12th Biology Chapter 1 Exercise Reproduction in Lower and Higher Plants Solutions Maharashtra Board

Class 12 Biology Chapter 1

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 1 Reproduction in Lower and Higher Plants Textbook Exercise Questions and Answers.

Reproduction in Lower and Higher Plants Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 1 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 1 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Insect pollinated flowers usually possess ………………
(a) sticky pollen with a rough surface
(b) large quantities of pollens
(c) dry pollens with a smooth surface
(d) light colored pollens
Answer:
(a) sticky pollen with a rough surface

Question 2.
In ovule, meiosis occurs in ………………
(a) Integument
(b) Nucellus
(c) Megaspore
(d) Megaspore mother cell
Answer:
(d) Megaspore mother cell

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 3.
The ploidy level is NOT the same in ………………
(a) Integuments and nucellus
(b) Root tip and shoot tip
(c) Secondary nucleus and endosperm
(d) Antipodals and synergids
Answer:
(c) Secondary nucleus and endosperm

Question 4.
Which of the following types require pollination but result is genetically similar to autogamy?
(a) Geitonogamy
(b) Xenogamy
(c) Apogamy
(d) Cleistogamy
Answer:
(a) Geitonogamy

Question 5.
If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?
(a) Endosperm
(b) Leaf cells
(c) Cotyledons
(d) Synergids
Answer:
(d) Synergids

Question 6.
In angiosperms, endosperm is formed by/ due to ………………
(a) free nuclear divisions of megaspore
(b) polar nuclei
(c) polar nuclei and male gamete
(d) synergids and male gametes
Answer:
(c) polar nuclei and male gamete

Question 7.
Point out the odd one.
(a) Nucellus
(b) Embryo sac
(c) Micropyle
(d) Pollen grain
Answer:
(d) Pollen grain

2. Very Short Answer Questions

Question 1.
The part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigmatic surface.

Question 2.
How many haploid cells are present in a mature embryo sac?
Answer:
6 cells, 2 synergids, 1 egg cell, 3 antipodals.

Question 3.
Even though each pollen grain has 2 male gametes why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?
Answer:
Angiosperms have phenomenon of double fertilization in which both the male gametes are utilized, one for fusion with egg cell to form zygote and other for fusion with secondary nucleus to form endosperm.

Question 4.
Megasporogenesis
Answer:
It is the process of formation of haploid megaspores from diploid megaspore mother cell.

Question 5.
What is hydrophily?
Answer:
Transfer of pollen grains in pollination process through agency of water is known as hydrophily.

Question 6.
The layer which supplies nourishment to the developing pollen grains.
Answer:
Tapetum

Question 7.
Parthenocarpy
Answer:
The condition in which fruit is developed without the process of fertilization is called parthenocarpy.

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 8.
Are pollination and fertilization necessary in apomixis?
Answer:
Apomixis is formation of embryos without formation of gametes hence there is no need of pollination and fertilization.

Question 9.
The part of pistil which develops into fruit and seed.
Answer:
Ovary develops into fruit and ovules into seed.

Question 10.
What is the function of filiform apparatus ?
Answer:
Filiform apparatus guides the pollen tube towards egg cell.

3. Short Answer Questions

Question 1.
How polyembryony can be commercially exploited?
Answer:

  1. Polyembryony is the development of more than one embryo inside the seed.
  2. When such polyembryonic seed germinate we get multiple seedlings from it.
  3. This condition increases the chances of survival of new plants.
  4. Nucellar embryos are genetically identical to parent plants hence we get uniform plants.
  5. In horticulture we can utilize these as rootstock for grafting, hence they have significant role in fruit breeding programmes e.g. Citrus, Mango.

Question 2.
Pollination and seeds formation are very crucial for the fruit formation, Justify.
Answer:

  1. After fertilization, ovary is transformed into fruit, where ovary wall becomes fruit wall, i.e pericarp.
  2. Mature ovules are transformed into seeds after fertilization.
  3. Fertilization is a process where male gametes unites with female gamete to form zygote which develops into embryo.
  4. In pollination process pollen grains carrying non-motile male gamete are transferred on stigma.
  5. Seeds have embryo which germinate into new plant hence the goal of reproduction to create offspring for next generation is achieved. Hence these are the crucial events for fruit formation.

Question 3.
Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?
Answer:

  1. Self incompatibility or self-sterility is a genetic mechanism that prevents germination of pollen on stigma of same flower. This favours cross pollination. E.g. Tobacco.
  2. In pollen-pistil interaction, when pollen grain is deposited on stigma, pistil has the ability to recognize and allow germination of right type of pollen.
  3. Special type of proteins on stigmatic surface determine compatibility or incompatibility.
  4. A physiological mechanism operates to ensure successful germination of compatible pollen.
  5. Compatible pollen absorbs water and nutrients from stigmatic surface that are absent in pollen and then pollen tube emerges which grow-s through style.

Question 4.
Describe three devices by which cross pollination is encouraged in Angiosperms by avoiding self-pollination?
Answer:

  1. Unisexuality, dichogamy, prepotency, heteromorphy and herkogamy are the outbreeding devices.
  2. Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.
  3. Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.
  4. Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. E.g. Calotropis where pollinia are situated below the stigma.

4. Long Answer Questions

Question 1.
Describe the process of double fertilization.
Answer:
Double fertilization:
(1) Out of the two male gametes produced by the male gametophyte in angiosperms, one unites with the female gamete and the other with the secondary nucleus. Since both the male gametes take part in fertilization and fertilization occurs twice, it is called double fertilization.

(2) During double fertilization, the pollen tube on reaching the ovule enters the embryo sac through micropyle and bursts in one of the synergids. Owing to this, the two male gametes contained in the pollen tube, are set free.
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 1

(3) Out of the two male gametes, one unites with the egg or female gamete and the other unites with the secondary nucleus of the embryo sac, forming a triploid or triple fusion nucleus, called the primary endosperm nucleus. The process involving the fusion of one of the male gametes with the egg nucleus, resulting in the formation of a diploid zygote is called syngamy.

(4) The reproductive process in which non-motile male nuclei are carried to the egg cell through a pollen tube is called siphonogamy.

(5) After fertilization, zygote develops into an embryo. Certain changes take place in the ovule leading to the development of a seed.

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 2.
Explain the stages involved in the maturation of microspore into male gametophyte.
OR
Describe the development of male gametophyte before pollination in angiosperms.
OR
Sketch and label male gametophyte in angiosperm.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 2

  1. Microspore or pollen grain is first cell of male gametophyte.
  2. The protoplast of pollen grain divides mitotically to form two unequal cells – a small thin walled generative cell and a large naked vegetative or tube cell.
  3. The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.
  4. The vegetative, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.
  5. In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.
  6. The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

Question 3.
Explain the development of dicot embryo.
Answer:
Development of embryo (dicot) in angio- sperm:
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 3
The oospore undergoes a transverse division to form a large basal cell towards the micropyle and a small apical or terminal cell towards the chalaza of the embryo sac. This two celled structure is called proembryo. The basal cell or suspensor initial undergoes repeated transverse divisions to form a multicellular structure called suspensor. The suspensor pushes the embryo towards the endosperm to draw its nutrition.

  1. The development of embryo from a zygote is called embryogenesis.
  2. The fusion of male gamete and an egg cell during fertilization results in the formation of a diploid zygote. The zygote develops a wall around it and is converted into oospore.
  3. The apical cell or embryonal initial of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
  4. From octant, the lower four cells form hypocotyl and radicle while four cells of upper side form plumule with two cotyledons.
  5. The lowermost cell of suspensor is hypophysis and by its further division forms part of radicle and root cap.
  6. The cells from upper side of octant divide repeatedly to form heart shaped which elongated further to form two lateral cotyledons.
  7. Enlargement of hypocotyl and cotyledon results into curved embryo which appears horse shoe shaped.

Question 4.
Draw a diagram of the L.S of anatropous ovule and list the components of embryo sac and mention their fate after fertilization.
Answer:
Components of Embryo sac.
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 4

  1. Mature embryo sac is 7-celled and 8 nucleate.
  2. Egg apparatus at micropylar end – with 2 synergids and egg cell.
  3. Central cell with secondary nucleus formed by 2 polar nuclei
  4. Antipodal cells at chalazal end – 3 cells.
  5. Pollen tube enters the synergids, Synergids guide the growth of pollen tube towards egg.
  6. Male gamete fuses with female gamete, i.e. syngamy to form zygote which develops into embryo.
  7. One male gamete fuses with secondarynucleus to form primary endosperm nucleus (PEN) which forms endosperm, nutritive tissue for embryo.

5. Fill in the Blanks

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 5
Question 1.
The ……………… collects the pollen grains.
Answer:
biotic agents

Question 2.
The male whorl, called the ……………… produces ………………
Answer:
androecium, pollen grains

Question 3.
The pollen grains represent the ………………
Answer:
male

Question 4.
The ……………… contains the egg or ovum.
Answer:
embryo sac

Question 5.
…………….. takes place when one male gamete and the egg fuse together. The fertilized egg grows vs into seed from which the new plants can grow.
Answer:
Fertilization

Question 6.
The ……………… is the base of the flower to which other floral parts are attached.
Answer:
thalamus

Question 7.
……………… is the transfer of pollen grains from anther of the flower to the stigma of the same or a different flower.
Answer:
Pollination

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 8.
Once the pollen reaches the stigma, pollen tube traverses down the ……………… to the ovary where fertilization occurs.
Answer:
style

Question 9.
The ……………… are coloured to attract the insects that carry the pollen. Some flowers also produce ……………… or ……………… that attracts insects.
Answer:
petals, fragrance, nectar

Question 10.
The whorl ……………… is green that protects the flower until it opens.
Answer:
Calyx.

6. Label the Parts of seed.

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 6
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 7

7. Match the following

Column I (Structure Before seed formation) Column II (Structure After seed formation)
A. Funiculus i. Hilum
B. Scar of Ovule ii. Tegmen
C. Zygote iii. Testa
D. Inner Integument iv. Stalk of Seed
v. Embryo

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

(a) A-v, B-i, C-ii, D-iv
(b) A-iii, B-iv, C-i, D-v
(c) A-iv, B-i, C-v D-ii
(d) A-iv, B-v C-iii, D-ii
Answer:
(c) A-iv, B-i, C-v D-ii

12th Std Biology Questions And Answers:

12th Biology Chapter 7 Exercise Plant Growth and Mineral Nutrition Solutions Maharashtra Board

Class 12 Biology Chapter 7

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 7 Plant Growth and Mineral Nutrition Textbook Exercise Questions and Answers.

Plant Growth and Mineral Nutrition Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 7 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 7 Exercise Solutions

1. Multiple choice questions

Question 1.
Which of the hormone can replace vernalization ?
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene
Answer:
(c) Gibberellins

Maharashtra Board Class 12 Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

Question 2.
The principle pathway of water translocation in angiosperms is ………………..
(a) Sieve cells
(b) Sieve tube elements
(c) Xylem
(d) Xylem and phloem
Answer:
(c) Xylem

Question 3.
Abscissic acid controls ………………..
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Answer:
(b) leaf fall and dormancy

Question 4.
Which is employed for artificial ripening of banana fruits?
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Gibberellin
Answer:
(b) Ethylene

Question 5.
Which of the following is required for stimulation of flowering in plants?
(a) Adequate oxygen
(b) Definite photoperiod
(c) Adequate water
(d) Water and minerals
Answer:
(b) Definite photoperiod

Question 6.
For short day plants, the critical period is ………………..
(a) light
(b) dark/night
(c) UV rays
(d) Both (a) and (c)
Answer:
(b) dark/night

Question 7.
Which of the following is NOT day neutral plant?
(a) Tomato
(b) Cotton
(c) Sunflower
(d) Soybean
Answer:
(d) Soybean

Question 8.
Essential macro elements are ………………..
(a) manufactured during photosynthesis
(b) produced by enzymes
(c) obtained from soil
(d) produced by growth hormones
Answer:
(c) obtained from soil

Question 9.
Function of Zinc is ………………..
(a) closing of stomata
(b) biosynthesis of 3-IAA
(c) synthesis of chlorophyll
(d) oxidation of carbohydrates
Answer:
(b) biosynthesis of 3-LAA

Question 10.
Necrosis means ………………..
(a) yellow spot on the leaves
(b) death of tissue
(c) darkening of green colour in leaves
(d) wilting of leaves
Answer:
(b) death of tissue

Question 11.
Conversion of nitrates to nitrogen is called ………………..
(a) ammonification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(d) denitrification

Question 12.
How many molecules of ATP are required to fix one molecule of nitrogen?
(a) 12
(b) 20
(c) 6
(d) 16
Answer:
(d) 16

2. Very short answer questions

Question 1.
Enlist the phases of growth in plants.
Answer:
The three phases of growth are phase of cell division, phase of cell enlargement and phase of cell maturation.

Question 2.
Give full form of IAA.
Answer:
Full form is Indole Acetic Acid.

