in dimensions to original dimensions in the direction of the applied force is called linear strain.
Linear strain = \(\frac{l}{\mathrm{~L}}\) …. (1)

iii. The ratio of change in dimensions to original dimensions in a direction perpendicular to the applied force is called lateral strain.
Lateral strain = \(\frac{\mathrm{d}}{\mathrm{D}}\) ….(2)

iv. Poisson’s ration,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 19

Question 27.
A wire of length 20 m and area of cross section 1.25 × 10^-4 m² is subjected to a load of 2.5 kg. (1 kg wt = 9.8 N). The elongation produced in wire is 1 × 10^-4 m. Calculate Young’s modulus of the material.
Solution:
L = 20 m, A = 1.25 × 10^-4m²,
F = mg = 2.5 × 9.8 N, l = 10^-4 m
Formula Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
To find: Young’s modulus (Y)
Calculation: From formula,
Y = \(\frac{2.5 \times 9.8 \times 20}{1.25 \times 10^{-4} \times 10^{-4}}\) = 3.92 × 10¹⁰Nm^-2
Answer: The Young’s modulus of the material is 3.92 × 10¹⁰ Nm^-2.

Question 28.
A wire of diameter 0.5 mm and length 2 m is stretched by applying a force of 2 kg wt. Calculate the increase in length of the wire, (g = 9.8 m/s², Y = 9 × 10¹⁰ N/m²)
Solution:
Given:
L = 2 m, F = 2 kg wt, d = 0.5 mm = 5 × 10^-4 m,
∴ r = \(\frac{\mathrm{d}}{2}\) = 2.5 × 10^-4 m.
Y = 9 × 10¹⁰ N/m², g = 9.8 m/s²
To find: Increase in length (l)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{MgL}}{\pi \mathrm{r}^{2} l}\)
Calculation:
From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 20
Answer:
The increase in length of the wire is 2.218 × 10^-3 m.

Question 29.
A brass wire of length 4.5 m with cross-section area of 3 × 10^-5 m² and a copper wire of length 5.0 m with cross section area 4 × 10^-5 m² are stretched by the same load. The same elongation is produced in both the wires. Find the ratio of Young’s modulus of brass and copper.
Solution:
L_B = 4.5 m, A_B = 3 × 10^-5 m²
L_C = 5 m, A_C = 4 × 10^-5 m² l_B = _C, F_B = F_C
To find: Ratio of Young’s modulus \(\left(\frac{Y_{B}}{Y_{C}}\right)\)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Calculation: For brass,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 21
= 1.2
Answer:
The ratio of Young’s modulus of brass and copper is 1.2 : 1.

Question 30.
The length of wire increases by 9 mm when weight of 2.5 kg is hung from the free end of wire. If all conditions are kept the same and the radius of wire is made thrice the original radius, find the increase in length.
Solution:
Given; l₁ = 9mm = 9 × 10^-3m,
M = 2.5 kg, r₂ = 3r₁,
Y₁ = Y₂ = Y (material is same)
To find: Increase in length (l₂)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 22
Answer:
The longitudinal strain produced in 1st wire is

Question 31.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in the figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.(Y_s = 2.0 × 10¹¹ Nm^-2, Y_B = 0.91 × 10¹¹ Nm^-2) (NCERT)
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 23
Solution:
Given: D = 0.25 cm = 0.25 × 10^-2 m,
L_S = 1.5 m, L_B = 1 m,
Y_S = 2.0 × 10¹¹ Nm^-2,
Y_B = 0.91 × 10¹¹ Nm^-2
To find: Elongations of brass wire (l_B)
Elongations of steel wire (l_S)

Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)

Calculation: Since, A = \(\frac{\pi \mathrm{D}^{2}}{4}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 24
Answer:
The elongation of the steel wire is 1.5 × 10^-4 m and that of brass wires is 1.32 × 10^-4 m.

Question 32.
One end of steel wire is fixed to a ceiling and load of 2.5 kg is attached to the free end of the wire. Another identical wire is attached to the bottom of load and another load of 2.0 kg, is attached to the lower end of this wire as shown in the figure. Compute the longitudinal strain produced in both the wires, if the cross-sectional area of wires is 10^-4m², (Y_steel = 20 × 10¹⁰N/m²)
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 25
Solution:
Given: M₁ = 2.5 kg, M₂ = 2kg, A = 10^-4m², Y_steel = 20 × 10¹⁰ N/m², L₁ = L₂ = L
To find: Longitudinal strain of 1^st wire (Strain₁).
Longitudinal strain of 2^nd wire (Strain₂)
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 26
Answer:
The longitudinal strain produced in 1^st wire is 1.225 × 10^-6 and in 2^nd wire is 2.205 × 10^-6

Question 33.
A steel wire having cross-sectional area 2 mm² is stretched by 10 N. Find the lateral strain produced in the wire. (Given: Y for steel = 2 × 10¹¹ N/m², Poisson’s ratio σ = 0.29)
Solution:
Given: A = 2 mm² = 2 × 10^-6 m²,
F = 10 N, Y_stee; = 2 × 10¹¹ N/m², σ = 0.29

To find: Lateral strain
Formulae:
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 27
Calculation: From formula (i),
longitudinal stress = \(\frac{10}{2 \times 10^{-6}}\)
= 5 × 10⁶ N/m²
From formula (ii),

From formula (iii),
lateral strain = σ × longitudinal strain
= 0.29 × 2.5 × 10^-5
∴ lateral strain = 7.25 × 10^-6
Answer:
Lateral strain produced in the wire is 7.25 × 10^-6.

Question 34.
A brass wire of radius 1 mm is loaded by a mass of 31.42 kg. What would be the decrease in its radius? (Y = 9 × 10¹⁰ N/m², Poisson’s ratio σ = 0.36)
Solution:
Given: R = 1 mm = 1 × 10^-3 m, M = 31.42 kg, Y = 9 × 10¹⁰ N/m², σ = 0.36
To find: Decrease in radius (r)

Formulae:

i. Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
ii. σ = \(\frac{\mathrm{Lr}}{l \mathrm{R}}\)

Calculation: From formula (i),
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 29
Answer:
The decrease in radius of the brass wire would be 3.92 × 10^-7 m

Question 35.
A metal cube of side 1 m is subjected to a force. The force acts normally on the whole surface of cube and its volume changes by 1.5 × 10^-5 m³. The bulk modulus of metal is 6.6 × 10¹⁰ N/m². Calculate the change in pressure.
Solution:
Given: L = 1 m, dV = 1.5 × 10^-5 m³,
K = 6.6 × 10¹⁰ N/m².
To find: Change in pressure (dP)
Formula: K = V\(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 30
Answer:
The change in pressure is 9.9 × 10⁵ N/m².

Question 36.
Determine the volume contraction of solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ P.a. (Bulk modulus of copper = 140 × 10⁹ pa)
Solution:
Given: L = 10 cm =0.1 m, ∆P = 7 × 10⁶ Pa, K = 140 × 10⁹ pa
To find: Volume contraction (dV)
Formula: K = V × \(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,
dV = \(\mathrm{L}^{3} \times \frac{\mathrm{dP}}{\mathrm{K}}\) = (0.1)³ × \(\frac{7 \times 10^{6}}{140 \times 10^{9}}\)
∴ dV = 5 × 10^-8 m³
Answer:
The volume contraction of solid copper cube is 5 × 10^-8 m³

Question 37.
Calculate the modulus of rigidity of a metal, if a metal cube of side 40 cm is subjected to a shearing force of 2000 N. The upper surface is displaced through 0.5 cm with respect to the bottom. Calculate the modulus of rigidity of the metal.
Solution:
Given: L = 40 cm = 0.4 m,
F = 2000 N = 2 × 10³N
l = 0.5 cm = 0.005 m, A = L² = 0.16 m²
To find: Modulus of rigidity (η)

Formulae:

i. θ = \(\frac{l}{\mathrm{~L}}\)
ii. η = \(\frac{\mathrm{F}}{\mathrm{A} \theta}\)

Calculation:

From formula (i),
θ = \(\frac{0.005}{0.4}\) = 0.0125
From formula (ii),
η = \(\frac{2 \times 10^{3}}{0.16 \times 0.0125}\) = 1 × 10⁶ N/m²
Answer:
The modulus of rigidity of the metal cube is 1 × 10⁶ N/m².

Question 38.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N forces producing only elastic deformation. Calculate the resulting strain. (Rigidity modulus of copper = 42 × 10⁹ N m^-2)
Solution:
Given: A = 15.2 × 19.14 × 10^-6 m²,
F = 44500 N, η = 42 × 10⁹ N m^-2
To find: Strain (θ)
Formula: η = \(\frac{\mathrm{F}}{\mathrm{A} \theta}\)

Calculation:
From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 31
Answer:
The strain produced in the piece of copper is 3.64 × 10^-3

Question 39.
A copper metal cube has each side of length 1 m. The bottom edge of cube is fixed and tangential force 4.2 × 10⁸ N is applied to top surface. Calculate the lateral displacement of the top surface, if modulus of rigidity of copper is 14 × 10¹⁰ N/m².
Solution:
Given: L = h = 1 m, F = 4.2 × 10⁸ N, η = 1.4 × 10¹¹ N/m²
To find: Lateral displacement (x)
Formula: η = \(\frac{\mathrm{F}}{\mathrm{A} / \theta}\) = \(\frac{\mathrm{Fh}}{\mathrm{Ax}}\)
Calculation: From formula,
x = \(\frac{\mathrm{Fh}}{\mathrm{A\eta}}\)
∴ x = \(\frac{4.2 \times 10^{8} \times 1}{(1 \times 1) \times 1.4 \times 10^{11}}\) = 3 × 10^-3 m
∴ x = 3 mm
Answer:
The lateral displacement of top is 3 mm.

Question 40.
The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.0 15 mm. Find the strain and shearing force. (Modulus of rigidity: η = 4.5 × 10¹⁰ N/m²)
Solution:
Given: A = 0.5 m × 0.5 m = 0.25 m²,
h = 1 cm = 10^-2m,
x = 0.015 mm = 15 × 10^-6m
η = 4.5 × 10¹⁰ N/m²
To find: Strain (θ). Shearing force (F)

Formulae:

i. θ = \(\frac{\mathrm{x}}{\mathrm{h}}\)
ii. F = ηAθ

Calculations:

Using formula (i),
θ = \(\frac{15 \times 10^{-6}}{10^{-2}}\) = 1.5 × 10^-3
Using formula (ii),
F = 4.5 × 10¹⁰ × 0.25 × 1.5 × 10^-3
= 1.688 × 10⁷ N
Answer:
Shearing force is 1.688 × 10⁷ N and strain is 1.5 × 10^-3

Question 41.
Explain the behaviour of metal wire under increasing load.
Answer:
Consider a metal wire suspended vertically from a rigid support and stretched by applying load to its lower end. The load is gradually increased in small steps until the wire breaks. The elongation produced in the wire is measured during each step. Stress and strain are noted for each load and a graph is drawn by taking tensile strain along X-axis and tensile stress along Y-axis. It is a stress-strain curve as shown in the figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 32

Proportional limit: The initial part of the graph is a straight line OA. This is the region in which Hooke’s law is obeyed and stress is directly proportional to stain. The straight line portion ends at A. The stress at this point is called proportional limit.
Yield point: If the load is further increased till point B is reached, stress and strain are no longer proportional and Hooke’s law is not valid. If the load is gradually removed starting at any point between O and B. The curve is retracted until the wire regains its original length. The change is reversible. The material of the wire shows elastic behaviour in the region OB. Point B is called the yield point. It is also known as the elastic limit.
Permanent Set: When the stress is increased beyond point B, the strain continues to increase. If the load is removed at any point beyond B, for example (at C), the material does not regain its original length. It follows the line CE. Length of the wire when there is no stress is greater than the original length. The deformation is irreversible and the material has acquired a permanent set.
Fracture point: Further increase in load causes a large increase in strain for relatively small increase in stress, until a point D is reached at which fracture takes place. The material shows plastic flow or plastic deformation from point B to point D. The material does not regain its original state when the stress is removed. The deformation is called plastic deformation.

Question 42.
Stress-strain curve for two materials A and B are shown in the figure. The graphs are drawn to the same scale.

Which material has greater Young’s modulus?
Which of the two is the stronger material? (NCERT)

Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 33
Answer:

For a given strain, stress for A is more than that of B. Hence, Young’s modulus (= stress/strain) is greater for A than that of B.
Material A is stronger than B because A can bear greater stress before the breaking of the wire.

Question 43.
Figure shows the stress-strain curve for a given material. What are

Young’s modulus and
approximate yield strength for this material? (NCERT)

Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 34
Answer:

The graph, implies stress of 150 × 10⁶ N m^-2 corresponds to a strain of 0.002. Therefore, Young’s modulus is,
The yield strength for the material is less than 300 × 10⁶ N m^-2, i.e.. 3 × 10⁸ N m^-2 and greater than 2.5 × 10⁸ N m^-2.

Question 44.
Explain the following terms:

Ductile
Malleable

Answer:

Metals such as copper, aluminium, wrought iron, etc. have large plastic range of extension. They lengthen considerably and undergo plastic deformation till they break. They are called ductile.
Metals such as gold, silver which can be hammered into thin sheets are called malleable.

Question 45.
Define strain energy.
Answer:
The elastic potential energy gained by a wire during elongation b a stretching force is called as strain energy.

Question 46.
A steel wire is acted upon by a load of 10 N. Calculate the extension produced in the wire, if the strain energy stored in the wire is 1.1 × 10^-3 J.
Solution:
Given: F = 10 N, Strain energy =1.1 × 10^-3 J,
To find: Extension (l)
Formula: W = \(\frac{1}{2}\) × F × l
Calculation:
From formula,
l = \(\frac{2 \mathrm{~W}}{\mathrm{~F}}\) = \(\frac{2 \times 1.1 \times 10^{-3}}{10}\) = 2.2 × 10^-4 m.
Answer:
The extension produced in the wire is 2.2 × 10^-4 m.

Question 47.
Calculate the strain energy per unit volume in a brass wire of length 3 m and area of cross-section 0.6 mm² when it is stretched by 3 mm and a force of 6 kg-wt is applied to its free end.
Solution:
Given: L = 3 m, F = 6 kg wt = 6 × 9.8N = 58.8N, A = 0.6 mm² = 0.6 × 10^-6 m², l = 3 mm = 3 × 10^-3 m
To find: Strain energy per unit volume (u)

Formulae:

i. Stress = \(\frac{F}{A}\)
ii. Strain = \(\frac{l}{L}\)
iii. u = \(\frac{1}{2}\) × Stress × Strain

Calculation:

From formula (i),
Stress = \(\frac{F}{A}\) = \(\frac{58.8}{0.6 \times 10^{-6}}\) = 98 × 10⁶ N/m²
From formula (ii),
Strain = \(\frac{l}{L}\) = \(\frac{3 \times 10^{-3}}{3}\) = 10^-3
From formula (iii),
u = \(\frac{1}{2}\) × Stress × Strain
= \(\frac{1}{2}\) × 98 × 10⁶ × 10^-3
u = 49 × 10³ J/m³
Answer:
Strain energy per unit volume in the brass wire is 49 × 10³ J/m³

Question 48.
A steel wire of diameter 1 × 10^-3 m is stretched by a force of 20 N. Calculate the strain energy per unit volume. (Y_steel = 2 × 10¹¹ N/m²)
Solution:
Given: d = 1 × 10^-3 m, r = 5 × 10^-4 m, y_steel = 2 × 10¹¹ N/m²
To find: Strain energy per unit volume
Formula: Strain energy per unit volume
= \(\frac{1}{2}\) × \(\frac{\text { (stress) }^{2}}{\mathrm{Y}}\)
Calculation:
From formula.
Strain energy per unit volume
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 36
Answer:
The strain energy per unit volume of the steel wire is 1621 J.

Question 49.
A uniform steel wire of length 3 m and area of cross section 2 mm² is extended through 3 mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. (Y_steel = 20 × 10¹⁰ N/m²)
Solution:
Given: L = 3 m, A = 2 mm² = 2 × 10^-6 m²,
l = 3 mm = 3 × 10^-3 m,
Y_steel = 20 × 10¹⁰ N/m²
To find: Energy stored (U)
Formula: W = \(\frac{1}{2}\) × F × l
Calculation: Since, Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 37
Answer:
The energy stored in the steel wire is 0.6 J.

Question 50.
Explain the following terms:
i. Hardness
ii. Strength
iii. Toughness
Answer:

i. Hardness:

Hardness is the property of a material which enables it to resist plastic deformation.
Hard materials have little ductility and they are brittle to some extent.
The term hardness also refers to stiffness or resistance to bending, scratching, abrasion or cutting.
It is the property of a material which gives it the ability to resist permanent deformation when a load is applied to it.
The greater the hardness, greater is the resistance to deformation.
Hardness of material is different from its strength and toughness.
The most well known example for hard material is diamond, while metal with very low hardness is aluminium.

ii. Strength:

If a force is applied to a body, it produces deformation in it.
Higher is the force required for deformation, the stronger is the material, i.e., the material has more strength.
Steel has high strength whereas plasticine clay is not strong because it gets easily deformed even by a small force.

iii. Toughness:

Toughness is the ability of a material to resist fracturing when a force is applied to it.
Plasticine clay is relatively tough as it can be stretched and deformed due to applied force without breaking.

Question 51.
Explain the concept of frictional force.
Answer:

Whenever the surface of one body slides over another, each body exerts a certain amount of force on other body.
These forces are tangential to the surfaces. The force on each body is opposite to the direction of motion between two bodies.
It prevents or opposes the relative motion between two bodies.
It is common experience that an object placed on any surface does not move easily when a small force is applied to it.
This is because of certain force of opposition acting between the surface of the object and the surface on which it is placed.
To initiate any motion between pair of surfaces, we need a certain minimum force. Also, after the motion begins, it is constantly opposed by some natural force.
This mechanical force between two solid surfaces in contact with each other is called as frictional force.

Question 52.
Explain few examples of friction.
Answer:

A rolling ball comes to rest after covering a finite distance on playground because of friction.
Our foot ware is provided with designs at the bottom of its sole so as to produce force of opposition to avoid slipping. It is difficult to walk without such opposing force. When we try to walk fast on polished flooring at home with soap water spread on it. There is a possibility of slipping due to lack of force of friction.
Relative motion between solids and fluids (i.e. liquids and gases) is also opposed naturally by friction, eg.: a boat on the surface of water experiences opposition to its motion.

Question 53.
Explain the origin of friction.
Answer:

If smooth surfaces are observed under powerful microscope, many irregularities and projections are observed.
Friction arises due to interlocking of these irregularities between two surfaces in contact.
The surfaces can be made extremely smooth by polishing to avoid irregularities but then case also, friction does not decrease but may increase.
Hence the interlocking of irregularities is not the real cause of friction.
According to modem theory, cause of friction is the force of attraction between molecules of two surfaces in actual contact in addition to the. force due to the interlocking between the two surfaces.
When one body is in contact with another body, the real microscopic area in contact is very small due to irregularities in contact.
Due to small area, pressures at points of contact is very high. Hence there is strong force of attraction between the surfaces in contact.
When the surfaces in contact become more and more smooth, the actual area of contact goes on increasing.
Due to this, the force of attraction between the molecules increases and hence the friction also increases.

Question 54.
State the following terms:

Cohesive force
Adhesive force

Answer:

When two Surfaces are of the same material, the force of attraction between them is called cohesive force.
When two surfaces are of different materials, the force of attraction between them is called adhesive force.

Question 55.
Explain the concept of static friction.
Answer:

Consider a wooden block placed on a horizontal surface as shown in the figure and small horizontal force F is applied to it.
The block does not move with this force as it cannot overcome the frictional force between the block and horizontal surface.
In this case the force of static friction is equal to F and balances it.
The frictional force which balances applied force when the body is static is called force of static friction. In other words, static friction prevents sliding motion.
If we keep increasing F, a stage will come
- For F < F_max, the force of static friction is equal to F.
- when for F = F_max, the object will start moving.
- For F ≥ F_max, the kinetic friction comes into play.
Static friction opposes impending motion i.e. the motion that would take place in absence of frictional force under the applied force.

Question 56.
Define limiting force of friction.
Answer:
Just before the body starts sliding over another body, the value of frictional force is maximum, it is called as limiting force of friction.

Question 57.
Explain the concept of kinetic friction.
Answer:

Once the sliding of block on the surface starts the force of friction decreases.
The force required to keep the body sliding steadily is thus less than the force required to just start its sliding.
The force of friction that comes into play when a body is in steady state of motion over another surface is called kinetic fòrce of friction.
Friction between two surfaces in contact when one body is actually sliding over the other body, is called kinetic friction or dynamic friction.

Question 58.
Why ball bearings are used in machines?
Answer:

For same pair of surfaces, the force of static friction is greater than the force of kinetic friction while the force of kinetic friction is greater than force of rolling friction.
As ball bearing undergo rolling friction and rolling friction is the minimum, ball bearings are used to reduce friction in parts of machines to increase its efficiency.

Question 59.
The coefficient of static friction between a block of mass 0.25 kg and a horizontal surface is 0.4. Find the horizontal force applied to it.
Solution:
Given: μ_s = 0.4, m = 0.25 kg, g = 9.8 m/s²
To find: Force (F_L)
Formula: F_L = μ_sN = μ_s(mg)
Calculation: From formula,
F_L = 0.4 × 0.25 × 9.8 = 0.98 N
Answer:
The horizontal force applied to it is 0.98 N.

Question 60.
Calculate the force required to move a block of mass 20 kg resting on a horizontal surface, if μ_s = 0.3 and g = 9.8 m/s².
Solution:
Given: m = 20 kg, μ_s = 0.3, g = 9.8 m/s²
To find. Force required (F)
Formula: F_S = μ_sN = μ_smg
Calculation: From formula,
F_S = 3 × 20 × 9.8 = 58.8N
Answer:
The force required to move the block is 58.8 N.

Question 61.
A block of mass 40 kg resting on a horizontal surface just begins to slide when a horizontal force of 120 N is applied to it. Once the motion starts, It can be maintained by a force of 80 N. Determine the coefficients of static friction and kinetic friction (g = 9.8 m/s²)
Solution:
Given: F_L= 120N, F_k = 80N, m = 40 kg, g = 9.8 m/s²

To find:

i. Coefficient of static friction (μ_s)
ii. Coefficient of kinetic friction (μ_k)

Formulae:

i. μ_s = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}\)
ii. μ_k = \(\frac{\mathrm{F}_{\mathrm{k}}}{\mathrm{N}}\)

Calculation:

From formula (i) we get.
N = mg = 40 × 9.8 = 392 N
∴ μ_s = \(\frac{F_{L}}{N}\) = \(\frac{120}{392}\) = 0.306
From formula (ii) we get,
∴ μ_k = \(\frac{F_{k}}{N}\) = \(\frac{80}{392}\) = 0.204
Answer:

The coefficient of static friction is 0.306.
The coefficient of kinetic friction is 0.204.

Question 62.
A 20 kg metal block is placed on a horizontal surface. The block just begins to slide when horizontal force of 100 N is applied to it. Calculate the coefficient of static friction. If coefficient of kinetic friction is 0.4, then find minimum force to maintain its uniform motion.
Solution:
Given: m = 20 kg, F_L = 100 N, μ_k = 0.4

To find:

i. Coefficient of static friction (μ_s)
ii. Minimum force required (F_k)

Formulae:

i. μ_s = \(\frac{F_{L}}{N}\) = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{mg}}\)
ii. μ_s = \(\frac{F_{k}}{N}\)

Calculation:

From formula (i),
μ_s = \(\frac{100}{20 \times 9.8}\) = 0.5102
From formula (ii),
F_k = μ_kN = μ_k × mg = 0.4 × 20 × 9.8
∴ F_k = 78.4 N
Answer:

The coefficient of static friction is 0.5102.
The minimum force required is 78.4 N.

Question 63.
The Mariana trench is located in the Pacific Ocean and at one place it is nearly 11 km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches the bottom? (K = 1.6 × 10¹¹ N/m²)
Answer:
Given: V = 0.32 m³, K = 1.6 × 10¹¹ N/m²,
P_W= 1.1 × 10⁸ Pa, P_atm = 1.01 × 10⁵ Pa
∴ dP = P_W – P_atm = 1.1 × 10⁸ – 1.01 × 10⁵ ≈ 1.1 × 10⁸ Pa
As, bulk modulus is given as,
K = V × \(\frac{d P}{d V}\)
∴ dV = \(\frac{\mathrm{V} \times \mathrm{dP}}{\mathrm{K}}\) = \(\frac{\left(1.1 \times 10^{8}\right) \times 0.32}{1.6 \times 10^{11}}\)
∴ dV = 2.2 × 10^-4 m³
The change in the volume of the ball when it reaches the bottom will be 2.2 × 10^-4 m³.

Question 64.
Two spheres A and B having same volume are dropped from same height in ocean. Sphere A is made up of brass and sphere B is made up of steel. Will there be same change in volume of spheres at a certain depth inside water? What will be the ratio of change in volumes of the two spheres at this depth?
Answer:

Brass and copper have different elastic moduli. Hence, there won’t be same change in the volume of spheres at a certain depth inside water.
As two spheres are dropped from same height and are at same depth in water pressure exerted on them remains same.
∴ dP_A = dP_B = dP
V_A = V_B = V
If dV_A and dV_B be the change in volume of two spheres A and B then,
dV_A = \(\frac{\mathrm{V} \mathrm{dP}}{\mathrm{K}_{\mathrm{A}}}\) and dV_B = \(\frac{\mathrm{V} \mathrm{dP}}{\mathrm{K}_{\mathrm{B}}}\)
Ratio of change in volume,

where, K_A and K_B are bulk modulus of material of spheres A and B respectively.

Question 65.
What is the basis of deciding the thickness of metallic ropes used in crane to lift the heavy weight?
Answer:

The thickness of the metallic ropes used in cranes is decided on the basis of the elastic limit of the material of the rope and the factor of safety.
To lift a load upto 10⁴ kg, the rope is made for a factor of safety of 10.
It should not break even when a load of (original load × factor of safety = 10⁴ × 10) 10⁵ kgf i.e., 10⁵ × 9.8 N is applied on it.
If ‘r’ is the radius of the rope, then maximum stress = \(\frac{10^{5}}{\pi r^{2}}\).
The maximum stress (ultimate stress) should not exceed the breaking stress (5 to 20 × 10⁸ N/m²) as well as elastic limit of steel Y_s(30 × 10⁷ N/m²).
In order to have flexibility, the rope is made up of large number of thin wires twisted together.

Question 66.
Multiple choice questions

Which one of the following substances possesses the highest elasticity?
(A) Rubber
(B) Glass
(C) Steel
(D) Copper
Answer:
(C) Steel

Question 1.
S.I unit of stress is
(A) Newton/ metre
(B) Newton/ metre²
(C) Newton²/metre
(D) Newton/metre³
Answer:
(B) Newton/ metre²

Question 2.
A wire is stretched to double of its length. The strain is
(A) 2
(B) 1
(C) zero
(D) 0.5
Answer:
(B) 1

Question 3.
A and B are two steel wires and the radius of A is twice that of B. If they are stretched by the same load, then the stress on B is ____.
(A) four times that of A
(B) two times that of A
(C) three times that of A
(D) same as that of A
Answer:
(A) four times that of A

Question 4.
Two wires of the same material have radii r_A and r_B respectively. The radius of wire A is twice the radius of wire B. If they are stretched by same load then stress on wire B is ___
(A) equal to that of A
(B) half that of A.
(C) two times that of A.
(D) four times that of A.
Answer:
(D) four times that of A.

Question 5.
The length of a wire increases by 1% by a load of 2 kg wt. The linear strain produced in the wire will be
(A) 0.02
(B) 0.001
(C) 0.01
(D) 0.002
Answer:
(C) 0.01

Question 6.
A wire of length ‘L’, radius ‘r’ when stretched with a force ‘F’ changes in length by l’. What will be the change in length of a wire of same material having length ‘2L’ radius ‘2r and stretched by a force ‘2F’?
(A) l/2
(B) l
(C) 2l
(D) 4l
Answer:
(B) l

Question 7.
A force of 100 N produces a change of 0.1% in a length of wire of area of cross section 1 mm². Young’s modulus of the wire is ____
(A) 10⁵ N/m²
(B) 10⁹ N/m²
(C) 10¹¹ N/m²
(D) 10¹² N/m²
Answer:
(C) 10¹¹ N/m²

Question 8.
A copper wire (Y = 1 × 10¹¹ N/m²) of length 6 m and a steel wire (Y = 2 × 10¹¹ N/m²) of length 4 m each of cross-section 10^-5 m² are fastened end to end and stretched by a tension of 100 N. The elongation produced in the copper wire is
(A) 0.2 mm
(B) 0.4 mm
(C) 0.6 mm
(D) 0.8 mm
Answer:
(C) 0.6 mm

Question 9.
When a force is applied to the free end of a metal wire, metal wire undergoes
(A) longitudinal and lateral extension.
(B) longitudinal extension and lateral contraction.
(C) longitudinal contraction and lateral extension.
(D) longitudinal and lateral contraction.
Answer:
(B) longitudinal extension and lateral contraction.

Question 10.
A force of 1 N doubles the length of a cord having cross-sectional area 1 mm². The Young’s modulus of the material of cord is
(A) 1 N m^-2
(B) 0.5 × 10⁶ N m^-2
(C) 10⁶ N m^-2
(D) 2 × 10⁶ N m^-2
Answer:
(C) 10⁶ N m^-2

Question 11.
In equilibrium the tensile stress to which a wire of radius r is subjected by attaching a mass ‘m’ is ____.
(A) \(\frac{\mathrm{mg}}{\pi \mathrm{r}}\)
(B) \(\frac{\mathrm{mg}}{2 \pi \mathrm{r}}\)
(C) \(\frac{\mathrm{mg}}{\pi \mathrm{r}^{2}}\)
(D) \(\frac{\mathrm{mg}}{2 \pi \mathrm{r}^{2}}\)
Answer:
(C) \(\frac{\mathrm{mg}}{\pi \mathrm{r}^{2}}\)

Question 12.
When load is applied to a wire, the extension is 3 mm, the extension in the wire of same length but half the radius by the same load is
(A) 0.75 mm
(B) 6 mm
(C) 1.5 mm
(D) 12.0 mm
Answer:
(D) 12.0 mm

Question 13.
The S.I. unit of compressibility is ____.
(A) \(\frac{\mathrm{m}^{2}}{\mathrm{~N}}\)
(B) Nm²
(C) \(\frac{\mathrm{N}}{\mathrm{m}^{2}}\)
(D) \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
Answer:
(A) \(\frac{\mathrm{m}^{2}}{\mathrm{~N}}\)

Question 14.
When the pressure applied to one litre of a liquid is increased by 2 × 10⁶ N/m². Its volume decreases by 1 cm³. The bulk modulus of the liquid is
(A) 2 × 10⁹ N/m²
(B) 2 × 10³ dyne/cm²
(C) 2 × 10³ N/m²
(D) 0.2 × 10⁹ N/m²
Answer:
(A) 2 × 10⁹ N/m²

Question 15.
A cube of aluminium of each side 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be
(A) 0.02
(B) 0.1
(C) 0.005
(D) 0.002
Answer:
(D) 0.002

Question 16.
A wire has Poisson’s ratio of 0.5. It is stretched by an external force to produce a longitudinal strain of 2 × 10^-3. If the original diameter was 2 mm, the final diameter after stretching is
(A) 2.002 mm
(B) 1.998 mm
(C) 0.98 mm
(D) 1.999 mm
Answer:
(B) 1.998 mm

Question 17.
The compressibility of a substance is the reciprocal of ___.
(A) Young’s modulus
(B) Bulk modulus
(C) Modulus of rigidity
(D) Poisson’s ratio
Answer:
(B) Bulk modulus

Question 18.
For which of the following is the modulus of rigidity highest?
(A) Aluminium
(B) Quartz
(C) Rubber
(D) Water
Answer:
(B) Quartz

Question 19.
An elongation of 0.2% in a wire of cross-section 10^-4 m² causes a tension of 1000 N. Then its Young’s modulus is
(A) 6 × 10⁸ N/m²
(B) 5 × 10⁹ N/m²
(C) 10⁸ N/m²
(D) 10⁷ N/m²
Answer:
(B) 5 × 10⁹ N/m²

Question 20.
Within the elastic limit, the slope of graph between stress against strain gives ____
(A) compressibility
(B) Poisson’s ratio
(C) modulus of elasticity
(D) extension
Answer:
(C) modulus of elasticity

Question 21.
Solids which break or rupture before the elastic limits are called
(A) brittle
(B) ductile
(C) malleable
(D) elastic
Answer:
(A) brittle

Question 22.
Substances which break just after their elastic limit is reached are ___.
(A) ductile
(B) brittle
(C) malleable
(D) plastic
Answer:
(B) brittle

Question 23.
Which of the following substances is ductile?
(A) glass
(B) high carbon steel
(C) Steel
(D) copper
Answer:
(D) copper

Question 24.
The Young’s modulus of a material is 10¹¹ Nm^-2 and its Poisson’s ratio is 0.2. The modulus of rigidity of the material is
(A) 0.42 × 10¹¹ N/m²
(B) 0.42 × 10¹⁴ N/m²
(C) 0.42 × 10¹⁶ N/m²
(D) 0.42 × 10¹⁸ N/m²
Answer:
(A) 0.42 × 10¹¹ N/m²

Question 25.
Two pieces of wire A and B of the same material have their lengths in the ratio 1:2, and their diameters are in the ratio 2:1. If they are stretched by the same force, their elongations will be in the ratio
(A) 2 : 1
(B) 1 : 4
(C) 1 : 8
(D) 8 : 1
Answer:
(C) 1 : 8

Question 26.
Young’s modulus of material of wire is ‘Y’ and strain energy per unit volume is ‘E’, then the strain is
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 42
Answer:
(C) \(\sqrt{\frac{2 \mathrm{E}}{\mathrm{Y}}}\)

Question 27.
Strain energy per unit volume is given by
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 43
Answer:
(A) \(\frac{1}{2} \times \frac{(\text { stress })^{2}}{\mathrm{Y}}\)

Question 28.
The strain energy per unit volume of the wire under increasing load is
(A) \(\frac{1}{2}\) × (stress)² × strain
(B) \(\frac{1}{2}\) × stress × (strain)²
(C) 0.5 × stress × strain
(D) 0.5 × (strain)² × \(\frac{1}{Y}\)
Answer:
(C) 0.5 × stress × strain

Question 29.
The energy stored per unit volume of a strained wire is
(A) \(\frac{1}{2}\) × (load) × (extension)
(B) \(\frac{1}{2}\) × \(\frac{Y}{(\text { strain })^{2}}\)
(C) \(\frac{1}{2}\) × Y × (strain)²
(D) Stress × strain
Answer:
(C) \(\frac{1}{2}\) × Y × (strain)²

Question 30.
If work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material, but with double the radius and half the length by 1 mm in joule is
(A) 1/4
(B) 4
(C) 8
(D) 16
Answer:
(D) 16

Question 31.
When the load on a wire is slowly increased from 3 to 5 kg wt, the elongation increases from 0.61 to 1.02 mm. The work done during the extension of wire is
(A) 0.16 J
(B) 0.016 J
(C) 1.6 J
(D) 16 J
Answer:
(B) 0.016 J

Question 32.
A body lies on a table. Its weight is balanced by the ___.
(A) frictional force
(B) normal force
(C) force causing motion on the body
(D) surface of the table
Answer:
(B) normal force

Question 33.
The friction that exists between the surface of two bodies in contact when one body is sliding over the other, is called ___.
(A) rolling friction
(B) friction
(C) kinetic friction
(D) static friction.
Answer:
(C) kinetic friction

Question 34.
Limiting force of static friction does NOT depend on
(A) actual area of contact.
(B) geometrical area of contact.
(C) interlocking between surfaces in contact.
(D) intermolecular forces of attraction between molecules of the two surfaces.
Answer:
(C) interlocking between surfaces in contact.

Question 35.
In case of coefficient of static friction (µ_s), kinetic friction (µ_k) and rolling friction (µ_r), which of the following relation is true
(A) µ_s > µ_k > µ_r
(B) µ_s > µ_r > µ_k
(C) µ_r > µ_s > µ_k
(D) µ_r > µ_k > µ_s
Answer:
(A) µ_s > µ_k > µ_r

Question 36.
A body of weight 50 N is placed on a smooth surface. If the force required to move the body on the surface is 30 N, the coefficient of friction is
(A) 0.6
(B) 0.4
(C) 0.3
(D) 0.8
Answer:
(A) 0.6

Question 37.
A wooden block of 50 kg is at rest on the table. A force of 70 N is required to just slide the block. The coefficient of static friction is
(A) \(\frac{6}{7}\)
(B) \(\frac{5}{7}\)
(C) \(\frac{7}{35}\)
(D) \(\frac{1}{7}\)
Answer:
(D) \(\frac{1}{7}\)

Question 38.
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:
(A) \(\frac{1}{2}\)Mgl
(B) \(\frac{1}{2}\)MgL
(C) Mgl
(D) MgL
Answer:
(A) \(\frac{1}{2}\)Mgl
Hint: elastic potential energy: \(\frac{1}{2}\) × F × L = \(\frac{1}{2}\)Mgl

Question 67.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1 π ms^-2, What will be the tensile stress that would be developed
(A) 6.2 × 10⁶ N m^-2
(B) 5.2 × 10⁶ N m^-2
(C) 3.1 × 10⁶ N m^-2
(D) 4.8 × 10⁶ N m^-2
Answer:
(C) 3.1 × 10⁶ N m^-2
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 44

Question 68.
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to:
(A) 10⁶ N m^-2
(B) 10⁴ N m^-2
(C) 10⁸ Nm^-2
(D) 10³ N m^-2
Answer:
(A) 10⁶ N m^-2
Hint: d = 6 mm = 0.006 m, l = 42 cm = 0.42 m, ∆l = 20 cm = 0.2 m, m = 0.02 kg, v = 20 m/s
Energy stored in the catapult,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 45

Question 69.
The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively),
(A) Plastic and ductile
(B) Ductile and brittle
(C) Brittle and ductile
(D) Brittle and plastic
Answer:
(C) Brittle and ductile
Hint: Ductile materials have a fracture strength lower than the ultimate Tensile strength (i.e., the points are far apart.) whereas in brittle materials, the fracture strength is equivalent to ultimate tensile strength (i.e., the points are close.)
∴ Material X is brittle and Y is ductile in nature.

Question 70.
The compressibility of water is ‘o’ per unit atmospheric pressure. The decrease in its volume ‘V’ due to atmospheric pressure ‘P’ will be
(A) \(\frac{\sigma \mathrm{V}}{\mathrm{P}}\)
(B) \(\frac{\sigma P}{\mathrm{~V}}\)
(C) σPV
(D) \(\frac{\sigma}{\mathrm{PV}}\)
Answer:
(C) σPV

Question 71.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount?
(A) 9 F
(B) 6 F
(C) 4 F
(D) F
Answer:
(A) 9 F
Hint: As material is same, Young’s modulus of two wires is same.
Also, volume of both wires is same.
V₁ = V₂
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 46

Question 72.
A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, \(\left(\frac{\mathrm{dr}}{\mathrm{r}}\right)\),is:
where negative sign indicates volume decreases with increase in pressure.
(A) \(\frac{\mathrm{mg}}{\mathbf{3} \mathrm{Ka}}\)
(B) \(\frac{\mathrm{mg}}{\mathrm{Ka}}\)
(C) \(\frac{\mathrm{Ka}}{\mathrm{mg}}\)
(D) \(\frac{\mathrm{Ka}}{3 \mathrm{mg}}\)
Answer:
(A) \(\frac{\mathrm{mg}}{\mathbf{3} \mathrm{Ka}}\)
Hint:
Bulk modulus is given as,
K = \(\left(\frac{-\mathrm{dP}}{\mathrm{dV} / \mathrm{V}}\right)\)
where negative sign indicates volume decreases with increase in pressure.
∴ Fractional decrease in volume will be,
\(\frac{\mathrm{dV}}{\mathrm{V}}\) = \(\frac{\mathrm{dP}}{\mathrm{K}}\)
If area of cross-section of cylinder is a, then,
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 47

Question 73.
Two metal wires ‘P’ and ‘Q’ of same length and material are stretched by same load. Their masses are in the ratio m₁ : m₂. The ratio of elongations of wire ‘P’ to that of’Q’ is
Maharashtra Board Class 11 Physics Important Questions Chapter 6 Mechanical Properties of Solids 48
Answer:
(C) m₂ : m₁
Hint:

Question 74.
The increase in energy of a metal bar of length ‘L’ and cross-sectional area ‘A’ when compressed with a load ‘M’ along its length is (Y = Young’s modulus of the material of metal bar)
(A) \(\frac{\mathrm{FL}}{2 \mathrm{AY}}\)
(B) \(\frac{\mathbf{F}^{2} \mathbf{L}}{\mathbf{2 A Y}}\)
(C) \(\frac{\mathrm{FL}}{\mathrm{AY}}\)
(D) \(\frac{F^{2} L^{2}}{2 A Y}\)
Answer:
(B) \(\frac{\mathbf{F}^{2} \mathbf{L}}{\mathbf{2 A Y}}\)
Hint:
U = \(\frac{1}{2}\) × F × l
= \(\frac{1}{2}\) × F × \(\frac{\mathrm{FL}}{\mathrm{AY}}\) = \(\frac{\mathrm{F}^{2} \mathrm{~L}}{2 \mathrm{AY}}\)

Maharashtra Board Class 12 Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry

July 24, 2024July 23, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 16 Green Chemistry and Nanochemistry

Question 1.
Define the following:
a. Atom economy.
Answer:
Atom economy : Atom economy is a measure of the amount of atoms from the starting materials that are present
in the final product at the end of chemical process.

Question 2.
How will you prevent the generation of waste or by-products?
Answer:
To prevent generating waste, there is the need to develop the zero waste technology (ZWT). ZWT in a chemical synthesis should result in waste product being zero or minimum. To use the waste product of one system as the raw material for other system is also the aim of ZWT.

For example :

Cement and brick industry can use the bottom ash of thermal power station as the raw material.
Thermal power station can use the effluent coming out from cleansing of machinery parts as coolant water.

Question 3.
(1) Calculate the atom economy of the following:
(At mass of C = 12, 11 = 1 ,0 = 16)

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 16 Green Chemistry and Nanochemistry 2
Formula weight of ethanol = 46
ethene 28
water= 18
% atom economy = \(\frac{28}{46}\) x 100 = 60.9%

(2) Calculate the atom economy of fermentation of sugar (glucose) to ethanol.

Answer:

Formula wt of glucose = 180
Formula Wi of ethanol =46
Relative massiwt of desired useful product in thc equation = 2 x 46 =92
% Atom economy = 92/180 x 100 = 51.1%

Question 4.
ExpLain less hazardous chemical synthesis with suitable example.
Answer:

To avoid formation of hazardous waste from chemical processes, the chemical reactions and synthesis routes should be designed to be as safe as possible.
Earlier used insecticide DDT (Dichlorodiphenyltrichloroethanc) was found to be harmful for human beings. So DDT has been replaced by benzenc hexachioride (BHC) as an insecticide, one of the y-isomer (gamma) of BHC is called gammexane or lindane.

Question 5.
How will you develop products that are less toxic or which require less toxic raw materials ?
Answer:

There is a need to design safer chemicals to prevent the workers in chemical industries from being exposed to toxic environment.
Adipic acid is extensively used in polymer industry. In synthesis of adipic acid, benzene is used as the starting material, but benzene is carcinogenic and being volatile organic compound (VOC) it pollutes the air and environment.
To overcome this health hazard Green technology developed by Drath and Frost, adipic acid is enzymatically synthesised from glucose.

Question 6.
How to apply the principle of green chemistry to achieve energy efficiency?
Answer:

Energy requirements during chemical synthesis is huge. To minimize the energy use it is better to carry out reactions at room temperature and pressure.
This can be achieved by applying the principle of green chemistry i.e. use of catalyst, use of micro-organisms or biocatalyst and use of renewable materials, etc.
The use of less energy can be achieved by improving the technology of heating system, use of microwave, etc.

Question 7.
Explain the use of renewable feed stocks.
Answer:

Industries use a lot of non-renewable feed stocks like petroleum. These resources are depleting fast and the future generation will be deprived. The excessive use of these resources have also put a burden on the environment.
If renewable resources like agricultural or biological products are used, this will ensure the sharing of resources by future generations. This practice will also not put a burden on the environment.
The products and waste produced are generally biodegradable and environmental friendly hence leading to a sustainable future.

Question 8.
Explain the need to design degradable chemicals.
Answer:

Environment protection is the prime concern which has lead to the need for designing chemicals that degrade and can be discarded easily. These chemicals and their degradation products should be non-toxic, non-bioaccumulative or should not be environmentally persistent.
This principle aims at waste product being automatically degradable to clean the environment. Thus the preference for biodegradable polymers and pesticides.
To make the separation and segregation easier for the consumer an international plastic recycle mark is printed on larger items.

Question 9.
Define the role of real time analysis in pollution prevention.
Answer:

There is a dire need to develop improvised analytical methods to allow for real time, in process monitoring and control prior to the formation of hazardous substances.
It is very much important for the chemical industries and nuclear reactors to develop or modify analytical
methodologies so that continuous monitoring of the manufacturing and processing unit is possible.

Question 10.
Define the role of safer chemistry in accident prevention.
Answer:
(1) It is needed to develop chemical processes that are safer and minimize the risk of accidents. It is important to select chemical substances used in a chemical reaction in such a way that they can minimize the occurrence of chemical accidents, explosions, fire and emissions.

(2) For example : Chemical process that works with the gaseous substances can lead to relatively higher possibilities of accidents including explosion as compared to the system working with nonvolatile liquid and solid substances.

Question 11.
Green chemistry plays an important role in sustainable development. Explain.
Answer:
Sustainable development is a development that protects the environment and the world’s resources. We can achieve sustainable development by adapting the twelve principles of green chemistry.

Green chemistry designs safer chemicals which are less toxic. It normally leads to low cost, use of less energy, environmentally friendly solvents and less production of waste. Green chemistry works on the principle of atom economy and minimum or no waste production. It encourages the use of renewable feed stocks and reduces the use of toxic and hazardous chemicals. It eliminates majorly stoichiometry reactions and prefers to use catalysis. It preserves the environment and safety requirements with added benefit of cost reduction.

Question 12.
How are nanomaterials classified ?
Answer:
Nanoparticles, nanowires and nanotubes can be classified according to dimensions. The nano structured materials may be large organic molecules, inorganic cluster compounds and metallic or semiconductor particles.

Question 13.
What are zero, one and two dimensional nanoscale system ?
Answer:

Zero-Dimensional Nanostructures : A zero-dimensional structure is one in which all three dimensions are in the nanoscale.
For example : Nanoparticles.
One-Dimensional Nanostructures : A one-dimensional nanostructure is one in which two dimensions are in the nanoscale. For example : Nanowires and Nano rods.
Two-Dimensional Nanostructures : A two-dimensional nanostructure is one in which one dimension is in the nanoscale. For example : Thin films.

Question 14.
State the different characteristic features of nanoparticles.
Answer:
The nanoparticle science is special as at such a small scale, different laws dominate than what we experience in our everyday life.

The characteristic features like optical properties, catalytical activities, have huge surface area and good thermal properties mechanical strength electrical conductivity vary than that of bulk material.

(1) Colour : At nanoscale this optical property behaves differently. Elemental gold has nice shining yellow colour, but nanoparticles of gold show red colour.

(2) Catalytic activity : Since the surface area of nanoparticles is large they show increased catalytic activity. They are usually heterogenous catalyst that means catalysts are solid form and the reactions occur on the surface of the catalyst. These catalysts can be easily separated and recycled. For example : Pd, Pt metal nanoparticles used in hydrogenation reactions. Ti02, ZnO are used in photocatalysis. Gold in bulk is unreactive but the nanoparticles of gold behave as very good catalyst for organic reactions.

(3) Surface area : High surface-to-volume ratio is a very important characteristic of nanoparticles. Bulk material if subdivided into a group of individual nanoparticles, the total volume remains the same, but the collective surface area is largely increased. With large surface area for the same volume, these small particles react much faster because more surface area provides more number of reaction sites, leading to more chemical reactivity. Explanation of increase in surface area with decrease in particle size.

(4) Thermal strength : The melting point of nanomaterial changes drastically with size.

For example : Sodium clusters (Nan) of 1000 atoms melts at 288 K, 10000 atoms melt at 303 K and bulk sodium melts at 371 K.

(5) Mechanical strength : The mechanic al strength of nano clusters increase the hardness of the metal.

For example : nanoparticles of copper and palladium clusters with diameter in the range of 5-7 nm have hardness up to 500 r. greater than the bulk metal.

(6) Electrical conductivity : At nanoscale level the electrical conductivity changes. For example : Carbon nanotubes behave as a conductor or semiconductor whereas carbon is nonconductor.

Question 15.
Describe the two methods of synthesising nanomaterials (nanoparticles).
Answer:
The two methods of synthesising nanomaterials :
(1) Bottom-up and
(2) Top down methods :

(1) Bottom-up method : Synthesis of nanoparticles in the bottom-up approach molecular components arrange themselves into more complex assemblies atom by atom, molecule by molecule and cluster by cluster from the bottom. Example : synthesis of nanoparticles by colloidal dispersion.

Maharashtra Board Class 12 Chemistry Solutions Chapter 16 Green Chemistry and Nanochemistry 7

(2) Top-down method : In the top-down approach, involves nanomaterials being synthesised from bulk material by breaking the material. The bulk solids are disassembled into finer pieces until they are constituted of only few atoms. Example : Nanoparticles are synthesised by colloidal dispersion.

Question 16.
Discuss the various analytical tools used for characterization of nanoparticles.
Answer:
The analytical tools used for characterization of nanoparticles are

U.V visible spectroscopy – It gives the preliminary confirmation of formation of nanoparticles.
X-ray Diffraction (XRD) – The information given by this tool is about particle size, crystal structure and geometry.
Scanning electron microscopy (SEM) : This is used to study the structure of surface of material that is the morphology of the material.
Transmission electron microscopy (TEM) gives information about the particles size.
(FTIR) Fourier transform infrared spectroscopy gives information about absorption of functional groups and binding nature of the nanomaterial.

Question 17.
Give evidence of use of nanoparticles by humans in ancient times with appropriate examples.
Answer:
There is enough evidence that nanomaterials have been produced and used by humans in ancient times. For example :

Gold and silver nanoparticles trapped in the glass matrix gives ruby red colour in some ancient glass paintings.
The decorative glaze or metallic film known as lustre found on some medieval pottery is due to certain spherical metallic nanoparticles.
Carbon black is a nanostructured material that is used in tyres of car to increase the life of tyre. (Discovery in 1900). Carbon nanotubes are made up of graphite sheets with nanosized diameter. They have highest strength.
Fumed silica, a component of silicon rubber, coatings, sealants and adhesives is also a nanostructured material.

Question 18.
Explain the different applications of nanoparticles.
Answer:
The contribution of nanochemistry in number of innovative products in various disciplines due to their unique physical, chemical, optical, structural, catalytic properties. Few applications are as follows :

Nanoparticles contribute to stronger, lighter, cleaner and smarter surfaces and systems. They are used in the manufacture of scratchproof eyeglasses, transport, sunscreen, crack resistant paints, etc.
Used in electronic devices like Magnetoresistive Random Access Memory (MRAM).
Nanotechnology plays an important role in water purification techniques. Silver nanoparticles are used in water purification system to get safe drinking water.
Self cleansing materials : Lotus is an example of self cleansing. Nanostructures on lotus leaves repel water which carries dirt as it rolls off. Lotus effect is the basis of self cleaning windows.

Question 19.
State the advantages of nanoparticles and nanotechnology.
Answer:

Nanotechnology has revolutionalized electronics and computing.
Nanotechnology has benefited the energy sector by making solar power more economical and energy storage more efficient.
Nanotechnology has transformed the medical field with the manufacture of smart dmgs which help cure the life threatening diseases like cancer and diabetes faster and without side effects.

Question 20.
State the disadvantages of nanoparticles and nanotechnology.
Answer:
Despite the benefits that nanotechnology offers to the world, it is accompanied by certain disadvantages and potential risks.

The standard of living has been raised by nanotechnology but at the same time it has increased the environmental pollution. The kind of pollution caused by nanotechnology is very dangerous for living organism, it is called nano pollution.

Nanoparticles can be potential health hazard depending on the size, chemical composition and shape. They can be inhaled and can be deposited in the human respiratory tract and in the lungs, causing lung damage.

Question 21.
Name the development that meets the needs of the present, without compromising the ability of future generation to meet their own need.
Answer:
Sustainable development

Question 22.
Give name of father of green chemistry.
Answer:
Paul T. Anastas

Question 23.
Environmentally safe chemistry is known as.
Answer:
Green chemistry

Question 24.
How many principles does green chemistry have ?
Answer:
Twelve

Question 25.
Which principle of green chemistry has its perspective largely towards petrochemicals?
Answer:
Use of renewable feedstocks.

Question 26.
Name the chemical which leachs out of plastic packaging materials.
Answer:
Phthalate

Question 27.
Name the materials having structural components with at least one dimension in the nanometer scale.
Answer:
Nanomaterials.

Question 28.
Name the class of nanomaterial i.e. nanotubes, fibres, nanowires belong to.
Answer:
Two dimensions are in the nanoscale.

Question 29.
Name the nanoparticles used in sunscreen.
Answer:
Zinc oxide (ZnO) and Titanium dioxide (TiO₂).

Question 30.
What is the colour of gold nanoparticles ?
Answer:
Red

Question 31.
Name the nanoparticles used as catalyst in hydrogenation reaction.
Answer:
Palladium and Platinum.

Question 32.
Name the two approaches used to synthesize nanomaterials.
Answer:
Bottom up and Top down.

Question 33.
Give the name of the wet chemical synthetic process for nanomaterials.
Answer:
Sol-gel process.

Question 34.
Give the steps involved in preparation of nanoparticle using sol-gel process.
Answer:
Hydrolysis, polycondensation, drying, thermal decomposition.

Question 35.
Name the analytical techniques used for characterisation of nanomaterials.
Answer:
u.v-visible spectroscopy, x-ray diffraction (XRD), scanning electron microscopy (SEM), Transmission electron microscopy (TEM), Fourier transform infrared spectroscopy (FTIR).

Question 36.
Name the technique used to analyse particle size, crystal structure and geometry of a nanoparticle.
Answer:
x-ray diffraction (XRD)

Question 37.
Name the analytical technique used to study the morphology (structure of surface) of a material.
Answer:
Scanning electron microscopy (SEM)

Question 38.
Which innovative material has been developed using the lotus effect ?
Answer:
Self cleansing material

Question 39.
Which are the sectors that are revolutionalized by nanoparticles ?
Answer:
Electronics, energy sector and medical fields.

Question 40.
What are the disadvantages of nanotechnology ?
Answer:
Nano pollution and lung damage.

Question 41.
Name the scientist who coined the word nanotechnology.
Answer:
Nario Taniguchi (University professor at Tokyo in 1974).

Question 42.
Select and write the most appropriate answer from the given alternatives for each subquestion:

1. The measure of the amount of atoms from the starting materials that are present in the useful product at the end of chemical process is known as
(a) catalyst
(b) atom economy
(c) design of safer chemicals
(d) design for efficient energy
Answer:
(b) atom economy

2. The atom economy of the following reaction is CH₃ – CH₂ – CH₂ – CH₂ – OH + NaBr + H₂SO₄ → CH₃ – CH₂ – CH₂ – CH₂ – Br + NaHSO₄ + H₂O
(a) 49.81%
(b) 49%
(c) 50%
(d) 100%
Answer:
(a) 49.81%

3. Green chemistry reduces risk by
(a) developing the process for reuse and recycle of solvents and chemicals
(b) inventing technologies to clean the environ-ment
(c) minimize the use of chemicals
(d) reducing or eliminating the use or generation of hazardous chemicals in chemical products and process
Answer:
(d) reducing or eliminating the use or generation of hazardous chemicals in chemical products and process

4. Chemical synthesis should be designed to mini-mizes the use of
(a) liquid fuels
(b) solid fuels
(c) gaseous fuels
(d) energy
Answer:
(d) energy

5. The chemistry that applies across the life cycle of a chemical product like design, manufacture and use is called
(a) eco-friendly chemistry
(b) green chemistry
(c) environmental chemistry
(d) inorganic chemistry
Answer:
(b) green chemistry

6. According to the principles of green chemistry the chemicals involved in the production must be
(a) non-hazardous
(b) toxic
(c) polluting
(d) highly toxic
Answer:
(a) non-hazardous

7. Which of the following is not one of the twelve principles of green chemistry ?
(a) using renewable feedstocks
(b) designing safer chemicals and products
(c) maximizing atom economy
(d) avoiding the use of catalysts
Answer:
(d) avoiding the use of catalysts

8. Chemical synthesis based on principle of green chemistry encourages the use of
(a) hazardous chemicals
(b) reactions with low atom efficiency
(c) catalyst
(d) high energy requirements
Answer:
(c) catalyst

9. The plastic bottles made of HDPE are used to store household cleaner and shampoo can be recycled to make
(a) carpets, furniture, new containers
(b) detergent bottles, fencing, floor tiles, pens
(c) custom-made products
(d) cables, mudflaps, panelling, roadway gutters
Answer:
(b) detergent bottles, fencing, floor tiles, pens

10. The plastic ketch-up bottles and syrup bottles made from polypropylene (pp) can be recycled to make
(a) battery cables, brooms, ice scrapers, rakes
(b) envelopes, floor tiles, lumber
(c) custom-made products
(d) carpet, furniture, new containers
Answer:
(a) battery cables, brooms, ice scrapers, rakes

11. The role of green chemistry aims to
(a) design chemical processes and products that maximize profits
(b) design safer chemicals and products by reduc¬ing or eliminating the use and generation of hazardous substances
(c) design processes and products that work efficiently
(d) utilize non-renewable feedstocks
Answer:
(b) design safer chemicals and products by reducing or eliminating the use and generation of hazardous substances

12. The study of phenomena and manipulation of materials of atomic, molecular and macromolecular scales where properties differ significantly from those at a large scale is called
(a) nanoscience
(b) nanochemistry
(c) nanotechnology
(d) nanomaterial
Answer:
(a) nanoscience

13. The term nanotechnology was first used by whom and when ?
(a) Richard Feynman, 1959
(b) Nario Taniguchi, 1974
(c) Eric Drexter, 1986
(d) Sumia Lijima, 1991
Answer:
(b) Nario Taniguchi, 1974

14. Which one of these statements is NOT true ?
(a) Gold at the nanoscale is red.
(b) A very highly useful application of nanochem¬istry is medicine.
(c) Sunscreen contains nanoparticles of zinc oxide (ZnO) and (SiO₂) silicon oxide.
(d) Silicon at nanoscale is not an insulator
Answer:
(c) Sunscreen contains nanoparticles of zinc oxide (ZnO) and (SiO₂) silicon oxide.

15. Which of the historical works mentioned below contain nanotechnology?
(a) Lycurgus cup
(b) Medieval stained glass windows in churches
(c) Damascus steel swords
(d) All of the above
Answer:
(d) All of the above

16. The nanometer scale is conventionally defined as
(a) 10 – 100nm
(b) 1 – 100nm
(c) 1 – 1000 nm
(d) 1 – 10000 nm
Answer:
(b) 1-100 nm

17. The material synthesized on the nanometer scale possess
(a) same bulk properties
(b) different bulk properties
(c) unique optical, magnetic, electrical properties
(d) no change in properties
Answer:
(c) unique optical, magnetic, electrical properties

18. Nanomaterials of zero dimension is
(a) one in which all three dimensions are in the nanoscale
(b) one in which two dimensions are in the nanoscale
(c) one in which one dimension is in the nanoscale
(d) None of the above
Answer:
(a) one in which all three dimensions are in the nanoscale

19. The science which deals with the design and synthesis of material on nanoscale with different size and shape is called
(a) nanoscience
(b) nanochemistry
(c) nanophysics
(d) nanotechnology
Answer:
(b) nanochemistry

20. Elemental has a shining yellow colour, but the colour of nanoparticles of gold is
(a) green
(b) yellow
(c) red
(d) blue
Answer:
(c) red

21. The surface area of nanoparticles
(a) is the same as in bulk
(b) increases with the same volume of the bulk
(c) decreases with the same volume of the bulk
(d) does not change with particle size
Answer:
(b) increases with the same volume of the bulk

22. The nanomaterial based catalyst are usually
(a) homogeneous catalyst
(b) heterogeneous catalyst
(c) good catalyst
(d) bad catalyst
Answer:
(b) heterogeneous catalyst

23. The catalyst used in photocatalysis is
(a) gold
(b) Raney Ni
(c) TiO₂
(d) AI₂O₃Answer:
(c) TiO₂

24. Nanosized copper clusters have the mechanical strength of
(a) less than the bulk copper wire
(b) 100% more than the bulk metal
(c) 200% more than the bulk metal
(d) 500% more than the bulk metal
Answer:
(d) 500% more than the bulk metal

25. The most common method used for synthesis of nanomaterials is
(a) sol-gels method
(b) only sol method
(c) only gel method
(d) colloidal dispersion method
Answer:
(a) sol-gels method

26. What is the information obtained from uv-visible spectroscopy when used for nanomaterials ?
(a) morphology of structure
(b) preliminary conformation of formation of nanoparticle
(c) particle size
(d) functional group present
Answer:
(b) preliminary conformation of formation of nanoparticle

27. What information of the nanoparticles is obtained from transmission electron microscopy technique ?
(a) structure
(b) functional group
(c) particle size
(d) geometry
Answer:
(c) particle size

28. The analytical tool used to study the structure of surface of nanoparticle i.e. morphology is
(a) Absorption spectroscopy
(b) Scanning electron microscopy
(c) Emission spectroscopy
(d) Nuclear magnetic resonance spectroscopy
Answer:
(b) Scanning electron microscopy

29. The constituents of carbon nanotubes are
(a) nanosized graphite sheets
(b) nanosized carbon black
(c) nanosized coal black
(d) None of the above
Answer:
(a) nanosized graphite sheets

30. Self cleansing windows are example of the
(a) Nanoparticle effect
(b) Crompton effect
(c) Lotus effect
(d) Tyndal effect
Answer:
(c) Lotus effect

31. Which highly effective and cost effective nano-particles are used for water purification ?
(a) gold nanoparticle
(b) copper nanoparticle
(c) silver nanoparticle
(d) silica nanoparticle
Answer:
(c) silver nanoparticle

32. The sectors revolutionalized by nanotechnology are
(a) electronics and computing
(b) energy
(c) medicine
(d) All of the above
Answer:
(d) All of the above

33. Name the body part that gets affected by the hazardous nano pollution.
(a) heart
(b) brain
(c) lungs
(d) eyes
Answer:
(c) lungs

34. The pollution caused by nanotechnology is known as
(a) air pollution
(b) nano pollution
(c) ground pollution
(d) environmental pollution
Answer:
(b) nano pollution.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 10 Halogen Derivatives

July 24, 2024July 23, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives

Question 1.
What are halogen derivatives of hydrocarbons?
Answer:
The replacement of hydrogen atom/s in aliphatic or aromatic hydrocarbons by halogen atom/s results in the formation of halogen derivatives of hydrocarbons.

Question 2.
How are halogen derivatives of hydrocarbons classified?
Answer:
Halogen derivatives of alkane are classified as :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 1

Monohalogen derivative (or alkyl halide) : It is a halogen derivative of an alkane in which one hydrogen atom is replaced by one halogen atom and it is also called alkyl halide. E.g. C₂H₅Br.
Poh halogen derivatives : These are halogen derivatives in which more than one hydrogen atoms of an alkane are substituted by corresponding number of halogen atoms.

They are classified as follows :
(i) Dihalogen derivatives : The compounds formed by the substitution of two hydrogen atoms of an alkane by two halogen atoms are called dihalogen derivatives.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 2

They are further classified as :

Vicinal dihalides (Two halogen atoms on vicinal or adjacent carbon atoms)
Geminal dihalides : CH₃ – CHCI₂ (Two halogen atoms on the same carbon atom)

(ii) Trihalogen derivatives : The compounds formed by the substitution of three hydrogen atoms of an alkane by three halogen atoms are called trihalogen derivatives.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 4

(iii) Tetrahalogen derivatives : The compounds formed by the substitution of four hydrogen atoms of an alkane by four halogen atoms are called tetrahalogen derivatives. E.g. CCI₄.

Question 3.
What are alkyl halides? How are they classified?
Answer:
The compound formed by the replacement of one hydrogen atom in an alkane by a halogen atom is called an alkyl halide. The halogen atom is bonded to sp³ hybridised carbon. Alkyl halides are classified into the following three classes depending on the type of the carbon to which halogen atom is bonded.

(1) Primary (1°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a primary carbon atom is called primary alkyl halide.

[Primary (1°) carbon atom i.e., the carbon atom which is attached to only one carbon atom.]
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 5

They are represented by the general formula R – ^1°CH₂– X.

(2) Secondary (2°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a secondary carbon atom is called secondary alkyl halide. [Secondary (2°) carbon i.e., the carbon atom which is attached to two other carbon atoms.]
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 6

They are represented by the general formula Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 7 (R and R’ can be same or different)

(3) Tertiary (3°) alkyl halide : Alkyl halide in which halogen atom is bonded to a tertiary carbon atom is called tertiary alkyl halide. [Tertiary (3°) carbon i.e., the carbon atom which is attached to three other carbon atoms.]
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 8

They are represented by the general formula Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 9 (R, R’ and R” may be same or different)

Question 4.
Explain the following :
(1) Alkyl halide or haloalkanes
(2) Allylic halides
(3) Benzylic halide
(4) Vinylic halide
(5) Haloalkyne
(6) Aryl halide or haloarenes.
Answer:
(1) Alkyl halide or haloalkanes : In alkyl halides or haloalkanes the halogen atom is bonded to sp3 hybridized carbon which is a part of saturated carbon chain.
Example :

(2) Allylic halides : In allylic halides, halogen atom is bonded to a sp³ hybridized carbon atom next to a carbon-carbon double bond.

Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 11

(3) Benzylic halide : In benzylic halides, halogen atom is bonded to a sp³ hybridized carbon atom which is further bonded to an aromatic ring.
Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 12

(4) Vinylic halides : In vinylic halides, halogen atom is bonded to a sp² hybridized carbon atom of aliphatic chain. Vinylic halide is a haloalkene.
Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 13

(5) Haloalkyne : In haloalkynes, halogen atom is bonded to a sp hybridized carbon atom.
Example :

(6) Aryl halides or haloarenes : In aryl halides, halogen atom is directly bonded to the sp² hybridized carbon atom of aromatic ring.
Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 15

Question 5.
Give the IUPAC names of the following :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 16

Question 6.
Draw the structures of the following compounds:
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 18

Question 7.
Write the structure of-
(a) 3-chloro-3-ethylhex-l-ene
(b) 1-Iodo-2, 3-dimethylbutane
(c) 1, 3, 5-tribromobenzene
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 22

Question 8.
Write structures of
(a) 2-iodo-3-methyl pentane
(b) 3-chiorolleNane
(c) 1-chloro-2, 2-dimethyl propane
(d) 1-chloro-4-ethyl cyclohexane.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 25

Question 9.
Write the possible isomers of monochloro derivatives of 2,3-Dimethylbutane and write their IUPAC names.
Answer:
The given parent hydrocarbon has molecular formula, C₆H₁₄. The monochloro derivative of this compound has molecular formula C₆H₁₃CI.
The parent hydrocarbon is,
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 25
Hence the structures of isomers of monochioroderivative are.

Question 10.
Write structures and IUPAC names of all possible isomers of C₅H₁₁Br and classify them as l°/2°/3°.
Answer:
C₅H₁₁Br is a monohalogen derivative.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 28

Question 11.
How are following compounds obtained from alcohols :
(1) ethyl chloride C₂H₅CI
(2) isopropyl chloride (CH₃ CHCI – CH₃)
(3) tert-butyl chloride (CH₃)₃ – CI?
Answer:
Alcohols in the presence of Lucas reagent which is a solution of concentrated HCI and ZnCI₂ form alkyl halides. Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 31

(3) Tertiary alcohols don’t need ZnCI₂ to react with HCI.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 32
The order of reactivity of alcohols with a given halo acid is 3° >2°> 1°.

Question 12.
How are following compounds prepared from alcohols:
(1) ethyl bromide (C₂H₅Br)
(2) isopropyl bromide (CH₃ – CHBr – CH₃)
(3) tert-butyl bromide (CH₃)₃C – Br?
Answer:
(1) Ethyl alcohol on heating with conc. hydrobromic acid (48%) forms ethyl bromide.
OR
When ethyl alcohol is treated with a mixture of NaBr and H₂SO₄, ethyl bromide is formed. Here HBr is generated in situ.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 33
(2) Isopropyl alcohol, on reaction with NaBr and dil. H₂SO₄ forms isopropyl bromide.

(3) Tertiary alcohol on reaction with sodium bromide and dil. H₂SO₄ forms tert-butyl bromide.

Question 13.
How is ethyl iodide obtained from ethyl alcohol?
Answer:
When ethyl alcohol is treated with sodium or potassium iodide in 95 % phosphoric acid, ethyl iodide is formed. Here HI is generated in situ.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 36

Question 14.
How will you prepare the following :

(1) Ethyl chloride (chloroethane) from ethyl alcohol using
(i) PCI₃
(ii) PCI₅ and
(iii) SOCI₂.
Answer:
(i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 37
(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 39

(2) Isopropyl chloride (2-chloropropane) from isopropyl alcohol using
(i) PCI₃
(ii) PCI₅
(iii) SOCI₂.
Answer:
When isopropyl alcohol is refluxed with phosphorus trichloride, isopropyl chloride is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 40

When isopropyl alcohol is refluxed with phosphorus pentachloride, isopropyl chloride is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 41

When isopropyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, isopropyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for the preparation of alkyl chloride.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 42

(3) Ethyl bromide (bromoethane) from ethyl alcohol.
Answer:
When ethyl alcohol is treated with a mixture of red phosphorus and bromine or hydrobromic acid (phosphorus tribromide is generated in situ), ethyl bromide is formed.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 43

(4) Ethyl iodide (do ethane) from ethyl alcohol.
Answer:
When ethyl alcohol is heated with a mixture of red phosphorus and iodine, (phosphorus triiodide is generated in situ), ethyl iodide is formed.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 44

Question 15.
Explain halogenation of methane.
Answer:
Halogenation : A reaction of alkanes with halogens (CI₂, Br₂, I₂) in the presence of appropriate conditions forming a mixture of alkyl halides.
(1) Chlorination:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 44
(2) When excess of chlorine is used, tetrachioro methane, a major product is obtained. When excess of methane is used, chioromethane, a major product is obtained. The order of reactivity of halogens towards alkane is
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 47
(3) lodination:

However, iodination reaction is a reversible reaction. HI being a strongest reducing agent reduces methyl iodide back to methane.

(4) Fluorination: A reaction of alkane with fluorine is explosive and also hydrofluoric acid is poisonous and corrosive. Hence, alkyl fluorides are not prepared by halogenation of alkane.

Question 16.
Predict the possible products of the following reaction :
(1) Bromination of propane
(2) Bromination of n-butane
(3) Bromination of 2-Methyl propane.
Answer:
(1) Bromination of propane
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 49
(2) Bromination of n-butane

(3) Bromination of 2-Methyl propane

Question 17.
How are following compounds prepared by halogenation of ethane :
(1) Chloroethane
(2) Bromoethane
(3) Iodoethane?
Answer:
(1) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorine in the presence of diffused sunlight or U.V. light or at high temperature, chloroethane is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 52

(2) Bromination of ethane : When ethane is heated with Br₂ in the presence of anhydrous AlBr₃, bromoethane is obtained.

(3) Iodination : When ethane is reacted with I₂ in the presence of suitable oxidising agents like-HgO or HIO₃ or dilute HNO₃ iodoethane is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 54

Question 18.
Direct iodination of alkanes is not possible.
Answer:
(1) Direct iodination of alkanes using iodine is highly reversible.
\(\mathrm{RH}+\mathrm{I}_{2} \rightleftharpoons \mathrm{RI}+\mathrm{HI}\)
(2) Hydroiodic acid HI being strong reducing agent, it reduces RI to alkane RH.
(3) The reaction takes place only in the presence of a suitable oxidizing agent like HgO, HIO₃ or dilute HNO₃ which decomposes HI. Hence, direct iodination of alkanes is not possible.

Question 20.
How are following compounds obtained from alkenes :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 56
Answer:
(1) Ethene on reaction with hydrogen chloride forms

(2) Ethene on reaction with hydrogen bromide forms

(3) Propene on reaction with hydrogen iodide forms

(4) but-2-ene on reaction with hydrogen iodide forms 2-iodobutane.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 60

Question 19.
State and explain Markovnikov’s rule.
Answer:
Markovnikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent gets attached to that carbon atom of the double bond which carries less number of hydrogen atoms.

Example : Addition of HBr’to unsymmetrical alkene like propene gives two products.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 61

Isopropyl bromide is the major product, since the negative part (Br^–) of HBr is attached to carbon atom of a double bond with less number of hydrogen atoms.

Question 20.
Explain peroxide effect.
OR
Write a note on the Kharasch-Mayo effect.
OR
Explain the addition of HBr to (unsymmetrical alkene) propane in the presence of benzoyl peroxide.
Answer:
The addition of HBr to an unsymmetrical alkene (propane) in the presence of benzoyl peroxide takes place in the opposite orientation to that of Markovnikov’s rule and this is known as Kharasch-Mayo effect or peroxide effect or Anti-Markovnikov addition.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 62

Question 21.
Write the structure of alkyl halide obtained by the action hydrogen bromide on 2-Methyiprop-1-ene in the presence of peroxide.
Answer:
In the presence of peroxide. HBr to 2-Methyl prop-I-cne forms l-Bromo-2-methylpropane.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 63

Question 22.
How are alkyl iodides prepared from alkyl chlorides/bromides?
Answer:
Alkyl iodide is prepared by treating alkyl chloride or alkyl bromide with sodium iodide, in the presence of dry acetone, sodium chloride or sodium bromide precipitates from the solution and can be separated by filtration. This reaction is known as Finkelstein reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 70

Question 23.
How are alkyl fluorides prepared with alkyl chlorides/alkyl bromides?
Answer:
When alkyl chloride or alkyl bromide is heated with metallic fluorides like AgF, CaF₂, CoF₂ or Hg₂F₂, alkyl fluoride is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 71

This reaction is known as Swarts reaction.

Question 24.
Explain the preparation of haloarenes using electrophilic substitution.
Answer:
When arene is treated with chlorine or bromine in dark at ordinary temperature in the presence of lewis acid as a catalyst like Fe, FeCI₃ or anhydrous AlCI₃, aryl chloride or aryl bromide is formed.

When toluene is brominated in dark at ordinary temperature in the presence of iron, a mixture of ortho and para bromo tolerene is obtained.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 72

Ortho and para isomers can be easily separated as there is large difference in melting points of ortho and para isomers.

Question 25.
Write a note on Sandmeyer’s reaction.
Answer:
Aryl halides are most commonly prepared by replacement of nitrogen of diazonium salt. The replacement of diazonium group by -Cl or -Br using cuprous salt is called Sandmeyer’s reaction. When a primary aromatic amine (like aniline) suspended in cold F1C1, is treated with sodium nitrite, a diazonium salt (benzene diazonium chloride) is formed. When diazonium salt is treated with cuprous chloride or cuprous bromide, aryl halide (chlorobenzene or bromobenzene) is formed.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 73

When benzene diazonium salt is mixed with potassium iodide, iodobenzene is formed.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 74

Question 26.
Define the following :
Answer:
(1) Monochromatic light : It consists of rays of single wavelength vibrating in different planes perpendicular to the direction of propagation of the light.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 77
(2) Plane polarized light : A light having oscillations only in one plane perpendicular to direction of propagation of light is known as plane polarized light.

(3) Optical isomerism : The steroisomerism in which the isomers have different spatial arrangements of groups/atoms around a chiral atom is called optical isomerism.

(4) Optical activity : The property of a substance by which it rotates plane of polarization of incident plane polarized light is known as optical activity.

(5) Optically active compound : The compound which rotate the plane of plane polarized light is called optically active compound.

(6) Enantiomers : The optical isomers which are non-superimposable mirror images of each other are called enantiomers or enantiomorphs or optical antipodes.
Example : 2-chlorobutane, lactic acid

(7) Chiral carbon atom : Carbon atom in a molecule which carries four different groups/atoms is called chiral carbon atom.

Chiral atom in a molecule is marked with asterisk (*)

For example : C in lactic acid
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 79

(8) Chiral molecule : When a molecule contains one chiral atom, it acquires a unique property i.e. it is non- superimposable with its mirror image is said to be chiral molecule.

(9) Chirality : The relationship between a chiral molecule and its mirror image is similar to the relationship between left and right hands. Therefore it is called handedness or chirality.

(10) Dextrorotatory substance or r/-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the right hand side (RHS) is called dextrorotatory substance (or isomer) and denoted by d or (+) sign.

(11) Laevorotatory substance or /-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the left hand side (LHS) is called laevorotatory substance (or isomer) and denoted by / or (-) sign.

(12) Racemic mixture or Racemate : A mixture containing equimolar quantities of dextro (d) and laevo (/) optical
isomers which is optically inactive due to molecules of one enantiomer is cancelled by equal and opposite optical rotation due to molecules of the other enantiomer is called a racemic mixture or racemate. It is represented as (dl) or (+).

Question 27.
Calculate the number of isomers for 2-chlorobutane.
Answer:
The number of optical isomers possible for a compound is 2ⁿ where n = number of asymetric carbon atoms.
As n = 1 for 2-ehlorobutane, 2ⁿ = 2¹ = 2. Hence, it has two optical isomers.

Question 28.
How many optical isomers are possible for C₅H₁₁ CI?
Answer:
The number of optical isomers : 3.

Question 29.
How many optical isomers are possible for glucose?
Answer:
The number of optical isomers : 16.

Question 30.
Draw the structures and indicate the chiral carbon atoms in
(1) Lactic acid
(2) 2-Chlorobutane.
Answer:
(1) In lactic acid structure, Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 80 the starred carbon atom is chiral carbon atom as it is attached to four different substituents, COOH, OH, CH₃ and H.
(2) In 2-chlorobutane structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, -CH₂– CH₃ (ethyl), CH₃ (methyl), Cl and H.

Question 31.
Identify chira! and achiral molecules.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 82

Question 32.
Complete the following reactions and explain optical activity of the products formed:
(i) Pent-1-ene with HBr
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 84
(ii) Pent-2-ene with HBr
Answer:

Question 33.
C₆H₁₂ (A) on treatment with HCI produced a compound Y. Which is optically active, what is structure A?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 86

Question 34.
A racemic mixture is optically inactive. Explain.
Answer:

A racemic mixture contains equimolar (or equimolecular) quantities of the dextrorotatory (d-) and laevorotatory (l-) isomers (enantiomers) of a compound.
The d-enantiomer rotates the plane of plane-polarized light to the right, while the l-enantiomer rotates the same to the left to the same extent.
The quantities of the d- and l-enantiomers being the same, both the rotations are of the same magnitude, but of opposite directions. Hence, they cancel each other. Hence, a racemic mixture is optically inactive.
It is represented as dl or ( + ). Example : ( ± ) lactic acid

Question 35.
Explain Fischer projection formula with illustration.
OR
Write a note on Fischer projection formula.
Answer:
Fischer projection formula or cross formula : The three dimensional (3-D) view of a molecule is presented on plane of paper. A Fischer projection formula can be drawn by visualizing the main carbon chain verical in the molecule. Each carbon on the vertical chain is represented by a cross.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 89

Conventionally the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines represent the bonds going below the carbon.

Question 36.
Explain Wedge formula with illustration
OR
Write a note on Wedge formula.
Answer:
Wedge formula : When a tetrahedral carbon is imagined to be present in the plane of paper all the four bonds at this carbon cannot lie in the same plane. The bonds in the plane of paper are represented by normal lines, the bonds projecting above the plane of paper are represented by solid wedges (or simply by bold lines) while bonds going below the plane of paper are represented by broken wedges (or simply by broken lines).
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 90

Question 37.
Give a laboratory test to confirm the presence of halogen in the original organic compound.
Answer:
Haloalkanes are of neutral type in aqueous medium. On warming with aqueous sodium or potassium hydroxide the covalently bonded halogen in haloalkane is converted to halide ion.

When this reaction mixture is acidified by adding dilute nitric acid and silver nitrate solution is added a precipitate of silver halide is formed which confirms presence of halogen in the original organic compound.

Question 38.
Define the following :
Answer:
(1) Mechanism of a reaction : It is a step by step description of exactly how the reactants are transformed into products in as much details as possible.
(2) Substitution reaction : When a group bonded to a carbon in a substrate is replaced by another group to get a product with no change in state of hybridization of that carbon, the reaction is called substitution reaction.

Question 39.
Describe the action of aqueous KOH (or NaOH) on :
(1) ethyl bromide
(2) isopropyl bromide
(3) tert-butyl chloride
(4) methyl bromide
(5) 2-chlorobutane.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 96
(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 98
(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 100

Question 40.
Describe the action of sodium ethoxide on
(1) ethyl bromide
(2) methyl bromide :
OR
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from alkyl halides?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained. In this reaction halide (-X) of alkyl halide is replaced by an alkoxy group (-OR). This reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (or unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 101

(1) Simple (symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.

e.g., When ethyl bromide is heated with sodium ethoxide, diethyl ether is formed.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 103

(2) Mixed (unsymmetrical) ether : When an alkyl halide and sodium alkoxide having different alkyl groups are heated, unsymmetrical ether is obtained.

When methyl bromide is heated sodium ethoxide, ethyl methyl ether is formed. Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 105
When ethyl bromide is heated with sodium meihoxide, ethyl methyl ether is formed.

Question 41.
What is the action of silver salt of carboxylic acid on alkyl halide?
Answer:
When an alkyl halide (R – X) is heated with silver salt of carboxylic acid (R -COOAg). an ester is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 107

Question 42.
Describe the action of alcoholic silver acetate on
(1) methyl bromide
(2) ethyl bromide.
Answer:
(1) Methyl bromide : When methyl bromide is heated with an alcoholic silver acetate, methyl acetate is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 108
(2) Ethyl bromide : When ethyl bromide is heated with an alcoholic silver acetate, ethyl acetate is formed.

Question 43.
What is the action of alcoholic silver propionate on ethyl bromide?
Answer:
When ethyl bromide is heated with an alcoholic silver propionate. ethyl propionate is formed.

Question 44.
Describe the action of excess of ammonia on (I) ethyl bromide (2) n.propyl bromide.
Answer:
(1) Ethyl bromide : When ethyl bromide is boiled under pressure with an excess of alcoholic ammonia, ethylamine (ethanamine) is formed. This is known as ‘ammonolysis of ethyl bromide.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 111

(2) n-propyl bromide : When n-propy1 bromide is boiled under pressure with an excess of ammonia, n-propyl amine (propanamine) is formed.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 112

Question 45.
What is ammonolysis? Give a suitable example for the reaction.
Answer:
When an alkyl halide is boiled under pressure with an excess of alcoholic solution of ammonia (NH₃), corresponding (primary amine) alkyl amine is formed. This reaction is known as ammonolysis of alkyl halide.

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 113

(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 96
(2) Isopropyl bromide : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl chloride : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 98
(4) Methyl bromide : When methyl bromide (bromomethane) is heated with aq. KOH, it is hydrolysed to methyl alcohol (methanol).

(5) 2-chlorobutane : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 100

Question 46.
Describe the action of aqueous alcoholic potassium cyanide on
(1) ethyl bromide
(2) methyl iodide.
Answer:
Ethyl bromide : When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 114
(2) Methyl iodide : When methyl iodide is boiled with alcoholic solution of potassium cyanide, methyl cyanide is formed.

Question 47.
Describe the action of alcoholic silver cyanide on
(1) ethyl bromide
(2) methyl chloride.
OR
Explain isocyanide reaction of
(1) ethyl bromide
(2) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl isocyanide is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 116
(2) Methyl chloride : When .methyl chloride is heated with alcoholic silver cyanide, methyl isocyanide is formed.

The above reactions (1) and (2) are called isocyanide reaction.

Question 48.
Describe the action of potassium nitrite on
(i) ethyl bromide,
(ii) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is treated with potassium nitrite, ethyl nitrite is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 118
(2) Methyl chloride : When methyl chloride is treated with potassium nitrite, methyl nitrite is formed.

Question 49.
Describe the action of silver nitrite on (1) ethyl chloride (2) n-propyl bromide.
Answer:
(1) Ethy chloride : When ethyl chloride is treated with silver nitrite, nitroethane is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 120
(2) n-Propyt bromide: When n-propyl bromide is treated with silver nitrate, nitropropane is obtained.

Question 50.
How will you convert (the following:

(1) Ethyl bromide to ethanol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 128

(2) Ethyl bromide to propane nitrile.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 129

(3) Ethyl bromide to ethyl amine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 130

(4) Ethyl bromide to ethyl acetate.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 131

(5) Ethyl bromide to ethyl isocyanide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 132

(6) Ethyl bromide to ethyl methyl ether.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 133

(7) Ethyl bromide to n-butane.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 134

(8) Ethyl bromide to Ethyl magnesium bromide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 135

Question 51.
Define the following :
Answer:
(1) Nucleophilic bimolecular reaction (SN²) : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends on the concentration of the substrate and the nucleophile is called a nucleophilic bimolecular reaction.