Maharashtra Board Class 12 Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

Question 3.
What does it mean by ‘open growth’?
Answer:
In plants the growth is indeterminate and takes place throughout the life at specific regions having meristems.

Question 4.
Plant stress hormone.
Answer:
Abscissic acid.

Question 5.
What is denitrification?
Answer:
Anaerobic bacteria can convert nitrates of soil back into nitrogen gas. That process performed by denitrifying bacteria is denitrification.

Question 6.
Bacteria responsible for conversion of nitrite to nitrate.
Answer:
Nitrobacter.

Question 7.
What is the role of gibberellins in rosette plants?
Answer:
In rosette plants like beet and cabbage, bolting, i.e. elongation of internodes before flowering is observed due to effect of gibberellins.

Question 8.
Vernalization
Answer:
The response of plant to the influence of low temperature on flowering in plants is called vernalization.

Question 9.
Photoperiodism
Answer:
The response of plant to the influence of light for initiation of flowering is known as photoperiodism.

Question 10.
What is grand period of growth?
Answer:
There are three phases of growth and the total time required for all phases to occur is called grand period of growth.

3. Short Answer Questions

Question 1.
(i) Differentiation
Answer:

  1. It is a process of maturation of cells derived from apical meristems.
  2. Differentiation is a permanent change in structure and function of cells that leads to its maturation.
  3. Cell undergoes major anatomical and physiological change during differentiation process.
  4. In hydrophytic plants parenchyma cells develop large schizogenous cavities which help them in aeration, buoyancy and mechanical support.

(ii) Redifferentiation
Answer:

  1. It is a process in which cells produced by de-differentiation lose their capacity of division and become mature.
  2. The cells mature to perform specific function.
  3. Interfascicular cambium is formed by process of dedifferentiation loses its capacity to divide.
  4. Secondary xylem and secondary phloem is formed form this cambium in vascular cylinder.

Question 2.
Arithmetic growth and Geometric growth
Answer:

Arithmetic growth Geometric growth
1. In arithmetic growth only one daughter cell continues to divide, while the other undergoes differentiation and maturation. 1. In geometric growth both the daughter cells continue to divide and redivide again and again.
2. Rate of growth is constant. 2 Rate growth is initially slow but later on rapid rate.
3. Linear curve is obtained. 3. Exponential curve is obtained.
4. Mathematical expression is
Lt = Lo + rt whereLt = length of time ‘t’
Lo = Length at time zero
rt = growth rate, t = time of growth
4. Mathematical expression is
Wt = Woe rt where,
Wt = final size,
Wo = initial size, r = growth rate, t = time of growth E = base of natural logarithm
5. e.g. Elongation of root 5. e.g. Divisions of zygote during embryo development.

Question 3.
Enlist the role and deficiency symptoms of: (a) nitrogen (b) phosphorus (c) potassium.
Answer:
(a) Nitrogen:
Role : Constituent of proteins as amino acids, nucleic acids, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.
Deficiency symptoms : stunted growth and chlorosis.

(b) Phosphorus:
Role : Constituent of cell membrane, certain proteins, nucleic acids and nucleotides, required for all phosphorylation reactions.
Deficiency symptoms : Poor growth, leaves dull green

(c) Potassium :
Role : Determination of anion – cation balance in cell, necessary for protein synthesis, involved in formation of cell membrane, opening and closing of stomata, activates enzymes, helps in maintenance of turgidity of cells.
Deficiency symptom : Yellow edges in leaves, premature death.

Maharashtra Board Class 12 Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

Question 4.
What is short day plant? Give any two examples.
Answer:
The plants which flower when the day length or light period is shorter than the critical photoperiod are called short day plants or SDP
SDPs usually flower during winter and late summer.
Examples – Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) and Cocklebur (Xanthium).

Question 5.
What is vernalization? Give its significance.
Answer:
A low temperature or chilling treatment that induces early flowering in plants is known as vernalization.

Significance:

  1. Due to chilling treatment crops can be produced earlier.
  2. Crops can be grown in areas where they do not grow naturally.

4. Long answer questions

Question 1.
Explain sigmoid growth curve with the help of diagram.
Answer:

  1. When growth occurs in plants three distinct phases of growth are noticed.
  2. Phase of cell formation is first phase where meristematic cells divide and new cells added.
  3. In phase of cell enlargement newly formed cells elongate and with turgidity there is cell enlargement.
  4. In phase of cell maturation cells get differentiated.
  5. When we compare the growth rate it differs in these three phases.
  6. In first phase or lag phase it is slow, while in log phase or exponential phase, growth rate accelerates and it reaches maximum.
  7. In stationary phase of maturation growth rate slows down and comes to steady state.
  8. When this changing rate of growth is plotted against time duration in a graph a sigmoid or S-shaped growth curve is obtained.

Question 2.
Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.
Answer:

  1. Effect of light duration on flowering of plants is known as photoperiodism.
  2. Depending on photoperiodic response, plants are categorised into three types – Short day plants, long day plants and day neutral plants.

1. Short day plants : Plants that flower under short day length conditions are called short day plants. Plants such as Dahlia, Xanthium, Soybean, Aster, Tobacco and Chrysanthemum are short day plants or SDR. Short day plants require a long uninterrupted dark period for flowering. Therefore, they are also called long night plants.

2. Long day plants : Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants or LDP Long day plants require a short dark or night period for flowering. Hence, they are also called short night plants. Plants such as radish, spinach, wheat, poppy, cabbage, pea, sugar beet, etc. are long day plants.

3. Day neutral plants : Plants in which the flowering is not affected by the day length period are called day neutral plants or DNP or photoneutral plants. Plants such as cucumber, sunflower, cotton, balsam, maize, tomato, etc. are day neutral plants.

Question 3.
Explain biological nitrogen fixation with example.
Answer:

  1. Conversion of atmospheric nitrogen into nitrogenous salts to make it available to plants for its update is described as nitrogen fixation.
  2. When living organisms are involved in nitrogen fixation process it is known as biological nitrogen fixation.
  3. The process is mainly carried out by prokaryotic organisms, i.e. different kinds of bacteria present in soil.
  4. The nitrogen fixing organisms are known as diazotrophs or nitrogen fixers and about 70% nitrogen is fixed by them.
  5. The nitrogen fixers are either free living bacteria or symbiotic associated with other higher plants e.g. Rhizobium.
  6. The cyanobacteria have specialized cells heterocysts which help in process of nitrogen fixation.
  7. Nitrogen fixation is high energy requiring process and 16 ATP molecules are needed for fixation of one molecule of nitrogen to ammonia.
  8. Soil bacteria like Nitrosomonas, Nitrosocyccus convert ammonia to nitrate and the Nitrobacter convert nitrite to nitrate. This is known as nitrification, biological oxidation.
  9. These bacteria are chemoautotrophic and utilize these processes for their metabolism.
  10. Fabaceae plants like pea, bean have root nodules which harbour symbiotic bacterium Rhizobium which fixes nitrogen. It is host specific, soil bacterium, Nitrogen is made available to host plant.

Maharashtra Board Class 12 Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

Question 4.
Write on macro and micro nutrients required for plant growth.
Answer:

  1. Plants absorb mineral nutrients from their surroundings.
  2. For a proper growth of plants about 35 to 40 different elements are required.
  3. Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO4, Sulphur as SO42- etc.
  4. Based on their requirement in quantity, they are classified as major nutrients or macronutrients and those needed in small amounts Eire minor or micronutrients.
  5. Macroelements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O, R Mg, N, K, S and Ca. – Ca pectate cell wall component, Mg component of chlorophyll.
  6. C, H, O are non-mineral major elements obtained from air and water e.g. CO2 is source of carbon, Hydrogen from water.
  7. Microelements are required in traces as they mainly have catalytic role as co-factors or activators of enzymes.
  8. Microelements may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
  9. The important micronutrients for plant growth are Mn, B, Cu, Zn, Cl.

12th Std Biology Questions And Answers:

12th Biology Chapter 13 Exercise Organisms and Populations Solutions Maharashtra Board

Class 12 Biology Chapter 13

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 13 Organisms and Populations Textbook Exercise Questions and Answers.

Organisms and Populations Class 13 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 13 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 13 Exercise Solutions

1. Multiple choice questions

Question 1.
Which factor of an ecosystem includes plants, animals and microorganisms?
(a) Biotic factor
(b) Abiotic factor
(c) Direct factor
(d) Indirect factor
Answer:
(a) Biotic factor

Question 2.
An assemblage of individuals of different species living in the same habitat and having functional interactions is ……………….
(a) Biotic community
(b) Ecological niche
(c) Population
(d) Ecosystem
Answer:
(a) Biotic community

Maharashtra Board Class 12 Biology Solutions Chapter 13 Organisms and Populations

Question 3.
Association between sea anemone and Hermit crab in gastropod shell is that of ………………..
(a) Mutualism
(b) Commensalism
(c) Parasitism
(d) Amensalism
Answer:
(b) Commensalism

Question 4.
Select the statement which explains best parasitism.
(a) One species is benefited.
(b) Both the species are benefited.
(c) One species is benefited, other is not affected.
(d) One species is benefited, other is harmed.
Answer:
(d) One species is benefited, other is harmed.

Question 5.
Growth of bacteria in a newly inoculated agar plate shows ………………….
(a) exponential growth
(b) logistic growth
(c) Verhulst-Pearl logistic growth
(d) zero growth
Answer:
(c) Verhulst-Pearl logistic growth

2. Very short answer questions.

Question 1.
Define the following terms
a. Commensalism
Answer:
The interaction between two species in which one species gets benefits and the other is neither harmed nor benefited is called commensalism.

b. Parasitism
Answer:
The interaction between two species in which one parasitic species derives benefit from the other host species by harming it is called parasitism.

c. Camouflage
Answer:
Camouflage is the disguising colouration or behaviour to merge with the surrounding so that prey or predator can remain hidden.

Question 2.
Give one example for each
a. Mutualism
b. Interspecific competition
Answer:
a. Lichen is composed of alga (cyanobacteria) and fungus. They cannot survive independently. Their association is mutualistic alga synthesises food by photosynthesis and fungus does the absorption of moisture.

b. Leopard and lion competing for a same prey. Sheep and cow competing for grazing in the same land.

Maharashtra Board Class 12 Biology Solutions Chapter 13 Organisms and Populations

Question 3.
Name the type of association:
a. Clown fish and sea anemone
b. Crow feeding the hatchling of Koel
c. Humming birds and host flowering plants
Answer:
a. Commensalism
b. Brood parasitism
c. Mutualism

Question 4.
What is the ecological process behind the biological control method of managing with pest insects?
Answer:

  1. Pest insects act as prey to predator birds or frogs.
  2. The biological control method consists of releasing the predators in the farms so that they can control the pest population in the natural way.
  3. This also eliminates the use of chemical pesticides.
  4. Frogs are natural predators of locust, therefore the population of this hazardous insect is controlled by frogs and the produce from agricultural farm can be saved.

Protocooperation:

  1. Protocooperation is a type of population interaction where two species interact with each other.
  2. Both are benefited but they have no need to interact with each other.
  3. They can survive and grow even in the absence of other species.
  4. Therefore this interaction is purely for the gain that they receive in such type of interaction.
  5. The interaction that occurs can be between different kingdoms.

3. Short answer questions.

Question 1.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown in to moist place.

Question 2.
If a marine fish is placed in a fresh water aquarium, will it be able to survive? Give reason.
Answer:
Marine fish has its own osmoregulation which is different from the osmoregulation seen in fresh water fish. In marine water, the ambient salinity is more than the concentration of ions in the body. But in fresh water reverse is the case. Therefore, marine fish has different machinery to cope up with high saline environment. Therefore, it cannot survive in fresh water as its osmoregulation is not possible in less saline waters.

Question 3.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown into moist place.

Maharashtra Board Class 12 Biology Solutions Chapter 13 Organisms and Populations

Question 4.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:

  1. Orchid is an epiphyte. It gets the support from the mango tree. But it does not cause any harm to the mango tree.
  2. Mango tree does not derive any benefit from this association. Therefore, this interaction is of type of commensalism.

Question 5.
Distinguish between the following:
a. Hibernation and Aestivation
Answer:

Hibernation Aestivation
1. Hibernation is winter sleep shown by some warm-blooded and some cold-blooded animals. 1. Aestivation is the type of summer sleep, shown by cold-blooded animals.
2. It is for the whole winter. 2. It is of short duration.
3. The animals look out for the warmer place to enter into hibernation. 3. Animals search for the moist, shady and cool place to sleep.
4. Metabolic activities of hibernators slowdown in this dormant stage. 4. Metabolic activities of aestivators remain low during aestivation period.
5. Hibernation helps in maintaining the body temperature and prevents any internal body damage due to low temperatures.

E.g. Bats, birds, mammals, insects, etc. show hibernation.

5. Aestivation helps in maintaining the body temperature by avoiding the excessive water loss and thus prevents any internal body damaged due to high temperatures.

E.g. Bees, snails, earthworms, salamanders, frogs, earthworms, crocodiles, tortoise, etc. show aestivation.

b. Ectotherms and Endotherms
Answer:

Ectotherms Endotherms
1. Ectotherms do not have ability to generate heat in the body. 1. Endotherms possess the ability to generate their own body heat.
2. Ectotherms depend on the environmental sources to heat their bodies. E.g sunlight. 2. Endotherms do not depend upon outside sources to generate heat.
3. Most ectotherms are confined to warmer parts of the world. 3. Endotherms inhabit coldest parts of the earth.
4. Body temperature of ectotherms fluctuate according to ambient temperature. 4. Body temperatures of endotherms remain constant and do not show fluctuations as per ambient temperatures.
5. Metabolic rate of ectotherms is low.

E.g. Amphibians and reptiles.

5. Metabolic rate of endotherms is high.

E.g. Mammals and birds

c. Parasitism and Mutualism
Answer:

Parasitism Mutualism
1. Parasitism is the relationship where only one organism receive benefits, while the other is harmed in return. 1. Mutualism is the relationship where both the organisms of distinct species are benefited.
2. Parasite cannot survive without host but if the host is overexploited then parasite too dies. 2. Both the species are dependent on each other for their benefits and survival.
3. Parasitism can be facultative or obligatory. 3. Mutualism is obligatory relationship.
4. Parasitism is a negative interaction. 4. Mutualism is a positive interaction.

Question 6.
Write a short note on
a. Adaptations of desert animals
Answer:

  1. Animals which are well-adapted to live in deserts are called xerocoles. These animals show adaptations for water conservation or heat tolerance.
  2. These animals show low basal metabolic rate. They obtain moisture from succulent plants and rarely drink water. E.g Gazella and Oryx.
  3. Desert animals like camel produce concentrated urine and dry dung.
  4. Many other hot desert animals are nocturnal, seeking out shade during the day or dwelling underground in burrows.
  5. Smaller animals from desert, emerge from their burrows at night.
  6. Mammals living in cold deserts have developed greater insulation through warmer body fur and insulating layers of fat beneath the skin.
  7. Few adaptations to desert life are unable to cool themselves by sweating so they shelter during the heat of the day. Many desert reptiles are ambush predators and often bury themselves in the sand, waiting for prey to come within range.
  8. Other animals have bodies designed to save water. Scorpions and wolf spiders have a thick outer covering which reduces moisture loss. The kidneys of desert animals concentrate urine, so that they excrete less water.

Maharashtra Board Class 12 Biology Solutions Chapter 13 Organisms and Populations

b. Adaptations of plants to water scarcity
Or
Adaptations in desert plants.
Answer:

  1. Thick cuticle on their leaf surfaces
  2. Stomata of desert plants is sunken that is it is in deep pits to minimize loss of water through transpiration.
  3. Desert plants also have a special photosynthetic pathway (CAM -Crassulacean acid metabolism) that enables their stomata to remain closed during daytime.
  4. Some desert plants like Opuntia, have their leaves reduced or they are modified to spines. Loss of leaf surface helps in prevention of transpiration.
  5. Photosynthetic function is taken over by the flattened stems called as phylloclade.

c. Behavioural adaptations in animals
Answer:

  1. Behavioural responses to cope with variations in their environment are shown by few animals.
  2. Desert lizards manage to keep their body temperature fairly constant by behavioural adaptations. They bask in the sun and absorb heat, when their body temperature drops below the comfort zone, but move into shade, when the ambient temperature starts increasing. Even snakes also show basking during winter months.
  3. Since they are ectothermic, this kind of behaviour saves them from extreme temperatures.
  4. Many smaller animals show burrowing behaviour to adapt to the temperature extremes.
  5. Some species burrow into the sand to hide and escape from the heat.
  6. Migrations shown by the birds and mammals are also behavioural responses for adapting to severe winter temperatures.

Question 7.
Define Population and Community.
Answer:
Population:
Group of organisms belonging to same species that can potentially interbreed with each other and live together in a well-defined geographical area by sharing or competing for similar resources, is called population.

Community:
Several populations of different species in a particular area makes a community.

4. Long answer questions.

Question 1.
With the help of suitable diagram, describe the logistic population growth curve.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 13 Organisms and Populations 1

  1. Naturally all populations of any species always have limited resources to permit exponential growth. Due to this there is always competition between individuals for limited resources. The most fit organisms succeed by survival and reproduction.
  2. A given habitat has enough resources to support a maximum possible number, but beyond a particular limit the further growth is impossible.
  3. This limit is called nature’s carrying capacity (K) for that species in that habitat.
  4. A population growing in a habitat with limited resources show following phases in a sequential manner, (a) A lag phase (b) Phase of acceleration (c) Phase of deceleration (d) An asymptote, when the population density reaches the carrying capacity.
  5. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth.
  6. Since resources for growth for most animal populations are finite and become limiting sooner or later, the logistic growth model is considered as a more realistic one.
  7. Logistic growth thus always shows sigmoid curve.

Maharashtra Board Class 12 Biology Solutions Chapter 13 Organisms and Populations

Question 2.
Enlist and explain the important characteristics of a population.
Answer:
Important characteristics of a population are as follows:
1. Natality:

  1. Natality is the birth rate of a population. Due to increased natality the population density rises.
  2. Natality is a crude birth rate or specific birth rate.
  3. Crude birth rate : Number of births per 1000 population/year gives crude birth rate. Crude birth rate is helpful in calculating population size.
  4. Specific birth rate : Crude birth rate is relative to a specific criterion such as age. E.g. If in a pond, there were 200 carp fish and their population rises to 800. Then, taking the current population to 1000, the birth rate becomes 800/200 = 4 offspring per carp per year. This is specific birth rate.
  5. Absolute Natality : The number of births under ideal conditions when there is no competition and the resources such as food and water are abundant, then it give absolute natality.
  6. Realized Natality : The number of births under different environmental pressures give realized natality. Absolute natality will be always more than realized natality.

2. Mortality:

  1. Mortality is the death rate of a population. It gives a measure of the number of deaths in a particular population, in proportion to the size of that population, per unit of time.
  2. Mortality rate is typically expressed in deaths per 1,000 individuals per year.
    A mortality rate of 9.5 (out of 1,000) in a population of 1,000 would mean 9.5 deaths per year in that entire population or 0.95% out of the total.
  3. Absolute Mortality : The number of deaths under ideal conditions when there is no competition, and all the resources such as food and water are abundant, then it gives absolute mortality.
  4. Realized Mortality : The number of deaths under environmental pressures come into play gives realized mortality.
  5. It must be remembered that absolute mortality will always be less than realized mortality.

3. Density:
The density of a population in a given habitat during a given period fluctuates due to changes in four basic processes, viz.

  1. Natality i.e. birth rate (The number of births during a given period in the population that are added to the initial density).
  2. Mortality i.e. death rate (The number of deaths in the population during a given period).
  3. Immigration i.e. number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
  4. Emigration i.e. the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.
  5. Natality and immigration increase in population density whereas mortality and emigration decrease it.

4. Sex ratio : Ratio of the number of individuals of one sex (male) to that of the other sex (female) is called sex ratio. In nature male, female ratio is always 1 : 1. This 1 : 1 ratio is called evolutionary stable strategy of ESS for each population.

5. Age distribution and age pyramid : This parameter is important for human population. Each population is composed of individuals of different ages. The age distribution is plotted for the population, the resulting structure is called an age pyramid. For making the age pyramid, the entire population is divided into three age groups as Pre-Reproductive (age 0-14 years), Reproductive (age 15-44 years) and Post-reproductive (age 45 -85+ years).

6. Growth : Growth of a population causes rise in its density. The size and density are dynamic parameters as they keep on changing with time, and various factors including food, predation pressure and adverse weather. From the density, one comes to know if the population is flourishing or declining.

12th Std Biology Questions And Answers:

12th Biology Chapter 5 Exercise Origin and Evolution of Life Solutions Maharashtra Board

Class 12 Biology Chapter 5

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 5 Origin and Evolution of Life Textbook Exercise Questions and Answers.

Origin and Evolution of Life Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 5 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 5 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Who proposed that the first form of life could have come from pre-existing non-living organic molecules?
(a) Alfred Wallace
(b) Oparin and Haldane
(c) Charles Darwin
(d) Louis Pasteur
Answer:
(b) Oparin and Haldane

Question 2.
The sequence of origin of life may be
(a) Organic materials – inorganic materials – Eobiont – colloidal aggregates – cell.
(b) Inorganic materials – organic materials – colloidal aggregates – Eobiont – cell.
(c) Organic materials – inorganic materials – colloidal aggregates – cell.
(d) Inorganic materials – organic materials – Eobiont – colloidal aggregates – cell.
Answer:
(b) Inorganic materials – organic materials- colloidal aggregates – Eobiont – cell.

Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life

Question 3.
In Hardy-Weinberg equation, the frequency of homozygous recessive individual is represented by-
(a) p²
(b) pq
(c) q²
(d) 2pq
Answer:
(c) q²

Question 4.
Select the analogous organs.
(a) Forelimbs of whale and bat
(b) Flippers of dolphins and penguin
(c) Thorn and tendrils of bougainvillea and Cucurbita
(d) Vertebrate hearts or brains
Answer:
(b) Flippers of dolphins and penguin

Question 5.
Archaeopteryx is known as missing link because it is a fossil and share characters of both
(a) Fishes and amphibians
(b) Annelida and Arthropoda
(c) Reptiles and birds
(d) Chordates and non-chordates
Answer:
(c) Reptiles and birds

Question 6.
Identify the wrong statement regarding evolution.
(a) Darwin’s variations are small and directional.
(b) Mutations are random and non- directional.
(c) Adaptive radiations leads to divergent evolution.
(d) Mutations are non-random and directional.
Answer:
(d) Mutations are non-random and directional

Question 7.
Gene frequency in a population remain constant due to ……………….
(a) Mutation
(b) Migration
(c) Random mating
(d) Non-random mating
Answer:
(c) Random mating

Question 8.
Which of the following characteristic is not : shown by the ape?
(a) Prognathous face
(b) Tail is present
(c) Chin is absent
(d) Forelimbs are longer than hind limbs
Answer:
(b) Tail is present

Question 9.
………………. can be considered as connecting link between ape and man.
(a) Australopithecus
(b) Homo habilis
(c) Homo erectus
(d) Neanderthal man
Answer:
(a) Australopithecus

Question 10.
The Cranial capacity of Neanderthal man was ……………….
(a) 600 cc
(b) 940 cc
(c) 1400 cc
(d) 1600 cc
Answer:
(c) 1400 cc

2. Very short answer questions

Question 1.
Define
(i) Gene pool
Answer:
The sum total of genes of all individuals of interbreeding population or Mendelian population is called gene pool.

(ii) Gene frequency
Answer:
The proportion of an allele in the gene pool as compared with other alleles at the same locus is termed as gene frequency.

(iii) Organic evolution
Answer:
Organic evolution can be defined as slow, gradual, continuous and irreversible changes through which the present-day complex forms of the life developed (or evolved) from their simple pre-existing forms.

(iv) Population
Answer:
All individuals of the same species form a group which is called a population.

(v) Speciation
Answer:
Formation of new species from the pre-existing single group of organisms is called speciation.

Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life

Question 2.
What is adaptive radiation?
Answer:
The process of evolution which results in transformation of original species to many different varieties is called adaptive radiation.

Question 3.
If the variation occurs in population by chance alone and not by natural selection and bring change in frequencies of an allele, what is it called?
Answer:
If the variation occurs in population by chance alone and not by natural selection to bring change in frequencies of an allele, it is called genetic drift.

Question 4.
State the Hardy-Weinberg equilibrium law.
Answer:
The Hardy-Weinberg equilibrium law states that at equilibrium point both the allelic frequency and genotypic frequency remain constant from generation to generation, in the diploid, sexually reproducing, large, free interbreeding population in which mating is random and there is absence of any other factors that change the allele frequency.

Question 5.
What is homologous organs?
Answer:
Homologous organs are those organs, which are structurally similar but perform different functions.

Question 6.
What is vestigial organ?
Answer:
Vestigial organs are imperfectly developed and non-functional organs which are in degenerate form, they may be functional in some related and other animals or in ancestor.

Question 7.
What is the scientific name of modern man?
Answer:
Homo sapiens sapiens is the scientific name of modern man.

Question 8.
What is coacervate?
Answer:
Coacervates are colloidal aggregations of hydrophobic proteins and lipids which grew in size by taking up material from surrounding aqueous medium.

Question 9.
Which period is known as ‘age of Reptilia’?
Answer:
Jurassic period from Mesozoic era is known as age of Reptilia.

Question 10.
Name the ancestor of human which is described as man with ape brain.
Answer:
Australopithecus, the ancestor of human which is described as man with ape brain.