Example :

(2) SN¹ reaction : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends only on the concentration of the substrate is called nucleophilic unimolecular or first order reaction or SN¹ reaction.

Question 52.
Explain, the mechanism of alkaline hydrolysis (reaction with aqueous KOH) of tert-butyl bromide (2-Bromo-2-methylpropane) with energy profile diagram.
OR
Explain only reaction mechanism for alkaline hydrolysis of tert-butyl bromide.
Answer:
(i) Consider alkaline hydrolysis of tert-butyl bromide (2-Bromo-2 methylpropane) with aqueous NaOH or KOH.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 137
(ii) Kinetics of the reaction : Due to steric hindrance of voluminous three methyl groups around carbon, nucleophile OH- cannot attack carbon atom directly. Hence, the reaction takes place in two steps.

Step I : This involves heterolytic fission of C – Br covalent bond in the substrate forming carbocation and Br^–. This is a slow process.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 138

Step II : This step involves attack of nucleophile OH- or carbocation forming C – OH bond and product tert-butyl alcohol. Since it involves ionic charge neutralisation, it is a fast step.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 139

Rate Determining Step (R.D.S.) : Since the first step is a slow step, it is R.D.S., and therefore the rate of the reaction depends on the concentration of only one reactant, (CH₃) C – Br.
Rate = R = k [(CH₃)₃ C – Br] where k is a rate constant of the reaction.

SN¹ reaction : The reaction between tert.butyl bromide and hydroxide ion to form tert.butyl alcohol follows a first-order kinetics. The rate of this reaction depends only on the concentration of one substance (tert-butyl bromide) and is independent of the concentration of alkali added. It is an unimolecular first (1st) order Nucleophilic Substitution reaction denoted as SN reaction.

Stereochemistry and mechanism of the reaction : The reaction takes place in two steps and both the steps involve formation of transition states (T.S.).

T.S. -1 for first step :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 140

In this transition state, C – Br bond is partially broken, so that carbon atom carries partial positive charge (+δ) and Br carries partial negative charge (-δ) which further breaks forming carbocation and Br . Tert-butyl cation (carbocation) has a planar structure and the CH₃ – C – CH₃ bond angle is 120°. It is the intermediate of the reaction. It is unstable. In this step, hybridisation of carbon atom changes from sp³ (tetrahedral geometry) to sp (planar geometry).

T.S. – II for second step :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 141

In this transition state, C – OH bond is partially fonned so that carbon atom carries partial positive charge (+ δ) and OH carries partial negative charge ( -δ) which further forms tert-butyl alcohol.

Formation of a racemic mixture : Since OH^– has equal probability of the attack on carbocation from frontside and from backside, the products obtained are equal. In case of optical active alkyl halide, a racemic mixture is obtained.

Question 53.
Discuss SN² mechanism of methyl bromide using aqueous KOH. Draw energy profile diagram.
OR
Discuss the mechanism of alkaline hydrolysis of methyl bromide or Bromomethane.
Answer:
(1) Consider alkaline hydrolysis of methyl bromide (Bromomethane). CH₃Br with aqueous NaOH or KOH.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 142

(2) Stereochemistry and Kinetics of the reaction iR.D.S.) : This hydrolysis reaction takes place only in one step which is a rate determining step i.e. R.D.S. The rate of hydrolysis reaction depends on the Concentration of CH3Br
and 0H which are present in the R.D.S. of the reaction.
Rate = R = k [CH₃Br] (OH]
where k is rate constant of the reaction.

SN² reaction : The reaction between methyl bromide and hydroxide ion to form methanol follows a second order kinetics, since the rate of the reaction depends on the concentrations of two reacting species, namely methyl bromide and hydroxide ion it is bimolecular second order (2nd) Nucleophilic Substitution reaction denoted by SN².

(3) Mechanism of the reaction :
(i) It is a single step mechanism. The reaction takes place in the following steps :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 143

(ii) Backside attack of the nucleophile : Nucleophile, OH^– attacks carbon atom of CH₃Br from back side i. e. from opposite side to that of the leaving group i.e. Br^– to experience minimum steric repulsion and electrostatic repulsion between the incoming nucleophile (OH^–) and leaving Br^–.

(iii) Transition state : When a nucleophile, OH^– approaches carbon atom of CH₃Br, the potential energy of the system increases until a transition state (T.S.) of maximum potential energy is formed in which C – Br bond is partially broken and C – OH bond is partially formed. The negative charge is equally shared by both incoming nucleophile- OH^– and outgoing, leaving group-Br^–. (Thus, the total negative charge is diffused.)

(iv) In CH₃Br, carbon atom is sp³-hybridized and CH₃Br molecule is tetrahedral. The hybridisation of carbon atom changes to sp² hybridisation. The transition state contains pentacoordinate carbon having three δ (sigma) bonds in one plane making bond angles of 120° with each other i.e., H₁; H₂ and H₃ atoms lie in one plane while two partial covalent bonds containing Br and OH lie collinear and on opposite sides perpendicular to the plane.

(v) Inversion of configuration : The transition state decomposes fast by the complete breaking of the C-Br bond and the new C-OH bond is formed on the other side. The breaking of C-Br bond and the formation of C-OH bond take place simultaneously. The energy required to break the C-Br bond is partly obtained from the energy released when C-OH bond is formed. The formation of product CH₃OH is accompanied by complete or 100% inversion of configuration forming again sp³-hybridized carbon atom giving tetrahedral CH₃OH molecule. But in this structure the positions of H₂ and H₃ atoms in the reactant (CH₃Br) and in product are on the opposite side. This inversion of configuration is called Walden inversion.

Question 54.
Discuss the factors influencing S_N1 and S_N2 mechanism.
Answer:
(1) Nature of substrate : S_N2 : The transition state (T.S.) of S_N2 mechanism + is pentacoordinate, it is crowded. As a result S_N2 mechanism is favoured in primary halides and least favoured in tertiary halides.

S_N1 : A planar carbocation intermediate is formed in S_N1 reaction. Bulky alkyl groups can be easily accommodated in planar carbocation and it has no steric crowding. As a result SN1 mechanism is favoured in tertiary halides and least favoured in primary halides.

The carbocation intermediate is stabilized by + effect of alkyl substituents and also by hyperconjugation y effect of alkyl substituents containing a-hydrogens. As a result, S_N1 mechanism is favoured in tertiary halides and least favoured in primary halides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 148

Thus, tertiary alkyl halides undergo nucleophilic substitution by S_N1 mechanism while primary halides follow S_N2 mechanism.

(2) Nucleophilicity of the reagent : A strong nucleophile attacks the substrate faster and favours S_N2 mechanism. The rate of S_N1 mechanism is independent of the nature of nucleophile. Nucleophile does not react in the 1st step (slow step) of S_N1. Nucleophile reacts fast after the carbocation intermediate is formed.

(3) Solvent polarity : (1) S_N1 reaction proceeds more rapidly in polar protic solvents than in aprotic solvent. Polar protic solvent decreases the rate of S_N2 reaction. (2) In S_N2 mechanism, rate depends on substrate as well as nucleophile. A polar solvent stabilizes nucleophile by solvation. Thus solvent deactivates the nucleophile by stabilizing it. Hence, aprotic solvents or solvent of low polarity will favour S_N2 mechanism.

Question 55.
How does relative reactivity for alkaline hydrolysis with respect to S_N2 and S_N1 vary in the following alkyl halides :
(1) Bromomethane
(2) Bromoethane
(3) 2-Bromopropane
(4) 2-Bromo-2-methylpropane ?
Answer:
(A) Relative reactivity for S_N2 mechanism decreases in the order of :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 153
(B) Relative reactivity for S_N1 mechanism decreases in the order of :
2-Bromo-2-methylpropane > 2-Bromopropane > Bromoethane > Bromomethane

Question 56.
Explain with reason the relative order of reactivity of l°/2°/3° alkyl halides by S_N1 mechanism.
Answer:
In alkaline hydrolysis of an alkyl halide by S_N1 mechanism, the formation of carbocation as an intermediate product is involved.

The increasing order of a stability of carbocation is,
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 155
The stability order for carbocation is 3° > 2° > 1°.
Therefore the increasing order of reactivity by S_N1 mechanism of alkyl halides is
(1°) primary < (2°) secondary < (3°) tertiary

Question 57.
Which one of the following is more easily hydrolysed in S_N1 and S_N2 reaction by aqueous KOH, C₆H₅ CHCIC₆H₅ and C₆H₅CH₂CI?
Answer:
In S_N1 reaction C₆H₅CHCI C₆H₅ will be more easily hydrolysed than C₆H₅CH₂CI
In S_N2 reaction C₆H₅ CH₂CI will be more easily hydrolysed than C₆H₅CHCIC₆H₅.

Question 58.
Choose the member that will react faster than the following pairs by S_N1 mechanism.
(1) l-bromo-2, 2-dimethyl propane or 2-bromopropane.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 159
Answer:
The reactivity of S_N1 reaction depends on the steric hindrance, in 2-bromopropane, a-carbon atom is attached to two methyl groups suffers greater steric hindrance to nucleophilic attack than l-bromo-2, 2-dimethyl propane. Hence, 2-bromopropane react faster by S_N1 mechanism.

(2) 2-Iodo-2-methyl butane or 2-iodio-3-methyl butane.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 160
Answer:
Since, 2-Iodo-2-methyl butane is a tertiary alkyl halide, it undergoes SN^-1 reaction faster than 2-iodo-3-methyl butane.

(3) 1-Chloro propane or 2-chloropropane.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 161
Answer:
Since, 2-chloropropane is a secondary alkyl halide, it undergoes SN^-1 reaction faster than 1-chloropropane.

(4) 2-Iodo-2-methyl butane or tert-butyl chloride.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 162
Answer:
Since, iodine is a better leaving group than chloride 2-iodo-2-methyl butane undergo SN^-1 reaction faster than tert-butyl chloride.

Question 59.
Write a note on elimination reaction.
OR
Explain dehydrohalogenation reaction.
Answer:
When alkyl halide having at least one β-hydrogen is boiled with alcoholic solution of potassium hydroxide (KOH), an alkene is formed due to elimination of hydrogen atom from β-carbon and halogen atom from α-carbon, is called dehydrohalogenation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 164

Tertiary butyl bromide when heated with alcoholic solution of potassium hydroxide forms isobutylene.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 165

This reaction is called β-elimination (or 1,2-elimination) reaction as it involves elimination of halogen and a β-hydrogen atom.

As hydrogen and halogen is removed in this reaction it is also known as dehydrohalogenation reaction.

Question 60.
Describe the action of alcoholic potassium hydroxide (aic. KOH) on
(1) ethyl bromide
(2) n-propyl bromide
(3) isopropyl bromide
(4) tert-butyl chlorIde.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 166

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 167

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 168

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 169

Question 61.
Describe the action of alc.KOH on 2-bromobutane.
When 2-bromobutane is boiled with alc.KOH on 2-bromobutane, a mixture of but-l-ene and but-2-ene is formed.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 170

Question 62.
Explain Saytzelf’s rule with suitable example.
Answer:
Saytzcff’s rule : In dehydrohalogenation reaction the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Hence the number of alkyl substituents on doubly bonded Carbon atoms increases, the stability of the alkene giving its major products.

Hence the increasing stability of alkenes is.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 171

There are two types of fi hydrogens (β₁ and β₂) therefore two alkenes are expected.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 172

Question 63.
What is a Grignard reagent ?
Answer:
Grignard reagent : An organometallic compound in which the divalent magnesium is directly linked to an alkyl group (R -) and a halogen atom (X), and has general formula R – Mg – X is called Grignard reagent. OR When alkyl halide is treated with magnesium in dry ether as solvent, it gives alkyl magnesium halide. It is known as Grignard reagent.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 185

The carbon-magnesium bond is highly polar and magnesium-halogen bond is in ionic in nature. Grignard reagent is highly reactive. It is an important reagent and used in the preparation of a large number of organic compounds.

Question 64.
How is Grignard reagent prepared ?
Answer:
Grignard reagent is an alkyl magnesium halide, R – Mg – X obtained by the reaction of alkyl halide R – X with magnesium (Mg) in dry ether.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 186

When an alkyl halide like CH₃I is added from a dropping funnel to a flask containing pieces of pure Mg in pure and dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH₃ – Mg – I is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 187

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesium iodide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 188

Question 65.
Write a note on Grignard reagent.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 166

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 167

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 169

Question 66.
Describe the action of water on
(1) methyl magnesium iodide
(2) ethyl magnesium iodide.
Answer:
(1) Methyl magnesium iodide : When methyl magnesium iodide is treated with water, methane is obtained
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 190
(2) Ethyl magnesium iodide : When ethyl magnesium iodide is treated with water, ethane is obtained.

Question 67.
Describe the action of ammonia on
(1) ethyl magnesium bromide
(2) n-propyl magnesium chloride.
Answer:
(1) Ethyl magnesium bromide : When ethyl magnesium bromide is treated with ammonia, ethane is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 192
(2) n-Propyl magnesium chloride : When n-propyl magnesium chloride is treated with ammonia, propane is formed.

Question 68.
Explain Wurtz reaction. OR Explain the action of sodium with alkyl halides.
Answer:
(1) When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher alkane is formed. This is called Wurtz reaction or Wurtz coupling reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 194
(2) In this reaction the alkyl radicals from two molecules of the reacting alkyl halide combine or couple to form the higher alkane.

(3) Thus, methyl bromide reacts with sodium in ether to form ethane (C₂H₆), while ethyl bromide under the same conditions forms n-butane (C₄H₁₀).
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 195

(4) If a mixture of two different alkyl halides is treated with Na in dry ether, then a mixture of alkanes is obtained called self coupling products. For example, a mixture of CH₃Br and C₂H₅Br gives propane along with C₂H₆ and C₄H₁₀.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 196

Question 69.
Explain the reaction of haloarene with alkyl halide and sodium metal.
Write a note on Wurtz-Fittig reaction.
Answer:
When an alkyl halide and an aryl halide is treated with sodium metal in dry ether the corresponding alkylarene (alkyl benzene) is formed. The reaction is known as Wurtz-Fittig reaction. This reaction allows alkylation of alkyl halides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 197

Question 70.
Describe the action of aryl halide on sodium metal.
Answer:
Aryl halide reacts with sodium metal in dry ether, biphenyl is formed. This reaction is known as Fittig reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 198

Question 71.
Identify the product A of following reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 200
Answer:

Question 72.
Explain the following substitution reactions of chlorobenzene :
(1) Halogenation
(2) Nitration
(3) Sulphonation.
Answer:
(1) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrous ferric chloride, a mixture of ortho and para-dichlorobenzene (major product) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 203
(2) Nitration : When chlorobenzene is heated with nitrating mixture (cone, nitric acid -I- cone, sulphuric acid) a mixture of l-chloro-4-nitro benzene (major product) and l-chloro-2-nitrobenzene is formed.

(3) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixture of 4-chlorobenzene sulphonic acid (major product) and 2-chlorobenzene sulphonic acid is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 205

Question 73.
Describe the action of the following on chlorobenzene :
(1) Methyl chloride in the presence of anhydrous AICI₃
(2) Acetyl chloride in the presence of anhydrous AICI₃.
Answer:
(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 206

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 207

Question 74.
Write a note on Friedel Craft’s reaction.
Answer:

(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 206

Question 75.
Convert 1-chlorobutane into the following compounds :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 221
Answer:
(1) 1-Chlorobutane to butan-l-ol :

(2) 1-Chlorobutane to 1-iodobutane :

(3) 1-Chlorobutane to n-butyl cyanide (CH₃– CH₂– CH₂– CH₂– CN) :

(4) l-Chlorobutane to

Question 76.
Predict the expected product of substitution reactions :
(1) Isobutyl chloride + sodium ethoxide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 227

(2) n-butyl chloride + sodium.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 228

(3) 1-chloropropane + aq. potassium hydroxide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 229

(4) Aniline + NaNO₂/HCl.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 230

Question 77.
Write the products:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 231

Question 78.
Identify A and B in the following :

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 235

Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 236
Answer:

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 239

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 241

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 243

Question 79.
State the uses of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or Chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freons
(6) DDT (p, p’-Dichlorodiphenyl trichloroethane).
Answer:
(1) Dichloromethane (CH₂CI₂) :

Dichloromethane dissolves wide range of organic compounds, hence it is used as solvent for many chemical reactions.
It is used as a solvent as a paint remover and degreaser.
It is used as propellant in aerosols and as a fumigant pesticide for grains and strawberries.
It is used to decaffinate tea or coffee.

(2) Trichloromethane or Chloroform (CHCI₃) :

Chloroform in the production of chlorofluoromethane, freon refrigerant R-22.
It is used as solvent in pharmaceuticals, pesticides, gums, fats, resins and dye industry.
It is a good source of dichlorocarbene species.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

Carbon tetrachloride is used in the manufacture of refrigerants.
It is used as a dry cleaning agent and as a pesticide for stored grains.
It is very useful solvent for oils, fats and resins. It serves as a source of chlorine.

(4) Iodoform (CHI₃) :

Iodoform is used as antiseptic, dressing of wounds and sores.
On small scale it is used as disinfectant.

(5) Freons :

Freons are widely used as propellants in aerosol, products of food, cosmetics and pharmaceutical industries.
Freons containing bromine in their molecules are used as fire extinguishers.
They are used in aerosol insecticides, solvent for cleaning clothes and metallic surfaces.
It is used as foaming agents in the preparation of foamed plastics and in production of certain fluorocarbons.
It is used as refrigerants and air conditioning purposes.

Question 80.
State the environmental effects of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freon.s (CCI₂F₂, CCI₃F, CHCIF₂)
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 247
Answer:
(1) Dichloromethane (CH₂CI₂) :

Higher levels of dichloromethane in air causes nausea, numbness in fingers and toes, dizziness.
Lower levels of dichloromethane causes impaired vision and hearing.
Direct contact with eyes can damage cornea.

(2) Trichloromethane or chloroform (CHCI₃) :

When chloroform is exposed to air in the presence of sunlight, it slowly oxidised to phosgene, a poisonous compound, therefore it is stored in dark, amber coloured bottles.
Chloroform vapour when inhaled for a short time causes dizziness, headache and fatigue and if inhaled for a long time affects central nervous system.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

Exposure to carbon tetrachloride causes eye irritation, damages nerve cells, vomiting sensation, dizziness, unconciousness or death. Long exposure to chloroform may affect liver.
When mixed with air it causes depletion of the ozone layer, which affects human skin leading to cancer.

(4) Iodoform (CHI₃) : Iodoform has a strong smell. It causes irritation to skin and eyes. It may cause respiratory irritation or breathing difficulty, dizziness, nausea, depression of central nervous system, visual disturbance.

(5) Freons (CCI₂F₂, CCI₃F, CHCIF₂) :

Freon as refrigerant causes ozone depletion.
Freons have low toxicity and low biological activity.
Freons from propane group are more toxic in nature.
Regular large inhalation of freon results in breathing problems, organ damage, loss of consciousness.

(6) DDT :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 248

DDT is not readily metabolised by animals.
It is deposited and stored in fatty tissues.
Exposure to high doses of DDT may cause vomiting, tremors or shakiness.
Laboratory animal studies showed adverse effect of DDT on liver and reproduction.
DDT is a pressistent organic pollutant, readily absorbed in soils and tends to accumulate in the ecosystem.
When dissolved in oil or other lipid, it is readily absorbed by the skin. It is resistant to metabolism.
There is a ban on use of DDT due to all these adverse effects.

Question 81.
What is the chemical name of freon?
Answer:
The chemical name of freon is Dichlorodifluoromethane.

Question 82.
What is the chemical name of DDT ?
Answer: The chemical name of DDT is p, p’-Dichlorodiphenyltrichloroethane.

Activity :
(1) Collect detailed information about Freons and their uses.
(2) Collect information about DDT as a persistent pesticide.
Reference books :
(1) Organic chemistry by Morrison, Boyd, Bhattacharjee, 7th edition, Pearson.
(2) Organic chemistry by Finar, Vol 1, 6th edition, Pearson

Multiple Choise Questions

Question 83.
Select and write the most appropriate answer from the given alternatives for each sub-question :

Question 1.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 250
Answer:
(b) CH₃ – CH₂ – CH₂ – I

Question 2.
The rate of S_N2 reaction depends on the concentra¬tion of
(a) only the substrate
(b) only the reagent
(c) both the substrate and the reagent
(d) neither the substrate nor the reagent
Answer:
(c) both the substrate and the reagent

Question 3.
In S_N2 reaction, the hydrolysis of alkyl halide shows
(a) the retention of configuration
(b) the inversion of configuration
(c) both retention and inversion of configuration
(d) no change in the configuration
Answer:
(b) the inversion of configuration

Question 4.
The one step exothermic reaction is
(a) S_N1
(b) S_N2
(C) S_N
(d) S2_NAnswer:
(b) S_N2

Question 5.
Which of the following is correct about S_N2 mechanism?
(a) Two step reaction
(b) Complete inversion of configuration
(c) Formation of carbonium ion
(d) Favoured by polar solvent
Answer:
(b) Complete inversion of configuration

Question 6.
Which of the following is not a nucleophile?
(a) Ammonia
(b) Ammonium ion
(c) Primary amine
(d) Secondary amine
Answer:
(d) Secondary amine

Question 7.
Which of the following undergoes nucleophilic substitution exclusively by S_N2 mechanism ?
(a) ethyl chloride
(b) isopropyl chloride
(c) chlorobenzene
(d) benzyl chloride
Answer:
(d) benzyl chloride

Question 8.
Which of the following is most reactive towards nucleophilic substitution reaction ?
(a) CH₂ = CH – CI
(b) CH₃CH = CHCI
(c) C₆H₅CI
(d) CICH₂– CH = CH₂Answer:
(d) CICH₂– CH = CH₂

Question 9.
The stability order of carbocation is
(a) 2° > 3° > 1°
(b) 3° > 2° > 1°
(c) 3° > 1° > 2°
(d) 1° > 3° > 2°
Answer:
(b) 3° > 2° > 1°

Question 10.
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 255
(a) ethane
(b) propane
(c) n-butane
(d) n-pentane
Answer:
(c) n-butane

Question 11.
Which of the following characteristic properties of the enantiomers is correct?
(a) The enantiomers possess same physical and chemical properties
(b) The enantiomers are optically active compounds
(c) The enantiomers have different optical rotations
(d) All of these
Answer:
(d) All of these

Question 12.
The optically inactive compound is
(a) glucose
(b) lactic acid
(c) isopropyl alcohol
(d) 2-bromo butane
Answer:
(c) isopropyl alcohol

Question 13.
A compound with the molecular formula CH₂OH(CHOH)₃CH₂OH has optically active forms
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 14.
A racemic mixture consists of
(a) equal amount of d and l isomers
(b) unequal amounts of d and / isomers
(c) unknown amounts of d and / isomers
(d) only d isomers
Answer:
(a) equal amount of d and l isomers

Question 15.
Which of the following compounds is not optically active ?
(a) Lactic acid
(b) Secondary butyl chloride
(c) n-propyl iodide
(d) Glucose
Answer:
(c) n-propyl iodide

Question 16.
Which of the following compounds shows optical activity ?
(a) n-butyl chloride
(b) isobutyl chloride
(c) sec-butyl chloride
(d) t-butyl chloride
Answer:
(c) sec-butyl chloride

Question 17.
The major product of the following reaction is
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 256
Answer:
(c)

Question 18.
The above reaction is known as
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 257
(a) Wurtz-Fittig reaction
(b) Friedel Craft’s reaction
(c) Sandmeyer’s reaction
(d) Swarts reaction
Answer:
(b) Friedel Craft’s reaction

Question 19.
Iodoform is used as
(a) an anaesthetic
(b) an antiseptic
(c) an analgesic
(d) an antibiotic
Answer:
(b) an antiseptic

Question 20.
p, p’-dichlorodiphenyl trichloroethane is used as
(a) insecticide
(b) anaesthetic
(c) antiseptic
(d) refrigerant
Answer:
(a) insecticide

Question 21.
The order of reactivity in nucleophilic substitution reaction is
(a) CH₃F < CH₃C1 < CH₃I < CH₃Br
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I
(c) CH₃F < CH₃Br < CH₃C1 < CH₃I
(d) CH₃I < CH₃Br < CH₃C1 < CH₃F
Answer:
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I

Question 22.
Racemate is
(a) optically active
(b) optically dextro rotatory
(c) optically inactive
(d) optically laevorotatory
Answer:
(c) optically inactive

Question 23.
The number of asymmetric carbon atoms in glucose are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 24.
The geometry of carbon lum ion is
(a) Tetrahedral
(b) planar
(c) linear
(d) pyramidal
Answer:
(b) planar

Question 25.
In its nucleophilic substitution reaction, aryl halide resembles
(a) Vinyl chloride
(b) allyl chloride
(e) Benzyl chloride
(d) ethyl chloride
Answer:
(a) Vinyl chloride

Question 26.
The weakest C-Cl bond is present in
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 258
Answer:
(d)

Question 27.
Which alkyl halide among the following com¬pounds has the highest boiling point ?
(a) (CH₃)₃CCI
(b) CH₃CH₂CH₂CH₂CI
(c) CH₃CH₂CH₂C1
(d) CH₃CH(CH₃)CH₂CI
Answer:
(b) CH₃CH₂CH₂CH₂CI

Question 28.
It is difficult to break C-Cl bond in CH₂ = CH – CI due to
(a) Hyper conjugation
(b) Resonance
(c) Electromeric effect
(d) Inductive effect
Answer:
(b) Resonance

Question 29.
Which one of the following when heated with metallic sodium will not give the corresponding alkane ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 259
Answer:
(c)

Question 30.
The most reactive alkyl halide towards SN2 reac¬tion is
(a) CH₃X
(b) R₃CX
(C) R₂CHX
(d) RCH₂X
Answer:
(a) CH₃X

Question 31.
The number of electrons surrounding the carbon- ium ion is
(a) 6
(b) 8
(c) 10
(d) 7
Answer:
(a) 6

Question 32.
The lowest stability of carbocation among the compounds
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 260
Answer:
(a)

Question 33.
Carbon atom in methyl carbocation contains how many pairs of electrons?
(a) 8
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 34.
The optically inactive compound is
(a) Glucose
(b) Lactic acid
(c) 2-Chlorobutane
(d) 2-Chloropropane
Answer:
(d) 2-Chloropropane

Question 35.
The hydrogen halide which does not obey Markownikv rule in presence of peroxide is
(a) HC1
(b) HBr
(c) HF
(d) HI
Answer:
(b) HBr

Question 36.
Which one of the following is NOT used to prepare alkyl halide from an alcohol ?
(a) SOCl₂
(b) PC1₃
(c) HC1 + ZnCl₂
(d) NaCl
Answer:
(d) NaCl

Question 37.
The total number of electrons present in the central carbon atom of a free radical is
(a) 7
(b) 8
(c) 9
(d) 6
Answer:
(a) 7

Question 38.
In which of the following pairs both are nucleophiles ?
(a) BF₃, AICI₃
(b) NO⁺₂, Cl^–
(c) CN^–, NH₃
(d) Br⁺, BC1₃Answer:
(c) CN^–, NH₃

Question 39.
Which one of the following alkane is NOT formed in Wurtz reaction ?
(a) Methane
(b) Ethane
(c) Propane
(d) Butane
Answer:
(a) Methane

Question 40.
Which of the following groups has highest priority according to R, S convention?
(a) CH₂OH
(b) COOH
(c) COCH₃
(d) COOCH₃Answer:
(d) COOCH₃

Question 41.
The halogen atom in aryl halides is
(a) o- and p-di reefing
(b) m-directing
(c) o, m and p-di reefing
(d) only m-directing
Answer:
(a) o- and p-di reefing

Question 42.
Chlorobenzene can be obtained by benzene diazonium chloride by
(a) Friedel Craft’s reaction
(b) Wurtz reaction
(c) Gatterman’s reaction
(d) Fittig reaction
Answer:
(c) Gatterman’s reaction

Question 43.
Which of the following carbocations is least stable ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 261

Answer:
(c)

Question 44.
But-l-ene on reaction with HCI in the presence of sodium peroxide yields
(a) n-butyl chloride
(b) isobutyl chloride
(c) secondary butyl chloride
(d) tertiary butyl chloride
Answer:
(c) secondary butyl chloride

Question 45.
Carbon tetrachloride is used as
(a) anaesthetic
(b) antiseptic
(c) dry cleaning agent
(d) fire extinguisher
Answer:
(c) dry cleaning agent

Question 46.
Identify the product D in the following sequence of reactions :
Maharashtra Board Class 12 Chemistry Solutions Chapter 10 Halogen Derivatives 263
(a) 2, 2-dimethyl butane
(b) 2, 3-dimethyl butane
(C) hexane
(d) 2, 4-dimethylpentane
Answer:
(b) 2, 3-dimethyl butane

Question 47.
The preparation of alkyl fluoride from alkyl chlor ide, in presence of metallic fluorides is known as
(a) Williamson’s reaction
(b) Finkeistein reaction
(c) Swarts reaction
(d) Wurlz reaction
Answer:
(c) Swarts reaction

Question 48.
UPAC name of the following compound is

(a) 3-Bromo-3, 4-dimetbyiheptane
(b) 3,4-dimethyl-3-bromoheptane
(c) 5-Bromo-4,5-dimethylheptane
(d) 4,5-dimethyl-5-bromoheptane
Answer:
(a) 3-Bromo-3, 4-dimetbyiheptane

Maharashtra Board Class 12 Chemistry Important Questions Chapter 9 Coordination Compounds

July 24, 2024July 23, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 9 Coordination Compounds Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 9 Coordination Compounds

Question 1.
What are double salts?
Answer:
Double salts are crystalline molecular or addition compounds containing more than one salt in simple molecular proportions soluble in water and in solution they ionise and exhibit all the properties of the constituent ions.

For example, K₂SO_4⁺A1₂(SO₄)_3^–24H₂O
\(\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{~K}_{(\mathrm{aq})}^{+}+2 \mathrm{Al}_{(\mathrm{aq})}^{3+}+4 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+24 \mathrm{H}_{2} \mathrm{O}_{0}\)

Question 2.
Define coordination compound.
Answer:
Coordination compound : It consists of a central metal ion or atom surrounded by atoms, molecules or anions called ligands by coordinate bonds, e.g. cisplatin Pt(NH₃)₂Cl₂, [Cu(NH₃)₄]SO₄.

Question 3.
Define Lewis bases and Lewis acids with respect to a coordination compound.
Answer:

Lewis bases : In a coordination compound the ligands being electron pair donors they are Lewis bases.
Lewis acids : The central metal atom or ion being electron acceptor behaves as a Lewis acid.
For example, in the coordination compound, [Cu(NH₃)₄]²⁺, NH₃ is a Lewis base and Cu²⁺ is a Lewis acid.

Question 4.
Define coordination sphere. Give example.
Answer:
Coordination sphere : A coordination entity consisting of a central metal atom or ion and the coordinating groups like neutral molecules or anions (ligands) written inside a square bracket is together called coordination sphere. This is a discrete structural unit. The ionisable groups (generally ions) called counter ions are written outside the bracket.

For example, in the coordination compound K₄[Fe(CN)₆], the coordination sphere is [Fe(CN)₆]^4- while K⁺ represents counter ion.

Question 5.
Define and explain charge number of a complexion.
Answer:
Charge number of a complexion : The net charge carried by a complexion or a coordination entity is called its charge number.

Explanation :
(i) Charge number is equal to the algebraic sum of the charges carried by central metal atom or ion and all the ligands attached to it.
(ii) E.g. consider anionic complex, [Fe(CN)₆]^4-.
Charge number of [Fe(CN)₆]^4- = Charge on Fe²⁺ ions + 6 x charge on CN^– = ( + 2) + 6( -1) = – 4 Hence charge number of [Fe(CN)₆]^4- is – 4.

Question 6.
Explain the oxidation state of a metal in a complex.
Answer:

The oxidation state of a metal atom or ion in the complex is the apparent charge carried by it in the complex.
It depends upon the atomic number and electronic configuration of the metal atom or ion.
The coordination number, the formula and geometry of a complex depend upon the oxidation state of the metal
atom or ion.

Question 7.
What is the charge on a monodentate ligand X in the complex, [NiX₄]^2-?
Answer:
The charge number of the complex ion is – 2. Nickel being divalent, its oxidation state is + 2. If the charge on monodentate ligand X is y, then Charge number = charge on Ni²⁺ charge on 6X – 2 = + 2 + 4 xy
∴ y = – 1
Hence the charge on ligand X is – 1.

Question 8.
Calculate the oxidation state of a metal in the following complexes :
(a) [Fe(NH₃)₆](NO₃)₃
(b) Ni(CO)₅.
Answer:
(a) [Fe(NH₃)₆](NO₃)₃ ⇌ [Fe(NH₃)₆]³⁺ + 3NO₃^–

NH₃ is a neutral ligand, and the charge number of complex ion is + 3.
If the oxidation state of Fe is x then,
+ 3 = x + 6(0)
∴ x = + 3
∴ The oxidation state of Fe is +3.

(b) Ni(CO)₅ is a neutral complex and CO is a neutral ligand. If the oxidation state of Ni is x, then zero = x + 5 x (zero)
∴ x = zero.
The oxidation state of Ni is zero.

Question 9.
Define and explain the term coordination number (C.N.) of a metal in the complex.
Answer:
Coordination number or legancy (C.N.) : The number of (monodentate) ligands which are directly bonded by coordinate bonds to central metal atom or ion in a coordination compound is called coordination number (C.N.) of the metal atom or ion.

Explanation :

The coordination number (C.N.) is a characteristic property of the metal and its electronic configuration.
C.N. takes the values from 2 to 10, of which 4 and 6 are very common.
The light transition metals show C.N. 4 and 6 while the heavier transition metals show C.N. 8.
The geometry and shape of a complex compound depends upon C.N. of the metal.

Question 10.
Mention primary valence, secondary valence and coordination number in the following complexes :
(a) [Cu(NH₃)JCI₂
(b) [Co(NH₃)₃CI₃]
(C) K4[Fe(CN)₆]
(d) [CoF₆]₃
(e) [Pt(NH₃)₂Cl₂]
(f) [Pt(NH₃)₂(P_y)₃CI₂]
(g) Cr(CO)₆
(h) [Ni(CN)₄]^2-
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 2

Question 11.
Classify the following complexes as homoleptic and heteroleptic complex :
(a) [Cu(NH₃)₄]SO₄;
(b) [Cu(en)₂(H₂O)CI]²⁺
(c) [Fe(H₂O)₅(NCS)]²⁺
(d) Tetraaminezinc(II) nitrate.

Question 12.
Summarise the rules of IUPAC nomenclature of coordination compounds.
Answer:
Following rules are followed for naming coordination compounds recommended by IUPAC :

In case of a complexion or a neutral molecule, name the ligand first and then the metal.
The names of anionic ligands are obtained by changing the ending -ide to -o and -ate to -ato.
The name of a complex is one single word. There must not be any space between different ligand names as well as between ligand name and the name of the metal.
After the name of the metal, write its oxidation state in Roman number which appears in parentheses without any space between metal name and parentheses.
If complex has more than one ligand of the same type, the number is indicated with prefixes, di-, tri-, tetra-, penta-, hexa- and so on.
For the complex having more than one type of ligands, they are written in an alphabetical order. Suppose two ligands with prefixes are tetraaqua and dichloro. While naming in alphabetical order, tetraaqua is written first and then dichloro.
If the ligand itself contains numerical prefix in its name, then display number by prefixes bis for 2, tris for 3, tetrakis for 4 and so forth. Put the ligand name in parentheses. For example, (ethylenediamine)3 or (en)3 would appear as tris (ethylenediamine) or tris(ethane-l, 2-diamine).
The metal in cationic or neutral complex is specified by its usual name while in the anionic complex the name of metal ends with ‘ate’.

Question 13.
State effective atomic number (EAN).
OR
State and explain effective atomic number (EAN). How is it calculated?
Answer:
Effective atomic number (EAN) : It is the total number of electrons present around the central metal atom or ion and calculated as the sum of electrons of metal atom or ion and the number of electrons donated by ligands.

It is calculated by the formula : EAN = Z – X + Y where.
Z = Atomic number of metal atom
X = Number of electrons lost by a metal atom forming a metal ion
Y = Total number of electrons donated by all ligands in the complex.

Generally the value of EAN is equal to the atomic number of the nearest inert element.

Explanation : Consider a complex ion [Co(NH₃)₆]³⁺
Oxidation state of cobalt is + 3 hence X = 3.
There are six ligands, hence Y = 2 x 6 = 12
Atomic number of cobalt. Z = 27
∴ EAN = Z – X + Y = 27 – 3 + 12 = 36.

Question 14.
Find effective atomic number (EAN) in the following complexes :
(1) [Ni(CO)₄]
(2) [Fe(CN)₆]^4-
(3) [Co(NH₃)₆]³⁺
(4) [Zn(NH₃)J²⁺
(5) [Pt(NH₃)₆]⁴⁺
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 4

Question 15.
What is effective atomic number (EAN) in the following complexes ?
(1) [Fe(CN)₆]^3-
(2) [CU(NH₃)₄]²⁺
(3) [Pt(NH₃)₄]²⁺
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 5

Question 16.
Calculate EAN in the following complexes :
(1) [Cr(H₂O)₂(NH₃)₂(en)]CI₃;
(2) [Ni(en)₂]SO₄;
(3) Na3[Cr(C₂O₄)₃].
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 6

Question 17.
Define in coordination compounds :
(1) Isomerism
(2) Isomers.
Answer:

Isomerism : It is the phenomenon in coordination compounds having same molecular formula but different physical and chemical properties due to different arrangements of the ligands around the central metal atom or ion in the space.
Isomers : The isomers are the coordination compounds having same molecular formula but different physical and chemical properties due to the difference in arrangements of the ligands in the space.

Question 18.
Mention the types of isomerisms in coordination compounds.
Answer:
There are two principal types of isomerisms in coordination compounds as follows :
(A) Stereoisomerism
(B) Structural isomerism (OR Constitutional isomerism)

(A) Stereoisomerism is further classified as :

Geometrical isomerism
Optical isomerism

(B) Structural isomerism is further classified as :

Ionisation isomerism
Linkage isomerism
Coordination isomerism
Solvate (or hydrate) isomerism

Question 19.
Why does stereoisomerism arise in the coordination compounds?
Answer:
In the coordination compounds (complexes) the ligands are linked to the central metal atom or ion by coordinate bonds which are directional in nature and hence give rise to the phenomenon of stereoisomerism.

In this isomerism, the different stereoisomers have different arrangements of ligands (atoms, molecules or ions) in space around the central metal atom or ion. Hence they have different physical and chemical properties and give rise to the phenomenon of stereoisomerism.

Question 20.
Define, in coordination compounds : (1) Stereoisomerism (2) Stereoisomers.
Answer:
(1) Stereoisomerism The phenomenon of isomerism in the coordination compounds arising due to different spatial positions of the ligands in the space around the central metal atom or ion is called stereoisomerism.

(2) Stereoisomers : The coordination compounds having same molecular formula but different stereoisomerism due to different spatial arrangements of the ligand groups in the space around the central metal atom or ion are called stereoisomers.

Question 21.
Define :
(1) Geometrical isomerism and
(2) Geometrical isomers.
Answer:
(1) Geometrical isomerism : The phenomenon of isomerism in the heteroleptic coordination compounds with the same molecular formula but different spatial arrangement of the ligands in the space around the central metal atom or ion is called geometrical isomerism.

(2) Geometrical isomers : The heteroleptic coordination compounds having same molecular formula but different geometrical isomerism due to different spatial arrangements of the ligands in the space around the central metal atom or ion are called geometrical isomers.

Question 22.
Define cis and trails isomers in the coordination compounds.
Answer:
(1) Cis-isomer : A heteroleptic coordination compound in which two similar ligands are arranged adjacent to each other is called cis-isomer. For example,

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 11

Gis-Diamminedichloroplatinum(II)

(2) Trans-isomer : A heteroleptic coordination compound in which two similar ligands are arranged diagonally opposite to each other is called trans-isomer. For example,

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 12

Trans-Diamminedichloroplatinum(II)

Question 23.
Write structures for geometrical isomers of Diamminebromochloroplatinum(II).
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 7

Question 24.
Explain the geometrical isomerism of the octahedral complex of the type [MA₄B₂] with a suitable example.
Answer:

Consider an octahedral complex of a metal M with coordination number six and monodentate ligands a and b having formula [MA₄B₂],
CA-isomer is obtained when both the B ligands occupy adjacent (1,2) positions.
Trans-isomer is obtained when the ligands B occupy the opposite (1,6) positions.
For example, consider a complex [CO(NH₃)₄CI₂]⁺. The structures of cis and trans isomers are

Question 25.
Explain the geometrical isomerism of the octahedral complex of the type [M(AA)₂B₂]^n± with a suitable example.
Answer:

Consider an octahedral complex of metal M with coordination number six and a bidentate ligand AA and monodentate ligand B having molecular formula [M(AA)₂B₂]^n± .
Bidentate ligand AA has two identical coordinating atoms.
Cis- isomer is obtained when two bidentate AA ligands as well as two ‘B’ ligands are at adjacent positions.
Trans-isomer is obtained when two AA ligands and two B ligands are at opposite positions.
For example, consider a complex [Co(en)₂CI₂]⁺.

Question 26.
Explain the geometrical isomerism of the octahedral complex of the type [MA4BC] with suitable example.
Answer:

Consider an octahedral complex of metal M with coordination number six and monodentate ligands A, B andC.
Cis-isomer is obtained when both the ligands B and C occupy adjacent (1,2) positions.
Trans-isomer is obtained when the ligands B and C occupy opposite positions.
For example, consider a complex [Pt(NH₃)₄BrCI] of the type [MA₄BC],

Question 27.
Define : (1) Optical isomerism (2) Optical isomers.
Answer:
(1) Optical isomerism : The phenomenon of isomerism in which different coordination compounds having same molecular formula have different optical activity is called optical isomerism.

(2) Optical isomers : Different coordination compounds having same molecular formula but different optical activity
are called optical isomers.

Question 28.
Explain : (1) Plane polarised light (2) Optical activity.
Answer:
(1) Plane polarised light : A monochromatic light having vibrations only in one plane is called a plane polarised light. This light is obtained by passing monochromatic light through NICOL prism.

(2) Optical activity : A phenomenon of rotating a plane of a plane polarised light by an optically active substance is
called optical activity. This substance is said to be optically active.

Question 29.
Explain : (1) Dextrorotatory substance (2) Laevorotatory substance. (1 mark each)
Answer:

Dextrorotatory substance : An optically active substance which rotates the plane of a plane polarised light to right hand side is called dextrorotatory or d isomer denoted by d.
Laevorotatory substance : An optically, active substance which rotates the plane of a plane polarised light to the left hand side is called laevorotatory or l isomer and denoted by l.

Question 30.
What are the conditions for the optical isomerism in coordination compounds?
Answer:

Optical isomerism is exhibited by those coordination compounds which possess chirality.
There should not be the presence of element of symmetry which makes the complex optically inactive.
The mirror images of the complex molecule or ion must be non-superimposable with the molecule or ion. B

Question 31.
What are enantiomers?
Answer:
Enantiomers : The two forms of the optical active complex molecule which are mirror images of each other are called enantiomers.

There are two forms of enantiomers, d form and l form.

Question 32.
Draw diagrams for the optical isomers of a complex, [Co(en)₃]³⁺.
Answer:
The complex [Co(en)₃]³⁺ has two optical isomers.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 14

Question 33.
Explain the optical isomerism in the octahedral complex with two symmetrical bidentate chelating ligands.
Answer:
The octahedral complexes of the type [M(AA)2Q2]”±, in which two symmetrical bidentate chelating ligands like AA and two monodentate ligands like a are coordinated to the central metal atom or ion exhibit optical isomerism and two optical isomers d and l can be resolved. For example, [Pt Cl2(en)2]2.

The cis-form is unsymmetrical and optically active while the trans-form is symmetrical and hence optically inactive. The optical isomers of cis-form (d and ) of this complex along with trans-form are shown below,
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 15

Question 34.
When are optical isomers called chiral?
Answer:
When the mirror images of optical isomers of the complex are nonsuperimposable they are said to be chiral. For example, [Co(en)₂(NH₃)₂]³⁺.

Question 35.
Define and explain ionisation isomerism.
Answer:
Ionisation isomerism : The phenomenon of isomerism in the metal complexes in which there is an exchange of ions between coordination (or inner) sphere and outer sphere is known as ionisation isomerism.

Explanation :

Ionisation isomers have same molecular formula but different arrangement of ions in the inner sphere and outer sphere in the complex,
Hence on ionisation, these ionisation isomers produce different ions in the solution. This ionisation isomerism is also called ion-ion exchange isomerism.

Examples:
(A) [CO(NH₃)₄CI₂]Br and (B) [Co(NH₃)₄CIBr] Cl

Ionisation :
(A) [CO(NH₃)₄CI₂] Br ⇌ [CO(NH₃)₄CI₂]⁺ + Br-
(B) [Co(NH₃)₄CIBr]Cl ⇌ [CO(NH₃)₄ClBr]⁺ + CP
In the isomers (A) and (B), there is an exchange of ions namely Br^– and CI^–.

Question 36.
Define : (1) Linkage isomerism (2) Linkage isomers.
OR
What is linkage isomerism ? Explain with an example.
Answer:
(1) Linkage isomerism : The phenomenon of isomerism in which the coordination compounds have same metal atom or ion and same ligand but bonded through different donor atoms or linkages is known as linkage isomerism.

(2) Linkage isomers : The coordination compounds having same metal atom or ion and ligand but bonded through different donor atoms or linkages are called linkage isomers.
For example : Nitro complex [CO(NH₃)₅NO₂]CI₂ = (Yellow) and nitrito complex [CO(NH₃)₅ONO]CI₂(Red)

Question 37.
Explain linkage isomers with NO₂^– group as a ligand.
Answer:
(1) Nitro group (_–NO₂^–) is an ambidentate ligand. NO₂^– group may link to central metal atom, through N or O.
(2) The two linkage isomers are, [CI: → Ag ← : NO₂]^– and [CI: → Ag ← O-NO]^–
Choloronitroargentate(I) ion and Chloronitritoargentate(I) ion

Question 38.
Write linkage isomers of a complex having constituents Co³⁺, 5NH3 and NO^–₂.
Answer:
(i) NO^–₂ is an ambidentate ligand which can be linked through N or O.
(ii) The linkage isomers are as follows :
(a) [CO(NH₃)₅(NO₂)]²⁺ Pentaamminenitrocobalt(III) ion
(b) [CO(NH₃)₅(ONO)]²⁺ Pentaamminenitritocobalt(III) ion

Question 39.
Define: (1) Coordination isomerism (2) Coordination isomers.
Answer:
(1) Coordination isomerism : The phenomenon of isomerism in the ionic coordination compounds having the same molecular formula but different complex ions involving the interchange of ligands between cationic and anionic spheres of different metal. ions is called coordination isomerism.

(2) Coordination isomers : The ionic coordination compounds having same molecular formula but different complexions duc to interchange of ligands between cationic and anionic spheres of different metal ions are called coordination isomers.

For example,

Question 40.
Give three examples of coordination isomers. (I mark each)
Answer:

[Cu(NH₃)₄] [PICI₄] and I Pt(NH₃)₄] [ICuCl4₄]
[Cr(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₄(CN)₂] [Cr(NH₃)₂(CN)₄]
[Cr(NH₃)₆] [Cr(SCN)₆] and [Cr(NH₃)₄(SCN)₂] [Cr(SCN)₄(NH₃)₂]

Question 41.
Define Solvate or Hydrate isomerism.
Answer:
Solvate ate or Hydrate isomerism : The phenomenon of isomerism in the coordination compounds arising due to the exchange of solvent or H₂O molecules inside the coordination sphere and outer sphere of the complex is known as solvate or hydrate isomerism.

Question 42.
Define solvate or hydrate isomers.
OR
What are hydrate isomers? Explain with examples.
Answer:
Solvate or Hydrate isomers : The coordination compounds having the same molecular formula but differ in the number of solvent or H₂O molecules inside the coordination sphere and outer sphere of the complexes are called solvate or hydrate isomers.

For example : [Cr(H₂O)₆] CI₃; [Cr(H₂O)₅CI]CI₂ H₂O; and [Cr(H₂O)₄CI₂] CI 2H₂O.

Question 43.
A coordination compound has the formula COCI₃6H₂O. Write the hydrate isomers of the complex.
Answer:
The possible hydrate isomers of the coordination compounds having molecular formula COCI₃6H₂O are as follows :
(1) [CO(H₂O)₆]CI₃;
(2) [CO(H₂O)₅CI]CI₂ H₂O
(3) [CO(H₂O)₄CI₂] Cl – 2H₂O
(4) [Co(H₂O)₃ CI₃] 3H₂O.

Question 44.
Explain the steps involved in describing the bonding in coordination compounds using valence bond theory.
Answer:

Vacant d-orbitais of metal ion form coordination bonds with ligands.
s, p orbitais along with vacant d-orbitais of metal ion take part in hybridisation.
The number of vacant hybrid orbitais formed is equal to number of hybridising orbitais which is equal to the number of ligand donor atoms or coordination number of the metal.
The metal-ligand coordination bonds are formed by the overlap between the vacant hybrid orbitais of metal and the filled orbitais of the ligands.
The hybrid orbitais used by the metal ion point in the direction of the ligands.
When inner (n – 1)d orbitais of metal ion are used in the hybridisation then the complex is called (a) inner orbital complex while when outer nd orbitais are used, complexes are called (b) outer orbital complexes. .

Question 45.
Explain the steps involved ¡n the metal-ligand bonding.
Answer:

Find the oxidation state of central metal ion in the complex.
Write the valence shell electronic configuration of metal ion.
From the formula of the complex determine the number of ligands and find the number of metal ion orbitais required for bonding.
Find the orbitais of metal ion available for hybridisation and the type of hybridisation involved.
Represent the electronic configuration of metal ion after hybridisation.
Exhibit filling of hybrid orbitais after complex formation.
Determine the nunther of unpaired electrons and predict magnetic property of the complex.
Find whether the complex is low spin or high spin (applicable for octahedral complexes with d⁴ or d⁸ electronic configuration.)

Question 46.
What are the salient features of valence bond theory (VBT)?
Answer:
The salient features of valence bond theory (VBT) are as follows :

According to this theory, a central metal atom or ion present in a complex provides a definite number of vacant orbitals (s, p, d and) to accommodate the electrons from the ligands for the formation coordinate bonds with the metal ion atom.
The number of vacant orbitals provided by the central metal atom or ion is the same as the coordination number of the metal. For example : Cu²⁺ provides 4 vacant orbitals in the complex. [Cu(NH₃)₄]²⁺.
The vacant orbitals of metal atom or ion undergo hybridisation forming the same number of hybridised orbitals, since the bonding with the hybrid orbitals is stronger.
Each ligand has one or more orbitals containing one or more lone pairs of electrons.
The shape or geometry of the complex depends upon the type of hybridisation of the metal atom.
When inner orbitals namely (n – 1) d orbitals in transition metal atom or ion hybridise, the complex is called inner complex and when outer orbitals i.e., nd orbitals hybridise then the complex is called outer complex.
When the central metal atom or ion in the complex contains one or more unpaired electrons the complex is
paramagnetic while if all the electrons are paired, the complex is diamagnetic.

Question 47.
What is the spin pairing process in the coordination compound?
Answer:
When the ligands approach the metal atom or ion for the formation of a complex, they influence the valence electrons of metal atom or ion. Accordingly the ligands are classified as (A) strong ligands and (B) weak ligands.

(A) Strong ligands :

They cause the pairing of unpaired electrons present in the metal atom or ion.
Spin pairing process :
- The process of pairing of unpaired electrons in metal atom or ion due to the presence of ligands in the complex is called spin pairing process.
- This spin pairing process decreases the number of unpaired electrons and hence decreases the paramagnetic character of the complex.
- The strong ligands also promote the outer ns electrons to the vacant inner (n – 1)d orbitals.

(B) Weak ligands : The weak ligands have no effect on the electrons in the valence shell of a metal atom or ion.
Strong ligands : CO, CN, ethylenediammine (en), NH₃, EDTA, etc.
Weak ligands : CI^–, I^–, OH^–, etc.

[Note : If a complex has n number of unpaired electrons then the magnetic moment, μ is given by ‘spin only’ formula μ = n(n + 2) B.M. where B.M. (Bohr Magneton) is the unit of magnetic moment. Hence from the magnitude of p, the number of unpaired electrons in the complex and its structure can be evaluated.]

Question 48.
Explain the structure of octahedral complex, [CO(NH₃)₆]³⁺ on the basis of valence bond theory.
Answer:
(1) Hexaamminecobalt(III) ion, [CO(NH₃)₆]³⁺ is a cationic complex, the oxidation state of cobalt is + 3 and the coordination number is 6.

(2) Electronic configuration : ₂₇CO [Ar]¹⁸ 3d⁷ 4s²
Electronic configuration : Co³⁺ [Ar]¹⁸ 3d⁶ 4s° 4p°

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 32

(3) Since NH₃ is a strong ligand, due to spin pairing effect, all the four unpaired electrons in 3d orbital are paired giving two vacant 3d orbitals.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 33

(4) Since the coordination number is Co³⁺ ion gets six vacant orbitais by hybridisation of two 3d vacant orbitais, One 4s and three 4p orbitais forming six d²sp³ hybrid orbitais giving octahedral geometly. It is an inner complex.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 34

(5) 6 lone pairs of electrons from 6NH₃ ligands are accommodated in the six vacant d²sp³ hybrid orbitals. Thus six hybrid orbitals of Co³⁺ overlap with filled orbitals of NH³ forming 6 coordinate bonds giving octahedral geometry to the complex.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 35

Since the complex has all electrons paired, it is diamagnetic.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 36

Question 49.
Explain the geometry of [CoF₆]^3- on the basis of valence bond theory.
Answer:
(1) Hexafluorocobaltate(III) ion, [CoF]^3- is an anionic complex, the oxidation state of cobalt is +3 and the coordination number is 6.
(2) Electronic configuration : ₂₇Co [Ar]¹⁸ 3d⁷ 4s² 4p° 4d°
Electronic configuration : Co³⁺ [Ar]18 3d⁶ 4s° 4p° 4d°

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 37

(3) Since F is a weak ligand, there is no spin pairing effect and Co³⁺ possesses 4 unpaired electrons.
(4) Since the coordination number is 6, the Co³⁺ ion gets six vacant orbitals by hybridisation of one 45 orbital, three 4p orbitals and two 4d orbitals forming six sp³d² hybrid orbitals giving octahedral geometry.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 38
(5) 6 lone pairs of electrons from 6F^– ligands are accommodated in the six vacant sp³d² hybrid orbitals. Thus six hybrid orbitals of Co³⁺ overlap with filled orbitals of F^– forming 6 coordinate bonds giving octahedral geometry to the complex. It is an outer complex.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 39

As the complex has 4 unpaired electrons it is paramagnetic.
Magnetic movement μ is,

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 40

Question 50.
Explain the structure of tetrachloronickelate(II) [NiCI]^2- on the basis of valence bond theory.
Answer:
(1) Tetrachloronickelate(II) ion is an anionic complex, oxidation state of Ni is +2 and the coordination number is 4.
(2) Electronic configuration : ₂₈Ni [Ar]¹⁸ 3d⁸ 4s² 4p°
Electronic configuration : Ni²⁺ [Ar]¹⁸ 3d⁸ 4s° 4p°

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 41

(3) Since the coordination number is 4, it gets 4 vacant hybrid orbitals by sp³-hybridisation of one 4s and three 4p orbitals giving tetrahedral geometry to the complex.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 42
(4) As Cl is a weak ligand, 2 unpaired electrons in Ni²⁺ remain undisturbed.
(5) 4 lone pairs of electrons from 40 ligands are accommodated in the vacant four sp³ hybrid orbitals. Thus four sp³ hybrid orbitals of Ni²⁺ overlap with filled orbitals of Cl^–forming 4 coordination bonds, giving tetrahedral geometry to the complex.
Since the complex has 2 unpaired electrons, it is paramagnetic.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 43
Magnetic moment \(\mu \text { is, } \mu=\sqrt{n(n+2)}=\sqrt{2(2+2)}=2.83 \text { B.M. }\)

Question 51.
Explain the structure of [Ni(CN)₄]^2- on the basis of valence bond theory.
Answer:
(1) Tetracyanonickelate (II) ion, INi(CN)₂]^2- is an anionic complex, oxidation state of Ni is + 2 and the coordination number is 4.
(2) Electronic configuration : ₂₂Ni [Ar]¹⁸ 3d⁸ 4s² 4p°
Electronic configuration: Ni²⁺ [Ai]¹⁸ 3d⁸ 4s° 4p°

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 45
(3) Since CN^– is a sa-ong ligand, one of the unpaired electrons in 3d orbital is promoted giving two paired electrons and one vacant 3d orbital.

(4) Since the coordination number is 4, Ni²⁺ gets 4 vacant hybrid orbitais by hybridisation of one 3d, one 4s and two 4p orbitais forming four dsp² hybrid orbitaIs. This has square planar geometry.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 47
(5) 4 lone pairs from 4CN^– ligands are accommodated in the vacant four dsp² hybrid orbitais. Thus four dps² hybrid orbitais of Ni²⁺ overlap with filled orbitais of CN forming 4 coordinate bonds giving square planar geometry to the complex. It is an inner complex.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 48
Since the complex ion has all electrons paired, it is diamagnetic.

Question 52.
What are the limitations of valence bond theory?
Answer:
In case of the coordination compounds, the valence bond theory has the following limitations :
(1) It cannot explain the spectral properties (colours) of the complex compounds.
(2) Even if the magnetic moments can be calculated from the number of unpaired electrons, it cannot explain the magnetic moment arising due to orbital motion of electrons.
(3) It cannot explain why the metal ions with the same oxidation state give inner complexes and outer complexes with different ligands.
(4) In every complex, it cannot explain magnetic properties based on geometry of the complex.
(5) Quantitative interpretations of thermodynamic and kinetic stabilities of the coordination compounds cannot be accounted.
(6) The complexes with weak field ligands and strong field ligands cannot be distinguished.
(7) It cannot predict the tetrahedral and square planar geometry of complexes with coordination number 4.
(8) The order of reactivity of inner complexes of d³, d⁴, d⁵ and d⁶ metal ions cannot be explained.
(9) It cannot explain the rates and mechanisms of reactions of the coordination compounds.

Question 53.
What are the assumptions of Crystal Field Theory (CFT)?
Answer:
Bethe and van Vleck developed Crystal Field Theory (CFT) to explain various properties of coordination compounds. The salient features of CFT are as follows :

In a complex, the central metal atom or ion is surrounded by various ligands which are either negatively charged ions (F^–, CI^–, CN^–, etc.) or neutral molecules (H₂O, NH₃, en, etc.) and the most electronegative atom in them points towards central metal ion.
The ligands are treated as point charges involving purely electrostatic attraction between them and metal ion.
- The central metal ion has five, (n – 1)d degenerate orbitals namely d_xy, d_yz, d_zx, d_{(x² – y²)} and d_z².
- When the ligands approach the metal ion, due to repulsive forces, the degeneracy of <i-orbitals is destroyed and they split into two groups of different energy, t_2g and e_g orbitals. This effect is called crystal field splitting which depends upon the geometry of the complex.
- The T-orbitals lying in the direction of ligands are affected to a greater extent while those lying in between the ligands are affected to a less extent.
- Due to repulsion, the orbitals along the axes of ligands acquire higher energy while those lying in between the ligands acquire less energy.
- Hence repulsion by ligands give two sets of split up orbitals of metal ion with different energies.
- The energy difference between two sets of d-orbitais after splitting by ligands is called crystal field splitting energy (CFSE) and represented by Δ₀ or by arbitrary term 10D_q. The value of Δ or 10D_q depends upon the geometry of the complex.
The electrons of metal ion occupy the split d-orbitais according to Hund’s rule. aufbau principle and those orbitais with minimum repulsion and the farthest away from the ligands.
CFI’ does not account for overlapping of orbitais of central metal ion and ligands, hence does not consider covalent nature of the complex.
From the crystal field stability energy, the stability of the complexes can he known.

Question 54.
What is crystal field splitting?
Answer:
The splitting of five degenerate d-orbitals of the transition metal ion into different sets of orbitals (to_2g and e_g) having different energies in the presence of ligands in the complex is called crystal field splitting.

Question 55.
What is crystal field stabilisation energy?
Answer:
Crystal field stabilisation energy (CFSE) It is the change in energy achieved by preferential filling up of the orbitals by electrons in the complex of metal atom or ion.

CFSE is expressed as a negative .quantity i.e., CFSE < 0. Higher the negative value more is the stability of the complex. m

Question 56.
Explain the factors affecting Crystal Field Splitting parameter (Δ₀).
Answer:
Crystal Field Splitting parameter (Δ₀) depends on. (a) Strength of the ligands and (b) Oxidation state of the metal.

(a) Strength of the ligands : Since strong field ligands like CN^–, en, etc. approach closer to the central metal ion, it results in a large crystal field splitting and hence Δ₀ has higher values.
(b) Oxidation state of the metal A metal ion with the higher positive charge draws the ligands closer to it which results in large separation of t_2g and e_g set of orbitals. The complexes involving metal ions with low oxidation state have low values of Δ₀. For example [Fe(NH₃)₆]³⁺ has higher Δ₀ than [Fe(NH₃)₆]²⁺. H

Question 57.
Explain the octahedral geometry of complexes using crystal field theory.
Answer:
(1) In an octahedral complex [MX₆]ⁿ⁺, the metal atom or ion is placed at the centre of regular octahedron while six ligands occupy the positions at six vertices of the octahedron.

(2) Among five degenerate d-orbitals. two orbitals namely d_{x² – y²} and d_z² are axial and have maximum electron density along the axes, while remaining three c-orbitals namely d_xy, d_yz and d_yz are planar and have maximum electron density in the planes and in-between the axes.

(3) Hence, when the ligands approach a metal ion, the orbitals d_{x² – y²} and d_z² experience greater repulsion and the orbitals d_zy, d_yz and d_zx experience less repulsion.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 55

(4) Therefore the energy of d_{x² – y²} and d_z² increases while the energy of d_xy, d_yz and d_zx decreases and five d-orbital lose degeneracy and split into two point groups. The orbitals d_xy, d_yz and d_yz form t_2g group of lower energy while d_{x² – y²} and d_z² form e group of higher energy.

Thus t_2g has three degenerate orbitals while e_g has two degenerate orbitals.

(5) Experimental calculations show that the energy of t_2g orbitals is lowered by 0.4Δ₀ or 4D_q and energy of e_g. orbital is increased by 0.6 Δ₀ or 6D_q Thus energy difference between t_2g and e_g orbitals is Δ₀ or 10D_g which is crystal field splitting energy.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 56

(6) CFSF increases with the increasing strength of ligands and oxidation stale of central metal ion.

Question 58.
Explain the tetrahedral geometry of complexes using crystal field theory.
Answer:
(1) In the tetrahedral complex, [MX₄]^n±, the metal atom or ion is placed at the centre of the regular tetrahedron and the four ligands, are placed at four corners of the tetrahedron.

(2) The ligands approach the central metal atom or ion in-between the three coordinates x, y and The orbitals d_xy, d_yz and d are pointed towards ligands and experience greater repulsion while the axial orbitals d_x² – _y² and d_z² lie in-between metal-ligand bond axes and experience comparatively less experience.

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 57
Fig. 9.12 (a) and (b) : Tetrahedral geometry hasing central metal atoll) (M) at the centre and four ligands (L) al the four corners

(3) Therefore energy of d_xy, d_yz and dd_zx orbitals is increased while that of d_x²y² and d_z² is lowered. Hence 5d-orbitals lose their degeneracy and split into two point groups, namely f2? of higher energy (d_xy, d_yzand d_zx) and e_g of lower energy (d_{x² – y²}and d_z²).

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 58

(4) The experimental calculations show that the energy of t_2g orbitals is increased by 0.4 Δ₀ or 4D_q and energy of e_g is lowered by 0.6 Δ₀ or 6D_g. Thus energy difference between t_2g and e_g orbitals is Δ₀ or 10D_g which is crystal field splitting energy (CFSE).

(5) This explains that the entry of each electron in e_g orbitals, stabilises the complex by 0.6 Δ₀ or 6D_q While the entry of each electron in t_2g orbitals destabilises the tetrahedral complex by 0.4 Δ₀ or 4D_g.

(6) In case of strong field ligands, the electrons prefer to pair up in e_g orbitals giving low spin (LS) complexes while in case of weak held ligands, the electrons prefer to enter higher energy t_2g orbitals giving more unpaired electrons and hence form high spin (HS) complexes.

Table 9.4 : Properties of complexes

Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 59

Question 59.
Give valence bond description for the bonding in the complex [VCI₄]^–. Draw box diagrams for free metal ion. Which hybrid prbitals are used by the metal? State the number of unpaired electrons.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 64
Since CI^– is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = 2
Type of hybridisation = sp³

Geometry of complex ion = Tetrahedral
The complex ion is paramagnetic.

Question 60.
Write a note on colour in coordination compounds.
Answer:

A large number of coordination compounds show wide range of colours due to d – d transition of electron and this can be explained by crystal field theory (CFT).
The complex absorbs the light in one visible region (400 nm to 700 nm) and transmits the light in different visible region giving complementary colour.
Consider an octahedral purple coloured complex of [Ti(H₂O)₆]³⁺ which absorbs green light and transmits purple colour. Similarly [Cu(H₂O)₆]²⁺ absorbs the light in the red region of radiation spectrum and transmits in the blue region, hence the complex appears blue.
The absorption of light arises due to d-d transition of electron from lower energy level (t_2g) to higher energy level (e_g) in octahedral complex.
The energy required for transition depends upon crystal field splitting energy Δ₀. If Δ₀ = ΔE, then the energy of an absorbed photon (hv) is \(\Delta E=h v=\frac{h c}{\lambda}\) where λ, v and c are wavelength, frequency and velocity of the absorbed light.
Higher the magnitude of Δ₀ or ΔE, higher is the frequency or lower is the wavelength of the absorbed radiation.
Since Δ₀ depends upon nature of metal atom or ion, its oxidation state, nature of ligands and the geometry of the complex, different coordination compounds have different colours.

Question 61.
Explain the purple colour of the complex, [Ti(H₂O)₆]³⁺ with the help of crystal field theory.
Answer:
(1) [Ti(H₂O)₆]³⁺ is an octahedral complex, oxidation state of titanium is +3 (Ti³⁺) and the coordination number is 6.

(2) Electronic configuration = ₂₂Ti [Ar]¹⁸ 3d² 4s²
Electronic configuration = Ti³⁺ + [Ar]¹⁸ 3d¹
OR

(3) According to crystal field theory, 3d orbitals undergo crystal field splitting giving higher energy e_g, two orbitals and lower energy t_2g, three orbitals.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 67
(4) The crystal field splitting energy (CFSE), Δ₀ is found to be 3.99 x 10^-19 J/ion from the spectrochemical studies.

(5) The absorption of radiation of wavelength λ or frequency v results in the transition of one unpaired electron from Photon energy t_2g orbital to e_g orbital.
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 68
(6) The wavelength of the absorbed radiation will be.

(7) Hence the complex [Ti(H₂O)₆]³⁺ absorbs the green radiation of wavelength 498 nm in the visible region and transmits the complementary purple light. Therefore the complex is purple coloured.

Question 62.
An octahedral complex absorbs the radiation of wavelength 620 nm. Find the crystal field splitting energy.
Answer:
Crystal field splitting energy Δ₀ is given by,
Maharashtra Board Class 12 Chemistry Solutions Chapter 9 Coordination Compounds 70

Question 63.
What are the applications of coordination compounds ?
Answer:
(1) In biology : Several biologically important natural compounds are metal complexes which play an important role in number of processes occurring in plants and animals.

For example, chlorophyll in plants is a complex of Mg²⁺ ions, haemoglobin in blood is a complex of iron, vitamin B₁₂ is a complex of cobalt.

(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more stable, more effective and can be assimilated easily.

For example, platinum complex [Pt(NH₃)₂CI₂] known as cisplatin is effectively used in cancer treatment. EDTA is used to treat poisoning by heavy metals like lead.

(3) To estimate hardness of water :

The hardness of water is due to the presence Mg²⁺ and Ca²⁺ ion in water.
The strong field ligand EDTA forms stable complexes with Mg²⁺ and Ca²⁺. Hence these ions can be removed by adding EDTA to hard water.

Similarly these ions can be selectively estimated due to the difference in their stability constants.

(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to supply continuously the metal ions in small amounts.

For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)₂] and K[Au(CN)₂] are used.

Multiple Choice Questions

Select and write the most appropriate answer from the given alternatives for each subquestion :

Question 1.
The coordination number of cobalt in the complex [Co(en)₂Br₂]CI₂ is
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(c) 6

Question 2.
EDTA combines with cations to form
(a) chelates
(b) polymers
(c) clathrates
(d) non-stoichiometric compounds
Answer:
(a) chelates

Question 3.
Which one of the following compounds can exhibit coordination isomerism?
(a) [Co(en)₂CI₂]Br
(b) [CO(NH₃)₆] [Cr(CN)₆]
(c) [Co(en)₃]CI₃
(d) [CO(NH₃)₅NO₂]CI₂Answer:
(b) [CO(NH₃)₆] [Cr(CN)₆]

Question 4.
Which of the following compounds can exhibit linkage isomerism?
(a) [Co(en)₃]CI₃
(b) [Co(en)₂CI₂]CI
(c) [Co(en)₂NO₂Br]CI
(d) [Co(NH₃)₅CI]Br₂Answer:
(c) [Co(en)₂NO₂Br]CI

Question 5.
Oxidation number of cobalt in K[COCI₄] is
(a) +1
(b) -1
(c) +3
(d) -3
Answer:
(b) -1

Question 6.
The correct structure of [Cr(H₂O)₆]³⁺ is ……………………
(a) octahedral
(b) tetrahedral
(c) square pyramidal
(d) trigonal bipyramidal
Answer:
(b) tetrahedral

Question 7.
Amongst the following ions which one has the highest paramagnetism?
(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [CU(H₂O)₆]²⁺
(d) [Zn(H₂O)₆]²⁺Answer:
(b) [Fe(H₂O)₆]²⁺

Question 8.
The geometry of [Ni(CN)₄]^3- and [NiCI₄]^-2 are
(a) both tetrahedral
(b) both square planar
(c) tetrahedral and square planar respectively
(d) square planar and tetrahedral respectively
Answer:
(d) square planar and tetrahedral respectively

Question 9.
The complex cis-[Pt(NH₃)₂CI₂] is used in treatment of cancer under the name.
(a) Aspirin
(b) Eqanil
(c) cisplatin
(d) transplatin
Answer:
(c) cisplatin

Question 10.
[CO(NH₃)₆]³⁺ is an orbital complex and is in nature.
(a) inner, paramagnetic
(b) inner, dimagnetic
(c) outer, paramagnetic
(d) outer, dimagnetic
Answer:
(b) inner, dimagnetic

Question 11.
The IUPAC name of [Ni(Co)₄] is
(a) tetra carbonyl nickel (O)
(b) tetra carbonyl nickel (II)
(c) tetra carbonyl nickelate (O)
(d) tetra carbonyl nickelate (II)
Answer:
(a) tetra carbonyl nickel (O)

Question 12.
The number of ions produced by the complex [CO(NH₃)₄CI₂] Cl is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 13.
The dimagnetic species is
(a) [Ni(CN)₄]^2-
(b) [NiCl₄]^2-
(c) [CoCI₄]^2-
(d) [CoF₆]^2-Answer:
(a) [Ni(CN)₄]^2-

Question 14.
Which one of the following is an inner orbital complex as well as diamagnetic in behaviour (Atomic no. Zn = 30, Cr = 24, Co = 27, Ni = 28)
(a) [Zn(NH₃)₆]²⁺
(b) [Cr(NH₃)₆]³⁺
(c) [CO(NH₃)₆]³⁺
(d) [Ni(NH₃)₆]²⁺Answer:
(c) [CO(NH₃)₆]³⁺

Question 15.
Among [Ni(Co)J, [Ni(CN)4]2A [NiClJ2- Species, the hybridisation states at the Nickel atom are respectively
(a) sp³, dsp², sp³
(b) sp³, dsp², dsp²
(c) dsp², sp³, sp³
(d) sp³, sp³, dsp²Answer:
(a) sp³, dsp², sp³

Question 16.
The strongest ligand in the following is
(a) CN^–
(b) Br^–
(c) HO^–
(d) F^–Answer:
(a) CN^–

Question 17.
Magnetic moment of (NH4)2 (MnBr4) is BM
(a) 5.91
(b) 4.91
(c) 3.91
(d) 2.91
Answer:
(a) 5.91

Question 18.
The complex which violates EAN rule is
(a) Fe(CO)₅
(b) [Fe(CN)₆]^3-
(c) Ni(CO)₄
(d) [Zn(NH₃)₄]CI₂Answer:
(b) [Fe(CN)₆]^3-

Question 19.
EDTA is a ligand of the type
(a) bidentate
(b) tridentate
(c) tetradentate
(d) hexadentate
Answer:
(d) hexadentate

Question 20.
The cationic complex among the following is
(a) K₃[Fe(CN)₆]
(b) Ni(CO)₄
(c) K₂HgI₄
(d) [CO(NH₃)₆]CI₂Answer:
(d) [CO(NH₃)₆]CI₂

Question 21.
If Z is the atomic number of a metal, X is number of electrons lost forming metal ion and Y is the number of electrons from the ligands then EAN is
(a) Z + X + Y
(b) X – Z + Y
(c) Z – X + Y
(d) X + Z – Y
Answer:
(c) Z – X + Y

Question 22.
Octahedral complex has hybridisation,
(a) dsp²
(b) d³sp³
(c) dsp³
(d) d²sp³Answer:
(d) d²sp³

Question 23.
Inner complex has hybridisation,
(a) d²sp³
(b) sp³d²
(c) sp³d
(d) sp³d³Answer:
(a) d²sp³

Question 24.
The number of unpaired electrons in [CO(NH3)6]3+ is
(a) 0
(b) 1
(c) 2
(d) 4
Answer:
(a) 0

Question 25.
The number of unpaired electrons in [NiClJ2- and [Ni(CN)4]2_ are respectively,
(a) 2, 2
(b) 2, 0
(c) 0, 0
(d) 1, 2
Answer:
(b) 2, 0

Question 26.
Among the following complexes, the highest magnitude of crystal field stabilisation energy will be for [Co(H₂O)₆]³⁺, [CO(CN)₆]^3-, [Co(NH₃)₆]³⁺, [CoF₆]^3-
(a) [Co(H₂O)₆]³⁺
(b) [CO(CN)₆]^3-
(c) [Co(NH₃)₆]³⁺
(d) [CoF₆]^3-Answer:
(b) [CO(CN)₆]^3-

Question 27.
The number of unpaired electrons in a low spin octahedral complex ion of d1 is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 28.
The number of unpaired electrons in a high spin octahedral complex ion of d⁷ is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(d) 3

Question 29.
Ligand used in the estimation of hardness of water is
(a) EDTA
(b) DBG
(c) chloride
(d) bromo
Answer:
(a) EDTA

Question 30.
Which of the following complexes will give a white precipitate on treatment with a solution of barium nitrate?
(a) [Cr(NH₃)₄SO₄] CI
(b) [CO(NH₃)₄CI₂] NO₂
(c) [Cr(NH₃)₄CI₂] SO₄
(d) [CrCI₃(H₂O)₄]CI
Answer:
(c) [Cr(NH₃)₄CI₂] SO₄

Question 31.
What is effective atomic number of Fe (z = 26) in [Fe(CN)₆]^4-?
(a) 12
(b) 30
(c) 26
(d) 36
Answer:
(d) 36

Question 32.
Cisplatin compound is used in the treatment of
(a) malaria
(b) cancer
(c) AIDS
(d) yellow fever
Answer:
(b) cancer

Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases

July 23, 2024July 22, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 10 Human Health and Diseases Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 10 Human Health and Diseases

Multiple Choice Questions

Question 1.
Infectious stage of Plasmodium is …………………
(a) trophozoite
(b) sporozoite
(c) cryptozoite
(d) metacercaria
Answer:
(b) sporozoite

Question 2.
After birth, antibodies are transferred from mother to infant through …………………
(a) colostrum
(b) placenta
(c) blood
(d) tissue fluid
Answer:
(a) colostrum

Question 3.
Which cells give rise to T-lymphocytes?
(a) Thymocytes
(b) Bone marrow cells
(c) Erythrocytes
(d) Leucocytes
Answer:
(a) Thymocytes

Question 4.
Where is antigen D is present?
(a) On Rhesus factor
(b) On the surface of RBCs
(c) On A-antigen
(d) On AB-antigen
Answer:
(b) On the surface of RBCs

Question 5.
Erythroblastosis foetalis is caused when mother is …………………
(a) Rh +ve
(b) with antibody ‘a’
(c) Rh -ve
(d) with antibody ‘b’
Answer:
(c) Rh -ve

Question 6.
Which of the following is NOT a parasitic vector insect?
(a) Mosquito
(b) Housefly
(c) Honey bee
(d) Head louse
Answer:
(c) Honey bee

Question 7.
Which is the proper sequence in the developmental stages of Plasmodium?
(a) Merozoites → Sporozoite → Trophozoites → Schizonts
(b) Trophozoites → Merozoites → Sporozoite → Schizonts
(c) Sporozoite → Merozoites → Trophozoites → Schizonts
(d) Schizonts → Merozoites → Sporozoite → Trophozoites.
Answer:
(c) Sporozoite → Merozoites → Trophozoites → Schizonts

Question 8.
There is no vaccination on this disease till today.
(a) Typhoid
(b) Tuberculosis
(c) Polio
(d) AIDS
Answer:
(d) AIDS

Question 9.
Charas, hashish, ganja are obtained from …………………
(a) Papaver somnijerum
(b) Erythroxylum coca
(c) Atropa belladorta
(d) Cannabis sativa
Answer:
(d) Cannabis sativa

Question 10.
………………. Plant is used to obtain cocaine alkaloid.
(a) Marijuana
(b) Papaver somntferum
(c) Cannabis sativa
(d) Coca
Answer:
(d) Coca

Question 11.
………………… fish is released in the waterbody to prevent the spread of Malaria and Filaria.
(a) Pomfret
(b) Tilapia
(c) Gambusia
(d) Gold fish
Answer:
(c) Gambusia

Question 12.
The carcinogen that can cause vaginal cancer is …………………
(a) Vinyl chloride
(b) Diethylstilboestrol
(c) Mustard gas
(d) Cadmium oxide
Answer:
(b) Diethylstilboestrol

Question 13.
Prostate cancer can be caused due to exposure to …………………
(a) cadmium oxide
(b) mustard gas
(c) asbestos
(d) Nickel and chromium compounds
Answer:
(a) cadmium oxide

Question 14.
Choose the correct definition of health …………………
(a) Health is not contracting any disorder or disease.
(b) State of complete physical, mental and social well-being.
(c) Health is complete absence of any disease.
(d) Health is feeling good all the time.
Answer:
(b) State of complete physical, mental and social well-being

Question 15.
The interval between infection and appearance of disease symptoms is called …………………
(a) inoculation
(b) penetration
(c) infection period
(d) incubation period
Answer:
(d) incubation period

Question 16.
What is injected in vaccination ?
(a) Half killed pathogen
(b) Dead pathogens
(c) Live pathogens
(d) Readymade antibodies
Answer:
(a) Half killed pathogen

Question 17.
Who among the following is considered as father of immunology ?
(a) Ferdinand Kohn
(b) Robert Koch
(c) Louis Pasteur
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 18.
Who coined the term antibiotics ?
(a) Charles Darwin
(b) Louis Pasteur
(c) Alexander Fleming
(d) Selman Waksman
Answer:
(d) Selman Waksman

Question 19.
Who coined the term antibody?
(a) Selman Waksman
(b) Alexander Fleming
(c) Paul Ehrlich
(d) Edward Jenner
Answer:
(c) Paul Ehrlich

Question 20.
Widal test is used for the diagnosis of …………………
(a) Malaria
(b) Typhoid
(c) Diabetes mellitus
(d) HIV/AIDS
Answer:
(b) Typhoid

Question 21.
Which bacterial genus out of the following is the common pathogen causing pneumonia ?
(a) Streptococcus sps
(b) Lactobacillus sps
(c) Pseudomonas sps
(d) Salmonella sps
Answer:
(a) Streptococcus sps

Question 22.
Given below are some statements. Which among them are symptoms of pneumonia ?
(i) Greenish, yellow sputum coughed out.
(ii) Hepatomegaly and hypoglycemia.
(iii) High fever with shaking chills.
(iv) Thickening of skin and underlying tissues.
(v) Stabbing chest pain with shortness of breath.
(vi) Mood swings and joint pains along with nausea and vomiting.
(a) (i), (ii), (iii), (iv)
(b) (i), (iii), (v), (vi)
(c) (i), (ii), (iv), (vi)
(d) (ii), (iii), (iv), (v)
Answer:
(b) (i), (hi), (v), (vi)

Question 23.
Which of the following is not the common way to prevent common cold ?
(a) Using hand sanitizers
(b) Blowing nose in open
(c) Staying away from people suffering from cold
(d) Sipping warm water
Answer:
(b) Blowing nose in open

Question 24.
Common cold is not cured by antibiotics because it is …………………
(a) caused by a virus
(b) caused by a Gram-positive bacterium
(c) caused by a Gram-negative bacterium
(d) not an infectious disease
Answer:
(a) caused by a virus

Question 25.
Motile zygote of Plasmodium occurs in …………………
(a) gut of female Anopheles
(b) salivary glands of Anopheles
(c) Human RBCs
(d) Human liver
Answer:
(a) gut of female Anopheles

Question 26.
Haemozoin is …………………
(a) a precursor of haemoglobin
(b) a toxin from Streptococcus
(c) a toxin from Plasmodium
(d) a toxin from Hemophilus
Answer:
(c) a toxin from Plasmodium

Question 27.
Vaccination against malaria is not possible because …………………
(a) they produce antibodies and antitoxins
(b) they do not produce antibodies and antitoxins
(c) antibodies resistant to vaccines are produced
(d) none of these
Answer:
(b) they do not produce antibodies and antitoxins

Question 28.
The active form of Entamoeba histolytica feeds upon …………………
(a) blood only
(b) erythrocytes, mucosa and submucosa of colon
(c) mucosa and submucosa of colon only
(d) food in intestine
Answer:
(b) erythrocytes, mucosa and submucosa of colon

Question 29.
Eating unwashed and raw green leafy vegetables grown along the railway tracks in Mumbai may cause …………………
(a) malaria
(b) influenza
(c) amoebic colitis
(d) ringworm
Answer:
(c) amoebic colitis

Question 30.
Which method of water purification can terminate amoebae ?
(a) Chlorination
(b) Sedimentation
(c) Filtration
(d) Boiling
Answer:
(d) Boiling

Question 31.
Which of the following measures should be taken to control amoebic dysentery ?
(A) Using insecticidal sprays to kill flies.
(B) Not allowing stagnant water to be accumulated over a long time.
(C) To avoid eating uncovered food.
(D) Drinking only boiled water.
(E) Eating plenty of fruits.
(a) (A) (C) (D)
(b) (A) (B) (D)
(c) (B) (C) (E)
(d) (A) (B) (E)
Answer:
(a) (A) (C) (D)

Question 32.
Which of the following should be avoided for endemic spread of amoebiasis ?
(a) Cleaning bathroom taps and toilet seats with disinfectants.
(b) Washing hands and using hand sanitizers.
(c) Proper sewage disposal and treatment.
(d) Eating uncovered roadside food.
Answer:
(d) Eating uncovered roadside food.

Question 33.
Name the disease in which the genital organs are grossly affected due to infective helminth.
(a) Ascariasis
(b) Ring worm
(c) Scabies
(d) Filariasis
Answer:
(d) Filariasis

Question 34.
Find the odd organism:
(a) Wuchereria bancrofti
(b) Brugia malayi
(c) Brugia timori
(d) Ascaris lumbricoides
Answer:
(d) Ascaris lumhricoides

Question 35.
When is hydrocele formed in a man ?
(a) When testis are not functioning properly.
(b) When scrotum is infected with filarial worms.
(c) When testis are injured due to accident.
(d) When there is water accumulation in testis.
Answer:
(b) When scrotum is infected with filarial worms

Question 36.
Which medicine is used for eradicating microfilariae from endemic areas ?
(a) Diethyle carbamacine
(b) Mebendazole
(c) Albendazole
(d) Rimfampcin
Answer:
(a) Diethyle carbamacine

Question 37.
Which of the following fungi are causative organisms of ringworm ?
(a) Microsporum
(b) Candida
(c) Thrush
(d) Tinea pedis
Answer:
(a) Microsporum

Question 38.
On which material present on the outer skin surfaces of the human body does the fungus causing infections feed on ?
(a) Melanin
(b) Keratin
(c) Lignin
(d) Suberin
Answer:
(b) Keratin

Question 39.
Which of the following statements is correct ?
(a) Fungus grows well on dry skin.
(b) Fungus cannot survive on the outside of the hair shafts.
(c) Fungus thrives well on the warm and moist skin.
(d) Nails can never show fungal infections.
Answer:
(c) Fungus thrives well on the warm and moist skin.