Short Answer Questions

Question 1.
Genetic drift.
Answer:

  1. Genetic drift is random, directionless fluctuation that takes place in allele frequency.
  2. It occurs by pure chance, in small sized population.
  3. Genetic drift becomes an evolutional factor as it can change the gene frequency.
  4. Sewall wright has given this concept and hence it is also known as Sewall wright effect.
  5. Due to genetic drift, some alleles of a population are lost or reduced by chance and some others may be increased.
  6. Some time, a few individuals become isolated from the large population and they produce new population in new geographical area.
  7. Genetic drift is also called founders’ effect because original drifted population becomes ‘founders’ in the new area.
    E.g. Non-adaptive character of huge horns in Antelope is fixed due to genetic drift.

Question 2.
Enlist the different factors that are responsible for changing gene frequency.
Answer:
Gene flow, genetic drift, gene mutations, chromosomal aberrations such as deletion, duplication, inversion and translocation, genetic recombinations, natural selection, isolation are some of the factors which are responsible for changing the gene frequency.

Question 3.
Draw a graph to show that natural selection leads to disruptive change.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life 1

Question 4.
Significance of fossils
Answer:

  1. Fossils are studied under palaeontology. They are used in reconstruction of phylogeny.
  2. Fossil study helps in studying various forms and structures of extinct animals.
  3. By understanding the structure of fossil, record of missing link between two groups of organisms can be deduced.
  4. By studying fossils various body forms and their evolution can be understood. They also help to understand the habit and habitat.
  5. Some fossils provide the evolutionary evidences such a connecting links.

Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life

Question 5.
Write the objections to Mutation theory of Hugo de Vries.
Answer:
Objections to Mutation Theory:

  1. Hugo de Vries observed the large and discontinuous variation. But these were chromosomal aberrations. Only gene mutations usually bring about minor changes.
  2. Rate by which mutations take place is very slow as compared to the requirement of evolution.
  3. Chromosomal aberrations are very unstable.
  4. The organisms with chromosomal aberration are usually sterile and thus chromosomal aberrations have little significance in evolution.

Question 6.
What is disruptive selection? Give example.
Answer:
Disruptive selection:

  1. The natural selection that disrupts the mean characters of the population, is called disruptive selection.
  2. Greater number of individuals acquire peripheral character value at both ends of the distribution curve. E.g. Finches with large size or small size, both will be selected.
  3. Extreme phenotypes are selected in evolutionary process and intermediate forms are eliminated.
  4. When distribution curve is plotted it shows two peaks for two extremes.
  5. Disruptive selection is rare because, nature always tries to balance the characters.
  6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

Example of disruptive selection:
African seed cracker finches are types of seed-feeder birds which have different sizes of beak. The seeds available to them were of small and large sized. Large beak sized birds feeds on large seeds while small beak sized birds feed on small seeds.

Such large and small birds thus thrive well. However, intermediate beak sized birds are unable to feed on either type of seeds so they starve and their population was decreased gradually. Natural selection eliminated them and thus the population of finches appear disrupted.

4. Match the columns

Question 1.

Column I Column II
(1) August Weismann (a) Mutation theory
(2) Hugo de Vries (b) Germplasm theory
(3) Charles Darwin (c) Theory of acquired characters
(4) Lamarck (d) Theory of natural selection

Answer:

Column I Column II
(1) August Weismann (b) Germplasm theory
(2) Hugo de Vries (a) Mutation theory
(3) Charles Darwin (d) Theory of natural selection
(4) Lamarck (c) Theory of acquired characters

5. Long Answer Questions

Question 1.
Would you consider wings of butterfly and bat as homologous or analogous and why?
Answer:
Wings of butterfly are made up of chitin. They neither have bones, nor muscles in the wings. The bat’s wings are actually patagium. They have muscles and bones just as those seen in all vertebrate limb series. Therefore, these two examples cannot be homologous. However, both the animals use the wings for flight. This is an indication that their function is similar but structure is different, hence they are analogous organs.

Question 2.
What is adaptive radiation? Explain with suitable example.
Answer:

  1. Adaptive radiation is the process of evolution which results in transformation of original species to many different varieties.
  2. The well-known example of adaptive radiation is Darwin’s Finches. When Charles Darwin went on his voyage to Galapagos islands, he noticed finches which is a variety of small birds.
  3. According to Darwin’s observations, the American main land species of finches was the original one which must have migrated to the different islands of Galapagos.
  4. Since environmental conditions here were different, they adapted in various ways to the differing environmental conditions of these islands.
  5. Original bird had a beak suited for eating seeds, but the changed feeding pattern has changed the shape of beaks too. Some birds also show altered beaks for insectivorous mode. Thus, this demonstrated adaptive radiation.
  6. Adaptive radiation in Australian Marsupials is also well studied. In Australia, there are many marsupial mammals who evolved from common ancestor.
  7. Adaptive radiation leads to divergent evolution.

Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life

Question 3.
By talking industrial melanism as one example, explain the concept of natural selection.
OR
Explain natural selection in action by quoting the example of industrial melanism.
Answer:
1. Industrial melanism is the best example of natural selection which was studied by Kettlewell. In U.K. there are two varieties of peppered moths, Biston betularia and Biston carbonaria.

2. Before industrialisation, in Great Britain, Biston betularia were more in number than Biston carbonaria. B. Betularia is greyish white while B.carbonaria is melanic form.

3. These nocturnal moths rest on tree trunk during day. White-winged moth can camouflage well with the lichen covered whitish barks of trees. They thus escaped the attention of the predatory birds. But at the same time melanic forms were visible due to white barks of the trees. Their number was thus reduced as they were preyed upon by birds.

4. Later there was an industrial revolution, which ultimately resulted in air pollution causing dark soot to settle on the barks of the trees. Lichens too were destroyed and the melanic forms were now at advantage. Melanic forms could camouflage with black tree trunks and their number increased. White-winged moth become clearly seen in changed colours of the trees and thus they were easily caught by predatory birds. This caused decrease in their number.

5. Natural selection thus acted in changed environmental conditions and helped in the establishment of a phenotypic traits. The changed traits were more adaptive and hence were selected. Natural selection encourages those genes or traits that assure highest degree of adaptive efficiency between population and its environment.

Question 4.
Describe the Urey and Miller’s experiment.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life 2
1. Urey and Miller performed an experiment to prove Oparin’s theory of chemical evolution.

2. They selected a spark discharge apparatus that consisted of closed system of glass having tungsten electrodes, flask for water boiling, a side tube connected to a vacuum pump, a cooling jacket and U-shaped trap.

3. The entire apparatus was first evacuated and made sterile and pre-biotic atmosphere was created in it.

4. The flask was filled with some water and mixture of methane, ammonia and hydrogen in the ratio of 1 : 2 : 2 were slowly passed through the stopcock, without allowing air.

5. Heat was supplied to the flask at very low temperature causing water to boil. The flask simulated the ocean present on primitive earth. Process of evaporation and precipitation was simulated by using heating mantle and condenser respectively.

6. Water vapours along with other gases were circulated continuously through continuous electric sparks. These sparks were given to the mixture for several days causing the gases to interact. This too simulated lightning.

7. Mixture of CH4, NH3 and H2 gases passed through a condenser and was condensed to liquid.

8. The liquefied mixture was collected in the U-shaped trap, present at the bottom of the apparatus. It was found that variety of simple organic compounds (urea, amino acids, lactic acid and sugars) were formed in the apparatus.

This experiment provides the evidence in support to the fact that simple molecules present in the earth’s early atmosphere combined to form the organic building blocks of life.

Question 5.
What is Isolation? Describe the different types of reproductive isolations.
Answer:
1. Isolation means separation of the population of a particular species into smaller units. The organisms belonging to these subunits are prevented from interbreeding due to some barrier. These barriers are called isolating mechanisms.

2. They prevent the genetic exchange and gene flow.

3. Due to isolating mechanisms in nature the divergence among organisms takes place gradually leading to speciation. The isolating mechanisms are of two types namely, geographical isolation and reproductive isolation.

I. Geographical Isolation : The barrier in the form of physical distance or geographical barrier is called geographical isolation. The original population gets divided into two or more groups by geographical barriers such as river, ocean, mountain, glacier, etc. Organisms cannot cross the barriers on their own and hence interbreeding is prevented between isolated groups.

The separated groups experience different environmental factors and they acquire new traits by mutations. The separated populations develop distinct gene pool and they do not interbreed. Each subgroup then evolves differently which results into formation of new species. E.g. Darwin’s Finches, African elephant, Loxodonta and Indian elephant, Elephas.

Maharashtra Board Class 12 Biology Solutions Chapter 5 Origin and Evolution of Life

II. Reproductive Isolation : Two populations may be occupying the same area, they may not be separated by geographical barrier, but then also they are reproductively isolated. Such reproductive isolation occurs due to change in genetic material, gene pool and structure of genital organs. Such differences prevent interbreeding between population. Such isolation later leads to speciation.

III. Different types of reproductive isolations : Reproductive isolation is of two types, viz. pre-zygotic and post-zygotic isolating mechanisms.

  1. Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.
  2. By various mechanisms the two groups remain isolated.
  3. In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
  4. Thus the populations remain isolated without the actual genetic exchange.

Question 6.
What is Genetic variations? Explain the different factors responsible for genetic variations.
Answer:
Genetic variations : The change in gene and gene frequencies is known as genetic variation. Genetic variations are caused by following factors:
(i) Mutations : Sudden permanent heritable change is called mutation. Mutation can occur in the gene, in the chromosome structure and in chromosome number. Mutation that occurs within the single gene is called point mutation or gene mutation. This leads to the change in the phenotype of the organism, causing variations.

(ii) Genetic recombination : In sexually reproducing organisms, during gamete formation, exchange of genetic material occurs between non-sister chromatids of homologous chromosomes. This is called crossing over. It produces new genetic combinations which result in variation. Fertilization between opposite mating gametes leads to various recombinations resulting into the phenotypic variations. These result in change in the frequencies of alleles.

(iii) Gene flow : Gene flow is movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Gene flow also alters gene frequency causing evolutionary changes.

(iv) Genetic drift : Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift. For example, when the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc. elimination of particular alleles from a population becomes possible. Smaller populations have greater chances for genetic drift. It results in the change in the gene frequency. Genetic drift is also an important factor for evolutionary change.

(v) Chromosomal aberrations : The structural, morphological change in chromosome due to rearrangement of genes is called chromosomal aberrations. Due to changes in the gene arrangement or gene sequence variations are caused.

6. Complete the chart

Era Dominating group of animals
1. Coenozoic ————–
2. ————- Reptiles
3. Palaeozoic ————-
4. ———— Lower Invertebrates

Answer:

Era Dominating group of animals
1. Coenozoic Mammals
2. Mesozoic Reptiles
3. Palaeozoic Insects, Fishes, Amphibians
4. Proterozoic Lower Invertebrates

12th Std Biology Questions And Answers:

12th Biology Chapter 6 Exercise Plant Water Relation Solutions Maharashtra Board

Class 12 Biology Chapter 6

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Plant Water Relation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 6 Exercise Solutions

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

  1. The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
  2. Passive absorption is the chief method of absorption (98%).
  3. There is no expenditure of energy in passive absorption.
  4. Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
  5. It occurs during daytime when there is active transpiration.
  6. Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
  7. Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
  8. Active absorption may be osmotic or non- osmotic type.
  9. For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

  1. Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
  2. The movement is against the gravity and described as ascent of sap.
  3. The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
  4. Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
  5. Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

  1. The loss of water in the form of vapour is called transpiration.
  2. Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
  3. Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
  4. Opening and closing of stomata is controlled by turgidity of guard cells.
  5. When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
  6. The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
  7. At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
  8. Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

  1. Removal of excess of water
  2. Helps in passive absorption of water and minerals
  3. Helps in ascent of sap – transpiration pull
  4. Maintains turgor of cells
  5. Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

  1. Root pressure theory is proposed by J. Pristley.
  2. For translocation of water, activity of living cells of root is responsible.
  3. Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
  4. Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
  5. Root pressure is an osmotic phenomenon.

Limitation of this theory:

  1. Not applicable to tall plants above 20 metres.
  2. Even in absence of root pressure ascent of sap is noticed.
  3. In actively transpiring plants, root pressure is not developed.
  4. In taller gymnosperms, root pressure is zero.
  5. Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

  1. Capillarity theory of water translocation is proposed by Bohem.
  2. Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
  3. Xylem vessels and tracheids are tubular elements having their lumen.
  4. In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
  5. As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

  1. The loss of water in the form of water vapour is called transpiration.
  2. About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
  3. For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
  4. Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
  5. The process is necessary evil because water which is important for plant is lost in the process.
  6. At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
    So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

  1. Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
  2. As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
  3. Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
  4. The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
  5. Root hair cell becomes flaccid and ready to absorb soil water.
  6. Water is passed on similarly in inner cortical cells.
  7. Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
  8. From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Osmotic Pressure (O.R) Turgor Pressure (T.P.)
1. The pressure exerted due to osmosis is called osmotic pressure. 1. The pressure exerted by turgid cell sap on cell membrane and cell wall, is called turgor pressure.
2. It is pressure caused by water when it moves by osmosis. 2. It is pressure caused by content of cell (cell sap).
3. It is generated by the osmotic flow of water through a semipermeable membrane. 3. It is maintained by osmosis.

Question 11.
How are the minerals absorbed by the plants ?
Answer:

  1. Soil is the chief source of minerals for the plants.
  2. Minerals get dissolved in the soil water.
  3. Minerals are absorbed by the plants in the ionic form mainly through roots.
  4. Absorption of minerals is independent of water.
  5. Absorbed minerals are pulled upwards along with xylem sap.
  6. Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation 1

  1. Water from soil is absorbed by plants with the help of root hairs.
  2. Root hairs are present in zone of absorption.
  3. Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
  4. Root hairs are nothing but cytoplasmic extensions of epiblema cell.
  5. Root hairs are long tube like structures of about 1 to 10 mm.
  6. They are colourless, unbranched and very delicate structures.
  7. A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
  8. The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

  1. Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
  2. The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
  3. Water molecules get adsorbed on cell wall of root hair (imbibition).
  4. They enter the root hair cell by diffusion through cell wall which is freely permeable.
  5. By process of osmosis they enter through plasma membrane which is semipermeable.
  6. The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
  7. The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
  8. The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
  9. From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
  10. The pathway of water is by apoplast and symplast.
  11. When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
  12. When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

Question 3.
Explain cohesion theory of translocation of water.
Answer:

  1. This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
  2. It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
  3. A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
  4. Ascent of sap occurs through lumen of xylem elements.
  5. Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
  6. Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
  7. This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
  8. In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
  9. Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
  10. Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
  11. Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

  1. Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
  2. Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
  3. The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
  4. When guard cells are flaccid that results in closure of stomata.
  5. According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
  6. Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
  7. During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
  8. According to proton transport theory, the movement is due to transport of H+ and K+ ions.
  9. Subsidiary cells are reservoirs of K+ ions. Starch is converted to malic acid which dissociate into malate and proton (H+) during day.
  10. Proton transported to subsidiary cells and K+ ions are taken from it. This forms potassium malate in guard cells.
  11. Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
  12. The uptake of K+ and Cl ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

Question 6.
Explain the active absorption of minerals.
Answer:

  1. Plants absorb minerals from the soil with their root system.
  2. MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
  3. The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
  4. The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
  5. Ions get accumulated in the root hair against the concentration gradient.
  6. These ions pass into cortical cells and finally reach xylem of roots.
  7. Along with the water these minerals are carried to other parts of plant.

12th Std Biology Questions And Answers:

11th Physics Chapter 14 Exercise Semiconductors Solutions Maharashtra Board

Class 11 Physics Chapter 14

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 14 Semiconductors Textbook Exercise Questions and Answers.

Semiconductors Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 14 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 14 Exercise Solutions 

1. Choose the correct option.

Question 1.
Electric conduction through a semiconductor is due to:
(A) Electrons
(B) holes
(C) none of these
(D) both electrons and holes
Answer:
(D) both electrons and holes

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 2.
The energy levels of holes are:
(A) in the valence band
(B) in the conduction band
(C) in the band gap but close to valence band
(D) in the band gap but close to conduction band
Answer:
(C) in the band gap but close to valence band

Question 3.
Current through a reverse biased p-n junction, increases abruptly at:
(A) Breakdown voltage
(B) 0.0 V
(C) 0.3V
(D) 0.7V
Answer:
(A) Breakdown voltage

Question 4.
A reverse biased diode, is equivalent to:
(A) an off switch
(B) an on switch
(C) a low resistance
(D) none of the above
Answer:
(A) an off switch

Question 5.
The potential barrier in p-n diode is due to:
(A) depletion of positive charges near the junction
(B) accumulation of positive charges near the junction
(C) depletion of negative charges near the junction,
(D) accumulation of positive and negative charges near the junction
Answer:
(D) accumulation of positive and negative charges near the junction

2. Answer the following questions.

Question 1.
What is the importance of energy gap in a semiconductor?
Answer:

  1. The gap between the bottom of the conduction band and the top of the valence band is called the energy gap or the band gap.
  2. This band gap is present only in semiconductors and insulators.
  3. Magnitude of the band gap plays a very important role in the electronic properties of a solid.
  4. Band gap in semiconductors is of the order of 1 eV.
  5. If electrons in valence band of a semiconductor are provided with energy more than band gap energy (in the form of thermal energy or electrical energy), then the electrons get excited and occupy energy levels in conduction band. These electrons can easily take part in conduction.

Question 2.
Which element would you use as an impurity to make germanium an n-type semiconductor?
Answer:
Germanium can be made an n-type semiconductor by doping it with pentavalent impurity, like phosphorus (P), arsenic (As) or antimony (Sb).

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 3.
What causes a larger current through a p-n junction diode when forward biased?
Answer:
In case of forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction.

Question 4.
On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer:
For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor.

Question 5.
Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?
Answer:

  1. In a p-type semiconductor, holes are majority charge carriers.
  2. When a p-type semiconductor is connected to terminals of a battery, holes, which are not actual charges, behave like a positive charge and get attracted towards the negative terminal of the battery.
  3. During transportation of hole, there is an indirect movement of electrons.
  4. The drift speed of these electrons is less than that in the n-type semiconductors. Mobility of the holes is also less than that of the electrons.
  5. As, electrical conductivity depends on the mobility of charge carriers, the conductivity of a n-type semiconductor is greater than that of p-type semiconductor even when both of these have same level of doping.

3. Answer in detail.

Question 1.
Explain how solids are classified on the basis of band theory of solids.
Answer:
i. The solids can be classified into conductors, insulators and semiconductors depending on the distribution of electron energies in each atom.

ii. As an outcome of the small distances between atoms, the resulting interaction amongst electrons and the Pauli’s exclusion principle, energy bands are formed in the solids.

iii. In metals, conduction band and valence band overlap. However, in a semiconductor or an insulator, there is gap between the bottom of the conduction band and the top of the valence band. This is called the energy gap or the band gap.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 1

iv. For metals, the valence band and the conduction band overlap and there is no band gap as shown in figure (b). Therefore, electrons can easily gain electrical energy when an external electric field is applied and are easily available for conduction.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 2

v. In case of semiconductors, the band gap is fairly small, of the order of 1 eV or less as shown in figure (c). Hence, with application of external electric field, electrons get excited and occupy energy levels in conduction band. These can take part in conduction easily.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 3

vi. Insulators, on the contrary, have a wide gap between valence band and conduction band of the order of 5 eV (for diamond) as shown in figure (d). Therefore, electrons find it very difficult to gain sufficient energy to occupy energy levels in conduction band.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 4

vii. Thus, an energy band gap plays an important role in classifying solids into conductors, insulators and semiconductors based on band theory of solids.

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 2.
Distinguish between intrinsic semiconductors and extrinsic semiconductors
Answer:

Intrinsic semiconductors Extrinsic semiconductors
1. A pure semiconductor is known as intrinsic semiconductors. The semiconductor, resulting
2. Their conductivity is low Their conductivity is high even at room temperature.
3. Its electrical conductivity is a function of temperature alone. Its electrical conductivity depends upon the temperature as well as on the quantity of impurity atoms doped in the structure.
4. The number density of holes (nh) is same as the number density of free electron (ne) (nh = ne). The number density of free electrons and number density of holes are unequal.

Question 3.
Explain the importance of the depletion region in a p-n junction diode.
Answer:
i. The region across the p-n junction where there are no charges is called the depletion layer or the depletion region.

ii. During diffusion of charge carriers across the junction, electrons migrate from the n-side to the p-side of the junction. At the same time, holes are transported from p-side to n-side of the junction.

iii. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

iv. The potential barrier thus developed, prevents continuous flow of charges across the junction. A state of electrostatic equilibrium is thus reached across the junction.

v. Free charge carriers cannot be present in a region where there is a potential barrier. This creates the depletion region.

vi. In absence of depletion region, all the majority charge carriers from n-region (i.e., electron) will get transferred to the p-region and will get combined with the holes present in that region. This will result in the decreased efficiency of p-n junction.

vii. Hence, formation of depletion layer across the junction is important to limit the number of majority carriers crossing the junction.

Question 4.
Explain the I-V characteristic of a forward biased junction diode.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 5

  1. Figure given below shows the I-V characteristic of a forward biased diode.
  2. When connected in forward bias mode, initially, the current through diode is very low and then there is a sudden rise in the current.
  3. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve.
  4. The corresponding voltage is called the knee voltage. It is about 0.7 V for silicon and 0.3 V for germanium.
  5. A diode effectively becomes a short circuit above this knee point and can conduct a very large current.
  6. To limit current flowing through the diode, resistors are used in series with the diode.
  7. If the current through a diode exceeds the specified value, the diode can heat up due to the Joule’s heating and this may result in its physical damage.

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 5.
Discuss the effect of external voltage on the width of depletion region of a p-n junction.
Answer:

  1. A p-n junction can be connected to an external voltage supply in two possible ways.
  2. A p-n junction is said to be connected in a forward bias when the p-region connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
  3. In forward bias connection, the external voltage effectively opposes the built-in potential of the junction. The width of depletion region is thus reduced.
  4. The second possibility of connecting p-n junction is in reverse biased electric circuit.
  5. In reverse bias connection, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased

11th Physics Digest Chapter 14 Semiconductors Intext Questions and Answers

Internet my friend (Textbookpage no. 256)

i. https://www.electronics-tutorials.ws/diode
ii. https://www.hitachi-hightech.com
iii. https://nptel.ac.in/courses
iv. https://physics.info/semiconductors
v. http://hyperphysics.phy- astr.gsu.edu/hbase/Solids/semcn.html

[Students are expected to visit above mentioned links and collect more information regarding semiconductors.]

11th Std Physics Questions And Answers:

11th Physics Chapter 8 Exercise Sound Solutions Maharashtra Board

Class 11 Physics Chapter 8

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 8 Sound Textbook Exercise Questions and Answers.

Sound Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 8 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 8 Exercise Solutions 

1. Choose the correct alternatives

Question 1.
A sound carried by air from a sitar to a listener is a wave of following type.
(A) Longitudinal stationary
(B)Transverse progressive
(C) Transverse stationary
(D) Longitudinal progressive
Answer:
(D) Longitudinal progressive

Question 2.
When sound waves travel from air to water, which of these remains constant ?
(A) Velocity
(B) Frequency
(C) Wavelength
(D) All of above
Answer:
(B) Frequency

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 3.
The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
(A) are longitudinal
(B) propagate isothermally
(C) propagate adiabatically
(D) are of long wavelength
Answer:
(C) propagate adiabatically

Question 4.
Speed of sound is maximum in
(A) air
(B) water
(C) vacuum
(D) solid
Answer:
(D) solid

Question 5.
The walls of the hall built for music concerns should
(A) amplify sound
(B) Reflect sound
(C) transmit sound
(D) Absorb sound
Answer:
(D) Absorb sound

2. Answer briefly.

Question 1.
Wave motion is doubly periodic. Explain.
Answer:
i. A wave particle repeats its motion after a definite interval of time at every location, making it periodic in time.
ii. Similarly, at any given instant, the form of a wave repeats itself at equal distances making it periodic in space.
iii. Thus, wave motion is a doubly periodic phenomenon, i.e., periodic in time as well as periodic in space.

Question 2.
What is Doppler effect?
Answer:
The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.

Question 3.
Describe a transverse wave.
Answer:
Transverse wave:
A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called transverse wave.
Example: Ripples on the surface of water, light waves.

Characteristics of transverse waves:

  1. All the particles of medium in the path of wave vibrate in a direction perpendicular to the direction of propagation of wave with same period and amplitude.
  2. When transverse wave passes through the medium, the medium is divided into alternate crests i.e., regions of positive displacements and troughs i.e., regions of negative displacement, that are periodic in time.
  3. A crest and an adjacent trough form one cycle of a transverse wave. The distance between any two successive crests or troughs is called wavelength ‘λ’ of the wave.
  4. Crests and troughs advance in the medium and are responsible for transfer of energy.
  5. Transverse waves can travel only through solids and not through liquids and gases. Electromagnetic waves are transverse waves, but they do not require material medium for propagation.
  6. When transverse waves advance through a medium, there is no change of pressure and density at any point of the medium, but the shape changes periodically.
  7. Transverse wave can be polarised.
  8. Medium conveying a transverse wave must possess elasticity of shape, i.e., modulus of rigidity.

Question 4.
Define a longitudinal wave.
Answer:
A wave in which particles of medium vibrate in a direction parallel to the direction of propagation of the wave is called longitudinal wave. Example: Sound waves.