Question 40.
Which of the following pair is viral diseases ?
(a) Common cold, AIDS
(b) Dysentery, Common cold
(c) Typhoid, Tuberculosis
(d) Ringworm, AIDS
Answer:
(a) Common cold, AIDS

Question 41.
Which one of the following glands is often referred in relation with AIDS ?
(a) Thymus
(b) Adrenal
(c) Thyroid
(d) Pancreas
Answer:
(a) Thymus

Question 42.
The first patient of AIDS was detected in India in …………………
(a) 1980
(b) 1986
(c) 1990
(d) 1996
Answer:
(b) 1986

Question 43.
Why is it said that for AIDS prevention is the only cure ?
(a) AIDS does not have any cure, once it is contracted.
(b) By prevention AIDS cannot be cured.
(c) AIDS can be cured by proper medication and vaccination.
(d) Only prevention helps as there is no cure for AIDS.
Answer:
(d) Only prevention helps as there is no cure for AIDS.

Question 44.
After a person is detected to be having AIDS by ELISA test, which is the next confirmatory test ?
(a) Western blot
(b) Southern blot
(c) PCR
(d) Northern blot
Answer:
(a) Western blot

Question 45.
What is full form of ELISA ?
(a) Enzyme Linked Inductive Assay
(b) Enzyme Linked Iron Sorbent Assay
(c) Enzyme Linked Immunosorbent Assay
(d) None of the above
Answer:
(c) Enzyme Linked Immunosorbent Assay

Question 46.
The possible ways of transmission of AIDS are …………………
(A) Intimate sexual contact
(B) Hugging and kissing
(C) Blood transfusion without properly checking it
(D) Eating from the same plate
(E) Sharing bed linen
(F) Transplacental infection from infected mother
(G) Sharing same tattoo gun and syringes
(a) (A), (C), (F), (G)
(b) (B), (D), (E), (G)
(c) (C), (D), (E), (F)
(d) (A), (B), (C), (D)
Answer:
(a) (A), (C), (F), (G)

Question 47.
Which one of the following statements is correct ?
(a) Benign tumours show the property of metastasis.
(b) Heroin accelerates body functions.
(c) Malignant tumours may exhibit metastasis.
(d) Patients who have undergone surgery are given cannabinoids to relieve pain.
Answer:
(c) Malignant tumours may exhibit metastasis.

Question 48.
Heroin or smack is chemically …………………
(a) diclofenac
(b) diacetyl morphine
(c) benzodiazepine
(d) amphetamines
Answer:
(b) diacetyl morphine

Question 49.
Ecstasy is a drug that is used in most of the Rev parties which is chemically a derivative of …………………
(a) Barbiturates
(b) Amphetamines
(c) Catecholamine
(d) Morphine
Answer:
(b) Amphetamines

Question 50.
From which plant is charas obtained?
(a) Cannabis sativa
(b) Erythroxylum coca
(c) Papaver somniferum
(d) Atropa belladonna
Answer:
(a) Cannabis sativa

Question 51.
Opium is obtained from the latex of the unripe fruits of …………………
(a) Cannabis sativa
(b) Thea siensis
(c) Papaver somniferum
(d) Erythroxylon coca
Answer:
(c) Papaver somniferum

Question 52.
Use of Cannabis products results in …………………
(a) depressed brain activity and feeling of calmness
(b) suppressed brain function and relief of pain
(c) stimulation of nervous system, increased alertness and activity
(d) alteration in perception, thoughts and feelings
Answer:
(d) alteration in perception, thoughts and feelings

Question 53.
Marijuana, ganja and LSD are …………………
(a) narcotics
(b) stimulants
(c) hallucinogens
(d) all of these
Answer:
(c) hallucinogens

Question 54.
What is the source of LSD ?
(a) Poppy seeds
(b) Datura plant
(c) Sugar
(d) Claviceps purpurea
Answer:
(d) Claviceps purpurea

Question 55.
Narcotics are …………………
(a) amphetamines and caffeine
(b) morphine and heroine
(c) LSD and cocaine
(d) barbiturates and benzodiazepine
Answer:
(b) morphine and heroine

Question 56.
Which is an incorrectly matched pair ?
(a) LSD – Ergot fungus
(b) Heroin – Opium
(c) Amphetamines – Depressant
(d) Benzodiazepine – Calmpose tablets
Answer:
(c) Amphetamines – Depressant

Question 57.
Choose the incorrect statement
(a) The excessive use of anabolic steroids cause severe acne.
(b) In both the sexes there is increased aggressiveness and mood swings, due to steroids.
(c) In females, anabolic steroids cause breast enlargement.
(d) In males, anabolic steroids cause enlargement of prostate gland.
Answer:
(c) In females, anabolic steroids cause breast enlargement.

Question 58.
Who are the first ones to note the danger signs of drug or alcohol abuse in the adolescents ?
(a) Alert parents and teachers
(b) Neighbours
(c) Relatives
(d) Doctors
Answer:
(a) Alert parents and teachers

Question 59.
Who can give professional help for the deaddiction?
(a) Highly qualified psychiatrist
(b) Parents
(c) Teachers
(d) Friends
Answer:
(a) Highly qualified psychiatrist

Match the columns

Question 1.

Column I	Column II
(a) Metchnikoff	(i) ABO Blood group system
(b) Fleming	(ii) Concept of immunity
(c) Edward Jenner	(iii) Phagocytic cells
(d) Karl Lands teiner	(iv) Lysozyme

Answer:

Column I	Column II
(a) Metchnikoff	(iii) Phagocytic cells
(b) Fleming	(iv) Lysozyme
(c) Edward Jenner	(ii) Concept of immunity
(d) Karl Lands teiner	(i) ABO Blood group system

Question 2.

Disease	Vector species
(a) Dengue	(i) Anopheles
(b) Malaria	(ii) Housefly
(c) Filaria	(iii) Culex
(d) Typhoid	(iv) Aedes

Answer:

Disease	Vector species
(a) Dengue	(iv) Aedes
(b) Malaria	(i) Anopheles
(c) Filaria	(iii) Culex
(d) Typhoid	(ii) Housefly

Classify the following to form Column B as per the category given in Column A

Question 1.
Benzyl penicillin, Chloromycetin, Mebendazole, Levamisole, Pyrimethamine Ampicillin, Ty21a vaccine, Sulfadoxine.

Column A (Disease)	Column B (Treatment)
(1) Pneumonia	————–
(2) Malaria	————–
(3) Ascariasis	————–
(4) Typhoid	————–

Answer:

Column A (Disease)	Column B (Treatment)
(1) Pneumonia	Benzyl penicillin, Ampicillin
(2) Malaria	Pyrimethamine, Sulfadoxine
(3) Ascariasis	Mebendazole, Levamisole
(4) Typhoid	Chloromycetin, Ty21a vaccine

Question 2.
Lung cancer, Pituitary, Spleen, Skin cancer, Cancer of adipose tissue, lymph nodes, Adrenal, Bone tumour.

Column (A Type of cancer)	Column B (Organs affected)
(1) Carcinoma	————–
(2) Sarcoma	————–
(3) Lymphoma	————–
(4) Adenocarcinoma	————–

Answer:

Column (A Type of cancer)	Column B (Organs affected)
(1) Carcinoma	Lung cancer, Skin cancer
(2) Sarcoma	Cancer of adipose tissue, Bone tumour
(3) Lymphoma	Spleen, Lymph nodes
(4) Adenocarcinoma	Pituitary, Adrenal

Very Short Answer Questions

Question 1.
By which process T-cells and B-cells are produced?
Answer:
T-cells and B-cells are produced from the stem cells called haemocytoblasts, in bone marrow of adults and in liver of the foetus, by the process of haematopoiesis and in the bone marrow in adult.

Question 2.
What is a hinge?
Answer:
Hinge is the region of Y-shaped structure, holding arms and stem of antibody where four polypeptide chains of antibody are held together by disulfide bonds (-s-s-) to form a ‘Y’-shaped structure.

Question 3.
What is epitope?
Answer:
Epitope is antigenic determinant, which is present on antigens.

Question 4.
What is paratope?
Answer:
Paratope is antigen binding site that is present on the antibodies.

Question 5.
Which antigen is present in Rh +ve person?
Answer:
Antigen D is present in Rh +ve person.

Question 6.
Give the role of flushing action of lachrymal secretions.
Answer:
The conjunctiva is freed from foreign particles by the flushing action of lachrymal secretions.

Question 7.
What happens when lachrymal secretion is absent in eyes?
Answer:
Eyes become susceptible to infection when lachrymal secretion is absent.

Question 8.
In tears which antibacterial substance is present?
Answer:
Lysozyme is the antibacterial substance present in the tears.

Question 9.
Which kind of immunity is provided by vaccination?
Answer:
Artificial acquired active and passive immunity is provided by vaccination.

Question 10.
Who was Edward Jenner?
Answer:
Edward Jenner was the British scientist who developed cowpox vaccine for the protection against small pox virus.

Question 11.
Mrunmayi is called universal blood acceptor. What is her blood group?
Answer:
Blood group of Mrunmayi is AB.

Question 12.
What are antigens?
Answer:
Different foreign substances that invade the body and are capable of stimulating an immune response are called antigens.

Question 13.
In which animal Rh factor was discovered at first?
Answer:
Rh factor was first discovered in Rhesus monkey for the first time.

Question 14.
What is elephantiasis?
Answer:
Elephantiasis is one of the symptoms of lymphatic filariasis, in which there is thickening of skin and underlying tissues due to presence of malarial parasite.

Question 15.
What is dermatophytosis ?
Answer:
Dermatophytosis is a clinical condition in which fungal infection of skin occur in humans, pets and cattle which is commonly called as ringworm.

Question 16.
What is sporozoite?
Answer:
Sporozoite is a developmental stage of Plasmodium produced by rupture of oocyst. Sporozoite can enter the bloodstream in human body and then infect hepatocytes or liver cells, where they multiply into merozoites.

Question 17.
Why does male mosquito not spread Malaria?
Answer:
Male mosquito feed only on plant sap and not blood of human beings; therefore it does not spread Malaria.

Question 18.
Where does Plasmodium reproduce asexually?
Answer:
Plasmodium reproduce asexually in the liver cells and red blood cells of infected human being.

Question 19.
Where does Plasmodium reproduce sexually?
Answer:
Plasmodium undergoes sexual reproduction by the process of fertilization and development in the intestine of mosquito.

Question 20.
Which fish can be used for mosquito control?
Answer:
Gambusia fish can be used for mosquito control.

Question 21.
What is arthralgia?
Answer:
Arthralgia means joint pains.

Question 22.
What do you mean by hepatomegaly?
Answer:
Hepatomegaly means enlargement of liver.

Question 23.
Which organism yields LSD?
Answer:
Ergot fungus, Claviceps purpurea yield LSD.

Question 24.
Enlist various types of barriers which prevent entry of foreign agents into the body.
Answer:

Epithelial surface
Antimicrobial substances in blood and tissues
Cellular factors in innate immunity
Fever
Acute phase proteins (APPs).

Question 25.
Which is the gastro-intestinal disease by which 15% Indian population is affected?
Answer:
Amoebiasis or amoebic dysentery is the gastro-intestinal disease by which 15% Indian population is affected.

Question 26.
What are the anti-helminthic drugs which are used in treatment of Ascariasis?
Answer:
Anti-helminthic drugs like Piperazine, Mebendazole, Levamisole, Pyrantel are used against Ascaris lumbricoidies.

Question 27.
Which are the diseases that can be avoided by eradication of mosquitoes in your area?
Answer:
Malaria, dengue, chikungunya and filariasis or elephantiasis can be avoided by eradication of mosquitoes.

Define the following

Question 1.
Serology
Answer:
A branch of immunology which deals with the study of antigen-antibody interactions is called serology.

Question 2.
Hygiene
Answer:
Hygiene is the science of health, which aims at preserving, maintaining and improving the health of an individual or the community as a whole.

Question 3.
Disease
Answer:
Disease is a condition of disrupted or deranged functioning of one or more organs or systems of the body caused due to infection, detective diet or heredity.

Question 4.
Immune system
Answer:
The system which protects us from various infectious agents is called immune system.

Question 5.
Resistance
Answer:
Resistance is an ability to ward off damage or disease through our defence mechanism.

Question 6.
Immunity
Answer:
The immunity is defined as the general ability of a body to recognize and neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.

Question 7.
Antibody
Answer:
The protective chemicals produced by immune cells in response to antigens is called antibodies.

Question 8.
Opsonisation
Answer:
The process of coating of bacteria to facilitate their subsequent phagocytosis by macrophages is called opsonisation.

Question 9.
Pathogen
Answer:
Pathogen are living agents such as viruses, ricketssia, bacteria, fungi, protozoans, helminth and certain insect larvae which are capable of causing diseases.

Question 10.
Parasite
Answer:
An organism that lives in or on another organism called host and takes its nourishment from it (host) is called parasite.

Question 11.
Pathogenicity
Answer:
The ability of an organism to enter a host and cause a disease is called pathogenicity.

Question 12.
Infectious disease
Answer:
The disease which is transmitted from infected person to another healthy person either directly or indirectly is called infectious disease or communicable disease.

Question 13.
Non-infectious disease
Answer:
The disease that cannot be transmitted from one infected person to another healthy person, either directly or indirectly is called non infectious or non-communicable disease.

Question 14.
Innate immunity
Answer:
Innate immunity is defined as the resistance to infections that an individual possesses due to his or her genetic make-up and thus it is inborn defence mechanism present naturally in the body.

Question 15.
Acquired immunity
Answer:
The resistance developed during lifetime is called acquired immunity.

Question 16.
APP proteins
Answer:
APP proteins or acute phase proteins are certain collection of plasma proteins which are suddenly increased by the infection caused after injury.

Question 17.
Incubation period
Answer:
Incubation period is the time interval from the invasion of a pathogen to the development of clinical manifestations or symptoms.

Name the following

Question 1.
Name the cell that produces lymphokines.
Answer:
Helper T-cells

Question 2.
Blood group systems in human beings.
Answer:
ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay blood group.

Question 3.
Types of sarcoma.
Answer:

Osteosarcoma (bone)
Myosarcoma (muscle)
Chondrosarcoma (cartilage)
Liposarcoma (adipose tissue)

Question 4.
Therapies used for treatment of cancer.
Answer:

Chemotherapy
Radiotherapy
Surgery
Immunotherapy
Supportive therapy

Question 5.
Name the term for the transmission of HIV from pregnant mother to foetus.
Answer:
Transplacental.

Question 6.
Factors that maintain good health.
Answer:

Balanced diet
Personal hygiene
Regular exercise
Right attitude of mind
Good habits.

Question 7.
Two examples of ascaricides.
Answer:
Mebendazole and Albendazole, etc.

Question 8.
Two vaccines for typhoid.
Answer:

Oral Ty21a
Injectable Typhoid polysaccharide vaccine or Typhium vi/ Typherix.

Question 9.
Parasites causing lymphatic filariasis.
Answer:

Wuchereria bancrojti
Brugiamalayi
Brugia timori.

Question 10.
Parasites causing Subcutaneous Filariasis.
Answer:

Loa loa
Mansonella spp.

Question 11.
Name the scientists who discovered AB blood group?
Answer:
Decastallo and Sturti

Distinguish between the following

Question 1.
Inborn immunity and acquired immunity.
Answer:

Inborn Immunity	Acquired Immunity
1. Inborn immunity or innate immunity is also called natural immunity.	1. Acquired immunity is also called adaptive immunity.
2. Innate immunity is present right from the birth.	2. Acquired immunity is not present at birth, but is acquired during lifetime of the individual.
3. Inborn immunity does not depend upon the previous exposure to a pathogen or foreign substance.	3. Acquired immunity always depends upon the previous exposure to a pathogen or foreign substance.
4. It is non-specific immunity as it can offer resistance to any pathogen.	4. It is specific immunity as it can offer resistance only to a particular pathogen.
5. Innate immunity consists of various types of barriers for defence against the pathogens.	5. Acquired immunity consists of various types of cells which are able to produce antibodies.
6. Inborn immunity shows immediate effect in the body.	6. Acquired immunity requires several days to become activated.
7. Inborn immunity is seen in all animals.	7. Acquired immunity is seen only in vertebrates.
8. Inborn immunity is genetic in nature and is heritable.	8. Acquired immunity is non-genetic in nature and is non-heritable.

Question 2.
Communicable and non-communicable diseases.
Answer:

Communicable diseases	Non-communicable diseases
1. Diseases transmitted from infected person to healthy person are called communicable or infectious diseases.	1. Diseases that are not passed from one person to other are non-communicable or non-infectious diseases.
2. Communicable diseases spread through pathogens.	2. Non-communicable diseases do not spread through pathogens.
3. Communicable diseases are not inherited from parental generation to offspring.	3. Non-communicable diseases like cancer can be from parental generation to offspring.
4. Vectors play the major role in spreading disease from one person to another.	4. Caused due to allergy, illness, malnutrition or abnormalities in cell proliferation, changes in lifestyle, environment play a significant role.
5. Treated by conventional methods using antibiotics and other drugs.	5. Treated conservatively for a long time or surgically.
6. Diseases are acute which develop suddenly due to infections. E.g. Pneumonia, Tuberculosis, AIDS, Typhoid, Cholera, Malaria.	6. Diseases are chronic which develop and persist for a long time. E.g. Cancer, Rickets, Allergies, Kwashiorkor, Diabetes, Heart disease, etc.

Question 3.
Ascariasis and Filariasis.
Answer:

Ascariasis	Filariasis
1. Only one species Ascaris lumbricoid.es cause ascariasis.	1. There are many species of nematode that can cause filariasis.
2. Ascaris causes the infection of alimentary canal.	2. Wuchereria bancrofti causes the infection of lymphatic system.
3. Ascaris does not cause swellings of upper and lower limbs.	3. Filariasis cause swellings of extremities.
4. Ascaris is caused due to faeco-oral transmission.	4. Filariasis is caused due to vector transmission (Culex mosquito).
5. Medicines for treatment of Ascariasis are Piperazine, Mebendazole, Levamisole, Pyrantel.	5. Medicines for treatment of filariasis are diethyl- carbamazine citrate.

Short answer questions

Question 1.
Despite constant exposure to variety of pathogens, why do most of us remain healthy?
Answer:

All human beings are exposed to various foreign bodies, including infectious agents like bacteria, viruses, etc. which are called pathogens.
But human body can resist almost all types of these pathogens.
For this purpose, there is immune system which protects us from various infectious agents.
There is resistance and prevention of the damage or disease, through our defence mechanisms.
Thus, despite constant exposure to variety of pathogens, most of us remain healthy.

Question 2.
What are the unique features of acquired immunity?
Answer:
Following are the unique features of acquired immunity:

Specificity : Production of specific antibody or T-lymphocyte against a particular antigen/ pathogen is called specificity.
Diversity : Ability to recognize vast variety of diverse pathogens or foreign molecules by immunity is called diversity.
Discrimination between self and non¬self : Acquired immunity can differentiate between own body cells (self) and foreign (non-self) molecules.
Memory : The first immune response upon encounter of a specific foreign agent and its elimination is retained as a memory. This results in quicker and stronger immune response when the same pathogen is encountered again.

Question 3.
Describe the polypeptide chains seen in the structure of an antibody.
Answer:

There are four polypeptide chains which make the antibody.
There are two heavy or H-chains and two light or L-chains.
Each chain has two distinct regions, the variable region and the constant region.
Variable regions carry the antigen binding site or paratope.
This part of antibody recognizes and bindsto the specific antigen to form an antigen- antibody complex.

Question 4.
Which are the antimicrobial substances in blood and tissues?
Answer:

There are more than 30 serum proteins, circulating in the blood in an inactive state, which forms the complement system.
‘Complement cascade’ is activated by presence of pathogens and thus they eliminate pathogens.
Cells which are affected by viruses secrete interferons which are a class of cytokines. These soluble proteins attack the pathogens.
Some leucocytes stimulate other cells to protect themselves from viral infection.

Question 5.
Is developing fever a bad sign or a good action? Explain.
Answer:

Fever is the innate immunity mechanism. When there is fever, the body rises.
This is in response to infection in a natural way.
Developing fever is a natural defence mechanism.
Fever helps to accelerate the physiological processes by which the invading pathogens are destroyed.
Fever stimulates the production of interferon and helps in recovery from viral infections.
Taking all these points into consideration, getting fever is a good action that innate immunity is working properly, however, entry of pathogen in the body causing illness is a bad sign.

Question 6.
With the help of a chart explain the compatibility of human blood groups.
Answer:
Different types of blood groups:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 1

Person with blood group A can donate blood to persons having blood group A or AB and receive blood from A or O.
Person with blood group B can donate blood to persons having blood group B or AB and receive blood from B or O.
Person with blood group O can donate blood to all the persons having blood group either A, B, AB or O, because blood group O is universal donor. But can receive blood only from person having blood group O.
Person with blood group AB can donate blood to only AB but can receive blood from all the persons having blood group either, A, B, O or AB because AB is universal recipient.

Question 7.
The blood group of Krutika is O Rh +ve. What would be the possible blood groups of her parents?
Answer:

Krutika has O blood group, therefore her genotype is I°I°.
Her parents can be of following combinations.
They may be having blood group A or B with heterozygous genotype respectively, i.e. IAI° and IBI°. If both of them are heterozygous, having either A or B blood group, Krutika can be of O type.
Other possibility is both the parents have to be O.
For being Rh positive, her at least one parent should be Rh positive. Rh negative is a recessive phenotype and hence needs double dose of these genes.

Question 8.
Can a person with blood group O Rh+ve donate blood to a patient with blood group O Rh-ve? Why?
Answer:
No. blood group O may be common to both donor and recipient but their Rh factor is different. A person with RH+ve blood cannot donate to a patient with Rh-ve blood group. In Rh+ve blood there is antigen D. This antigen D when enters the body of recipient, there are production of anti-RH antibodies in his or her body. These anitibodies with cause agglutination of the Rh +ve blood which will be given to the patient. This agglutination will cause clots inside the vital organ of the recipient and the death may follow.

Question 9.
Why do we suffer from common cold repetitively in our life, but other viral diseases like Influenza or Small pox only once?
Answer:
Influenza infection causes production of antibodies in our body, once the virus attacks us. Therefore, second encounter with the virus may not cause effect. But in case of common cold, large number of different virus families are responsible for developing infection of common cold. Many a times different allergens are also inducing agents for common cold. Thus we may suffer from common cold again and again.

Question 10.
What are the symptoms of pneumonia?
Answer:

Main symptoms of infectious pneumonia are cough producing greenish or yellow sputum or phlegm and a high fever with chills.
Shortness of breath, stabbing chest pain, coughing up blood, headaches, sweaty and clammy skin, loss of appetite, fatigue, blueness of the skin, nausea, vomiting, mood swings and joint pains or muscle aches are some other symptoms.

Question 11.
What is the diagnosis and treatment of filariasis? How can we control this disease ?
Answer:
I. Diagnosis and Treatment : For the patient, diethyl-carbamazine citrate is the drug used for twice a day for three weeks. Thereafter for five days every six months the same treatment is repeated. This becomes effective against filarial worms.

II. Prevention and Control:

Mosquito eradication should be done for controlling filariasis.
In the areas with mosquitoes, avoid mosquito bite by using mosquito nets and insect repellents.

Question 12.
What are the signs and symptoms of filariasis?
Answer:
Signs and symptoms of filariasis:

As the lymphatic drainage does not take place, there is oedema with thickening of skin and underlying tissue.
Extremities like legs, arms, breasts, scrotum, etc. are affected by nematode causing lymphatic filariasis, i. e. Wuchereria bancrofti.
Lymph vessels and lymph nodes are enlarged and swollen.
Elephantiasis is seen in which limbs are swollen like legs of elephant.
Lymphoedema, i.e. accumulation of lymph fluid is seen in tissue causing swelling.
Hydrocele condition develops in which testis are enlarged due to accumulation of lymphatic fluid in testis.

Question 13.
What are the various ways in which mosquitoes can be eradicated from any area?
Answer:
Eradication of mosquitoes:

Removal of all stagnant water pools around the houses.
If such water bodies are there, they should be sprayed with insecticides.
But better option which is eco-friendly is releasing mosquito eating fish like Gambusia or Tilapia.
Use of mosquito repellent plants like Citronella. Use of coils and repellent creams.
Fumigation of the area to kill the mosquito.
Aedes sps. breed in man-made containers, especially plastic and cement tanks. Care should therefore be taken to dispose such containers properly. Water should not be accumulated in them.

Question 14.
What precautions will you take if you are travelling in an area which has lot of mosquitoes?
Answer:

Avoiding areas where mosquitoes are in more concentration.
Carrying mosquito repellent creams or coils.
Use of mosquito nets and other fumigation devices.
Wearing full clothing in light colours.
Staying indoors when mosquitoes are swarming, especially in the evening.
Taking anti-malarial pills as a precautionary measure.

Question 15.
Deaddiction may be difficult but not impossible. Collect information about NGOs, working in the field of deaddiction.
Answer:
There are many NGOs that work in the field of deaddiction. In different cities, there are different organizations. Even the central Government has started helpline number 1800-11-0031 for those drug and alcohol addicts who need help to come out of these addictions. Muktangan in Pune is one such reputed organization which does the great work in the field of deaddiction.

Question 16.
What are the most common warning signs of drug and alcohol abuse among youth ?
Answer:
The most warning signs of addictions are as follows:

Drop in academic performance, absenteeism from school or college.
No interest in personal hygiene and hobbies.
Withdrawal from the society, increased tendency of isolation and depression.
Aggressive and rebellious behaviour resulting into strained relationships with family and friends.
Fatigue and change in sleeping and eating habits.
Fluctuations in weight, appetite, deteriorating health, etc.

Question 17.
What are the preventive measures for malaria?
Answer:
Preventive measures of malaria :

Transmission of malarial parasite can be reduced by preventing mosquito bites. Therefore, mosquitoes should be controlled or totally eradicated.
This can be done by using of mosquito nets and insect repellents.
Mosquito control measures such as spraying insecticides inside houses and draining stagnant water where mosquitoes lay their eggs.
The mosquito larvae can be eradicated by releasing Gambusia fish which can feed upon these larvae.
Vaccine against malaria is also under preparation.

Question 18.
How does Entamoeba histolytica causes amoebiosis ?
Answer:

Amoebiosis is spread through ingestion of the cyst form of Entamoeba histolytica. This is a commensal organism.
Cyst is a semi-dormant and hardy structure found in faeces of infected person.
Non-encysted amoebae are called trophozoites. The trophozoites die quickly after leaving the body but may also be present in faeces.
Trophozoites are rarely the source of new infections.
The infection may remain asymptomatic for many days as Amoeba can remain latent in the gastrointestinal tract.

Question 19.
Describe the signs and symptoms of amoebiasis.
Answer:
Amoebiasis shows following common symptoms:

Diarrhoea, flatulence, stool with mucus and abdominal pains (cramps) are common.
Stool sticky with mucus and blood.
Amoebae form cysts in the liver, in such case there is hepatomegaly, i.e. enlargement of liver.
Liver shows amoebic liver abscess accompanied with fever and pain in right side of the abdomen.

Question 20.
How can amoebiasis be prevented?
Answer:
Prevention of amoebiasis is to be done at two levels, viz. at home and at endemic level.
1. Prevention of the spread of amoebiasis at the home level:

Washing hands with soap and water after using the toilet or changing a baby’s diaper and before handling and eating food.
Cleaning bathrooms and toilets properly with germicides.
Avoiding raw vegetables when in endemic areas where they are grown in soil fertilized by human faeces.
Boiling and purifying the drinking water.

2. Prevention of the spread of amoebiasis at endemic level:

Avoiding consumption of street foods especially in public places.
Following good sanitary practice, as well as using proper sewage disposal or treatment.
E. histolytica cysts are usually resistant to chlorination; therefore sedimentation and filtration of water supplies are necessary to reduce the incidence of infection.
Avoiding shared towels or face washers.

Question 21.
Describe the symptoms of ascariasis.
Answer:

After infection by Ascaris lumbricoides, there is appearance of eggs in stools in 60 – 70 days.
In larval ascariasis, symptoms are seen in 4-16 days after infection.
The final symptoms are gastrointestinal discomfort, colic and vomiting, fever and appearance of live worms in faeces.
Some patients may have pulmonary symptoms. Inflammation of alveolar walls is seen. This is known as pneumonitis.
Some may show neurological disorders during migration of the larvae.
Loss of appetite which reflects in weight loss.
A bolus of worms may obstruct the intestine.
Larvae that migrate may also cause eosinophilia, i.e. increase in number of eosinophils.

Question 22.
What are the preventive measures against ascariasis?
Answer:

Prevention of ascariasis can be done by adopting the following measures :
Use of proper toilet facilities.
Safe disposal of excreta.
Protection of food from dirt and soil.
Washing of vegetables before cooking and avoiding eating raw, unwashed vegetables and fruits.
Hand washing and use of safe food. Observing personal hygiene.
Use of pharmaceutical drugs such as Mebendazole and Albendazole can kill Ascaris.

Question 23.
Discuss the clinical manifestation of AIDS.
Answer:
There are four stages of clinical manifestations or symptoms of AIDS.

Stage I : This is initial infection with the virus and formation of antibodies, usually 2-8 weeks after initial infection.
Stage II : In this stage the person is asymptomatic carrier. Incubation takes place with a period ranging for 6 months to 10 years.
Stage III : This is called AIDS related complex (ARC). In this stage, one or more of the following clinical signs are seen. E.g. Recurrent fever for longer than one month, fatigue, unexplained diarrhoea, night sweats, shortness of breath, loss of more than 10 per cent body weight, etc.
Stage IV : This is the end stage in which patient shows full blown AIDS. Thus it is called the end stage of HIV infection. Life threatening opportunistic infections (like pneumonia, tuberculosis, Kaposi sarcoma, etc.) are easily caught during this period.

Question 24.
Through which modes HIV infection does not spread?
Answer:

HIV does not spread through casual contact such as hugging, etc.
Insect bite such as mosquito bites does not transmit HIV.
Participation in sports is not the mode by which HIV transmits.
Contact between articles used by AIDS patient, a hand shake with him or her does not transmit HIV
HIV infections do not occur through swimming pool or by sharing clothes, utensils, etc.

Question 25.
Write about laboratory diagnosis and treatment of AIDS.
Answer:
I. Laboratory diagnosis :

There are two tests for diagnosis of AIDS.
First test is ELISA (Enzyme-Linked Immunosorbent Assay) which is used to detect the HIV antibodies.
The second confirmatory test is Western Blot, which is used to weed out any false positive results. It is a highly specific test.
It is based on detecting specific antibody to viral core protein and envelope glycoprotein.

II. Treatment of AIDS:

AIDS cannot be cured.
Antiretroviral drugs are used to reduce the viral load and prolong the life of HIV patient. E.g. Antiretroviral therapy (ART) uses drugs such as TDF (tenofovir), EFV (Efavirenz), Lamivudine (3TC), etc.

Question 26.
Describe the structure of HIV with a suitable diagram.
Answer:

Human Immunodeficiency Virus or HIV is spherical and 100 to 140 nm in diameter.
It has centrally located two ss RNA molecules along with reverse transcriptase enzymes.
There are coverings of two layers of proteins. The outer layer formed by matrix protein (pi7) while in inner layer is of capsid protein (p24).
An additional layer of lipids is seen over the matrix protein layers. This layer is impregnated with glycoprotein GP120 and GP41.

Question 27.
What are the modes of transmission of HIV or AIDS?
Answer:
The transmission of HIV occurs through following routes:

Sexual relations, mainly unsafe sexual contact including oral, vaginal and anal sex.
Through blood and blood products either by blood transfusions or sharing needles and syringes.
Transplacental From pregnant mother to her foetus through placenta. Nursing mother can also transmit HIV to her baby through lactation.
Spreading the virus is very rare in case of accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs.
HIV is seen in urine, tears, saliva, breast milk and vaginal secretions but unless these body fluids enter the injuries and wounds, transmission is not easy.

Question 28.
What are the measures of prevention and control of AIDS ?
Answer:

Preventive measures : AIDS has no cure, hence prevention is the best choice. The
following steps help in preventing this dreadful disease.
High risk group people should be educated about HIV transmission. They should never donate blood.
Use of disposable needles and syringes should be done with proper disposal.
Risky sexual habits should be avoided.
Tooth brushes, razors, other articles that can become contaminated with blood should not be shared.
Blood should be screened before receiving it.
Routine screening of blood and semen donors, organ donors (kidney, liver, lung, cornea), and patients undergoing haemodialysis must be done.
Pregnant women or those women who are contemplating pregnancy should be regularly screened.

Question 29.
Explain the ill-effects of opioids and cannabinoids on health.
OR
What are harmful effects of drug abuse?
Answer:

Opioids bind to specific opioid receptors which are present in central nervous system and gastrointestinal tract. Some opioids e.g. heroin act like depressants and slow down all the body functions.
Cannabinoids have the capacity to interact with receptors present in the brain. Inhalation or ingestion of cannabinoids such as marijuana, hashish, charas and ganja have adverse effect on cardiovascular system.
Cannabinoid like LSD causes hallucinations.
All these substances are addictive and hence cause adverse effects on the body and health.

Question 30.
Give the adverse effects of opioids, cannabinoids and morphine on human health.
Answer:
1. Opioids:

Opioids bind to specific opioid receptors present in the central nervous system and in gastrointestinal tract.
They are depressants and slow down the body functions.

2. Cannabinoids:

Cannabinoids interact with receptors in the brain.
They affect cardiovascular system of the body.

3. Morphine:

Morphine is an effective sedative and pain killer when used for medicinal purpose.
When abused it affects physical, physiological and psychological functions.

Give reasons

Question 1.
Vaccines are safe.
Answer:
During their manufacture, vaccines are rigorously tested. Many rounds of study, examination and research are carried out before they are used for general public. Extensive research and evidences are gathered to check their safety. Sometimes, some vaccines produce side effects but these are rare and mild. Hence vaccines are considered to be safe.

Question 2.
Innate immunity is also known as non-specific immunity.
Answer:
Innate immunity is non-specific because it does not depend on previous exposure to foreign substances. It is inborn capacity of the body to resist the pathogen that causes the disease. It is natural immunity and hence it remains non-specific, trying to protect the body in case of any invasion of foreign body.

Question 3.
Vaccination is important for preventing pneumonia.
Answer:
Vaccinations for pneumonia are available against Haemophilus influenzae and Streptococcus pneumoniae. If given earlier in life, they reduce the chances of catching pneumococcal infections. The deaths can be prevented which are common due to lung infections. Since it is a common and chronic j disease for all age groups, for the prevention one must take vaccination.

Question 4.
Common cold is the most frequent ! infectious disease in humans.
Answer:
Common cold is caused by virus which is abundantly present in congested city environments. The upper respiratory tract is infected due to these rhinoviruses and corona viruses. Average adult contracts such infections 2 to 4 times in a year while children capture it 6 to 12 times in a year. Therefore, it is said to be the most frequent infectious disease of human beings.

Question 5.
Typhoid is food and water-borne disease.
Answer:
Typhoid is caused due to Salmonella typhi which isa Gram-negative bacterium, transmitted from a patient or carrier to another healthy person through contaminated food or water. Flying insects, mostly houseflies transmit the bacteria from faeces to the food. Poor hygienic habits and improper public sanitation system spreads typhoid. Therefore, it is said to be food and water-borne disease.

Question 6.
Malignant tumour is more dangerous than benign tumour.
Answer:
Malignant tumour cells can show metastasis and hence can spread far and wide in the body, affecting other healthy cells and tissues. They are difficult to cure by any therapy as one does not know exact location of the cancerous cells. They also produce variety of symptoms depending on their i location. On the contrary, benign tumours can be treated surgically. Since they are covered by a cyst like membrane, the cancerous cells do not spread from them. Thus malignant tumour can be lethal as against the benign tumour.

Question 7.
Prevention is better than cure for AIDS.
Answer:
Till this time, there is no preventive vaccination for AIDS. There is also no cure for AIDS. The medicines are also costly and may not give complete cure. The only way to remain away from AIDS is the complete awareness about it. Thus it should be prevented by not allowing HIV to enter our body. Once HIV finds the entrance, the cure is impossible. Therefore, it is said that prevention is better them cure for AIDS.

Write short notes

Question 1.
Cellular factors in innate immunity.
Answer:

Phagocytic cells ingest and destroy the pathogens.
This is natural defence against the invasion of pathogenic microorganisms and other foreign particles in blood and tissues.
Phagocytic cells are of two types, viz. microphages and macrophages. They can remove foreign particles that enter the body.
Natural killer (NK) cells is a class of lymphocytes which carry out important and non-specific defence against viral infections and tumours.

Question 2.
Acute phase proteins (APPs).
Answer:

Acute phase proteins are involved in innate immune mechanism.
When there is an infection or injury, it leads to a sudden increase in concentration of certain plasma proteins, which are called acute phase proteins or APPs.
These include C Reactive Protein (CRP), Mannose binding protein, Alpha-1-acid glycoprotein, Serum Amyloid P etc.
APPs enhance host resistance, prevent tissue injury and promote repair of inflammatory lesions.

Question 3.
Rh factor.
Answer:
Rh factor:

Rh factor is the term adapted from Rhesus monkey.
In rhesus monkey, there is antigen D on the surface of their RBCs.
Landsteiner and Wiener discovered this antigen and termed it as Rh factor.
Persons having Rh factor or D antigen are called Rh positive while those lacking D antigen or Rh factor are called Rh negative.

Question 4.
Erythroblastosis foetalis.
Answer:

Erythroblastosis foetalis is condition in which there is destruction of the erythrocytes of the foetus. It is the haemolytic disease of the newborn (HDN).
This is caused in foetus, if mother is Rh -ve and father is Rh +ve. Rh +ve is the dominant allele, the foetus becomes Rh +ve, when its father is RH +ve.
Rh +ve blood groups have D antigen which induces a strong immunogenic response when introduced into Rh -ve individuals.
During foetal life, there is connection between mother and foetus through placenta, therefore Rh +ve antigen D from the foetus enters maternal circulation.
This triggers formation of anti-Rh antibodies in mother. Subsequently Rh+ve foetus receives anti-Rh antibodies produced by mother.
This causes agglutination reaction resulting into haemolysis in foetus. In order to prevent HDN, Rh -ve mother is injected with the anti-Rh antibody during all her pregnancies if her husband is Rh +ve.

Question 5.
Common cold.
Answer:

The common cold (nasopharyngitis or rhinopharyngitis) is a viral infectious disease of the upper respiratory system. The causative organisms are rhinoviruses and coronaviruses.
Symptoms include cough, sore throat, runny nose and fever.
There is no known treatment, however, symptoms usually resolve spontaneously in 7 to 10 days.
The best prevention for the common cold is to stay away from infected people and places where infected individuals have been.
Hand washing with plain soap and water is recommended. Also alcohol-based hand sanitizers provide very little protection.

Question 6.
Life cycle of Plasmodium.
Answer:

Anopheles Female mosquito which is a carrier carries sporozoites. When it bites the human, these sporozoites enter human circulation.
Sporozoites undergo asexual reproduction through fission or schizogony in the liver cells or erythrocytes of the human.
It forms merozoites. The cells formed within erythrocytes function as gametocytes. They undergo gamogony.
Upon biting such person, the gametocytes enter into female Anopheles, fertilization occurs in its gut.
Diploid zygote transforms into oocyst. Oocyst forms large number of haploid sporozoites through meiosis (sporogony).
Sporozoites migrate to salivary glands and are ready to infect new human host.
Again Sporozoite → Merozoite → Trophozoite → Schizont sequence is carried on for plasmodial stages in human body.
The sexual life cycle of Plasmodium occurs in mosquito body which acts as a vector. While its asexual phase takes place in human body.

Question 7.
Pneumonia.
Answer:

Pneumonia is an inflammatory condition of alveoli in the lungs causing formation of fluid in the lungs. This condition is called consolidation and exudation.
Causes of pneumonia are infection due to bacteria, viruses, fungi or parasites, chemical burns or physical injury to the lungs.
Influenza virus, adenovirus, para influenza and Respiratory Syncytial Virus (RSV) are some viruses that can cause pneumonia. Bacteria like Streptococcus pneumoniae and fungal pathogens e.g. Pneumocystis jirovecii and Pneumocystis carinii can also spread infection of pneumonia. Chemical burns of physical injury to lungs also cause similar infection.
Main symptoms of infectious pneumonia are cough producing greenish or yellow sputum or phlegm and a high fever with chills.
Shortness of breath or dyspnea, stabbing chest pain, coughing up blood, headaches, sweaty and clammy skin, loss of appetite, fatigue, blueness of the skin, nausea, vomiting, mood swings and joint pains or muscle aches are some other symptoms.
Preventive vaccination against pneumonia is available. Medicines such as Benzyl penicillin, Ampicillin and Chloramphenicol are effective to prevent pneumonia.

Question 8.
Ringworm.
Answer:

Ringworm or Dermatophytosis is a clinical condition caused by Trichophyton and Microsporum fungal infection of the skin. This infection is seen in humans and pets.
Dermatophytes are the fungi that feed on keratin. Keratin is the material found in the outer layer of skin, hair and nails.
These fungi attack various parts of the body. Infections on the body forms enlarged raised red rings. These patches have intense itching. Infection on the skin of the feet may cause athlete’s foot and jock itch.
When the nails are infected it causes onychomycosis. During this the nails thicken, discolour and finally crumble and fall off.
Prevention of ringworm infection is to be done by avoiding sharing of clothing, sports equipment, towels or sheets. Clothes should be washed in hot water with fungicidal soap after suspected exposure to ringworm. One should not walk barefoot but use appropriate footwear.
Diagnosis of ringworm is done by physical examination and treatment is done with uses drugs like nystatin, fluconazole, itraconazole, etc.

Question 9.
Dengue.
Answer:

Dengue is a viral disease causing high fever. It is a painful, debilitating vector-borne disease.
There are four closely related dengue viruses that cause infection.
Vector of Dengue virus is female Aedes mosquito. The mosquito takes up the dengue virus when it sucks blood of a person suffering from dengue.
The spread of dengue is not directly from one person to another person.

Question 10.
Performance enhancers.
Answer:

Performance enhancers are certain drugs used by sportspersons to enhance their performance during competitions.
Narcotic analgesics, anabolic steroids, diuretics and certain hormones are misused by such sportspersons to increase muscle strength and bulk. It also promotes aggressiveness and improve overall performance.
Use of anabolic steroids cause side effects.
Females show masculinization, increased aggressiveness, mood swings, depression, abnormal menstrual cycles, excessive hair growth on the face and body, enlargement of clitoris, deepening of voice.
Males show acne, increased aggressiveness, mood swings, depression, and reduction of size of the testicles, decreased sperm production, kidney and liver dysfunction, breast enlargement, premature baldness, enlargement of the prostate gland.
These effects may be permanent with prolonged use. Using such drugs is illegal and punishable.

Chart Based Questions

Question 1.
Complete the chart of ABO blood group system and answer the questions given below:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 3
Questions:
(i) Which blood group from the above table is called universal acceptor?
(ii) Which blood group from the above table is called universal donor?
Answer:

(i) Blood group AB is called universal acceptor.
(ii) Blood group O is called universal donor.

Question 2.
Complete the table

Plasmodium species	Incubation period	Pattern of fever
————	————–	High fever after 48 hours.
————–	28 days	————–
————–	17 days	————-
————	————-	High fever at irregular intervals between 22 to 48 hours.

Answer:

Plasmodium species	Incubation period	Pattern of fever
P. vivax	14 days	High fever after 48 hours.
P. malariae	28 days	High fever after 72 hours interval
P. ovale	17 days	High fever after 48 hours interval
P. falciparum	12 days	High fever at irregular intervals between 22 to 48 hours.

Question 3.
Complete the following table
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 5
Answer:

Question 4.
Complete the following table:

Carcinogen	Organ affected
N-nitrosodimethlene	————–
Aflatoxin	———-
————–	Vagina
————–	Urinary bladder
————–	Prostate
————–	Skin and lungs

Answer:

Carcinogen	Organ affected
N-nitrosodimethlene	Lungs
Aflatoxin	Liver
Diethylstilboestrol	Vagina
2-naphthylamine and 4-aminobiphenyl	Urinary bladder
Cadmium oxide	Prostate
Soot, coal tar (2-4 benzopyrene)	Skin and lungs

Diagram Based Questions

Question 1.
Label the given diagram
Img 7 Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 7
Answer:

Antigen binding site.
Variable region of heavy chain
Varible region of light chain
Constant region of light chain
Constant region of heavy chain
Disulphide bond
Hinge
Light chain
Heavy chain

Question 2.
Observe the given diagram and answer the following questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 8
(1) What is I and II in the above diagram?
Answer:
I is a virus which is trying to cause infection, II are two antigen molecules which are trying to attack the virus.

(2) What structures are responsible for antigen and antibody complex? Identify them in the above diagram.
Answer:
(a) is epitope which is antigen determinant and
(b) is a paratope which is part of the antibody. Epitope and paratope are specific to each other and hence they form a complex.

(3) What is the study of antigen-antibody interactions called?
Answer:
The study of antigen-antibody interactions is called serology.

Question 3.
Fill in the blanks after observing the diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 9
On the right side of the diagram, the stages of plasmodium are passed in the body of ……………….. Whereas on the left side of the diagram, those take place in the body of ………………, ………………. is the stage that is dormant in the liver of human host. From this ………………… and then ………………… is the stage in the erythrocytes, which rupture and gives rise to …………….. Microgamete and macrogamete fuse with each other to form …………………… which later gives rise to ookinete which forms ………………… This enters the salivary glands of mosquito …………………. phase of Plasmodium occurs in mosquito body, whereas …………………… phase is in human body.
Answer:
On the right side of the diagram, the stages of plasmodium are passed in the body of human. Whereas on the left side of the diagram, those take place in the body of mosquito. Hypnozoite is the stage that is dormant in the liver of human host. From this schizont and then merozoites Trophozoite is the stage in the erythrocytes, which rupture and gives rise to gamerocyte. Microgamete and macrogamete fuse with each other to form zygote which later gives rise to ookinete which forms sporozoites. This enters the salivary glands of mosquito. Sexual phase of Plasmodium occurs in mosquito body, whereas asexual phase is in human body.

Question 4.
Observe the given diagram and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 10
(1) Enlist the stages of Entamoeba histolytica you see in the above diagram.
Answer:
Trophozoite, pre-cystic form, cyst, binucleate cyst, quadrinucleate cyst, Metacycstic amoeba, amoebulae are the different stage of Entamoeba histolytica that are seen in the above diagram.

(2) Where are these stages passed?
Answer:
These stages are passed in the lumen of intestine of the host human being.

(3) How does Entamoeba come out of the body of the host?
Answer:
Encysted Entamoeba pass out with the faecal matter of the host.

Question 5.
Observe the given diagram and answer the following questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 11

(1) In life cycle of Ascaris at what stage do they enter the human body?
Answer:
When there is development of infective larva inside the egg of Ascaris, it enters the human body.

(2) How do they enter the human body and through which organ do they enter?
Answer:
Ascaris eggs are deposited in the faeces. They mix in the soil, if faeces is exposed in open. From there, it can enter into nearby water body or it may contaminate vegetables or other food stuffs. Such unhygienic food or unclean hands pass these eggs in the body of human through the mouth.

(3) What are the vital organs affected by the Ascaris during its development within the body of host human ?
Answer:
Ascaris can affect trachea, lungs, heart, brain and eyes too.

(4) In which organ do they copulate and produce fertilized eggs?
Answer:
Adult male and female copulate in the intestine of the host human.

Question 6.
Observe the given diagram and answer the following questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 12

(1) In which host stage II is completed?
Answer:
Stage II is completed in human.

(2) What happens in step 1 and step 2?
Answer:
Humans are infected at step 1 when mosquito bites human and larvae enter blood stream. In step 2 adult Wuchereria worms are formed in lymphatics.

(3) Describe the events in step 3.
Answer:
Mosquito carries the blood as it bites the human in step 4 and ingests microfilariae in human blood. Later the microfilariae start growing in the midgut of mosquito.

(4) In which host stage I is completed?
Answer:
Stage I is completed in mosquito.

Question 7.
Sketch and label the diagram of pathogen that causes typhoid.
Answer:
Pathogen of typhoid is Salmonella typhi.
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 13

Question 8.
Sketch and label disease causing agents of pneumonia.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 14

Question 9.
Sketch and label benign and malignant tumours.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 15

Question 10.
Sketch and label structure of HIV.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 10 Human Health and Diseases 16

Long answer questions

Question 1.
Describe different ways in which epithelial surface offers the innate immunity.
Answer:

Skin and mucous covering, when intact protects the body against the invasion by pathogens. The healthy skin has bactericidal activity due to the salts present in drying sweat.
Sebaceous glands in the skin produce secretions and long chain of fatty acids. These are bactericidal and fungicidal.
Respiratory tract is provided with mucosa which prevents entry of microorganisms to a large extent.
The inhaled particles are arrested through hair in the nasal passage. The particles that pass beyond nasal passage are caught by mucus lining the epithelium. They are swept back to pharynx. Then they are either swallowed or coughed out.
The cough reflex is an important defence mechanism of respiratory tract.
There is saliva in the mouth which has inhibitory effect on microorganisms. Gastric secretions has acidity and hence microorganisms are destroyed in stomach.
The flushing action of urine eliminates bacteria from the urethra. Semen too has antibacterial substances, e.g. Spermine and zinc.

Question 2.
Explain ABO blood group system in human being with a suitable chart.
Answer:

In ABO system, the blood groups are determined by the antigen present on the surface of red blood cells.
The blood groups are of four types, viz. A. B, AB and O.
In person with blood group A there is antigen ‘A’ on the surface of their red blood cells (RBCs) and antibodies ‘b’ in their plasma.
In person with blood group B there is antigen ‘B’ on the surface of their red blood cells (RBCs) and antibodies ‘a’ in their plasma.
In person with blood group AB there are both antigens ‘A’ and ‘B’ on the surface of their RBCs and no antibodies in their plasma.
In person with blood group ‘O’ there are no antigens ‘A’ and ‘B’ on the surface of their RBCs but have both ‘a’ and ‘b’ antibodies in their plasma.
During blood transfusion compatibility of blood has to be taken into consideration.
Person with ‘O’ blood group is called universal donor while the person with ‘AB’ blood group is called universal recipient. Individuals with blood group O can donate blood to anyone, while those individuals with blood group AB can receive blood from any person.

Question 3.
What are the main causes of cancer?
Answer:
Causes of Cancer : Following carcinogenic factors are responsible for causing cancer.

Chemicals : Many induce development of cancer. E.g. nicotine, caffeine, polycyclic hydrocarbons and products of combustion of coal and oil. Sex hormone and steroids, if given or secreted in excess, can cause cancer. E.g. Breast cancer.
Radiation : Radiations such as X-rays, gamma-rays, cosmic rays, ultra-violet rays are carcinogenic.
Viruses : Virus possessing oncogenes (v-onc genes) are carcinogenic. E.g. EBV (Epstein-barr virus), HPV (Human papiloma virus) are oncogenic viruses.
Oncogenes : Cellular oncogenes (c-onc genes) or proto-oncogenes can cause cancer. They are present in normal cells but if activated they lead to oncogenic transformation of cells.
Addiction : Addictive substances like cigarette smoke, tobacco lead to cancer of mouth, lips and lungs. Alcohol can cause cancer of oesophagus, stomach, intestine and liver. Drugs like marijuana or anaerobic steroids can also cause cancer.

Question 4.
What are the different ways of treating cancer?
Answer:
Cancer treatment consists of combination of a number of therapies which are follows:
(1) Chemotherapy : Chemotherapy means giving certain anticancer drugs. These drugs check cell division by inhibiting DNA synthesis. But these are more toxic to cancerous cell than to normal cells. Chemotherapy shows side effects such as hair loss or anaemia.

(2) Radiotherapy : In addition to chemotherapy, radiations are given. The cancer cells are bombarded with the radiations from radioactive materials such as cobalt, iridium and iodine. The X-rays, gamma rays and charge particles are used to destroy the cancerous tissue or cells. They cause minimum damage to the surrounding normal tissue or cells.

(3) Surgery : Entire cancerous tissue or cells are removed surgically. E.g. breast tumour or uterine tumour. After removing the cancerous tissue, additionally other treatments are also given.

(4) Immunotherapy : For tackling with tumour, patients are given biological response modifiers such as a-interferon which activates their immune system to destroy the tumour.

(5) Supportive therapy : With supportive therapy, patient’s quality of life is increased. To treat symptoms of cancer and side effects of cancer treatments, this therapy is used. This therapy varies depending upon condition of individual patient.

Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination

July 23, 2024July 22, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 9 Control and Co-ordination Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 9 Control and Co-ordination

Multiple choice questions

Question 1.
The supporting cells that produce myelin sheath in the CNS are ……………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(a) Oligodendrocytes

Question 2.
Human brain develops to its full size at an age of year/s.
(a) 1
(b) 6
(c) 12
(d) 18
Answer:
(b) 6

Question 3.
Telencephalon is the other name of ……………….
(a) pons varolii
(b) medulla oblongata
(c) cerebrum
(d) cerebellum
Answer:
(c) cerebrum

Question 4.
Olfactory tracts merge in olfactory area of lobe.
(a) Frontal
(b) Parietal
(c) Occipital
(d) Temporal
Answer:
(d) Temporal

Question 5.
………………. is the largest commissure of the human brain.
(a) Corpora striata
(b) Corpora quadrigemina
(c) Habencular commissure
(d) Corpus callosum
Answer:
(d) Corpus callosum

Question 6.
Grey matter of the brain shows large collection of ……………….
(a) dendrons
(b) cytons
(c) axons
(d) synapsis
Answer:
(b) cytons

Question 7.
Masses of grey matter in white matter of the cerebrum are called ……………….
(a) corpora striata
(b) corpus callosum
(c) paracoel
(d) basal ganglia
Answer:
(d) basal ganglia

Question 8.
Parietal and temporal lobes of cerebrum are separated by sulcus.
(a) lateral
(b) parieto occipital
(c) central
(d) median longitudinal
Answer:
(a) lateral

Question 9.
……………… area is motor speech area
(a) Acoustic
(b) Wernike’s
(c) Somato sensory
(d) Broca’s
Answer:
(d) Broca’s

Question 10.
Maxillary nerve is a branch of nerve.
(a) Occulomotor
(b) Trochlear
(c) Trigeminal
(d) Facial
Answer:
(c) Trigeminal

Question 11.
Spinal accessory is the cranial nerve.
(a) IV
(b) VI
(c) IX
(d) XI
Answer:
(d) XI

Question 12.
Rotation of eye ball is controlled by ……………….
(a) Optic nerve
(b) Pathetic nerve
(c) Auditory nerve
(d) Hypoglossal nerve
Answer:
(b) Pathetic nerve

Question 13.
The spinal nerves emerge out of vertebral column through ……………….
(a) intervertebral foramina
(b) neural canal
(c) central canal
(d) foramen magnum
Answer:
(a) intervertebral foramina

Question 14.
The neuro transmitter is removed by an enzyme called ……………….
(a) noradrenaline
(b) acetylcholine
(c) hyaluronidase
(d) cholinesterase
Answer:
(d) cholinesterase

Question 15.
The reflex action originates in ……………….
(a) sensory neuron
(b) motor neuron
(c) receptor organ
(d) effector organ
Answer:
(c) receptor organ

Question 16.
Cytons of neurons are located in dorsal root ganglion.
(a) afferent
(b) efferent
(c) adjustor
(d) association
Answer:
(a) afferent

Question 17.
Wall of carotid arteries contain ……………….
(a) thermoreceptors
(b) mechanoreceptors
(c) baroreceptors
(d) statoacoustic receptors
Answer:
(c) baroreceptors

Question 18.
The electronegativity inside the membrane is due to ……………….
(a) less anions than cations
(b) less cations than anions
(c) bicarbonates
(d) carbonates
Answer:
(b) less cations than anions

Question 19.
The neuro transmitters stimulate ……………….
(a) presynaptic membrane
(b) cyton
(c) axon terminals
(d) postsynaptic membrane
Answer:
(d) postsynaptic membrane

Question 20.
………………. is an extero-receptor.
(a) Thermoreceptor
(b) Baroreceptor
(c) Proprioreceptor
(d) Enteroreceptor
Answer:
(a) Thermoreceptor

Question 21.
The are described as windows for brain.
(a) sensory neurons
(b) motor neurons
(c) effectors
(d) sense organs
Answer:
(d) sense organs

Question 22.
Otolith organ is formed of ……………….
(a) cochlea and vestibule
(b) sacculus and utriculus
(c) semicircular canals
(d) ear ossicles
Answer:
(b) sacculus and utriculus

Question 23.
Olfactory bulbs are extensions of brain’s ……………….
(a) cerebrum
(b) limbic system
(c) RAS
(d) pons varolii
Answer:
(b) limbic system

Question 24.
Gustatory senses are noted by ……………….
(a) retina
(b) skin
(c) nose
(d) tongue
Answer:
(d) tongue

Question 25.
………………. is attached to the eardrum.
(a) Malleus
(b) Incus
(c) Stapes
(d) Cochlea
Answer:
(a) Malleus

Question 26.
Eustachian tube is present in ……………….
(a) external ear
(b) internal ear
(c) heart
(d) middle ear
Answer:
(d) middle ear

Question 27.
The internal ear is a fluid filled structure called ……………….
(a) cochlea
(b) vestibule
(c) labyrinth
(d) otolith
Answer:
(c) labyrinth

Question 28.
The space within cochlea is known as ……………….
(a) scala vestibule
(b) scala tympani
(c) aqueous chamber
(d) scala media
Answer:
(d) scala media

Question 29.
Thermoregulatory centre in the body is ……………….
(a) hypothalamus
(b) cerebellum
(c) spinal cord
(d) pituitary
Answer:
(a) hypothalamus

Question 30.
Which of the following is a sensory nerve?
(a) Vagus
(b) Auditory
(c) Facial
(d) Lumbar
Answer:
(b) Auditory

Question 31.
Chemical transmission in synapse occurs due to ……………….
(a) cholesterol
(b) ADH
(c) acetylcholine
(d) cholinesterase
Answer:
(c) acetylcholine

Question 32.
Voluntary muscular coordination is under the control of ……………….
(a) medulla
(b) pons
(c) hypothalamus
(d) cerebrum
Answer:
(d) cerebrum

Question 33.
All involuntary vital activities are under the control of ……………….
(a) medulla oblongata
(b) cerebellum
(c) cerebral hemispheres
(d) pons Varolii
Answer:
(a) medulla oblongata

Question 34.
Cerebellum is controlling centre for ……………….
(a) muscular strength
(b) memory
(c) equilibrium
(d) muscular coordination
Answer:
(c) equilibrium

Question 35.
Which receptors are present in the retina?
(a) Chemoreceptors
(b) Thermoreceptors
(c) Photoreceptors
(d) Baroreceptors
Answer:
(c) Photoreceptors

Question 36.
Breathing is controlled by ……………….
(a) trachea
(b) medulla oblongata
(c) lungs
(d) hypothalamus
Answer:
(b) medulla oblongata

Question 37.
Corpus callosum is a nerve fibre bridge which connects ……………….
(a) two cerebral hemispheres
(b) cerebrum and cerebellum
(c) cerebellum and medulla
(d) midbrain and hindbrain
Answer:
(a) two cerebral hemispheres

Question 38.
Centre for thirst and hunger are located in ……………….
(a) cerebrum
(b) cerebellum
(c) hypothalamus
(d) medulla
Answer:
(c) hypothalamus

Question 39.
Gyri in the brain are present in ……………….
(a) cerebral cortex
(b) olfactory lobes
(c) medulla oblongata
(d) hypothalamus
Answer:
(a) cerebral cortex

Question 40.
Which of the following is a structure of mesencephalon?
(a) Inferior colliculi
(b) Thalamus
(c) Cerebellum
(d) Pons varolii
Answer:
(a) Inferior colliculi

Question 41.
Third ventricle lies in ……………….
(a) midbrain
(b) forebrain
(c) cerebellum
(d) medulla oblongata
Answer:
(b) forebrain

Question 42.
Medulla oblongata encloses ……………….
(a) third ventricle
(b) fourth ventricle
(c) first ventricle
(d) second ventricle
Answer:
(b) fourth ventricle

Question 43.
Loss of memory may result from injury to the ……………….
(a) corpora quadrigemina
(b) pons varolii
(c) cerebellum
(d) cerebrum
Answer:
(d) cerebrum

Question 44.
Terminal non-nervous part of spinal cord is ……………….
(a) funiculus
(b) filum terminale
(c) cauda equina
(d) conus terminalis
Answer:
(b) filum terminale

Question 45.
Which part of the pituitary is neurohaemal organ?
(a) Pars distalis
(b) Infundibulum
(c) Pars nervosa
(d) Pars intermedia
Answer:
(c) Pars nervosa

Question 46.
Development of secondary sexual characteristics in female is under the control of ……………….
(a) growth hormone
(b) TSH
(c) estrogen
(d) progesterone
Answer:
(c) estrogen

Question 47.
Hypersecretion of STH in children causes ……………….
(a) cretinism
(b) gigantism
(c) dwarfism
(d) myxoedema
Answer:
(b) gigantism

Question 48.
Milk secretion in lactating woman is controlled by ……………….
(a) LH
(b) prolactin
(c) relaxin
(d) oestrogen
Answer:
(b) prolactin

Question 49.
ADH is secreted by
(a) adrenal gland
(b) thyroid
(c) hypothalamus
(d) pancreas
Answer:
(c) hypothalamus

Question 50.
BMR is increased by the administration of ……………….
(a) insulin
(b) GH
(c) thyroxine
(d) testosterone
Answer:
(c) thyroxine

Question 51.
The largest endocrine gland in the body is ……………….
(a) pituitary
(b) adrenal
(c) liver
(d) thyroid
Answer:
(d) thyroid

Question 52.
Diabetes insipidus is caused by the deficiency of ……………….
(a) calcitonin
(b) oxytocin
(c) atrial natriuretic factor
(d) vasopressin
Answer:
(d) vasopressin

Question 53.
Simple goitre is caused by the deficiency of ……………….
(a) TSH
(b) thyrocalcitonin
(c) thyroxine
(d) iodine
Answer:
(d) iodine

Question 54.
Exopthalmic goitre is also known as ……………….
(a) Grave’s disease
(b) Gull’s disease
(c) Simple goitre
(d) Cushing’s disease
Answer:
(a) Grave’s disease

Question 55.
Cushing’s syndrome is developed due to ……………….
(a) hyposecretion of ACTH
(b) hypersecretion of corticoids
(c) hyposecretion of thyroxine
(d) hypersecretion of thyroxine
Answer:
(b) hypersecretion of corticoids

Question 56.
Pituitary gland is under the control of ……………….
(a) thyroid
(b) adrenal
(c) pineal
(d) hypothalamus
Answer:
(d) hypothalamus

Question 57.
FSH is secreted by ……………….
(a) pituitary gland
(b) thyroid gland
(c) ovary
(d) adrenal gland
Answer:
(a) pituitary gland

Question 58.
ICSH stimulates ……………….
(a) ovary
(b) Leydig cells
(c) seminiferous tubules
(d) kidney
Answer:
(b) Leydig cells

Question 59.
Which of the following secrete LH ?
(a) Pituitary
(b) Thyroid
(c) Ovary
(d) Adrenal
Answer:
(a) pituitary

Question 60.
TSH regulates secretion.
(a) thyroxine
(b) MSH
(c) androgens
(d) insulin
Answer:
(a) thyroxine

Question 61.
Deficiency of thyroxine in adults cause ……………….
(a) cretinism
(b) myxoedema
(c) diabetes
(d) Cushing’s disease
Answer:
(b) myxoedema

Question 62.
Osmotic pressure and blood pressure are maintained by ……………….
(a) glucocorticoids
(b) aldosterone
(c) TRF
(d) MSH
Answer:
(b) aldosterone

Question 63.
Hormone secreted by corpus luteum is ……………….
(a) aldosterone
(b) progesterone
(c) testosterone
(d) cortisol
Answer:
(b) progesterone

Question 64.
………………. is also called hypophyseal stalk.
(a) Infundibulum
(b) Median eminence
(c) Pars intermedia
(d) Sphenoid bone
Answer:
(a) infundibulum

Question 65.
………………. is like a collar around hypophyseal stalk.
(a) Pars distalis
(b) Pars nervosa
(c) Pars intermedia
(d) Pars tuberalis
Answer:
(d) Pars tuberalis

Question 66.
Herring bodies are the parts of ……………….
(a) hypothalamo-hypophyseal tracts
(b) pituicytes
(c) hypothalamo-hypophyseal portal system
(d) pituitary cleft
Answer:
(a) hypothalamo-hypophyseal tracts

Question 67.
Corticotropin is the other name of ……………….
(a) ACTH
(b) STH
(c) Aldosterone
(d) ADH
Answer:
(a) ACTH

Question 68.
Adrenal failure leads to ……………….
(a) Acromegaly
(b) Simmond’s disease
(c) Midget
(d) Addison’s disease
Answer:
(d) Addison’s disease

Question 69.
Prolactin inhibiting factor is secreted by ……………….
(a) Hypophysis
(b) Hypothalamus
(c) Thyroid
(d) Mammary glands
Answer:
(b) Hypothalamus

Question 70.
Which one of the following is not applicable to prolactin ?
(a) Mammotropin
(b) Lactogenic hormone
(c) Somatotropin
(d) Luteotropin
Answer:
(c) Somatotropin

Question 71.
………………. is a gonadotropic hormone.
(a) STH
(b) LTH
(c) ACTH
(d) FSH
Answer:
(d) FSH

Question 72.
Rhythmic integrated contractions of jejunum are controlled by ……………….
(a) coherin
(b) insulin
(c) glucagon
(d) ADH
Answer:
(a) coherin

Question 73.
Thyroid gland is derived from of embryo.
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) ecto-endoderm
Answer:
(c) endoderm

Question 74.
Deficiency of thyroxine in infants causes ……………….
(a) Cretinism
(b) Grave’s disease
(c) Myxoedema
(d) Exophthalmos
Answer:
(a) Cretinism .

Question 75.
………………. is a hypercalcemic hormone.
(a) PTH
(b) TCT
(c) Thyroxine
(d) ACTH
Answer:
(a) PTH

Question 76.
………………. is a middle layer of adrenal cortex.
(a) Zona fasciculata
(b) Zona pellicida
(c) Zona glomerulosa
(d) Zona reticularis
Answer:
(a) Zona fasciculata

Question 77.
Decrease in the blood calcium level is ……………….
(a) hyperglycemia
(b) hypercalcemia
(c) hypoglycemia
(d) hypocalcemia
Answer:
(d) hypocalcemia

Question 78.
………………. stimulates RBC production.
(a) Aldosterone
(b) Cortisol
(c) Epinephrine
(d) Parathormone
Answer:
(b) Cortisol

Question 79.
Chemicals which are released at the synaptic junction are called ……………….
(a) hormones
(b) neurotransmitters
(c) cerebrospinal fluid
(d) lymph
Answer:
(b) neurotransmitters

Question 80.
Potential difference across resting membrane is negatively charged. This is due to differential distribution of the following ions.
(a) Na⁺ and K⁺ ions
(b) Ca⁺⁺ and Cl^– ions
(c) Ca⁺⁺ and Mg⁺⁺ ions
(d) Mg⁺⁺ and Cl^– ions
Answer:
(a) Na⁺ and K⁺ ions

Question 81.
Which of the following is not involved in Knee-jerk reflex?
(a) Muscle spindle
(b) Motor neuron
(c) Brain
(d) Inter neurons
Answer:
(c) Brain

Question 82.
An area in the brain which is associated with strong emotions is ……………….
(a) Cerebral cortex
(b) Cerebellum
(c) Limbic system
(d) Medulla
Answer:
(c) Limbic system

Question 83.
Which is the vitamin present in Rhodopsin?
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D
Answer:
(a) Vitamin A

Question 84.
Wax gland present in the ear canal is modified ……………….
(a) Sweat gland
(b) Vestibular gland
(c) Cowper’s gland
(d) Sebaceous gland
Answer:
(d) Sebaceous gland

Question 85.
The part of internal ear responsible for hearing is ……………….
(a) cochlea
(b) semicircular canal
(c) utriculus
(d) sacculus
Answer:
(a) cochlea

Question 86.
The organ of corti is a structure present in ……………….
(a) external ear
(b) middle ear
(c) semi circular canal
(d) cochlea
Answer:
(d) cochlea

Question 87.
Select the right match of endocrine gland and their hormones among the options given below.
A. Pineal i. Epinephrine
B. Thyroid ii. Melatonin
C. Ovary iii. Estrogen
D. Adrenal medulla iv. Tetraiodothyronine
(a) A-iv, B-ii, C-iii, D-i
(b) A-ii, B-iv, C-i, D-iii
(c) A-iv, B-ii, C-i, D-iii
(d) A-ii, B-iv, C-iii, D-i
Answer:
(d) A-ii, B-iv, C-iii, D-i

Question 88.
Listed below are the hormones of anterior pituitary origin. Tick the wrong entry.
(a) Growth hormone
(b) FSH
(c) Oxytocin
(d) ACTH
Answer:
(c) Oxytocin

Question 89.
Mary is about to face an interview. But during the first five minutes before the interview she experiences sweating, increased rate of heartbeat, respiration, etc. Which hormone is responsible for her restlessness?
(a) Estrogen and progesterone
(b) Oxytocin and vasopressin
(c) Adrenaline and noradrenaline
(d) Insulin and glucagon
Answer:
(c) Adrenaline and noradrenaline

Question 90.
The steroid responsible for balance of water and electrolytes in our body is ……………….
(a) Insulin
(b) Melatonin
(c) Testosterone
(d) Aldosterone
Answer:
(d) Aldosterone

Question 91.
Thymosin is responsible for ……………….
(a) raising the blood sugar level
(b) raising the blood calcium level
(c) increased production of T lymphocytes
(d) decrease in blood RBC
Answer:
(c) increased production of T lymphocytes

Question 92.
In the mechanism of action of a protein hormone, one of the second messengers is ……………….
(a) Cyclic AMP
(b) Insulin
(c) T₃
(d) Gastrin
Answer:
(a) Cyclic AMP

Question 93.
Leydig cells produce a group of hormones called ……………….
(a) androgens
(b) estrogen
(c) aldosterone
(d) gonadotropins
Answer:
(a) androgens

Question 94.
Corpus luteum secretes a hormone called ……………….
(a) prolactin
(b) progesterone
(c) aldosterone
(d) testosterone
Answer:
(b) progesterone

Question 95.
Cortisol is secreted from ……………….
(a) pancreas
(b) thyroid
(c) adrenal
(d) thymus
Answer:
(c) adrenal

Question 96.
A hormone responsible for normal sleep – wake cycle is ……………….
(a) epinephrine
(b) gastrin
(c) melatonin
(d) insulin
Answer:
(c) melatonin

Question 97.
Match the pairs and choose the correct answer among the following options.
A. Epinephrine
i. Increase in muscle growth
B. Testosterone
ii. Decrease in blood pressure
C. Glucagon
iii. Decrease in liver glycogen content
D. Atrial natriuretic factor
iv. Increase in heartbeat
(a) A-ii, B-i, C-iii, D-iv
(b) A-iv, B-i, C-iii, D-ii
(c) A-i, B-ii, C-iii, D-iv
(d) A-i, B-iv, C-ii, D-iii
Answer:
(b) A-iv, B-i, C-iii, D-ii

Question 98.
Blood calcium level is a resultant of how much dietary calcium is absorbed, how much calcium is lost in the urine, how much bone dissolves releasing calcium into the blood and how much calcium from blood enters tissues. Several factors play an important role in these processes. Mark the one which has no role.
(a) Vitamin D
(b) Parathyroid hormone
(c) Thyrocalcitonin
(d) Thymosin
Answer:
(d) Thymosin

Question 99.
All the following tissues in mammals except one consists of a central ‘medullary’ region surrounded by a cortical region. Mark the wrong entry.
(a) Ovary
(b) Adrenal
(c) Liver
(d) Kidney
Answer:
(c) Liver

Match the columns

Question 1.
Match the hormones with their source

Column A	Column B
(1) Glucagon	(i) Neurohypophysis
(2) Adrenaline	(ii) Islets of Langerhans
(3) Somato tropins	(iii) Adenohypophysis
(4) ADH	(iv) Medulla

Answer:

Column A	Column B
(1) Glucagon	(ii) Islets of Langerhans
(2) Adrenaline	(iv) Medulla
(3) Somato tropins	(iii) Adenohypophysis
(4) ADH	(i) Neurohypophysis

Question 2.
Match the layer of adrenal with its hormone.

Column A (Layers of adrenal cortex)	Column B (Hormones)
(1) Zona glomerulosa	(A) Cortisols
(2) Zona fasciculata	(B) Androgens
(3) Zona reticularis	(C) Aldosterone

Answer:

Column A (Layers of adrenal cortex)	Column B (Hormones)
(1) Zona glomerulosa	(C) Aldosterone
(2) Zona fasciculata	(A) Cortisols
(3) Zona reticularis	(B) Androgens

Question 3.
Match the disorder with the gland associated with it.

Column A (Disorder)	Column B (Associated Gland)
(1) Addison’s disease	(A) Hypothalamus
(2) Grave’s disease	(B) Pituitary
(3) Diabetes insipidus	(C) Thyroid
(4) Acromegaly	(D) Adrenal

Answer:

Column A (Disorder)	Column B (Associated Gland)
(1) Addison’s disease	(D) Adrenal
(2) Grave’s disease	(C) Thyroid
(3) Diabetes insipidus	(A) Hypothalamus
(4) Acromegaly	(B) Pituitary

Classify the following to form Column B as per the category given in Column A

Question 1.

Column A	Column B
(1) Forebrain	————–
(2) Midbrain	————–
(3) Hindbrain	————–

Answer:

Column A	Column B
(1) Forebrain	Olfactory lobes, Corpus callosum
(2) Midbrain	Superior colliculi, Iter
(3) Hindbrain	Pons varolii, Vermis

Question 2.
Types of nerves
Occulomotor, Facial, Optic, Vagus, Abducens, Vestibulocochlear

Column A	Column B
(1) Sensory	————–
(2) Motor	————–
(3) Mixed	————–

Answer:

Column A	Column B
(1) Sensory	Optic, Vestibulocochlear
(2) Motor	Occulomotor, Abducens
(3) Mixed	Facial, Vagus

Question 3.
Hormones
Estrogen, Glucagon, Epinephrine, Relaxin, Somatostatin, Nor-Adrenalin

Column A	Column B
(1) Ovary	————
(2) Pancreas	————
(3) Adrenal Medulla	————

Answer:

Column A	Column B
(1) Ovary	Estrogen, Relaxin
(2) Pancreas	Glucagon, Somatostatin
(3) Adrenal Medulla	Epinephrine, Nor-Adrenaline

Question 4.
Disorders
Dwarfism, Myxoedema, Addison’s disease, Cushing’s disease, Gigantism, Goitre

Column A	Column B
(1) Pituitary	————
(2) Thyroid	————
(3) Adrenal Cortex	————

Answer:

Column A	Column B
(1) Pituitary	Dwarfism, Gigantism
(2) Thyroid	Myxoedema, Goitre
(3) Adrenal Cortex	Addison’s disease, Cushing’s disease

Very short answer questions

Question 1.
What is the need for the control and coordination in multicellular animals?
Answer:
Multicellular animals need control and coordination to maintain constancy of internal environment, i.e. homeostasis.

Question 2.
How do plants carry out control and coordination?
Answer:
Plants carry out control and coordination by sending chemical signals and bringing about various types of movements.

Question 3.
What is the type of nervous system of Earthworm?
Answer:
Earthworm is an annelid having ventral, ganglionated nervous system. It consists mainly of nerve ring, nerve cord and peripheral segmentally arranged nerves.

Question 4.
What kind of nervous system is seen in Hydra, earthworm and cockroach?
Answer:
In Hydra, the nervous system is in the form of nerve net, while in earthworm and cockroach the nervous system is ganglionated.

Question 5.
What is a gland?
Answer:
An organized collection of secretory epithelial cells capable of producing of some secretion is called a gland.

Question 6.
What do you mean by discrete endocrine glands ?
Answer:
The glands which are exclusively endocrine in function are discrete endocrine glands.

Question 7.
What is genu and splenium?
Answer:
Genu is anterior and splenium is posterior fold of corpus callosum.

Question 8.
Which is the largest basal nucleus of brain ?
Answer:
Corpus striatum is the largest basal nucleus of the brain.

Question 9.
What is EEG?
Answer:
EEG is electro encephalography. It is used to detect the electrical changes taking place in the brain.

Question 10.
Mention the major sulci present in the cerebral hemispheres.
Answer:
The sulci present in the cerebral hemisphere are – central sulcus, parieto-occipital sulcus and lateral sulcus.

Question 11.
What is the difference in ‘tract’ and ‘nerve’?
Answer:
A bundle of axons within CNS is called a ‘tract’ while the one outside the CNS is called ‘nerve’.

Question 12.
What is a choroid plexus? State its locations.
Answer:
Network of blood capillaries associated with the brain is called choroid plexus, which are anterior choroid plexus present on the roof of epithalamus and posterior choroid plexus located on the roof of medulla oblongata.

Question 13.
What is a synapse ?
Answer:
The interconnection between two neurons or neuron with motor organ is called synapse.

Question 14.
What is a polarised membrane?
Answer:
The cell membrane of a neuron at resting stage is called polarised membrane. In such membrane, the outer side of the cell membrane is more electropositive due to more Na⁺ ions.

Question 15.
What is summation effect?
Answer:
Many weak stimuli given in a quick succession may produce a response due to summation or additive effect of stimuli, which is known as summation effect.

Question 16.
What is synaptic delay?
Answer:
The time required for a nerve impulse to cross a synapse during transmission of a nerve impulse is called synaptic delay, which is about 0.3-0.5 milliseconds.

Question 17.
What is refractory period?
Answer:
The time interval between two consecutive nerve impulses is called refractory period, during which a nerve cannot be stimulated. Nerve is stimulated only after completion of this period.

Question 18.
What is a synaptic cleft?
Answer:
Small intercellular space between two successive neurons which is about 20¬30 nm in width, is called synaptic cleft.

Question 19.
What is synaptic transmission?
Answer:
The process by which the impulse from the pre-synaptic neuron is conducted to the post-synaptic neuron or cell is called synaptic transmission.

Question 20.
What is a pre-synaptic neuron?
Answer:
The neuron carrying an impulse to the synapse is the pre-synaptic neuron.

Question 21.
What is transmission terminal and generator region?
Answer:
The pre-synaptic membrane of a neuron is transmission terminal while the post- synaptic membrane is called generation region.

Question 22.
What is synaptic fatigue?
Answer:
The synaptic fatigue is the time lag or halting of the transmission of nerve impulse, temporarily at synapse due to exhaustion of its neurotransmitter.

Question 23.
What is the width of synaptic cleft?
Answer:
The width of synaptic cleft is about 20-30 nm.

Question 24.
How much is the resting potential of axon?
Answer:
The resting potential of axon is -70 mV.

Question 25.
How does Na⁺ – K⁺ pump work?
Answer:
During repolarisation of certain part of a nerve, Na⁺ – K⁺ Pump, pumps out 3 Na⁺ ions for every 2 K⁺ ions they pump into the cell.

Question 26.
What is blood-brain barrier?
Answer:
The barrier that keeps a check on passage of ions and large molecules from the blood to the brain tissue is called blood-brain barrier. Endothelial cells lining the blood capillaries help in this process along with the astrocytes.

Question 27.
Enlist meninges of human brain.
Answer:
Dura mater, arachnoid mater or membrane and pia mater are the meninges of human brain.

Question 28.
What is the source of CSF?
Answer:
CSF is secreted by the choroid plexuses of brain and ependymal cells lining the ventricles of brain and central canal of spinal cord.

Question 29.
What is the function of tympanic membrane?
Answer:
Tympanic membrane vibrates on receiving the sound waves and then transfers the vibrations to malleus, the first of the three ear ossicles.

Question 30.
Mention the role of semicircular canals in ear.
Answer:
Semicircular canals help in balancing the equilibrium of the body.

Question 31.
What is the cause for diabetes insipidus?
Answer:
Diabetes insipidus is caused by the deficiency of ADH.

Question 32.
Give role of Parathormone.
Answer:
Parathormone regulates the calcium balance in the body. It increases blood calcium level by increasing reabsorption of calcium from bones.

Question 33.
What is meningitis?
Answer:
Infection of meninges is called meningitis.

Question 34.
Where is general motor area located? What is the function of this area?
Answer:
General motor area is located on precentral gyrus of frontal lobe. It is concerned with all motor activities.

Question 35.
What is the role of Wernicke’s area?
Answer:
Wernicke’s area is the sensory speech area responsible for understanding and formulating written and spoken language.

Question 36.
What is the location of Wernicke’s area?
Answer:
Wernicke’s area is located in the area of contact between temporal, parietal and occipital lobes of cerebrum.

Question 37.
What is the role of Broca’s area?
Answer:
The role of Broca’s area which is the motor speech area that translates thoughts into speech and controls movement of tongue, lips and vocal cords.

Question 38.
What is the location of Broca’s area?
Answer:
Broca’s area is present in the frontal lobe of cerebrum.

Question 39.
Which areas are present in the post central gyrus?
Answer:
General sensory areas concerned with sensation of temperature, touch, pressure, pain and speech are present in the post central gyrus.

Question 40.
Which areas are located on temporal lobe?
Answer:
The areas concerned with sense of taste, sense of hearing and sense of smell are located on temporal lobe.

Question 41.
What is the main function of occipital lobe?
Answer:
Occipital lobe carries sensory visual area which is concerned with the sense of sight. Association visual area is concerned with perception, analysis and storing of information obtained by sight.

Question 42.
What is neuroendocrine system?
Answer:
Neuroendocrine system is an integrated and coordinated system consisting of nervous and endocrine system which brings about the coordination of the body.

Question 43.
What is the quantity of cerebrospinal fluid in adult human being?
Answer:
Cerebrospinal fluid is about 120 ml in adult human being.

Question 44.
How many neurons are present in the brain?
Answer:
There are about 30,000 million neurons in the brain.

Question 45.
Which part of the brain forms 80-85% of the brain?
Answer:
Cerebrum forms 80-85% of the brain.

Question 46.
Which sense is poorly developed in human beings and what is the reason for this?
Answer:
Sense of smell is poorly developed in human beings and it is due to less developed olfactory lobes.

Question 47.
Where are lateral ventricles situated and what is its roof called?
Answer:
Lateral ventricles are the cavities filled with cerebrospinal fluid, present inside the cerebrum and its roof is called pallium.

Question 48.
What is foramen of Monro?
Answer:
Foramen of Monro is a narrow opening that connects lateral ventricle present in cerebrum with third ventricle present in diencephalon.

Question 49.
What is the roof and floor of diencephalon called respectively?
Answer:
Roof of diencephalon is epithalamus and floor of diencephalon is hypothalamus.

Question 50.
Which hormones are produced from pineal gland? Name their functions.
Answer:
Serotonin and melatonin are hormones of pineal gland which are concerned with metabolic activities and regulation of biological rhythm respectively.

Question 51.
What is Habenular commissure?
Answer:
Habenular commissure connects the lateral walls of diencephalon or thalami with each other.

Question 52.
What structure is present on the anterior side of the hypothalamus?
Answer:
Optic chiasma or crossing of two optic nerves is seen on the anterior side of the hypothalamus.

Question 53.
Name the parts of corpora quadrigemina and give their functions.
Answer:
The upper two lobes of corpora quadrigemina are called superior colliculi which receive impulse from optic nerves while the lower two lobes are called inferior colliculi which receive auditory stimuli. Both colliculi control and coordinate head movements.

Question 54.
What is arbour vitae?
Answer:
Arbor vitae is white, branching tree-like processes of white matter that are sent into grey cortex of cerebellum.

Question 55.
Which cranial nerve is called a dentist nerve?
Answer:
Trigeminal, the Vth cranial nerve is called a dentist nerve.

Question 56.
What is central canal?
Answer:
Central canal is the narrow central cavity present inside the spinal cord.

Question 57.
Which cranial nerves originate from the midbrain?
Answer:
Cranial nerves (III) Occulomotor, (IV) Pathetic and (VI) Abducens originate from the midbrain.

Question 58.
What is the other name for the auditory cranial nerve?
Answer:
Vestibulocochlear.

Question 59.
What is the difference between unconditioned and conditioned reflexes?
Answer:
Unconditioned reflexes are inborn or hereditary and permanent while the conditioned reflexes are temporary, learnt or acquired during lifetime.

Question 60.
Give examples of unconditional reflexes.
Answer:
Blinking of eyes, withdrawing of hand upon pricking, suckling to breast by infant, swallowing, knee jerk, sneezing and coughing are some of the unconditional reflexes.

Question 61.
What do you mean by accommodation power of the lens?
Answer:
The ability of the lens by which the light ray from far and near objects is focused on the retina is called accommodation power of the lens. The lens makes fine adjustments to bring such sharp focus on retina.

Question 62.
Where is pituitary located?
Answer:
Pituitary is located inside a depression called sella turcica which is present in sphenoid bone of the skull.

Question 63.
What is the difference between Lorain dwarf and Frohlic dwarf?
Answer:
Lorain dwarf is mentally normal while Frohlic dwarf is mentally abnormal.

Question 64.
How is Simmond’s disease caused?
Answer:
Simmond’s disease is caused due to hyposecretion of GH during adulthood.