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 5.
State Newton’s formula for velocity of sound.
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Question 6.
What is the effect of pressure on velocity of sound?
Answer:
Effect of pressure:
i. Let v be the velocity of sound in air when the pressure is P and density is ρ.

ii. Using Laplace’s formula, we can write,
v = \(\sqrt{\frac {γP}{ρ}}\) ….(1)

iii. If V be the volume of a gas having mass M then, ρ = \(\frac {M}{V}\)

iv. Substituting ρ in equation (1), we get,
v = \(\sqrt{\frac {γPV}{M}}\) ….(2)

v. But according to Boyle’s law,
PV = constant (at constant temperature)
Also, M and γ are constant.
∴ v = constant

vi. Hence, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

vii. For a gaseous medium, PV= nRT.
Substituting in equation (2), we get,
v = \(\sqrt{\frac {γnRT}{M}}\)

viii. Thus, even for a gaseous medium obeying ideal gas equation, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

Question 7.
What is the effect of humidity of air on velocity of sound?
Answer:
Effect of humidity:
i. Let vm and vd be the velocities of sound in moist air and dry air respectively.
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 1

ii. Humid air contains a large proportion of water vapour. Density of water vapour at 0 °C is 0.81 kg/m³ while that of dry air at 0°C is 1.29 kg/m³. So, the density ρm of moist air is less than the density ρd of dry air i.e., ρm < ρd.

iii. Thus \(\frac {v_m}{v_d}\) > 1
∴ vm > vd

iv. Hence, sound travels faster in moist air than in dry air. It means that velocity of sound increases with increase in moistness (humidity) of air.

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 8.
What do you mean by an echo?
Answer:
An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound.

Question 9.
State any two applications of acoustics.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Medical applications of acoustics:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Underwater applications of acoustics:
i. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
ii. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
iii. Motion and position of submerged objects like submarine can be measured with the help of this system.

Applications of acoustics in environmental and geological studies:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics. This is useful in studying the properties of the Earth.

Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 10.
Define amplitude and wavelength of a wave.
Answer:
i. Amplitude (A): The largest displacement of a particle of a medium through which the wave is propagating, from its rest position, is called amplitude of that wave.
SI unit: (m)

ii. Wavelength (λ): The distance between two successive particles which are in the same state of vibration is called wavelength of the wave.
SI unit: (m)

Question 11.
Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of π/2.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 2
i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, FI and J, E and F,
iii. Point having phase difference of π/2: A and B; B and C; D; D and F; F and H; H and I; J and I

Question 12.
Define the relation between velocity, wavelength and frequency of wave.
Answer:
i. A wave covers a distance equal to the wavelength (λ) during one period (T).
Therefore, the magnitude of the velocity (v) is given by,
Magnitude of velocity = \(\frac {Distance covered}{Corresponding time}\)

ii. v = \(\frac {22}{7}\) i.e., v = λ × (\(\frac {1}{T}\)) …………….. (1)

iii. But reciprocal of the period is equal to the frequency (n) of the waves.
∴ \(\frac {1}{T}\) = n …………… (2)

iv. From equations (1) and (2), we get
v = nλ
i.e., wave velocity = frequency × wavelength.

Question 13.
State and explain principle of superposition of waves.
Answer:
Principle:
As waves don’t repulse each other, they overlap in the same region of the space without affecting each other. When two waves overlap, their displacements add vectorially.

Explanation:
i. Consider two waves travelling through a medium arriving at a point simultaneously.

ii. Let each wave produce its own displacement at that point independent of the others. This displacement can be given as,
y1 = displacement due to first wave.
y2 = displacement due to second wave.

iii. Then according to superposition of waves, the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.
∴\(\vec{y}\) = \(\vec{y_1}\) + \(\vec{y_2}\)

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 14.
State the expression for apparent frequency when source of sound and listener are
i) moving towards each other
ii) moving away from each other
Answer:
i. Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

ii. Apparent frequency heard by the listener is given by,
n = n0(\(\frac {v±v_L}{v±v_s}\))
Where upper signs (+ ve in numerator and -ve in denominator) indicate that source and observer move towards each other. Lower signs (-ve in numerator and +ve in denominator) indicate that source and listener move away from each other.

iii. If source and listener are moving towards each other, then apparent frequency is given by,
n = n0(\(\frac {v+v_L}{v-v_s}\)) i.e., apparent frequency increases.

iv. If source and listener are moving away from each other, then apparent frequency is given by,
n = n0(\(\frac {v-v_L}{v+v_s}\)) i.e., apparent frequency decreases.

Question 15.
State the expression for apparent frequency when source is stationary and listener is
1) moving towards the source
2) moving away from the source
Answer:
Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

i. If listener is moving towards source then apparent frequency is given by,
n = n0(\(\frac {v+v_L}{v}\)) i.e., apparent frequency increases.

ii. If listener is receding away from source then apparent frequency is given by,
n = n0(\(\frac {v-v_L}{v}\)) i.e., apparent frequency decreases.

Question 16.
State the expression for apparent frequency when listener is stationary and source is.

(i) moving towards the listener
(ii) moving away from the listener
Answer:
Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

i. If source is moving towards observer then apparent frequency is given by,
n = n0(\(\frac {v}{v-v_s}\)) i.e., apparent frequency increases.

ii. If source is receding away from observer then apparent frequency is given by,
n = n0(\(\frac {v}{v+v_s}\)) i.e., apparent frequency decreases.

Question 17.
Explain what is meant by phase of a wave.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 3
i. The state of oscillation of a particle is called the phase of the particle.

ii. The displacement, direction of velocity and oscillation number of the particle describe the phase of the particle at a place.

iii. Particles r and t (q and u or v and s) have same displacements but the directions of their velocities are opposite.

iv. Particles having same magnitude of displacements and same direction of velocity are said to be in phase during their respective oscillations. Example: particles v and p.

v. Separation between two particles which are in phase is wavelength (λ).

vi. The two successive particles differ by ‘1’ in their oscillation number i.e., if particle v is at its nth oscillation, particle p will be at its (n + 1)th oscillation as the wave is travelling along + X direction.

vii. In the given graph, if the disturbance (energy) has just reached the particle w, the phase angle corresponding to particle is 0°. At this instant, particle v has completed quarter oscillation and reached its positive maximum (sin θ = +1). The phase angle θ of this particle v is \(\frac {π^c}{2}\) = 90° at this instant.

viii. Phase angles of particles u and q are πc (180°) and 2rcc (360°) respectively.

ix. Particle p has completed one oscillation and is at its positive maximum during its second oscillation.
∴ phase angle = 2πc + \(\frac {π^c}{2}\)
= \(\frac {5π^c}{2}\)

x. v and p are the successive particles in the same state (same displacement and same direction of velocity) during their respective oscillations. Phase angle between these two differs by 2πc.

Question 18.
Define progressive wave. State any four properties.
Answer:
i. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
Properties of progressive waves are:
Amplitude, wavelength, period, double periodicity, frequency and velocity.

Question 19.
Distinguish between traverse waves and longitudinal waves.
Answer:

Longitudinal wave Transverse wave
1. The particles of the medium vibrate along the direction of propagation of the wave. 1. The particles of the medium vibrate perpendicular to the direction of propagation of the wave.
2. Alternate compressions and rarefactions are formed. 2. Alternate crests and troughs are formed.
3. Periodic compressions and rarefactions, in space and time, produce periodic pressure and density variations in the medium. There are no pressure and density, variations in the medium.
4. For propagation of a longitudinal wave, the medium must be able to resist changes in volume. For propagation of a transverse wave, the medium must be able to resist shear or change in shape.
5. It can propagate through any material medium (solid, liquid or gas). It can propagate only through solids.
6. These waves cannot be polarised. These waves can be polarised.
7. eg.: Sound waves eg.: Light waves

Question 20.
Explain Newtons formula for velocity of sound. What is its limitation?
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton’s formula and the experimental value.

2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

3. Solve the following problems.

Question 1.
A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m for this wave, calculate
(i) the frequency
(ii) the period.
Answer:
Given: v = 340 m/s, λ = 1.7 m
To find: frequency (n), period (T)
Formulae:
i. n = \(\frac {v}{λ}\)
ii. T = \(\frac {1}{n}\)
Calculation: From formula, (i)
n = \(\frac {340}{1.7}\)
∴ n = 200 Hz
From formula, (ii)
T = \(\frac {1}{n}\) = \(\frac {1}{2×10^2}\)
= 5 × 10-3
…….. (using reciprocal Table)
∴ T = 0.005 s

Question 2.
A tuning fork of frequency 170 Hz produces sound waves of wavelength 2m. Calculate speed of sound.
Answer:
Given: n = 170 Hz, λ = 2 m
To find: velocity of sound (v)
Formula: v = nλ
Calculation: From formula,
v = 170 × 2
∴ v = 340 m/s

Question 3.
An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?
Answer:
Given: t = 0.45 s, v = 1500 m/s,
To Find: depth (d)
Formula: speed (v) = \(\frac {distance}{time}\)
Calculation:
For an echo distance travelled by the sound wave = 2 × (distance between echo sounder and shoal) (d)
v = \(\frac {2 × d}{t}\)
∴ d = \(\frac {1500 × 0.45}{2}\) = 337.5 m

Question 4.
A girl stands 170 m away from a high wall and claps her hands at a steady rateso that each clap coincides with the echo of the one before.
a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.
Answer:
i. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.

ii. The girl is at a distance of 226.67 m from the wall when she produces 45 claps per minute.
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 5.
Sound wave A has period 0.015 s, sound wave B has period 0.025. Which sound has greater frequency?
Answer:
Given: TA = 0.015 s, TB = 0.025 s
To find: greater frequency (n)
Formula: n = \(\frac {1}{T}\)
Calculation: From formula,
nA = \(\frac {1}{T_A}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ nA = 66.67
…. (using reciprocal table)
nB = \(\frac {1}{T_B}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ nB = 40 Hz
…. (using reciprocal table)
∴ nA > nB

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 6.
At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?
Answer:
Given:
vair = 1.75 VS.T.P = \(\frac {7}{4}\) vS.T.P
TS.T.P = 273 K
To find: temperature Tair
Formula: v ∝ √T
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 4

Question 7.
A man standing between 2 parallel eliffs fires a gun. He hearns two echos one after 3 seconds and other after 5 seconds. The separation between the two cliffs is 1360 m, what is the speed of sound?
Answer:
distance (s) = 1360 m,
time for first echo = 3 s,
time for second echo = 5 s
To Find : speed of sound (v)
Formula : speed = \(\frac {distence}{time}\)
Calculation:
Time for first echo = 3 s
∴ time taken by sound to travel given distance t1
= \(\frac {3}{2}\) = 1.5 s
Time for second echo = 5 s
∴ time taken by sound to travel given distance t2
= \(\frac {5}{2}\) = 2.5 s
∴Total time taken by sound to travel given distance, T = 1.5 + 2.5 = 4 s
From formula,
v = \(\frac {1360}{4}\)
∴v = 340 m/s

Question 8.
If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1 : 1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?
Answer:
Let v1 and v2 be the velocity of sound on day 1 and day 2 respectively.
\(\frac {v_1}{v_2}\) = \(\frac {1}{1.1}\)
We know, v ∝ \(\frac {1}{√ρ}\)
Let ρ1 and ρ2 be the density of air on day 1 and day 2 respectively.
∴ \(\sqrt{\frac {ρ_2}{ρ_1}}\) = \(\frac {1}{1.1}\)
∴ \(\frac {ρ_2}{ρ_1}\) = (\(\frac {1}{1.1}\))²
∴ ρ1 = 1.1² ρ2 = 1.21 ρ²
From above equation, we can conclude,
ρ1 > ρ2
∴ v2 > v1 i.e., the velocity of sound is greater on the second day than on the first day.
We know, speed of sound in moist air (vm) is greater than speed of sound in dry air (vd).
∴ We can conclude, air is moist on second day and dry on the first day.

Question 9.
A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.
Answer:
Given: vs = 15 m/s, n0 = 250 Hz, v = 340 m/s
To find: Frequency (n)
Formula: n = n0(\(\frac {v}{v-v_s}\))
Calculation: As the source approaches listener, apparent frequency is given by,
n = 250 (\(\frac {340}{340-15}\)) = \(\frac {3400}{13}\)
∴ n = 261.54 Hz

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 10.
The sound emitted from the siren of an ambulance has frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.
Answer:
Given: vs = 30 m/s, n0 = 1500 Hz, v = 340 m/s
To find: Difference in apparent frequencies (nA – n’A)
Formulae:
i. When the ambulance moves towards he stationary observer then nA = n0(\(\frac {v}{v-v_s}\))

ii. When the ambulance moves away from the stationary observer then, n’A = n0(\(\frac {v}{v+v_s}\))

Calculation:
From formula (i), icon’ 340
nA = 1500(\(\frac {340}{340-30}\))
∴ nA = 1645 Hz
From (ii)
n’A = 1500(\(\frac {340}{340+30}\))
∴ nA = 1378 Hz
Difference between nA and n’A
= nA – n’A = 1645 – 1378 = 267 Hz

11th Physics Digest Chapter 8 Sound Intext Questions and Answers

Can you recall? (Textbook page no. 142)

i. What type of wave is a sound wave?
ii. Can sound travel in vacuum?
iii. What are reverberation and echo?
iv. What is meant by pitch of a sound?
Answer:
i. Sound wave is a longitudinal wave.

ii. Sound cannot travel in vacuum.

iii. a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
b. An echo is the repetition of the original sound because of reflection by some surface.

iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.