Question 65.
How is Addison’s disease caused?
Answer:
Addison’s disease is caused by the hyposecretion of ACTH that leads to adrenal failure.

Question 66.
How is Cushing’s disease caused?
Answer:
Hypersecretion of corticoids causes Cushing’s disease.

Question 67.
What is the main difference between diabetes mellitus and diabetes insipidus?
Answer:
Diabetes mellitus is caused due to deficiency of insulin while diabetes insipidus is caused due to deficiency of ADH.

Question 68.
Which cells secrete coherin? What is the function of coherin?
Answer:
Coherin is the hormone secreted by hypothalamic neurons that brings about prolonged, rhythmic integrated contractions of the jejunum.

Question 69.
What is the difference between Gull’s disease (Myxoedema) and Grave’s disease?
Answer:
Gull’s disease is caused due to hyposecretion of thyroxine while Grave’s disease is caused due to hypersecretion of thyroxine.

Question 70.
In which part thyroid gland stores its hormones?
Answer:
The lumen of thyroid follicles store the thyroid hormones in the form of thyroglobulins.

Question 71.
What are thymosins?
Answer:
Thymosins are hormones secreted by thymus gland which promote the production of antibodies.

Question 72.
Which hormone is secreted by the heart?
Answer:
Heart secretes ANF or Atrial Natriuretic Factor.

Name the following

Question 1.
Name the region consisting of nerve fibres that connects cerebrum and medulla oblongata.
Answer:
Pons Varolii.

Question 2.
Give the names of cranial nerve number VIth and VIIth.
Answer:
VTth cranial nerve is Abducens and Vllth cranial nerve is Facial.

Question 3.
Name the three sulci present on the cerebral hemispheres.
Answer:
Central sulcus, lateral sulcus and parieto¬occipital sulcus.

Question 4.
Name the band of nerve fibres that connect cerebrum, cerebellum and spinal cord.
Answer:
Crura cerebri.

Question 5.
Name the second largest part of the brain.
Answer:
Cerebellum.

Question 6.
Name the three branches of trigeminal nerve.
Answer:
Ophthalmic, Maxillary and Mandibular.

Question 7.
Name the nerve which arises from ventral side of medulla and supplies the tongue.
Answer:
Hypoglossal.

Question 8.
Name of emergency hormones secreted by sympathetic nervous system.
Answer:
Adrenaline and nor-adrenaline are the emergency hormones secreted by sympathetic nervous system.

Question 9.
Name the disorders caused by hyposecretion of thyroxine in children and adults.
Answer:

Hyposecretion of thyroxine in children : Cretinism.
Hyposecretion of thyroxine in adults : Myoxedema.

Question 10.
Name the dual exocrine as well as endocrine gland. Name the hormones secreted by it.
Answer:
Pancreas is the dual gland, exocrine as well as endocrine, it secretes hormones like insulin, glucagon and somatostatin.

Question 11.
Name the four peptide hormones secreted by endocrine cells of alimentary canal.
Answer:
Gastrin, secretin, cholecystokinin and GIPT or Gastric inhibitory peptide.

Question 12.
Name the disorder caused by the under secretion of thyroxine in children.
Answer:
Cretinism is the disorder seen in children due to under secretion of thyroxine.

Give functions of the following

Question 1.
Meninges and CSF.
Answer:
Functions of meninges:

Meninges give protection to the brain and the spinal cord.
They are also nutritive in function.
Cerebrospinal fluid acts as a shock absorber. It protects the brain from mechanical injuries and from desiccation.
It also maintains constant pressure inside and outside the CNS and regulates the temperature.

Functions of CSF:

CSF helps in exchange of nutrients and wastes between the blood and the brain tissue.
CSF supplies oxygen to the brain.

Question 2.
The functions of forebrain.
Answer:
(A) Functions of cerebrum:

The cerebrum controls the voluntary activities.
The cerebrum perceives various sensory stimuli received through vision, taste, smell, sound, touch, speech, etc.
It is the centre of memory, will-power, intelligence, reasoning and learning.
The cerebrum is the centre for emotions, thoughts and feelings, pain, pleasure, fear, fatigue, pressure, temperature, etc.
It is also the centre for micturition, defecation, weeping, laughing, etc

(B) Function of olfactory lobes: Sensation of smell.

Diencephalon acts as a relay centre for motor and sensory impulses between spinal cord, brainstem and various areas of cerebral cortex.
Diencephalon consists of epithalamus, thalami and hypothalamus. Therefore it acts as a centre for homeostasis and higher centre of autonomous nervous system.
Hypothalamic nuclei secrete neurohormones which influence the pituitary gland.
Diencephalon regulates heartbeats, blood pressure and water balance.
Anterior choroid plexus which is located in the diencephalon secretes cerebrospinal fluid.
Hypothalmic regions control many involuntary functions such as hunger, thirst, thermo-regulation, fear, anger, sleep, sexual desire, etc.

Question 3.
Write various functions of hindbrain.
Answer:
Functions of hindbrain:
(i) Cerebellum:

Cerebellum is a primary centre for the control of equilibrium, posture, balancing and orientation.
Neuromuscular activities are regulated by the cerebellum.
Coordination of walking, running, speaking, etc. is under the control of hindbrain.

(ii) Pons:

Activities of two cerebellar hemispheres are coordinated by pons.
Nerve fibres cross over in this area and thus the right side of the brain controls the left part of the body and vice versa.
Pons controls the consciousness of the brain.
Breathing centre is located in pons along with medulla.

(iii) Medulla oblongata:

Medulla oblongata controls all the involuntary activities such as heartbeats, respiration, vasomotor activities.
Peristalsis and reflex actions such as coughing, sneezing, swallowing, etc. are also under the control of medulla oblongata.
Medulla oblongata is essential for all the vital functions of the body.

Question 4.
Spinal cord.
Answer:
Functions of spinal cord:

Spinal cord is the main pathway for conduction of sensory and motor nerve impulses.
The sensory impulses travel from the body
to the brain and the motor impulses travel from the brain to the body.
Spinal reflexes are controlled by spinal cord.
The spinal cord reduces the load on the brain by taking appropriate actions in a reflex way.

Question 5.
Pituitary gland.
Answer:
Functions of pituitary:

Pituitary secretes seven main hormones viz. ACTH, TSH. GH or STH, LTH or Prolactin, MSH and gonadotropins such as FSH, LH or ICSH.
These hormones are secreted upon receiving proper releasing factor from hypothalamus.
These hormones bring about many coordinating functions in the body. Almost all endocrine glands are under the control of these hormones.
The neurohypophysis part stores two hormones, viz. vasopressin and oxytocin which are secreted by hypothalamus.
Important functions such as growth and reproductive processes, pregnancy, childbirth, lactation, etc. are under the control of pituitary.
Neurohypophysis acts as a neurohaemal organ and stores the hormones for some time.

Question 6.
Functions of hormones secreted by the thyroid gland.
Answer:

Thyroxine is the main metabolic hormone in the body.
Thyroxine maintains basal metabolic rate (BMR) by increasing glucose oxidation. It brings about calorigenic effect by energy production.
It also controls normal protein synthesis.
The physical growth, development of gonads and development of mental faculties is under the control of thyroxine.
It controls tissue differentiation during metamorphosis particularly in amphibia.
Body weight, respiration rate, heart rate, blood pressure, temperature, digestion, etc. is regulated by thyroxine.
Another hormone of thyroid, i.e. calcitonin regulates calcium metabolism of the body.

Question 7.
Give significance of relaxin and inhibin.
Answer:
1. Relaxin : Relaxin relaxes the cervix of the pregnant female and ligaments of pelvic girdle during parturition.
2. Inhibin : Inhibin inhibits the FSH and GnRH production.

Question 8.
Enlist hormones secreted by GI tract and state their role.
Answer:
Gastrointestinal tract:
In the gastrointestinal mucosa, certain ells are endocrine in function. These cells produce hormones which play vital role in digestive processes and flow of digestive juices.

Gastrin : This hormone stimulates gastric glands to produce gastric juice.
Secretin : This hormone is responsible for secretion of pancreatic juice and bile from pancreas and liver.
Cholecystokinin CCK/Pancreozymin PZ : This hormone stimulates gall bladder to release bile and stimulates the pancreas to release its enzymes.
Entero-gastrone/Gastric inhibitory peptide (GIP) : It slows gastric contractions and inhibits the secretion of gastric juice.

Distinguish between the following

Question 1.
Electrical and chemical synapses:
Answer:

Electrical synapse	Chemical synapse
1. The gap between the successive neurons in electrical synapse is very less [3.8 nm],	1. The gap between two successive neurons in chemical synapse is larger than electrical synapse [10-20 nm].
2. Transmission across the gap is faster in electrical synapse.	2. Transmission across the gap in chemical synapse is relatively slower than electrical synapse.
3. Electrical synapse is less common.	3. Chemical synapse is more common.
4. Electrical synapse is found in those places of the body requiring instant response.	4. Chemical synapse is found almost everywhere and connects neuron to neuron, muscles or glands.

Question 2.
Sympathetic and parasympathetic nervous system.
Answer:

Sympathetic nervous system	Parasympathetic nervous system
1. Sympathetic nervous system is formed by 22 pairs of sympathetic ganglia, 2 sympathetic cords which run parallel to vertebral column.	1. Parasympathetic nervous system has nerve fibres which run along with cranial and spinal nerves.
2. Sympathetic nervous system works through neurotransmitter, adrenaline.	2. Parasympathetic nervous system works through release of acetylcholine.
3. Sympathetic nervous system enhances all the involuntary functions.	3. Parasympathetic nervous system retards all the involuntary functions.
4. It brings about fight, fright and flight responses.	4. It brings about relaxation, comfort, pleasure, etc.
5. The pre-ganglionic nerve fibres are short and the post-ganglionic nerve fibres are long in sympathetic nervous system.	5. The pre-ganglionic nerve fibres are long and the post-ganglionic nerve fibres are short in parasympathetic nervous system.

Question 3.
Unconditional reflexes and Conditional reflexes
Answer:

Unconditional reflexes	Conditional reflexes
1. Unconditional reflexes are inborn.	1. Conditional reflexes are not inborn, they require training.
2. Unconditional reflexes are permanent.	2. Conditional reflexes are temporary.
3. They never disappear and need no previous experience.	3. They may disappear after sometime and need proper training for developing it.
4. Unconditional reflexes are heritable.	4. Conditional reflexes are non-heritable.
5. Sneezing, coughing, blinking of eye, etc. are unconditional reflexes.	5. Cycling, driving, playing games, etc. are due to conditional reflexes.

Question 4.
Epithalamus and hypothalamus.
Answer:

Epithalamus	hypothalamus
1. Epithalamus is the roof of diencephalon.	1. Hypothalamus is the floor of diencephalon.
2. Epithalamus shows pineal stalk to which pineal gland is attached.	2. Hypothalamus shows infundibulum to which pituitary gland is attached.
3. Epithalamus is non-nervous in nature.	3. Hypothalamus is the higher centre of autonomous nervous system.
4. Epithalamus has anterior choroid plexus which secretes cerebrospinal fluid.	4. Hypothalamus has neurons which secrete two endocrine hormones.
5. Epithalamus controls biological rhythm.	5. Hypothalamus controls homeostasis of the body.

Question 5.
Dura mater and pia mater.
Answer:

Dura mater	pia mater
1. Dura mater is the outermost meninx.	1. Pia mater is the innermost meninx.
2. Dura mater lies on the innermost side of skull or cranium.	2. Pia mater lies on outermost side of the brain.
3. Dura mater is tough, thick and fibrous.	3. Pia mater is thin and highly vascular.
4. Dura mater is mainly protective in function.	4. Pia mater is mainly nourishing in nature.
5. Below dura mater is subdural space.	5. Above pia mater is sub-arachnoidal space.

Question 6.
Cerebrum and cerebellum
Answer:

Cerebrum	Cerebellum
1. The cerebrum is the larger part forming 85% of the brain. It has four lobes.	1. The cerebellum is the smaller part forming 11%. of the brain. It has three lobes.
2. The cerebrum coordinates the functions of the sensory and motor areas.	2. The cerebellum coordinates the equilibrium of muscular movements during walking and running.
3. The cerebrum plays an important role in receiving the sensory impulses such as touch, pain, heat, cold, etc.	3. The cerebellum plays an important role in maintaining the posture and balance of the body.
4. The cerebrum is concerned with higher mental faculties such as memory, will and intelligence.	4. The cerebellum is concerned with muscular mechanism.

Question 7.
Cranial nerves and Spinal nerves
Answer:

Cranial nerves	Spinal nerves
1. Nerves arising from the brain are cranial nerves.	1. Nerves arising from the spinal cord are spinal nerves.
2. There are 12 pairs of cranial nerves.	2. There are 31 pairs of spinal nerves.
3. Cranial nerves are of three types, viz sensory, mixed and motor.	3. All spinal nerves are of mixed type.
4. Cranial nerves are responsible for cerebral reflexes.	4. Spinal nerves are responsible for spinal reflexes.

Question 8.
Extero and entero receptors.
Answer:

Exteroceptors	Interoceptors
1. Receptors receiving stimuli from outer environment of the body are called exteroceptors.	1. Receptors receiving stimuli from inside the body are called interoceptors.
2. These are somatic receptors.	2. These are visceral receptors.
3. Exteroceptors keep the body informed about . changes in the environment like temperature, pressure, touch, etc.	3. Interoceptors keep the homeostasis in the body by receiving stimuli from inside the body.
4. Mechanoreceptors, thermoreceptors, chemical receptors, photoreceptors and statoacoustic receptors are the different types of exteroceptors.	4. Propioceptors, enteroceptors, baroceptors are the different types of interoceptors.

Question 9.
Adenohypophysis and neurohypophysis.
Answer:

Adenohypophysis	Neurohypophysis
1. Adenohypophysis is the anterior lobe of pituitary.	1. Neurohypophysis is the posterior lobe of pituitary.
2. There is portal system between adenohypo-physis and hypothalamus which has blood sinusoids.	2. There is axonal knobs and blood vessels that connect neurohypophysis and hypothalamus.
3. Adenohypophysis forms 75% of pituitary.	3. Neurohypophysis forms 25% of pituitary.
4. Adenohypophysis has three parts, pars tuberalis, pars distalis and pars intermedia.	4. Neurohypophysis has three parts, median eminence, infundibulum and pars nervosa
5. Adenohypophysis has chromophil (acidophil and basophil) and chromophobe cells.	5. Neurohypophysis has axonic fibres and pituicytes.
6. Adenohypophysis secretes seven different hormones after receiving an appropriate message from hypothalamus through releasing factors.	6. Neurohypophysis does not produce hormones on its own. It is a neurohaemal organ as it receives and stores two hormones from hypothalamus.

Question 10.
FSH and LH
Answer:

FSH	LH
1. FSH is follicle stimulating hormone essential for the development of ovary.	1. LH is luteinizing hormone responsible for ovulation in females.
2. FSH stimulates ovary (follicular cells) to produce estrogen.	2. LH stimulates ovary (corpus luteum) to produce progesterone.
3. FSH in males is responsible for the spermatogenesis.	3. LH in females is responsible for the development of corpus luteum.
4. Negative feedback mechanism exists between amounts of FSH and estrogen in females.	4. Negative feedback mechanism exists between amounts of LH and progesterone in females.
5. FSH is indirectly responsible for the development of secondary sexual characters in females.	5. LH is indirectly responsible for maintenance of pregnancy in females.

Question 11.
Glucocorticoids and mineralcorticoids
Answer:

Glucocorticoids	Mineralocorticoids
1. Glucocorticoids control carbohydrate metabolism.	1. Mineralocorticoids regulate mineral concentration.
2. These are secreted by the cells of zona fasciculata.	2. These are secreted by the cells of zona glomerulosa.
3. These also regulate protein and fat metabolism.	3. These regulate salt-water balance.
4. Cortisol is the main glucocorticoid.	4. Aldosterone is the main mineralocorticoid.

Give scientific reasons

Question 1.
Number of gyri is related to the degree of intelligence.
Answer:

Gyri are the ridges in the folds present on the cerebral cortex.
The number of folds increase the surface area of the cerebral cortex.
Cerebral cortex has sensory and motor areas such as Wernicke’s area, Broca’s area, etc. If the surface area is greater, the neurons in these areas are also more in number.
Greater the number of neurons greater is the intelligence.
The number of gyri therefore is said to be related to the degree of intelligence.

Question 2.
A drunken person cannot maintain balance of the body.
Answer:

Cerebellum is the primary centre for controlling equilibrium and balance of the body.
Alcohol has an adverse effect on the neurons of cerebellum.
Consciousness of brain is also controlled by cerebellum.
When a person is drunk, the alcohol in his or her blood affects the activities of cerebellum and hence the person cannot maintain the balance of the body.

Question 3.
We are able to understand the smell of the first showers of rain or the sudden changes in the climate.
Answer:

We are able to understand any smell because of our olfactory mucosa and olfactory lobes of the brain.
Volatile substances are received by the olfacto-receptors in the nose.
Nerve impulse generated are carried by olfactory nerve and transmitted to brain where the impulse is interpreted.
The characteristic earthy smell is due to a compound ‘geosmin’.
Geosmin is produced by some species of Streptomyces [ gram positive soil bacterium],
Similarly, sudden change in the climate is easily noticed in the form of temperature change in the surrounding.
This change is detected by caloreceptors of the skin. From these receptors the signal is transmitted to CNS where the change is perceived.

Question 4.
We are able to hear the chirping of the birds and recognize the sound of the bird.
Answer:

The phonoreceptors of the body receive the sound waves and transfer the nerve impulses to the auditory areas of the brain.
The interpretation, of the sound is a combined effort of sensory and association areas [auditory areas] of temporal lobes of the brain.
This is how we are able to hear the chirping of the birds and recognize the sound of the bird.

Question 5.
We can see and enjoy the beautiful colours of the nature after the sunrise.
Answer:

The light receptor are present in the retina of eyes.
These are photosensitive rod and cone cells. Cone cells are sensitive to bright light-and colours.
On receiving the light rays, these cells convert them into nerve impulses.
The cones are of three types, which contain their own characteristic photo-pigments that respond to red, green and blue lights.
Various combinations of these cones and their photo pigments produce sensation of different colours.
These impulses are carried to the visual cortex of the occipital lobe where the image is interpreted.
In this manner, we can see and enjoy the beautiful colours of the nature after the sunrise.

Question 6.
Cerebellum is well developed in humans.
Answer:

Our posture is upright and mode of locomotion is bipedal.
While standing, walking and running, our body has to be in a state of balance.
Cerebellum controls balancing, posture, body equilibrium and orientation.
Thus to control static as well as dynamic equilibrium of the body, cerebellum is well developed.

Question 7.
Mr. Sharma suffered from a stroke and the right side of his body was paralysed. However his response was normal for knee jerk reflex with either leg. Explain how and why?
Answer:

Stroke is damage to the brain due to interruption of its blood supply.
Due to this brain functions and cerebral reflexes are severely affected.
However, spinal reflexes remain largely unaffected.
In the above case. Mr. Sharma suffered from a stroke and developed paralysis. But his response to the knee jerk was normal as it is controlled by spinal cord.

Question 8.
The word hormone is a misnomer.
Answer:

Hormone is adapted from the Greek word ‘hormein’ means to excite.
But the hormones secreted by the endocrine glands and cells carry out two fold functions, viz. excitatory and inhibitory.
The word hormone is therefore a misnomer.

Question 9.
Pituitary may be considered as the coordinator of endocrine orchestra but not master endocrine gland.
Answer:

The pituitary gland was formerly considered as a master endocrine gland, because all other endocrine glands are under the control of pituitary hormones.
But now it is known that the pituitary gland itself is under the control of hypothalamus through hypo thalamo – hypophysial axis.
Through various releasing factors and release inhibiting factors, the secretions of pituitary are regulated by hypothalamus.
Pituitary in turn controls growth, secretion and maintenance of glands such as adrenal cortex, thyroid and gonads.

Question 10.
Pituitary gland plays an important role in pregnancy and childbirth.
Answer:
Pituitary gland plays an important role in pregnancy and childbirth for the following reasons:

Luteinizing hormone secreted by the pituitary gland stimulates ovulation, formation of corpus luteum and the synthesis of progesterone which are necessary for pregnancy.
Lactogenic hormone secreted by the pituitary gland promotes breast development during pregnancy and stimulates the secretion of milk after childbirth.
Oxytocin secreted by hypothalamus and sent by pituitary gland stimulates the contraction of the uterus during childbirth.

Question 11.
Fall and rise in blood calcium stimulates secretion of parathyroid.
Answer:

Normal calcium level in the blood is called calcemia.
Increase in the blood calcium is called hypercalcemia while decrease in the level is called hypocalcemia.
Decrease in the blood calcium, stimulates the parathyroid glands to secrete PTH.
PTH stimulates osteoclasts of bone to start bone resorption. This will help to cause increase in the blood calcium level.
Increase in the blood calcium above normal will retard the secretion of PTH. Hence further increase in the blood calcium is stopped.

Question 12.
Pancreas is both exocrine as well as endocrine gland.
Answer:

Pancreas is heterocrine gland i.e. both exocrine and endocrine gland.
The exocrine part is pancreatic acini. The cells of acini secrete pancreatic juice containing digestive enzymes like trypsinogen, chymotrypsinogen, etc.
Endocrine part of pancreas is made up of groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

Alpha (α) cells (20%) secrete glucagon.
Beta (β)) cells (70%) secrete insulin.
Delta (δ) cell (5%) secrete somatostatin. PP cells or F cells (5%) secrete pancreatic polypeptide (PP).
Hence, pancreas is said to be exocrine as well as endocrine gland.

Question 13.
Patient suffering from hypothyroidism shows increased level of TSH. Why?
Answer:

Hypothyroidism means decrease in the secretion of T₃ and T₄
Decrease in T₄, i.e. thyroxine in the blood triggers negative feedback mechanism.
The hormone receptors in hypothalamus detect this change and secrete TRE
TRF stimulates the pituitary to secrete TSH
Thus, Patient suffering from hypothyroidism shows increased level of TSH due to negative feedback control.

Question 14.
We are advised to use iodized salt.
OR
Why do we use iodized salt?
Answer:

Iodine is needed for synthesis of thyroid hormone.
If there is deficiency of iodine in the diet, it causes enlargement of thyroid gland leading to simple goitre.
This disease is common in hilly areas hence it is also called endemic goitre.
Addition of iodine to table salt prevents this disease.
Therefore we must use iodised salt.

Write short notes on the following

Question 1.
Nervous system of Hydra.
Answer:

Hydra, a cnidarian shows the diffused nervous system in the form of nerve net.
It is the most primitive nervous system.
There are two nerve nets in the mesoglea-one connected towards the epidermis and second towards the gastro-dermis.
Hydra lacks sensory organs, but the sensory cells scattered in the body wall.
The nerve impulse shows no polarity or direction. As all neurons are interconnected the response is seen throughout the body.

Question 2.
Nervous system of Planaria.
Answer:

Planaria is a flatworm and belongs to the phylum Platyhelminthes.
It is the most primitive animal with a Central Nervous System (CNS) located on the ventral side of body.
Nervous system consists of a mass of cerebral or cephalic ganglion appearing like an inverted U-shaped brain.
Ventrally from below the ganglia arise a pair of Ventral Nerve Cords (VNC) or long nerve cords. These are interconnected to each other by transfer nerve or commissure in a ladder like manner.
The PNS include sensory cells arranged in lateral cords in the body.

Question 3.
Resting potential of a nerve fibre.
Answer:

The neurons have a property of excitability.
The transmission of the nerve impulse along the long nerve fibre is a result of electrical changes across the neuronal membrane during conduction of an excitation.
The plasma membrane separates the outer [extra cellular fluid] and inner solutions [intra cellular fluid/cytoplasm] of different chemical compounds.
The external tissue fluid has both Na⁺ and K⁺
On the inside there is an excess of K⁺ along with large amount of negatively charged protein molecules and nucleic acid.
This condition of a resting nerve is also called a polarised state and it is established by maintaining an excess of Na+ on the outer side.

Question 4.
Electrical synapse.
Answer:

In this type of synapse gap between the neighbouring neurons is very narrow.
This electrical link is formed between the pre-and post-synaptic neurons.
At the gap junction, the two cells are within almost 3.8 nm distance of each other.
Transmission across the gap is faster.
Electrical synapses are found in those places of the body requiring fastest response as in the defence reflexes.

Question 5.
Velocity of nerve impulse.
Answer:

The rate of transmission of impulse is higher in long and thick nerves.
It is higher in warm blooded animals [homeotherms] than in cold blooded animal [poikilotherms].
The velocity of transmission is higher in voluntary fibres (100- 120 m/second in man) as opposed to autonomic or involuntary nerves (10-20 m/second).
It is faster in medullated nerve fibre [up to 150 m/s], as the impulse has to jump from one node of Ranvier to the next as compared with non-medullated nerve fibre [10-25 m/s]

Question 6.
Meninges of CNS.
Answer:

Meninges are the connective tissue membranes that cover the brain and the spinal cord. There are three meninges, viz. dura mater, arachnoid mater and pia mater that cover the Central nervous system.
The outermost tough, thick and fibrous meninx is dura mater. It is protective in function as it is attached to the inner side of the cranium.
The middle, thin and vascular membrane formed of reticular connective tissue is called arachnoid mater. It carries out nutritive function and also gives protection to the brain.
The innermost highly vascular and thin membrane is pia mater. It lies in contact with CNS. It is nutritive in function.
There is subdural space between the dura mater and arachnoid mater. It is filled with serous fluid.
Besides, there is a sub-arachnoidal space lying between arachnoid mater and pia mater. It is filled with cerebrospinal fluid.

Question 7.
Ventricles in human brain.
Answer:

Ventricles are the cavities present in different parts of the brain.
There are four ventricles in the human brain. All the ventricles are connected with each other.
They are filled with cerebrospinal fluid.
Paracoel or lateral ventricles-I and-II are present inside the cerebral hemispheres.
The diencephalon has ventricle-III.
Ventricle-III is in connection with lateral ventricles by foramen of Monro.

Question 8.
Sympathetic nervous system.
Answer:

Sympathetic Nervous System is formed by 22 pairs of sympathetic ganglia. These ganglia are linearly arranged on two sympathetic cords. Sympathetic nerve cords run on either side of the vertebral column.
Sympathetic nerve cords are connected to CNS by rami communicans of spinal nerve fibres.
This system works during stress, pain, anger, fear or emergency. It is supposed to bring about fight, flight or fright reactions.
Action of sympathetic nervous system is dependent on adrenaline or noradrenaline. This neurotransmitter is secreted by sympathetic nervous system as an emergency hormone.

Question 9.
Parasympathetic nervous system.
Answer:

Parasympathetic nervous system consists of nerve fibres of some cranial nerves, sacral nerves and parasympathetic ganglia.
These parasympathetic ganglia are present on the sides of visceral organs like heart, lungs, stomach, kidney, etc.
Parasympathetic ganglia gives out parasympathetic fibres which innervates these involuntary organs.
Parasympathetic nervous system works through release of acetylcholine which acts as neurotransmitter. It is an inhibiting neurotransmitter which affects visceral organs.
This system works during rest and brings about relaxation, comfort, pleasure, etc.

Question 10.
Parkinson’s disease.
Answer:

Degeneration of dopamine-producing neurons in the CNS causes Parkinson’s disease.
Symptoms develop gradually over the years.
Symptoms are tremors, stiffness, difficulty in walking, balance and coordination.
Seen in old age and is incurable.

Question 11.
Alzheimer’s disease.
Answer:

Alzheimer’s disease is the most common form of dementia.
Its incidence increases with the age.
Symptoms include the loss of cognitive functioning, thinking, remembering, reasoning and behavioural abilities. It interferes with the person’s daily life and activities.
It occurs due to loss of cholinergic and other neurons in the CNS and accumulation of amyloid proteins.
There is no cure for Alzheimer’s, but treatment slows down the progression of the disease and may improve the quality of life.

Question 12.
Types of reflex actions.
Answer:

The reflex actions are of two types, viz. cerebral and spinal.
Cerebral reflex actions are controlled by the brain.
Spinal reflex actions are controlled by the spinal cord.
In man, most of the reflex actions are controlled by the spinal cord.

Question 13.
Pavlov’s experiment about conditional reflex.
Answer:

Conditional reflex was demonstrated by Pavlov while performing experiments with dogs.
Pavlov offered some food to dog and noticed that the dog starts salivating after smelling and seeing the food.
Simultaneously he rang the bell so that the dog associated the food with the sound of bell. This experiment was repeated many a times by him.
Later he only rang the bell and did not give any food to the dog. But still the dog salivated.
This shows that the dog was conditioned to the sound of bell. The dog learnt that there is a relation between food and sound of bell. This is called conditional response.

Question 14.
Mechanism of vision.
Answer:

The light rays of visible wavelength pass through the cornea and the lens and are focused on the retina of the eye.
The sight is possible due to conjugated proteins present in the rods and the cones.
These are photo pigments which are composed of opsin (a protein) and retinal (Vitamin A derivative).
The light induces dissociation of retinal from the opsin, which causes a change in the structure of the opsin.
This causes the change in the permeability of the retinal cells.
It generates action potential which is carried via bipolar cells and ganglion cells and further conducted by the optic nerves to the visual cortex (vision centre) of the brain.
The neural impulses are analyzed and the image formed on the retina is thus recognized.

Question 15.
Mechanism of hearing.
Answer:

The external ear receives the sound waves and sends to the tympanic membrane. The tympanum vibrates transmitting the vibrations to the chain of three ossicles and then to the oval window.
The vibrations are further passed on to the fluid of cochlea.
The waves in the perilymph and endolymph induces movements in the basilar membrane.
The hair cells of organ of Corti bend and are pressed against the tectorial membrane.
Due to this pressure, the nerve impulses are generated and are sent to the afferent nerve fibres.
The impulses are carried by the auditory nerves to the auditory centre of the brain, where the impulses are analyzed and the sound is perceived.

Question 16.
Types of endocrine systems.
Answer:
There are 3 types of endocrine systems.

Discrete endocrine system : The glands exclusively endocrine in function are called discrete endocrine glands.
Mixed endocrine system : The glands endocrine as well as exocrine in function are called mixed endocrine glands.
Diffused endocrine system : Some endocrine cells scattered in a particular region/gland form diffused endocrine system.

Question 17.
Thymus.
OR
Functions of Thymosin.
Answer:

Thymus is located on the dorsal side of the heart and the aorta.
It consists of many lobules.
The thymus plays major role in the development of the immune system.
Thymus secretes thymosin which plays an important role in the differentiation of T-lymphocytes. These cells built cell mediated immunity.
The thymosin also promotes the production of antibodies providing humoral immunity.
The degeneration of thymus gland occur in old individuals leading to decreased production of thymosin thereby weakening of immune response.

Question 18.
The role of heart and kidney in hormone secretion.
Answer:
(I) Kidney:

Kidney produces renin, erythropoietin and calcitriol (calcitriol is the active form of vitamin cholecalciferol (D3).
Renin along with angiotensin helps in maintaining the blood pressure in the renal artery by vasoconstriction.
Erythropoietin stimulates erythropoiesis.
Calcitriol helps in absorbing calcium from the stomach.

(II) Heart:

Heart walls secrete Atrial natriuretic hormone /ANE
ANF increases sodium excretion [natriuresis] along with water.
It acts along with kidneys and reduces blood pressure by lowering blood volume.

Question 19.
Hormones secreted by adrenal gland.
Answer:

The adrenal cortex secretes corticoids. Corticoids is a group of several hormones that control several vital body functions.
Corticoids are of two types, viz. mineralocorticoids and glucocorticoids.
Small amounts of androgenic steroids are also secreted by the adrenal cortex which have the role in the growth of axial and pubic hairs and facial hairs during puberty.
The mineralocorticoids regulate the electrolyte balance while the glucocorticoids are involved in carbohydrate metabolism.
Adrenal medulla secretes adrenaline or epinephrine and noradrenaline or norepinephrine.

Question 20.
Role of mineralocorticoids.
Answer:

The mineralocorticoids regulate ionic and osmotic balance, by regulating the amounts of electrolyte and water.
Aldosterone is the main mineralocorticoid that acts on the renal tubules.
Aldosterone stimulates the re-absorption of Na⁺ and water and excretion of K⁺ and phosphate ions.
The aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

Question 21.
Secretions of adrenal medulla and their role.
Answer:

The adrenal medulla secretes two catecholamine hormones, viz. adrenaline (epinephrine) and noradrenaline (nor-epinephrine).
Adrenaline and noradrenaline increase alertness, dilation of pupils, piloerection, sweating, etc.
Both the hormones increase the rate of heartbeat, strength of heart contraction and rate of respiration.
These hormones also stimulate the breakdown of glycogen, lipids and proteins thereby increasing blood glucose level.
All the above reactions are useful for survival during emergency situations and in stress condition. Therefore, they are called emergency hormones or 3 F hormones of fright, fight or flight.

Question 22.
Cortisols and their role.
Answer:
I. Cortisol : Cortisol is the main glucocorticoid hormone.

II. Role of cortisol:

Cortisol stimulates many metabolic reactions such as gluconeogenesis, lipolysis and proteinolysis.
It inhibits cellular uptake and utilization of amino acids.
Cortisol also plays an important role in maintaining the cardiovascular system and kidney functions.
It is also involved in anti-inflammatory reactions and suppresses the immune response.
Cortisol stimulates the RBC production.

Question 23.
Disorders of adrenal cortical hormones.
Answer:

Disorders of adrenal cortical secretions are of two types, viz. hyposecretion and hypersecretion.
Hyposecretion of corticosteroids causes Addison’s disease.
The symptoms of Addison’s disease are general weakness, weight loss, low body temperature, feeble heart action, low BR acidosis, excessive loss of Na+ and Cl in urine, impaired kidney functioning and kidney failure, etc.
Hypersecretion of corticoids causes Cushing’s disease.
The symptoms of Cushing’s disease are alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, etc.

Question 24.
Hormones of adenohypophysis.
Answer:
I. Somatotropic Hormone (STH) or Growth Hormone (GH):

The secretion of GH is under dual control of hypothalamus through GHRF (Growth hormone releasing factor) and GHIF (Growth hormone inhibiting factor).
The GH brings about general growth of the body.
The principal actions of GH Eire promotion of linear growth in the skeleton, increase in the size of the muscles and connective tissue.
GH enhances the protein synthesis. The lipolysis in adipose tissue to release more fatty acids is also stimulated by GH.
The growth of bones by absorption of calcium takes place due to GH.

II. Thyroid Stimulating Hormone (TSH) or Thyrotropin:

TSH is regulated by TRF (Thyrotropin releasing factor) from hypothalamus’
For inhibition there is a negative feedback between thyroxine level in the blood and secretion of TSH.
TSH stimulates thyroid glands to increase uptake of iodine for synthesis of thyroxine. It brings breakdown of colloid to release thyroxine.

III. Adrenocorticotropic Hormone (ACTH) Corticotropin :

ACTH stimulates growth of adrenal cortex Eind stimulates it to secrete glucocorticoids Eind mineralocorticoids.
The regulation of ACTH secretion is under the control of hypothalamic CRF (Corticotropin releasing factor) and the negative feedback mechanism between plasma level of cortisol and ACTH.

IV. Prolactin (PRL) or Luteotropic hormone (LTH):
1. Secretion of prolactin is under duad control by hypothalamus by two factors such as PRF (Prolactin releasing factor) and PIF (Prolactin inhibiting factor) of hypothalamus.

2. Prolactin performs many functions therefore it has many terms as follows :

Development of mammary glands (Mammo tropin).
Milk secretion by mammary glands (Lactogenic hormone).
Maintenance of corpus luteum so that- it keeps on secreting progesterone during pregnancy (Luteotropin).

3. It may be inhibiting the chances of pregnancy during lactation period.

V. Gonadotropic Hormones (GTH) or Gonadotropins:
1. There are two types of gonadotropins, viz. FSH and LH or ICSH.

2. The secretions of gonadotropin are regulated by gonadotropin releasing factor (GHRF) of hypothalamus.

(3) They are regulated by negative feedback by sex hormones such as testosterone and estrogen.
(a) Follicle Stimulating Hormone (FSH):

FSH in female stimulates development of Graafian follicles. It helps in the formation of ovum by stimulating oogenesis.
It also stimulates ovarian follicular cells for secretion of female sex hormones, estrogen.
Under influence of estrogen, development of secondary sexual characters occurs in female.
In males, FSH stimulates germinal epithelium of seminiferous tubules for spermatogenesis and helps in the production and maturation of sperms.
Deficiency of FSH leads to infertility in both the sexes.

(b) Luteinizing hormone (LH) in females and Interstitial cell stimulating hormone (ICSH) in males:

LH brings about ovulation, i.e. rupture and release of ovum from the mature Graafian follicle. The empty Graafian follicle is transformed into corpus luteum.
Corpus luteum is a secondary endocrine source which secretes gestational hormone progesterone. Progesterone is a pregnancy stabilizing hormone.
In males, ICSH stimulates interstitial cells of Leydig which in turn secretes male sex hormone, the testosterone.
Testosterone develops secondary sexual characters in males.
High level of progesterone in female signals negative feedback, to pituitary to stop secretion of LH.
In males, high level of testosterone in blood gives negative feedback signal to Inhibit the secretion of FSH.

Short answers questions

Question 1.
Give one point of distinction between nervous coordination and hormonal coordination.
Answer:

The activity of nervous coordination is quick, immediate and fast as it sends the electrical signals.
Hormonal coordination is slow and long lasting as it takes place through the action of hormones.

Question 2.
Write about types of nerves.
Answer:
Nerve is a group of neurons enclosed in a connective tissue sheath epineurium. It is classified as:

Sensory nerve : A nerve having all sensory neurons is called sensory nerve. It carries information from sense organs to CNS. It is also called afferent nerve.
Motor nerve : A nerve having all motor neurons is called motor nerve. It carries information from CNS to effector organs. It is also called efferent nerve.
Mixed nerve : A nerve having with both sensory and motor neurons is called a mixed nerve. Sensory neurons in it carry nerve impulses from sense organs to CNS while motor neurons carry nerve impulses from CNS to effector organs.

Question 3.
What is the composition of neural tissue?
Answer:
Neural tissue is derived from embryonic ectoderm. It consists of two types of cells.

Neurons or nerve fibres : These are structural and functional units of nervous tissue. They conduct nerve impulses and coordinate various body activities.
Neuroglial cells : These are supportive cells which protect neurons throughout CNS and PNS. They perform other functions like secretion of myelin sheath, phagocytosis, production of CSF, etc. e.g Schwann cells, astrocytes, satellite cells, etc.

Question 4.
Explain the structure of cyton.
Answer:

Cyton is the main body of a neuron or nerve fibre.
The cyton has a distinct central nucleus with a nucleolus and neuroplasm.
Cytoplasm surrounds the nucleus around which there are neurofibrils, Nissl’s granules and other cell organelles.
Nissl’s granules are rich in ribosomes and proteins.
Neurofibrils play an important role in transmission of nerve impulse.
The cytons are generally found inside the brain, spinal cord (CNS) and in the ganglia.
Cytons within CNS form ‘nuclei’ while those present outside CNS in nerves form ‘ganglia’.

Question 5.
Enlist the various connective tissue layers in a nerve along with their location.
Answer:
Connective tissue layers in a nerve are:

Endoneurium : Covers each nerve fibre.
Perineurium : Covers each nerve bundle having a number of neurons.
Epineurium : Covers many nerve bundles to form a peripheral nerve.

Question 6.
What is a synapse?
Answer:

Synapse is a microscopic functional gap between two successive neurons.
In this telodendrites of pre-synaptic neuron are in close proximity with dendrites of post-synaptic neuron.
This gap is also called synaptic cleft.
During transmission of nerve impulse, the synapse get filled with neurotransmitters like acetyl choline.

Question 7.
What are the three divisions of nervous system? What are their chief functions?
Answer:

The three divisions of nervous system are central nervous system, peripheral nervous system and autonomous nervous system.
The central nervous system (CNS) consists of brain and spinal cord. The brain and spinal cord are the coordinators for all nervous functions.
The peripheral nervous system (PNS) is constituted by several nerves given by the central nervous system to all the body parts. All these nerves carry impulses to the CNS and bring back the responses from them. They are divided into cranial nerves and spinal nerves.
The autonomous nervous system controls all the internal organs and is not under voluntary control.

Question 8.
Does this CSF remain enclosed inside the ventricles? What can be the outcome of such a situation?
Answer:

CSF is present within the CNS as well as around it.
This fluid communicates with each other on the roof of medulla oblongata through 3 apertures, viz. Foramina of Luschka and foramen of Magendie.
This communication ensures constant pressure of CSF within as well as outside the CNS.
In the absence of this communication, there would be a pressure difference within as well as outside the CNS which will result in disturbances in the activities of CNS. Moreover, intercranial pressure would rise.

Question 9.
Enlist the different parts of the brain.
Answer:

There are three divisions of the brain, viz. forebrain (prosencephalon), midbrain mesencephalon) and hindbrain (rhombencephalon).
Forebrain is divided into cerebrum (telencephalon) and diencephalon (thalamencephalon). Underdeveloped of factory lobes (rhinencephalon) can also be seen in the anterior region.
Midbrain consists of corpora quadrigemina and crura cerebri.
Hindbrain has cerebellum (metencephalon) and brain stem. It is divided into pons varolii and medulla oblongata (myelencephalon).

Question 10.
Describe functional areas of cerebral cortex.
Answer:
Functional areas of cerebrum:

There are three functional areas in cerebrum viz., sensory, association and motor area.
In sensory area, sensory receptors bring the sensory inputs. These inputs are analysed in sensory area.
The sensory speech area is located in parietal lobe. It is called Wernicke’s area.
Association area forms the major portion of the cerebrum. It processes, analyses and stores the information given by the inputs. Power of reasoning, will, understanding, memory, etc. are the faculties present in the cerebral cortex.
Motor area is present in the frontal lobe lying anterior to the premotor area. In the lower part of the motor area just above the lateral sulcus lies the Broca’s area or motor speech area. The Broca’s area controls the movements necessary for speech.

Question 11.
Explain in detail the regions associated with the diencephalon.
Answer:
Diencephalon is a part present between forebrain and midbrain. It has three regions epithalamus, thalami and hypothalamus.
(1) Epithalamus : Epithalamus is the roof of diencephalon. It is highly vascular and non- nervous. It forms anterior choroid plexus that secretes cerebrospinal fluid. Pineal body is attached to epithalamus with the help of pineal stalk. Pineal body secretes serotonin and melatonin.

(2) Thalami : The lateral parts of diencephalon which are interconnected with the habencular commissure are called thalami. From thalami all sensory impulses (except olfactory impulses) pass on to the cerebrum.

3. Hypothalamus : Hypothalamus is the floor of the brain. Pituitary gland is attached to this floor by an infundibular stalk. Hypothalamus has many hypothalamic nuclei which are scattered in the white matter.

Question 12.
What is EBG? What information can be obtained from the EEG?
Answer:

EEG stands for electroencephalography.
It refers to the recordings of the brain’s spontaneous electrical activities in certain period of time.
These are recorded using multiple electrodes.
EEG is non-invasive method and measures voltage fluctuations resulting from ionic current within the neurons.
The basic concepts involved in this are similar to ECG.
It is used to diagnose conditions like epilepsy, sleep disorders, encephalopathies, coma, etc.

Question 13.
Find out how different functional areas of the brain can be mapped?
Answer:
Functional areas and status of the brain can be mapped by several imaging techniques available such as-

MRI : Magnetic Resonance Imaging
CT : Computed Tomography
PET : Positron Emission Tomography

Question 14.
Which are silent areas of the brain?
Answer:

Silent areas of the brain refer to association areas of the brain.
One such area is in the prefrontal cortex of brain.
These are the areas of the brain in which pathogenic conditions may occur without producing symptoms.
Injury to these areas is not accompanied by symptoms related to sensory and motor functions.

Question 15.
Is nervous tissue without lymphatic vessels?
Answer:

CSF is the lymph of CNS.
CSF is continuously generated by the ependymal cells lining the ventricles and central canal and simultaneously drained out of the brain into the blood stream.
There are no lymphatic vessels in the nervous system.
But the CSF is drained into peripheral blood circulation with the help of lymph vessels associated with meninges mainly the dura mater.

Question 16.
Explain the structure of spinal cord.
Answer:
Structure of spinal cord:

Spinal cord is a 42 to 45 cm long, 2 cm thick and hollow tube, extending from medulla oblongata to lumbar region.
It lies in the neural canal of vertebral column.
At the other end, it tapers down and is called conus medullaris. The posterior most end is called filum terminale which appears as a thread-like structure.
Beyond the second lumbar vertebra, it forms a horse tail-like structure called cauda equina. Cauda equina is a bunch of dorsal and ventral roots of last pair of spinal nerves.
There are two swellings on the spinal cord. The upper is cervical swelling and lower is lumbar swelling. Accordingly there are two plexuses, the cervical plexus supplying nerves to hands and the lumbar plexus supplying nerves to legs.
31 pairs of spinal nerves arise from spinal cord.

Question 17.
What is the significance of reflex action?
Answer:
Significance of reflex action :

Reflex action helps the animals to adjust quickly to the changing environment.
Reflex action is for quick actions necessary for survival. The life may have been in danger in the absence of reflex action.
Most of the reflexes are spinal reflexes, i.e. reflexes controlled by spinal cord. Thus the brain is not involved in these actions. This prevents overloading of the brain and brain fatigue.
Some reflexes are inborn and hence training or learning is not required for these.

Question 18.
During extraction of a tooth, the dentist gives an injection of Anaesthesia to the patient before extraction. Is the action potential generated? How does the local anaesthesia work? What is the effect of pain killer on the nervous system?
Answer:

Teeth are innervated by branches of trigeminal nerve [Vth cranial nerve]
Extraction of tooth stimulates this nerve which then carries the impulse [action potential) to the pain centre of the brain where the pain is perceived.
To avoid this, dentists give anaesthesia, to numb the nerve.
Action potential is not generated due to anaesthesia.
Hence the pain is not perceived.
Similarly, some common pain killers act on the nerve endings and pain centres of the brain, preventing generation of action potential.

Question 19.
Give a list of psychological disorders.
Answer:

Autism spectrum disorder.
Bipolar disorder.
Depression.
Anxiety disorder.
ADHD (Attention Deficit Hyperactivity Disorder).
Stress related disorders.

Question 20.
What are endocrine glands?
Answer:

Endocrine glands are ductless glands which are capable of secreting hormones.
The hormones are poured directly into the bloodstream as the endocrine glands do not have duct.
Hormones regulate the function of target tissue or organ.
They either have excitatory effect or have an inhibitory effect.

Question 21.
What are the main endocrine glands in human body?
Answer:
The main endocrine glands in human body are as follows:

Pituitary or hypophysis
Hypothalamus
Thyroid
Parathyroid
Adrenal or suprarenal
Islets of Langerhans in pancreas
Endocrine parts of gonads, i.e. testis and ovary.
Pineal gland and thymus are also endocrine glands of less importance.

Question 22.
What are the common properties of hormones?
OR
State properties of hormones.
Answer:

Hormones are specifically produced in response to a certain stimulus.
Depending on nature and intensity of the stimulus, the rate of secretion of a hormone varies from low to very high.
Hormones are produced in one organ and show their effect on distant ‘target’ organ. The source and the target region may be distantly located.
Hormones are directly poured in blood circulation and always carried through blood.
Hormones are always bound to specific carrier proteins while being transported through the blood.
Hormones have a high degree of target specificity.
Every hormone acts basically by modifying some aspect of cellular metabolism.
The excessive secretions or deficiencies- of hormones may lead to serious disorders. Such disorders are called hyper – and hypo- disorders, respectively.

Question 23.
What are the disorders caused due to hyposecretion and hypersecretion of GH or STH?
Answer:
(1) Hypersecretion is excessive secretion. In children, the hypersecretion of GH causes gigantism. In adults, it causes Acromegaly.

(2) Hyposecretion, i.e. lesser secretion of GH in children cause dwarfism. The person is also referred to as midget. There are two types of dwarfs, viz. Frohlic dwarf who are mentally abnormal and Lorain dwarf who are mentally normal.

(3) Hyposecretion of GH in adults cause Simmond’s disease.

Question 24.
What are the disorders caused due to hyposecretion and hypersecretion of ACTH?
Answer:

Hyposecretion of ACTH leads to Addison’s disease, i.e. adrenal failure. This results in affected carbohydrate metabolism leading to weakness and fatigue.
The hypersecretion leads to excessive growth of adrenal cortex. This causes Cushing’s disease.

Question 25.
Write an account of hormones secreted by the thyroid gland.
Answer:

Thyroid secretes triiodothyronine or T₃, tetraiodothyronine or thyroxine or T₄ and thyr ocalcitonin.
The thyroid gland synthesize, store and discharge these hormones.
T₃ and T₄ are iodinated derivatives of amino acid tyrosine which are secreted by thyroid follicular cells and stored in follicles. They have similar function. The secretion of T₃ and T₄ are regulated by Thyroid stimulating hormone (TSH) or thyrotropin of pituitary gland in negative feedback manner. T₃ is more active and T₄ is more potent hormone.
Thyrocalcitonin is secreted by the parafollicular cells.
Thyrocalcitonin regulates blood calcium level. It stimulates bones to take up Ca⁺⁺ from the blood and deposit it in the form of calcium phosphates in the bones, thereby decreasing blood Ca⁺⁺ level. Increased calcium level of blood stimulates ‘C’ cells to secrete thyrocalcitonin and vice versa.

Question 26.
Describe adrenal glands with respect to morphology, histology and secretions.
Answer:

A pair of adrenal or suprarenal glands are located just on the upper border of kidneys.
The adrenal gland shows two distinct regions, viz. thicker outer cortex and thinner inner medulla.
The adrenal cortex consists of three distinct regions. The outer zona glomerulosa, the middle zona fasciculata and inner zona reticularis.
Adrenal cortex produces corticoids. Corticoids is a collective term for many hormones, such as glucocorticoids, mineralocorticoids and steroid sex hormones.
Adrenal medulla secretes adrenaline or epinephrine and noradrenaline or norepinephrine.

Question 27.
Why are reproductive organs called dual in function?
Answer:

A pair of testes in males and a pair of ovaries in female both secrete hormones which are essential for sexual characters and function.
Besides this, they also produce male and female gametes respectively. Therefore they are said to be dual in function.

Question 28.
What are male hormones? What is their source and functions?
OR
Write a short note on the functions of androgens.
Answer:

Androgens are male hormones. The most significant androgen is the testosterone.
Interstitial cells of Leydig present in the testis of mature man produce androgen. Androgens are steroid in chemical nature.
Androgens regulate and stimulate the development, maturation and functions of the male reproductive organs like seminiferous tubules, epididymis, vas deferens, seminal vesicles, prostate glands and urethra.
Androgens are made sex hormones. They produce secondary sexual characteristics. Low pitch of voice is produced due to changes in the vocal cords which take place due to testosterone, etc. They stimulate muscular growth and growth of facial and axillary hairs.
The mental make up of a man like aggressiveness is due to testosterone.
They stimulate seminiferous tubules for the process of spermatogenesis.

Question 29.
What are female sex hormones? What role do they play?
Answer:
(1) The ovaries secrete two steroid hormones viz., estrogen and progesterone.

(2) The estrogen is secreted by the developing ovarian follicles. It has many roles in stimulation of female reproductive functions and growth of ovaries, fallopian tubes, uterus and vagina.

(3) It also controls female secondary sexual characteristics like high pitch of voice, development of mammary glands, broadening of pelvis, growth of pubic hairs and deposition of subcutaneous fats to produce feminine stature.

(4) The estrogen also regulates female sexual behaviour.

(5) The empty Graafian follicle after ovulation is converted into a structure called corpus luteum which secretes a hormone known as progesterone. Progesterone is a gestational hormone which is essential for maintaining the pregnancy. It also acts on the mammary glands and stimulates them for lactation, milk synthesis and ejection.

Question 30.
Why is pancreas called a dual gland?
Answer:

Pancreas is called a dual gland because it is exocrine as well as endocrine in nature.
The exocrine pancreas secretes digestive enzymes through acini.
The endocrine pancreas secretes hormones through its endocrine cells called Islets of Langerhans.

Question 31.
What are the hormones of pancreas? Describe the functions of pancreatic hormones.
OR
Pancreas plays an important role in controlling diabetes mellitus. Explain.
Answer:

Islets of Langerhans consists of three types of cells known as a-cells, β-cells and δ-cells.
α-cells secrete glucagon while β-cells secrete insulin. δ-cells secrete somatostatin.
The glucagon is a hyperglycemic hormone. It is a peptide hormone which acts mainly on the liver cells. Here it stimulates hepatocytes for glycogenolysis (i.e. breakdown of glycogen) leading to increased level of blood glucose (i.e. hyperglycemia).
It also stimulates gluconeogenesis (i.e. formation of glucose from non-carbohydrate sources). This in turn brings rise in blood glucose level or hyperglycemia.
Glucagon reduces the cellular glucose uptake and utilisation.
Insulin is also a peptide hormone, which plays a major role in maintenance of blood glucose level.
Insulin stimulates hepatocytes and adipocytes for cellular glucose uptake and utilization.

Therefore glucose from the blood decreases causing hypoglycemia. This hormone helps in the conversion of glucose to glycogen (i.e. glycogenesis) that occurs in target cells.

Question 32.
How is blood glucose level maintained?
Answer:
The blood glucose level is maintained by the joint but antagonistic action of insulin and glucagon.
Insulin is hypoglycemic hormone while glucagon is hyper/glycemic hormone.

When there is excess sugar in the blood, more insulin is secreted by the pancreatic islets. This causes the conversion of blood glucose into glycogen. This process is known as glycogenesis. This causes decline in the level of glucose in the blood.

When there is less blood glucose level then the glucagon is secreted. It causes stored glycogen to be converted into glucose. This process is called glycogenolysis.

Question 33.
What happens when there is insufficiency or deficiency of insulin in the body?
Answer:

Due to insufficiency, of insulin level there is prolonged hyperglycemia. This leads to diabetes mellitus.
In this diabetic condition cells are unable to utilize glucose. Therefore, in a diabetic person blood glucose levels are high. The glucose is excreted in urine.
The harmful compounds like ketone bodies are formed leading to ketosis.
Diabetes can be treated by taking insulin injections or tablets (insulin therapy) or with hypoglycemic drugs.

Question 34.
Where are parathyroid glands located? What are their functions?
OR
Write a short note on the functions of Parathyroid hormone (PTH).
Answer:

Parathyroid glands are located on the back side of the thyroid gland.
There are two pairs of parathyroid glands. One pair of parathyroid is in each lobe of thyroid.
Parathyroid glands secrete a peptide hormone known as parathromone or parathyroid hormone (PTH).
The level of Ca⁺⁺ in the blood regulates the secretion of PTH.
PTH is hypercalcemic hormone, it increases blood calcium level. Thus the calcium balance is maintained by TCT and PTH.

Question 35.
What are the gastrointestinal hormones? Explain the function of each.
Answer:

There are scattered endocrine cells in different parts of alimentary canal.
These cells secrete four peptide hormones which are gastrin, secretin, cholecystokinin (CCK) and gastric inhibitory peptide (GIP).
Gastrin stimulates gastric glands for the secretion of hydrochloric acid and pepsinogen.
The secretin acts on exocrine pancreas and stimulates secretion of water and bicarbonate ions to form pancreatic juice.
CCK acts on pancreas and gall bladder and stimulates the secretion of pancreatic enzymes and bile juice respectively.
GIP inhibit gastric secretion and motility.

Question 36.
Name the hormone secreted by the heart. What is its function?
Answer:

The atrial wall of the heart secrete a peptide hormone known as atrial natriuretic factor (ANF).
When the blood pressure increases, ANF hormone is secreted.
It causes dilation of the blood vessels.
Blood then can easily flow with lesser resistance and hence BP decreases.

Question 37.
What are the hormones of kidney? What function do they carry out ?
Answer:

The juxtaglomerular cells of the kidney produce a peptide hormone known as erythropoietin.
Erythropoiesis stimulates bone marrow for the production of RBCs. It thus stimulates the process of erythropoiesis.
Hormone calcitriol from kidney helps in the formation of bones.

Question 38.
Give importance of hypothalamus.
Answer:

Hypothalamus is the controlling centre for hypophysis.
Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of hypophysis. E.g. Adrenocorticotropin hormone releasing factor or CRF; Thyrotropin releasing factor or TSHRF; GHRF and GHRIE i.e. Growth hormone releasing and release inhibiting factor, etc.
Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.
Hormones like vasopressin and oxytocin are secreted by neurosecretory cells of hypothalamus.
Hypothalamus can register the internal changes in the body as it is a part of diencephalon and thus it accordingly brings about coordination in the body through endocrine system.

Question 39.
Write a brief account of releasing factors secreted by hypothalamus.
Answer:
(1) In the hypothalamus, there are several groups of neurosecretory cells which form different nuclei.

(2) These hypothalamic nuclei are supraoptic, paraventricular, dorso-median and ventromedian nuclei. These neurosecretory cells produce releasing and inhibiting factors.

(3) The hypothalamic neurohormones regulating the release of pituitary hormones are called releasing factors. The following are some of J the important releasing factors:

CRF or corticotropin releasing factor or ACTH releasing factor releases secretion of Adrenocorticotropin hormone (ACTH).
TRF or TSHRF (Thyroid stimulating hormone releasing factor) stimulates release of TSH.
FSH RF (Follicle stimulating hormone ; releasing factor) stimulates release of FSH.
GHRF (Growth hormone releasing factor) GHRIF (Growth hormone release inhibiting factor) act on release and regulation of growth hormone.
PRF (Prolactin releasing factor) and PRIF ; (Prolactin release inhibiting factor) act on release and regulation of prolactin.
MSHRF (Melanocyte stimulating hormone releasing factor) and MSH RIF (Melanocyte stimulating hormone release inhibiting factor) act on release and regulation of MSH.

Question 40.
Hormones are called chemical messengers and regulators. Explain.
Answer:

Hormones bring about coordination in the body with the help of nervous system.
Endocrine system and nervous system together form neuro-endocrine system.
This system works in tune with the external and internal environmental changes.
The hormones are either excitatory or inhibitory. They bring about the actions accordingly to keep the body in homeostasis or equilibrium.
Almost all endocrine glands are controlled by negative feedback inhibition. Some glands are auto-regulatory. Therefore, the concentration of hormones cannot be in excess or in deficiency.
Almost all the functions such as metabolism, growth, reproduction, etc. are under the control of hormones. Therefore hormones are called regulators and messengers.

Chart based/Table based questions

Question 1.
Draw a flow chart of – steps in generation and conduction of a nerve impulse.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 1

Question 2.
Enlist the names of following cranial nerves: I, II, VII, XII
Answer:

Number	NAME	NATURE
I	Olfactory	Sensory
II	Optic	Sensory
VII	Facial	Mixed
XII	hypoglossal	Motor

Question 3.
Enlist the names of following cranial nerves : III, W VI, XI
Answer:

Number	NAME	NATURE
III	Occulomotor	Motor
IV	Trochlear/Pathetic	Motor
VI	Abducens	Motor
XI	Spinal accessory	Motor

Question 4.
Complete the table.

Number	Type	No. of Pairs	Region
————	Cervical	————-	—————
T₁– T₁₂	————-	12 pairs	————-
L₁ – L₅	————-	5 pairs	Lower back
————-	Sacral	————-	Pelvic
————–	Coccygeal	—————	Tall region

Answer:

Number	Type	No. of Pairs	Region
C₁ – C₈	Cervical	8 pairs	Neck
T₁-T₁₂	Thoracic	12 pairs	Thorax / Upper back
L₁ -L₅	Lumber	5 pairs	Lower back
S – S₅	Sacral	5 pairs	Pelvic
Co₁	Coccygeal	1 pair	Tall region

Question 5.
Write types of neuroglial cells of CNS and PNS in tabular form.
Answer:

CNS – glial cells	PNS – glial cells	Functions
Oligodendrocytes [cells with few branches]	Schwann cells	Secrete myelin sheath
Astrocytes [star-shaped and most abundant glial cells in CNS]	Satellite cells	Protect, cushion and supply nutrients to nearby neurons. Help in maintaining blood-brain barrier.
Microglia [small cells with few branches]	Macrophages	Phagocytosis
Ependymal cells lining the ventricles of brain [mostly columnar]	Ependymal cells lining central canal of spinal cord	Secrete cerebrospinal fluid

Question 6.
Enlist the various receptors found at various location in the body.
Answer:

Receptors	Types	locations
Mechanoreceptors	Thermoreceptors	Skin
	Tango [touch and pressure] receptors	Skin
	Tactile [light touch] receptors	Skin
Chemoreceptors	Gustato receptors	tongue
Chemoreceptors	Olfacto receptors	Olfactory mucosa
Photoreceptors	Rod and cone cells	Retina of eye
Phonoreceptors	Organ of Corti	Cochlea of internal ear
Stato receptors	Cristae and maculae	Semicircular canals, utricle, saccule of internal ear

Question 7.
Sketch the concept maps for mechanism of vision and mechanism of hearing.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 2

Question 8.
Complete the table given below by putting a tick (✓) in the boxes where applicable.

Action	Reflex	Voluntary
1. Touching a hot object	———-	———–
2. Releasing saliva on smelling food	———–	————
3. Applying a car’s brakes in an emergency	————	————–
4. Blinking of eyes when a small insect touches the eye	———–	————

Answer:

Action	Reflex	Voluntary
1. Touching a hot object		✓
2. Releasing saliva on smelling food	✓
3. Applying a car’s brakes in an emergency	can be a conditioned feclex too.	✓
4. Blinking of eyes when a small insect touches the eye	✓	————

Question 9.
Complete the following table

Action	Reflex	Voluntary
1. Optic nerve	———–	———–
2. Facial	———–	————
3. Hypoglossal	————	————–
4. Trigeminal	———–	————
5. Auditory
6. Glosso-pharyngeal

Answer:

Action	Reflex	Voluntary
1. Optic nerve	Sensory	Sense of vision and light
2. Facial	Mixed	Facial expression, movement of neck, tongue, etc. and saliva secretion
3. Hypoglossal	Motor	Movement of tongue
4. Trigeminal	Mixed	Sensation of touch, taste and jaw movements
5. Auditory	Sensory	Hearing and equilibrium
6. Glosso-pharyngeal	Mixed	Taste, pharyngeal contractions ‘and saliva secretion

Diagram based questions

Question 1.
Sketch and label – nerve net of Hydra.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 4

Question 2.
Sketch and label – nervous system of Planaria
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 5

Question 3.
Sketch and label – depolarization and repolarization along nerve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 6

Question 4.
Sketch and label ultrastructure of synapse.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 7

Question 5.
Sketch and label – lateral view of brain.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 8

Question 6.
Sketch and label – functional areas of Brain?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 9

Question 7.
Sketch and label – ventral view of human brain.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 10

Question 8.
Sketch and label – ventricles of brain.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 11

Question 9.
Sketch and label T.S. of spinal cord.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 12

Question 10.
Sketch and label – formation of spinal nerve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 13

Question 11.
Sketch and label – mechanism of hormone action.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 14

Question 12.
Sketch and label – V.S. of pituitary.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 16

Question 13.
Sketch and label – morphology of thyroid
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 17

Question 14.
Sketch and label – histology of thyroid
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 18

Question 15.
Sketch and label – parathyroid glands
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 19

Question 16.
Sketch and label thymus.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 20

Question 17.
Sketch and label – adrenal gland.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 21

Long answer questions

Question 1.
Explain transmission of nerve impulse across a synapse.
OR
Explain how is impulse transmitted through a synapse?
Answer:

The nerve impulse travels along the axon of the pre-synaptic neuron to the axon terminal.
Pre-synaptic neurons or axons have several synaptic knobs at their ends or terminals.
These knobs have membranous sacs, called synaptic vesicles having neurotransmitter molecules.
When an impulse reaches a synaptic knob, voltage sensitive Ca⁺⁺ channels open and calcium ions (Ca⁺⁺) diffuse inward from the extracellular fluid.
The increased calcium concentration inside the cells, initiates a series of events that help to fuse the synaptic vesicles with the cell membrane of pre-synaptic neuron, where they release their neurotransmitters by exocytosis.
The neurotransmitters bind to the receptors of the post-synaptic cell,
This action is either excitatory (stimulating) or inhibitory (slowing down/stopping).
Once the impulse has been transferred across the synapse, the enzyme like acetyl cholinesterase destroys the
neurotransmitter and the synapse is ready to receive a new impulse.

Question 2.
Explain transmission of nerve impulse along the axon.
OR
Describe the conduction of a nerve impulse in the neuron.
Answer:
1. Before conduction of nerve impulse, the cell membrane is in the polarized state.

2. When a stimulus is applied at a site on the polarised membrane, the membrane at that site becomes freely permeable to Na⁺.

3. This leads to a rapid influx of Na⁺ followed by the reversal of the polarity at that site, i.e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged.

4. The polarity of the membrane at that site [site A] is thus reversed and hence depolarised. The electrical potential difference across the plasma membrane at the site of stimulation is called the action potential, which is in fact termed as a nerve impulse.

5. At sites immediately ahead [site B], the axon membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site A to site B.
Maharashtra Board Class 12 Biology Important Questions Chapter 9 Control and Co-ordination 22

6. On the outer surface current flows from site B to site A to complete the circuit of current flow. Hence, the polarity at the site is reversed, and an action potential is generated at site B. Thus, the impulse (action potential) generated at site A arrives at site B.

7. The sequence is repeated along the length of the axon and consequently the impulse is conducted.

8. The rise in the stimulus-induced permeability to Na⁺ is extremely short-lived. It is quickly followed by a rise in permeability to K⁺.

9. Within a fraction of a second, K⁺ diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation and the fibre becomes once more responsive to further stimulation.

Question 3.
Explain the structure of cerebrum. Structure of cerebrum
Answer:
Answer: Structure of cerebrum:

Cerebrum is the largest part of the brain. It forms 80-85% volume of the brain.
A median longitudinal fissure divides the cerebrum into two cerebral hemispheres. These hemispheres are interconnected by a thick band of transverse nerve fibres called corpus callosum.
The outer part of cerebrum is called cerebral cortex while the inner part is called cerebral medulla.
The roof of cerebrum is called pallium. Pallium is highly folded forming sulci and gyri. Sulci are depressions while gyrl are ridges. The gyri increase the surface area of cerebral cortex.
The ventro-lateral walls of cerebrum are thickened and are called corpora striata.
The cerebral cortex has three deep sulci, the central, lateral and parieto-occipital.
These sulci divide the cerebral hemisphere into four lobes. These are frontal, parietal, occipital and temporal lobes. A fifth median lobe called insula or insular cortex is folded deep within the lateral sulcus.
The central sulcus separates frontal and parietal lobes, the lateral sulcus separates parietal and temporal lobes and the parieto¬occipital sulcus separates parietal and occipital lobe.

Question 4.
Describe the structure and function of midbrain.
Answer:
1. Structure of midbrain:

Midbrain is the middle part of the brain situated between the forebrain and the hindbrain. It is present between the pons varolii and diencephalon.
It has two distinct regions : Corpora quadrigemina and crura cerebri.
Corpora quadrigemina consists of two pairs of lobes, viz., superior colliculi and inferior colliculi. These are located in the thick wall of midbrain.
Crura cerebri are thick bands of longitudinal nerve fibres, present on the floor of midbrain.

2. Functions of midbrain:

Inferior colliculi control and coordinate auditory reflexes.
Superior colliculi control head and eye movements.
Crura cerebri connect the cerebrum to cerebellum and spinal cord.

Question 5.
Give an account of structure of hindbrain.
Answer:
Structure of hindbrain:

Hindbrain includes cerebellum, pons varolii and medulla oblongata.
Cerebellum is 11% of the total brain and is the second largest part of the brain.
It has three lobes, median vermis and lateral two cerebral hemispheres. It has outer grey and inner white matter.
Cerebral cortex shows sulci and gyri. The inner white matter of cerebellar medulla shows arbor vitae or branching tree-like processes.
Pons is the part that connects the two cerebral hemispheres. It has outer white and inner grey matter. Pons is made up of nerve fibres which form bridges between cerebrum and medulla oblongata.
Medulla oblongata is the last part of the hindbrain which continues further as a spinal cord. It has outer white and inner grey matter.
Its roof shows has posterior choroid plexus.
Eight pairs of cranial nerves arise from medulla oblongata.

Question 6.
Describe T.S. of spinal cord.
Answer:

Externally there are three meninges covering spinal cord Duramater, arachnoid membrane and pia mater.
Dorsoventrally there are two fissures, the shallow dorsal or posterior fissure and the deeper ventral or anterior fissure.
From dorsal fissure a dorsal septum extends inside.
Neurocoel or central canal is situated in the centre of spinal cord.
The central canal is filled with cerebro¬spinal fluid and is lined by cuboidal epithelial cells called layer of ependyma.
There is inner grey and outer white matter in the spinal cord. This grey matter is in the shape of ‘H’ with two dorsolateral horns and two ventro-lateral horns.
Dorsal horns form dorsal roots and ventral horns form ventral roots.
White matter is divided into three columns, viz., the dorsal funiculi, ventral funiculi and lateral funiculi on either side.
Ascending and descending tracts of nerve fibres arise from dorsal and ventral roots of the spinal cord. Ascending tracts are sensory while descending tracts are motor in nature.

Question 7.
What are the different types of reflexes?
Answer:
1. Based on the location of their action : Thereflexes are divided into somatic reflexes and visceral reflexes.

When effector is located in body structures such as skeletal muscles, it is called a somatic reflex.
When the effector is located in the visceral organs such as glands or smooth muscles then it is called a visceral reflex.

2. Based on the basis of number of neurons : Reflexes are of two types, viz. monosynaptic reflexes and polysynaptic reflexes.

Simple or monosynaptic reflexes are those in which one sensory and one motor neuron are involved in the reflex action.
Polysynaptic or complex reflexes are those when more than two neurons are involved in the reflex action.

3. Based on inheritance and experience of learning : The reflexes are subdivided into unconditional or inborn and conditional or acquired.

Unconditional or inborn reflexes are inborn or hereditary. They are permanent, never disappear and need no previous experience, e.g. blinking of eyes, suckling, swallowing, knee jerk, sneezing, coughing, etc.
Conditional or acquired reflexes are acquired during life by experience or learning. They are based on individual learning or experience. These are not heritable, temporary and may disappear or reappear, e.g. driving, cycling, etc.

Question 8.
Explain the mechanism of reflex action.
Answer:

Mechanism of reflex action: There are series of sequential events in which reflex action is completed:
Stimulus is picked up by any receptors, e.g. pricking of a needle in the hand, causes skin to be a receptor.
Sensory impulse is formed in grey matter of spinal cord. It receives sensory impulse, interprets it and generates motor impulse.
The cyton of motor neuron present in the ventral horn of grey matter and axon conducts motor impulse from spinal cord to effector organ. This is further carried by dendrites innervating the skin.
Impulse is carried to the association neuron by axon of sensory neuron, when impulse reaches the end of the axon there is a synapse.
Transmission takes place by releasing acetylcholine from the synaptic buttons at the end of the axon.
It fills the synaptic gap and helps in chemical transmission of the impulse from axon of one neuron to dendron of the other neuron. Once the impulse reaches the dendrites of association neuron; axonic button releases an enzyme, acetylcholine esterase which neutralizes the acetylcholine and again a synaptic gap is formed. This mechanism helps to receive new impulse or avoid the mixing of different impulses.
The association neuron receives sensory impulse, interprets it, analyses it and generates motor impulse. Motor impulse again travels through synapse between association neuron and motor neuron.
Impulse travels through motor neuron and reaches the effector organ like skeletal muscles or the gland. The effector organ gives a proper response like contraction of the muscles or secretion by the gland.

Question 9.
Define receptors. Enlist different types of receptors.
Answer:
1. Receptors : Receptors are specialized cells, tissues or organs present in the body which receive different stimuli.

2. Types of receptors:

Receptors are of two types, viz. exteroceptors and interoceptors.-
Exteroceptors receive stimuli directly from the external environment. They are somatic in nature.
Interoceptors are located inside the body and are visceral in nature. They respond to internal changes in the body.

The various types of exteroceptors and interoceptors, their location and functions have been summarized in the table given below:

Types of Exteroceptors	Location	Function
1. Mechanoreceptors	Touch corpuscles in skin	Tangoreceptors Pressure Tactile receptors- Touch
2. Thermoreceptors	Skin	Frigido receptors (cold) Heat receptors (warmth)
3. Chemoreceptors	Tongue, nasal mucosa	Gustatoreceptors – Taste Olfactoreceptors- Smell
4. Statoacoustic receptors	Internal ear	Cochlea – Hearing Semicircular canals-Balance and equilibrium
5. Photoreceptors	Retina of the eye	Rods and cones interpret images Rods-black and white image. Sensitive to dim light. Cones – Coloured image. Sensitive to bright light.

Question 10.
Describe the different parts of human eye.
OR
Describe briefly the structure of eye.
Answer:

Human eyes are a pair of organs located in sockets of the skull called orbits.
Eyeball is spherical and has three layers.
Sclera is the outer layer of dense connective tissue with anterior transparent cornea.
Choroid is the middle layer. It is bluish in colour containing many blood vessels. The anterior region is thick and forms the ciliary body. Posterior 2/3rd region is thinner.
Iris is the forward segment of the ciliary body which is pigmented and opaque. This part is the visible coloured portion of the eye.
Lens is present anteriorly inside the iris and is held in position by the ligaments of ciliary body.
The aperture surrounded by the iris in front of the lens is known as pupil. The movement of the pupil is regulated by the muscle fibres of iris.
The innermost layer of the eye is the retina having three sub-layers formed by ganglion cells, bipolar cells and photoreceptor cells, which are sensitive to light.
There are two types of photoreceptor cells, viz. rods and cones containing light sensitive proteins. They are termed as photo pigments, rhodopsin which is a derivative of vitamin A (in rods) and iodopsin (in cones).
The cones are responsible for daylight or photopic vision and colour vision. The rods function in dim light giving scotopic vision.
The cones are of three types, each containing its own characteristic photopigments that respond to red, green and blue lights.
The optic nerve leaves the eye at a point slightly away from the median posterior pole of the eyeball. In this region, the rods and cones are absent therefore this region is known as blind spot. Macula lutea, a yellowish pigmented spot is present lateral to the blind spot.
Fovea is a central pit present beside it. Fovea is a thinned out portion of the retina where only the cones are densely packed and therefore have greatest visual acuity (resolution).
A space between the cornea and the lens is called aqueous chamber. It contains a thin watery fluid known as aqueous humor.

Question 11.
Describe the internal structure of human ear.
OR
Ear is one of the important sense organs known for its role in hearing and balancing. Describe those structures present in the internal ear which helps in these functions.
Answer:

The ears are the auditory sensory organs, also involved in maintaining equilibrium of the body.
The ear is composed of three divisions namely the outer ear, middle ear and internal ear.
The external ear consists of the pinna and external auditory meatus (canal). The pinna is for the collection of sound waves coming from the environment. The external auditory canal is the circular tube leading inside up to the eardrum or tympanic membrane.
The tympanic membrane or eardrum is formed of connective tissues having outer skin cover and inner mucus membrane.
The middle ear consists of chain of three ossicles called malleus, incus and stapes, The malleus is attached to the tympanic membrane and the stapes is connected to the oval window of the internal ear. They help in the transmission of sound waves from external auditory canal to internal ear.
Connecting middle ear with the pharynx is eustachian tube which helps in equalizing the air pressure on either side of the tympanic membrane.
The internal ear is fluid filled structure called labyrinth. It has two parts, bony and the membranous labyrinth.
The outer bony labyrinth is formed by the series of channels in which the membranous abyrinth containing endolymph fluid is present.
The membranes consist of coiled cochlea, the reissner’s membrane and basilar membranes. These membranes divide the surrounding perilymph filled bony labyrinth into an upper scala vestibule and a lower scala tympani.
The space within cochlea which is known as scala media is filled with endolymph. The scala vestibule ends at the oval window at the base of cochlea.
The scala tympani terminates at the round window which opens to the middle ear. The organ of corti is located on the basilar membrane. It contains the hair cells which act as auditory receptors.
The hair cells are columnar cells present in rows. The basal ends of the hair cells are in close contact with the afferent nerve fibres while their apical end contains numerous cilia. A thin elastic membrane projects above the rows of the hair cells called tectorial membrane.
Above the cochlea, the internal ear also contains vestibular apparatus. It consists of three semicircular canals and the otolith organ formed of the sacculus and utriculus. The semicircular canals lie in different plane at right angles to each other and are suspended in the perilymph.
The bases of canals are swollen and are called ampullae, which contain a projecting ridge known as crista ampullaris which contain hair cells.
The sacculus and utriculus also have projecting ridge called macula. The crista and macula are the specific receptors of vestibular apparatus. They are responsible for maintenance of body posture and the balance.

Question 12.
Write an account of position and structure of pituitary gland.
Answer:
Pituitary gland (Hypophysis):
I. Position : Pituitary or Hypophysis is located on the ventral side of brain below the hypothalamus. Infundibulum or hypophyseal stalk attaches pituitary to hypothalamus just behind the optic chiasma. It is well protected in sella turcica which is a depression of the sphenoid bone of the skull.

II. Morphological structure of pituitary- gland: The pituitary gland shows two distinct regions : Anterior lobe or adenohypophysis and posteriorlobe or neuro-hypophysis.
(1) Adenohypophysis or Anterior lobe : It is the largest lobe of the gland and forms about 75% of pituitary gland. It develops as an outgrowth called Rathke’s pouch from the roof of embryonic buccal cavity. It has three divisions, viz. pars tuberalis, pars distalis and pars intermedia.

(i) Pars tuberalis : Tubular region present below the hypothalamus is known as pars tuberalis. It is like a collar around the infundibulum. It is non-secretory in nature.
(ii) Pars distalis : The largest anterior region which is secretory in nature is called pars distalis. It is made up of loose cords of epitheloid secretory cells which are separated by reticular connective tissue containing blood sinusoids. It is connected to the hypothalamus by portal system formed by blood sinusoids.
(iii) Pars intermedia : The narrow cleft between the pars distalis and pars nervosa of neuro – hypophysis is called the intermediate part or pars intermedia. It is reduced, less developed and non-functional in human being.

(2) Neuro-hypophysis or Posterior lobe : The posterior lobe of the pituitary which is attached to hypothalamus by infundibular stalk is called neuro-hypophysis. It is smaller and constitute 25% of pituitary. It has the following three parts:

Median eminence : The swollen median part of the hypothalamus where infundibulum gets attached is called median eminence.
Infundibulum : Infundibulum is the hypophyseal stalk that helps in attachment of pituitary gland to the hypothalamus. It contains mainly the axonic fibres of neurosecretory cells present in hypothalamus. It forms the major connection of hypothalamo-hypophysis axis.
Pars nervosa : The lowermost, larger region of neuro-hypophysis that contains axons is called pars nervosa. It acts as a neurohaerhal organ and contains specialized cells called pituicytes.

Question 13.
In a person, Pars distalis part of the Pituitary gland is not producing hormones in sufficient quantity. Explain the effects it will produce with respect to the different hormones.
OR
Give names and functions of hormones secreted by adenohypophysis.
Answer:
Pars distalis of the pituitary gland produces following hormones:
GH, ACTH, TSH, FSH, LH/ICSH, LTH and MSH. If these hormones are produced in deficient quantities, following disorders are seen.
1. GH:
(a) Hyposecretion of GH in childhood leads to dwarfism. Frohlich dwarf or Lorain dwarf may be produced based on mental capacity. Hyposecretion in adulthood causes Simmonds’s disease.
(b) Hypersecretion of GH in childhood causes gigantism and in adulthood it causes acromegaly.

2. TSH :
(a) Hyposecretion of TSH leads to thyroid atrophy.
(b) Hypersecretion of TSH causes excessive functioning of thyroid gland.

3. ACTH :
(a) Hyposecretion of ACTH causes Addisons’ disease, in which adrenal gland shows failure of functions.
(b) Hypersecretion of ACTH causes Cushing’s disease, in which the adrenal cortex undergoes excessive growth.

4. FSH : Hyposecretion of FSH leads to infertility in both the sexes. Hypersecretion of FSH in females cause disturbances in menstrual cycle.

5. LH/ICSH : Hyposecretion of LH in females will cause lack of ovulation. Hyposecretion of ICSH in males cause reduction in masculinity. Sperm production may be affected. Hypersecretion of LH/ICSH can cause disturbances in reproductive cycles.

6. LTH : Corpus luteum is not maintained due to lesser amount of LTH. Lactogenesis will also hamper if there is hyposecretion of LTH.

Question 14.
Describe the hormones of neuro¬hypophysis.
Answer:
Hormones of neuro-hypophysis : Neuro-hypophysis does not secrete any hormone itself but stores the hormones which are secreted by hypothalamic neurons. It stores and releases the following hormones, viz. ADH, Oxytocin and coherin.
1. Anti Diuretic Hormone (ADH) or Vasopressin:

ADH brings about anti-diuretic action and also increases blood pressure.
It is a regulatory hormone which plays a major role in osmoregulation.
It increases the permeability of distal convoluted tubule or collecting tubules of uriniferous tubules of kidney.
Higher ADH levels decrease the urine output and helps for water conservation. It helps in the absorption of water from the ultrafiltrate thus regulates the water balance of body fluids.
ADH also controls constriction of arterioles and increases blood pressure in kidney which facilitates ultra filtration. Therefore it is also called vasopressin.
ADH is regulated by increase or decrease of osmotic pressure of blood in a feedback manner.
The osmotic pressure is detected by osmoreceptors in the hypothalamus.

2. Oxytocin (Birth hormone):

Oxytocin helps in parturition.
It is a powerful stimulant of contraction of uterine myometrium at the end of gestation due to which the labour is initiated.
It also stimulates myoepithelial cells of mammary glands for milk ejection during lactation.
It also helps in fertilization by powerful contractions of the uterine musculature to drive the sperms upward towards fallopian tubes.
Oxytocin also excites musculature of gallbladder, ureters, urinary bladder, intestine, etc. for proper functioning of these organs.

3. Coherin : Coherin induces prolonged, rhythmic integrated contractions of the jejunum.

Question 15.
Describe the morphology of thyroid gland.
Or
With the help of a suitable diagram describe the structure of thyroid gland.
Answer:
Morphology of thyroid gland:

Thyroid is the largest endocrine gland in the body.
It weighs about 25 to 30 g and measures about 5 cm in length and 3 cm in width.
It is located in the neck region anteriorly just below the larynx and situated ventrolaterally to the trachea.
The thyroid is derived from the endoderm of the embryo.
The thyroid can vary in size as per age, sex and diet.
It is reddish brown, bilobed and highly vascular
The two lobes are joined by connective tissue called isthmus which is located at 2nd to 4th tracheal cartilage.
Therefore, the right and left lobe of thyroid are seen on both sides of the trachea.
The gland is H-shaped having butterfly-like appearance.
The structural and functional units of thyroid gland are the thyroid follicles.
From the outer surface there lies a connective tissue capsule. A number of septa arise from the connective tissue which are called trabeculae. They divide the gland into lobules. Each lobule has follicles which store hormone. The number of follicles are about three million.

Question 16.
Describe neurohormonal regulation of pituitary and thyroid gland.
Answer:
Pituitary gland is directly under the influence of neurohormones of hypothalamus while thyroid is indirectly influenced.
I. Neurohormonal regulation of pituitary:

Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
Hypothalamus forms the hypothalamohypophysial axis through which transportation of neuro-hormones take place.
Pituitary secretes a variety of hormones which influence other endocrine glands of the body. E.g. GH, PRL, TSH, ACTH, Gonadotropins

II. Neurohormonal regulation of thyroid :

Hypothalamus secretes TRF [Thyrotropin releasing factor] which influences the anterior pituitary to release TSH.
TSH in turn stimulates thyroid follicles to produce and release two thyroid hormones – T₃ and T₄. (Thyroxin)
Increase in T₃ and T₄ triggers negative feedback mechanism that stops the secretion of TRF.

(4) As the pituitary does not get the signal in the form of TRF TSH secretion is stopped.

Question 17.
Name the hormones secreted by the adrenal cortex and state their role.
Answer:
Adrenal cortex secretes 3 types of corticoids – mineralocorticoids, glucocorticoids and sex corticoids.
I. Mineralocorticoids:

The mineralocorticoids regulate ionic and osmotic balance, by regulating the amounts of electrolyte and water.
Aldosterone is the main mineralocorticoid that acts on the renal tubules.
Aldosterone stimulates the re-absorption of Na+ and water and excretion of K⁺ and phosphate ions.
The aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

II. Glucocorticoids:

Cortisol is the main glucocorticoid. Cortisol stimulates many metabolic reactions such as gluconeogenesis, lipolysis and proteinolysis.
It inhibits cellular uptake and utilization of amino acids.
Cortisol also plays an important role in maintaining the cardiovascular system and kidney functions.
It is also involved in anti-inflammatory reactions and suppresses the immune response.
Cortisol stimulates the RBC production.

III. Sex corticoids (Gonadocorticoids).

Sex corticoids, Androgens and estradiols are produced by the adrenal cortex.
In males, they have a role in development and maintenance of external sex characters.
Excess sex corticoids in female causes adrenal virilism and hirsutism (excess hair on face)
Excess sex corticoids in males causes gynaecomastia i.e. enlarged breast.