Activity (Textbook page no. 144)

i. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B.
ii. Determine the wavelength and amplitude of each of the two waves P and Q shown in figure below.
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 5
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 6

Wave Wavelength (λ) Amplitude (A)
A 4 m 2 m
B 2 m 4 m
Wave Wavelength (λ) Amplitude (A)
P 6 units 3 units
Q 4 units 2 units

11th Std Physics Questions And Answers:

11th Biology Chapter 14 Exercise Human Nutrition Solutions Maharashtra Board

Class 11 Biology Chapter 14

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 14 Human Nutrition Textbook Exercise Questions and Answers.

Human Nutrition Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 14 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 14 Exercise Solutions

1. Choose correct option

Question A.
Acinar cells are present in ……………..
a. liver
b. pancreas
c. gastric glands
d. intestinal glands
Answer:
b. pancreas

Question B.
Which type of teeth are maximum in number in human buccal cavity?
a. Incisors
b. Canines
c. Premolars
d. Molars
Answer:
d. Molars

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question C.
Select odd one out on the basis of digestive functions of tongue.
a. Taste
b. Swallowing
c. Talking
d. Mixing of saliva in food
Answer:
c. Talking

Question D.
Complete the analogy:
Ptyalin: Amylase : : Pepsin : …………….. .
a. Lipase
b. Galactose
c. Proenzyme
d. Protease
Answer:
d. Protease

2. Answer the following questions

Question A.
For the school athletic meet, Shriya was advised to consume either Glucon-D or fruit juice but no sugarcane juice. Why it must be so?
Answer:
Sugarcane juice contain disaccharides. Disaccharides take time to digest i.e. breaking into monosaccharides, Glucon — D and fruit juices contain monosaccharide. Therefore, for instant supply of energy during athletic meet Glucon – D or fruit juices are preferred and not sugarcane.

Question B.
Alcoholic people may suffer from liver disorder. Do you agree? Explain your answer.
Answer:

  1. Liver disorder in alcoholic people may occur after years of heavy drinking.
  2. Most of the alcohol in the body is broken down in the liver by an enzyme called alcohol dehydrogenase, which transforms ethanol into a toxic compound called acetaldehyde (CH3CHO).
  3. ver consumption of alcohol leads to cirrhosis (distorted or scarred liver) and eventually to liver failure.
    Therefore, alcoholic people may suffer from liver disorder.

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question C.
Digestive action of pepsin comes to a stop when food reaches small intestine. Justify.
Answer:
Pepsin acts in acidic medium thus it is active in stomach. There is alkaline condition in the small intestine. pH of small intestine is very high for pepsin to work. Therefore, pepsin gets denatured in the small intestine.

Question D.
Small intestine is very long and coiled. Even if we jump and run, why it does not get twisted? What can happen if it gets twisted?
Answer:

  1. Mesentery is a tissue that is located in the abdomen. It attaches the small intestine to the wall of the abdomen and keeps it in place and therefore it does not get twisted while running and jumping.
  2. If small intestine gets twisted, the affected spot may block the food, liquid passing through it. It may sometimes cut off the blood flow if the twist is very severe. If this happens the surrounding tissue may die and can cause serious problems.

3. Write down the explanation

Question A.
Digestive enzymes are secreted at appropriate time in our body. How does it happen?
Answer:

  1. The digestive enzymes and juices are produced in sequential manner and at a proper time.
  2. These secretions are under neurohormonal control.
  3. Sight, smell and even thought of food trigger saliva secretion.
  4. Tenth cranial nerve stimulates secretion of gastric juice in stomach.
  5. Even the hormone gastrin brings about the same effect.

B. Explain the structure of tooth. Explain why human dentition is considered as thecodont, diphydont and heterodont.
Answer:

  1. Structure of tooth:
    • A tooth consists of the portion that projects above the gum called crown and the root that is made up of two or three projections which are embedded in gum.
    • A short neck connects the crown with the root.
    • The crown is covered by the hardest substance of the body called enamel which is made up of calcium phosphate and calcium carbonate.
    • Basic shape of tooth is derived from dentin which is a calcified connective tissue.
    • The dentin encloses the pulp cavity. It is filled with connective tissue pulp. It contains blood vessels and nerves.
    • Pulp cavity has extension in the root of the tooth called root canal.
    • The dentin of the root of tooth is covered by cementurn which is a bone like substance that attaches the root to the surrounding socket in the gum.
  2. Human dentition is described as thecodont, diphyodont and heterodont.
  3. It is called the codont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
  4. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
  5. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.
    Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 7

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question C.
Explain heterocrine nature of pancreas with the help of histological structure.
Answer:
Pancreas:

  1. Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
  2. Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
  3. Pancreatic juice is collected and carried to duodenum by pancreatic duct.
  4. The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
  5. Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
  6. Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
  7. It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
  8. Glucagon and insulin together control the blood-sugar level.
  9. Somatostatin hormone inhibits glucagon and insulin secretion.

4. Write short note on

Question A.
Position and function of salivary glands.
Answer:
Salivary Glands:

  • There are three pairs of salivary glands which open in buccal cavity.
  • Parotid glands are present in front of the ear.
  • The submandibular glands are present below the lower jaw.
  • The glands present below the tongue are called sublingual.
  • Salivary glands are made up of two types of cells.
  • Serous cells secrete a fluid containing digestive enzyme called salivary amylase.
  • Mucous cells produce mucus that lubricates food and helps swallowing.

Question B.
Jaundice
Answer:

  1. Jaundice is a disorder characterized by yellowness of conjunctiva of eyes and skin and whitish stool.
  2. It is a sign of abnormal bilirubin metabolism and excretion.
  3. Jaundice develops if excessive break down of red blood cells takes place along with increased bilirubin level than the liver can handle or there is obstruction in the flow of bile from liver to duodenum.
  4. Bilirubin produced from breakdown of haemoglobin is either water soluble or fat soluble.
  5. Fat soluble bilirubin is toxic to brain cells.
  6. There is no specific treatment to jaundice.
  7. Supportive care, proper rest are the treatments given to the patient.
    [Note: Treatment ofjaundice will depend on the underlying cause of it. For example, hepatitis-induced jaundice would require treatment which includes antiviral or steroid medications ]

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 5.
Observe the diagram. This is histological structure of stomach. Identify and comment on significance of the layer marked by arrow.
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 1
Answer:
The layer marked in the diagram represents glandular epithelium of mucosa.
Significance of the glandular epiihelium of mucosa:
Goblet cells of the epithelial layer of a mucous membrane secrete mucus which lubricates the lumen of the alimentary canal. This helps in movement of food through the gastrointestinal tract.

Question 6.
Find out pH maxima for salivary amylase, trypsin, nucleotidase and pepsin and place on the given pH scale
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 2
Answer:
Salivary amylase = 6.8
Trypsin = 8
Nucleotidase = 7.5
Pepsin = 2

Question 7.
Write the name of a protein deficiency disorder and write symptoms of it.
Answer:

  1. Kwashiorkor is a protein deficiency disorder.
  2. This protein deficiency disorder is found generally in children between one to three years of age.
  3. Children suffering from Kwashiorkor are underweight and show stunted growth, poor brain development, loss of appetite, anaemia, protruding belly, slender legs, bulging eye, oedema of lower legs and face, change in skin and hair colour.

Question 8.
Observe the diagram given below label the A, B, C, D, E and write the function of A, C in detail.
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 3
Answer:
A- Bile duct, B- Stomach, C- Common hepatic duct, D- Pancreas, E- Gall Riadder

Functions: Bile duct: It carries hile from the gall bladder and empties it into the tipper part of the small intestine. Common hepatic duct: It drains bile from the liver. It helps in transportation of waste from liver and helps in digestion by releasing bile.
[Note: Labels (A) and (O) have been modified for the better understanding of the students]

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Practical / Project : Here are the events in the process of digestion. Fill in the blanks and complete the flow chart.
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 4
Answer:
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 5
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 6

11th Biology Digest Chapter 14 Human Nutrition Intext Questions and Answers

Can you recall? (Textbook Page No. 161)

Question 1.
What is nutrition?
Answer:

  1. Nutrition is the sum of the processes by which an organism consumes and utilizes food substances,
  2. WHO defines nutrition as the intake of food, considered in relation to the body’s dietary needs.
  3. The term nutrition includes the process like ingestion, digestion, absorption, assimilation and egestion.

Question 2.
Enlist life processes that provide us energy to perform different activities.
Answer:
The life processes which are essential and provide us energy are nutrition and respiration.

Think about it (Textbook Page No. 161)

Question 1.
Our diet includes all necessary nutrients. Still we need to digest it. Why is it so?
Answer:

  1. Digestion is a very important process of converting complex, noil-diffusible and non-absorbable food substances into simple, diffusible and assimilable substances.
  2. Our diet includes all necessary nutrients, which are in the form of complex substances like carbohydrates, proteins, fats and vitamins.
  3. These complex substances are converted into simple, diffusible and assimilable substances through the process of digestion.
    Hence, there is a need for digestion of food.

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Human Digestive System (Textbook Page No. 161)

Question 1.
Label the diagram
Answer:
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 8

Do you know? (Textbook Page No. 162)

Question 1.
Who controls the deglutition?
Answer:
The process of swallowing is called deglutition. Medulla oblongata controls the deglutition.

Question 2.
Is deglutition voluntary or involuntary?
Answer:

  • Deglutition consists of three phases: oral phase, pharyngeal phase and oesophagal phase.
  • The oral phase is voluntary whereas the pharyngeal and oesophagal phases are involuntary.
    [Source: Goya!, R. K., & Mashimo, H. (2006,.). Physio!o’ of oral, pharyngeal, and esophageal motility. GI Motility online.]

Use your brain power (Textbook Page No. 165)

Question 1.
Draw a neat labelled diagram of human alimentary canal and associated glands in situ.
Answer:
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 8

Question 2.
Write a note on human dentition.
Answer:

  1. Human dentition is described as thecodont, diphyodont and heterodont.
  2. It is called thecodont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
  3. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
  4. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 3.
Muscularis layer in stomach is thicker than that in intestine. Why is it so?
Answer:
Muscularis layer in stomach is thicker than that of intestine because food is churned and gastric juices are mixed in the stomach whereas in intestine only absorption takes place.

Question 4.
Liver is a vital organ. Justify.
Answer:

  1. Kupffer cells of liver destroy toxic substances, dead and worn-out blood cells and microorganisms.
  2. Bile juice secreted by liver emulsifies fats and makes food alkaline.’
  3. Liver stores excess of glucose in the form of glycogen.
  4. Deamination of excess amino acids to ammonia and its further conversion to urea takes place in liver.
  5. Synthesis of vitamins A, D, K and BI2 takes place in liver.
  6. It also produces blood proteins like prothrombin and fibrinogen.
  7. During early development, it acts as haemopoietic organ.
    Therefore, liver is a vital organ.

Internet my friend: (Textbook Page No. 171)

Question 1.
Collect the different videos of functioning of digestive system,
Answer:
[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 9

Find out (Textbook Page No. 162)

Question 1.
What will be the dental formula of a three years old child?
Answer:
The dental formula of a three-year-old child will be: I \(\frac{2}{2}\), C \(\frac{1}{1}\), M \(\frac{2}{2}\) = \(\frac{2,1,2}{2,1,2}\)
i. e. 5 × 2 = 10 teeth in each jaw = 20 teeth.
As a child has 20 teeth by the age of three.

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 2.
What is dental caries and dental plaque? How can one avoid it?
Answer:

  • Dental caries are tooth decay or cavities caused by acids secreted by bacteria. Dental caries may be yellow or black in color.
  • Dental plaques also known as tooth plaque is a soft, sticky film which forms on the teeth regularly. It is colourless to pale yellow in colour.
  • Tooth decay and dental plaque can be prevented by brushing teeth twice a day with a fluoride containing tooth paste.
  • Rinsing mouth thoroughly with a mouth wash and use of dental floss or interdental cleaners to clean teeth daily can help to avoid dental caries and dental plaque.

Internet my friend (Textbook Page No. 162)

Question 1.
Find out the role of orthodontist and dental technician.
Answer:
a. Orthodontics is a specialization in dental profession. Orthodontist straightens the crooked teeth, locates problem in patients’ teeth and their overall oral development. They might use X-rays, plaster molds or dental appliances like retainers and space maintainers to correct the problems,

b. Dental technicians are the ones which improves patients’ appearance, ability to chew and speech. They make dentures, crowns, bridges and dental braces.

Question 2.
What is a root canal treatment?
Answer:

  • Root canal treatment is also known as endodontic treatment.
  • It is a dental treatment of removing infection from inside of a tooth.
  • Root canal is hollow section of tooth which contains the nerve tissue, blood vessels and other cells, this is also known as pulps.
  • Crown and root are a part of tooth. Crown is present above the gum while root is embedded in the gum.
    e. Pulp which is present inside the root canal nourishes the tooth and provides moisture to the surrounding material.
  • The nerves present inside the pulp sense hot cold temperatures as pain.
  • First step of a root canal treatment is removal of dead pulp tissues by making a hole on the surface of tooth.
  • In second step, the dentist cleans and decontaminates the area and fills the hollow area with adhesive cement in order to seal the canal completely.
  • The tooth is dead after the therapy and the patient no longer feel any pain but the tooth becomes more fragile than ever.
  • The last step of root canal is adding a crown or filling. Until the crown or filling is complete, patient is not supposed to chew or bite using that tooth. After the crown or filling patient can use that tooth as before.

Find out (Textbook Page No. 163)

Question 1.
You must have heard about appendicitis. It is inflammation of appendix. Find more information about this disorder.
Answer:

  1.  Appendicitis is a condition where there is inflammation of appendix.
  2. Appendix is a vestigial organ. It is a linger shaped pouch that projects from colon on the lower right side of the abdomen.
  3. Appendicitis pain is very severe. It initially starts from the navel and then moves.
  4. It occurs in the people of age group between 10 to 30.
  5. Surgical removal is the standard treatment for appendicitis.
  6. Symptoms: Nausea and vomiting, loss of appetite, low grade fever, constipation, abdominal bloating, severe pain in the right side of the abdomen.
  7. Appendicitis is caused when there is blockage in the lining of the appendix that results in infection. The bacteria multiply rapidly and causes inflammation and it is then filled with pus.
  8. If not treated properly appendix can rupture which can lead to further complications.
    [Students can use above answer for reference and find more information about appendicitis.]

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 2.
What is heartburn? Why do we take antacids to control it?
Answer:
Heart burn is a problem created when stomach contents (acid) are forced back up to oesophagus. It causes a burning pain in lower chest.

Antacids are bases and help to treat heartburn by neutralizing the stomach acid. The key ingredients of antacids are calcium carbonate, magnesium hydroxide, aluminium hydroxide or sodium bicarbonate.

Activity (Textbook Page No. 163)

Make a model of human digestive system in a group.
Answer:
[Students are expected to perform this activity on their own.]

Always Remember (Textbook Page No. 166)

Question 1.
Food remains for a very short time in mouth but action of salivary amylase continues for further IS to 30 minutes till gastric juice mixes with food in the stomach. Why do you think it stops after the food gets mixed with gastric juice?
Answer:

  1. The gastric juices are mixed with food in the stomach.
  2. The pH of the stomach is 1.0-2.0 which is very acidic. Such high level of acidity leads to denaturation of salivary amylase’s protein structure.
  3. On the other hand, pH 6.8 is required for salivary amylase to carry out the activity which is not found in stomach. Thus, activity of salivary amylase is stopped when food is mixed with gastric juice.

Internet my friend (Textbook Page No. 167)

Question 1.
How are bile pigments formed?
Answer:

  1. When old and worn out red blood cells are destroyed by macrophages in liver, the globin portion of hemoglobin is split off and heme is converted to biliverdin.
  2. Most of this biliverdin is converted to bilirubin, which gives bile its major pigmentation.
    [Source http://www.biologydiscussion.com/human-physiology/digestive-system/bile-pigments/bile-pigments-origin-and-formation-digestive-juice-human-biology/81803]

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Think about it (Textbook Page No. 167)

Question 1.
How can I keep my pancreas healthy? Can a person live without pancreas?
Answer:

  1. Pancreas can be kept healthy by:
    • Eating proper balanced and low-fat diet, with plenty of whole grains, fruits and vegetables.
    • Regular exercise and maintaining a healthy weight.
    • Limiting alcohol consumption and avoid smoking.
    • Adequate intake of water.
    • Regular checkups.
  2. The pancreas is a gland that secretes digestive enzymes and insulin which is needed for a person to survive.
  3. Without pancreas the person will develop diabetes and will have to take insulin for the rest of the life.
  4. Without pancreas the body’s ability to absorb nutrients also decreases.
    Hence, though a person can survive without pancreas he may have to remain dependent on the medicines for survival.

Do it yourself? (Textbook Page No. 167)

Question 1.
You have studied the representation of enzymatic actions in the form of reactions.
Write the reactions of pancreatic enzymes.
Answer:
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 10

Do it yourself (Textbook Page No. 168)

Question 1.
Observe the following reactions and explain in words.
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 11
Answer:

  1. Maltase acts on maltose to form glucose.
  2. Sucrase acts on sucrose to form glucose and fructose.
  3. Lactase acts on lactose to form glucose and galactose.
  4. Dipeptidase acts on dipeptides to form amino acids.
  5. Emulsified fats are converted into fatty acids and glycerol by lipase.

Use your brain power (Textbook Page No. 168)

Question 1.
Make a flow chart for digestion of carbohydrate.
Answer:
Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition 12

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 2.
What is a proenzyme? Enlist various proenzymes involved in process of digestion and state their function.
Answer:
Proenzymes are synthesized in cells as an inactive precursor that undergo some modification before becoming catalytically active.
The various proenzymes involved in process of digestion are as follows:

  • Pepsinogen: Pepsinogen when converted into its active form pepsin acts on proteins to form peptones and proteoses.
  • Trypsinogen: Trypsinogen when converted to it active form trypsin converts proteins, proteoses and peptones to polypeptides.
  • Chymotrypsinogen: Chymotrypsinogen when converted to active form chymotrypsin it converts polypeptides to dipeptides.

Question 3.
Differentiate between Chyme and Chyle.
Answer:

No. Chyme Chyle
a. Chyme is a semi-fluid acidic mass of partially digested food. Chyle is an alkaline slurry which contains various nutrients ready for absorption.
b. Chyme leaves stomach and enters the small intestine. Chyle leaves small intestine and enters large intestine.

Question 4.
Digestion of fats take place only after the food reaches small intestine. Give reason.
Answer:
Digestion of fats takes place in small intestine because the presence of fats in small intestine stimulates the release of pancreatic lipase from pancreas and bile from liver. Pancreatic lipases hydrolyze fat molecules into fatty acids and monoglycerides and bile brings about emulsification of fats. Therefore, digestion of fats occur when food reaches small intestine.

Observe and Discuss (Textbook Page No. 169)

Question 1.
Action of digestive juice in your group.
Answer:

Digestive juices

Action

Saliva Saliva contains salivary amylase which breaks down starch into maltose.
Gastric juice HC1 breaks converts inactive pepsinogen into its active form pepsin. Pepsin then breakdown proteins into peptones and proteoses.
Pancreatic juice Pancreatic amylase acts on glycogen and starch and converts those into disaccharides. Enterokinase converts trypsinogen into trypsin (active form).
Trypsin converts proteins, proteoses, peptones to polypeptides.
Chymotrypsin converts polypeptides to dipeptides.
Nucleases digest nucleic acids to pentose sugar.
Intestinal enzymes Maltase converts maltose to glucose.
Sucrase converts sucrose to glucose and fructose.
Lactase converts lactose to glucose and galactose.
Dipeptidases converts dipeptides to amino acids.
Lipase converts emulsified fats into fatty acids and monoglycerides.
Bile juice It brings about emulsification of fats.

Can you recall? (Textbook Page no. 170)

Question 1.
What is balanced diet?
Answer:
Balanced diet is a diet which contains proper amount of carbohydrates, fats, vitamins, proteins and minerals to maintain a good health.

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 2.
Explain the terms undernourished, over-nourished and malnourished in details.
Answer:

  • Undernourished: When supply of nutrients is less than the minimum amount of nutrients or food required for good health is called undernourished.
  • Over-nourished: The intake of nutrients is excessive. In over-nourished the amount of nutrients exceeds the amount required for normal growth.
  • Malnourished: Malnourished is a condition where a person’s diet does not contain right amount of nutrients.

Do you know? (Textbook Page No. 170)

Question 1.
What is gross calorific value?
Answer:
The amount of heat liberated by complete combustion of lg food in a bomb calorimeter is termed as gross calorific (gross energy) value.

Question 2.
What is physiological value?
Answer:
The actual energy produced by 1 g food is its physiological value.

Question 3.
Name the following
Energy content of food in animals is expressed in terms of?
Answer:
Heat Energy

Question 4.
Complete the following table representing Gross calorific value and physiological value of food component.

Food Component

Gross calorific value (Kcal/g)

Physiological value (Kcal/g)

Fats (A) 9.0
(B) 5.65 4.0
Carbohydrates (C) (D)

Answer:

Food Component

Gross calorific value (Kcal/g)

Physiological value (Kcal/g)

Fats 9.45 9.0
Proteins 5.65 4.0
Carbohydrates 4.1 4.0

Find out (Textbook Page No. 171)

Question 1.
Find out the status of nialnutrition among children in Maharashtra and efforts taken by the government to overcome the situation. Search for various NGOs working in this field.
Answer:
93,783 children have been diagnosed with severe acute malnutrition and 5.7 lakh with moderate acute malnutrition in Maharashtra.
Steps taken by government to overcome malnutrition:

  1. Promotion of infant and young child feeding practices.
  2. Management of malnutrition at community and facility level by trained service providers.
  3. Treatment of children with severe acute malnutrition at special units called the Nutrition Rehabilitation Centres (NRCs), set up at public health facilities.
  4. A special program to combat micronutrient deficiencies of Vitamin A, Iron and Folic acid.
  5. The initiatives like Mother and Child protection card, village health and nutrition days, are taken by the government for addressing the nutrition concerns in children, pregnant women and lactating mothers.

Various NCOs working in this field:

  1. Akshay Patra
  2. Fight Hunger Foundation,
  3. Feeding India,
  4. No Hungry child
    [Source: http://pib.nic.in/newsite/PrintRelease.aspx?relid=l 13725; https://yourstory.com/2016/10/world- food-day-ngosj
    [Note: Students can use above answer as reference and find more information from the internet.]

Maharashtra Board Class 11 Biology Solutions Chapter 14 Human Nutrition

Question 2.
Are jaundice and hepatitis same disorders?
Answer:
Jaundice and Hepatitis are two different disorders.

Jaundice: Jaundice occurs when the rate of bilirubin production exceeds the rate of its elimination. It causes yellowing of skin and eyes.

Hepatitis: It is a disease where there is inflammation of liver. It may be caused because of infection, over alcohol consumption, immune system disorder etc.

Do you know (Textbook Page No. 171)

Question 1.
Alcoholism causes different disorders of liver like steatosis (fatty liver), alcoholic hepatitis, fibrosis and cirrhosis. Collect more information on these disorders and try to increase awareness against alcoholism in society. Collect information about NGOs working against alcoholism.
Answer:
Steatosis (fatty liver): Steatosis is accumulation of fat in the liver. Treatment can help but it cannot be cured. Major risk factors are obesity and Diabetes type II, it is also associated with excessive alcohol consumption. Fatigue, weight loss and abdominal pain are some symptoms. It is a benign condition but in very smaller number of patients it can lead to liver failure. Treatment involves diet and exercise to reduce obesity.

Alcoholic hepatitis: Alcoholic Hepatitis is liver inflammation caused by excessive consumption of alcohol. It occurs in people who drink heavily for many years. Symptoms like yellowing of skin and eye, accumulation of fluid in stomach which leads to increase in stomach size. Treatments like completely stopping of alcohol consumption, hydration and nutrition care are carried out. Administration of steroid drugs reduces liver inflammation.

Fibrosis: There is significant scarring of liver tissue in this condition. Fibrosis itself does not cause any symptoms. Diagnosis includes doctor’s evaluation, blood tests and imaging tests, liver biopsy. Treatments include stopping the consumption of alcohol. There are no such effective drugs for curing of fibrosis.

Cirrhosis: It is a chronic liver damage caused due to various reasons which leads to irreversible scarring of liver and liver failure. Causes of cirrhosis are chronic alcohol abuse and hepatitis. Patients may experience fatigue, weakness and weight loss. In later stages, patients may develop jaundice, abdominal swelling and gastrointestinal bleeding. In advanced stage, a liver transplant is required.

NGOs working against alcoholism:

  1. Muktangan Rehabilitation Centre
  2. Anmol Jeevan Foundation
  3. Sankalp Rehabilitation Trust
  4. Kripa Foundation
  5. Harmony Foundation
  6. Hands for you Rehab Centre

11th Std Biology Questions And Answers:

11th Physics Chapter 12 Exercise Magnetism Solutions Maharashtra Board

Class 11 Physics Chapter 12

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 12 Magnetism Textbook Exercise Questions and Answers.

Magnetism Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 12 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 12 Exercise Solutions 

1. Choose the correct option.

Question 1.
Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at r is always proportional to
(A) \(\frac {1}{r^2}\)
(B) \(\frac {1}{r^3}\)
(C) \(\frac {1}{r}\)
(D) Not necessarily \(\frac {1}{r^3}\) at all points
Answer:
(B) \(\frac {1}{r^3}\)

Question 2.
Magnetic meridian is the plane
(A) perpendicular to the magnetic axis of Earth
(B) perpendicular to geographic axis of Earth
(C) passing through the magnetic axis of Earth
(D) passing through the geographic axis
Answer:
(C) passing through the magnetic axis of Earth

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 3.
The horizontal and vertical component of magnetic field of Earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be
(A) 30°
(B) 45°
(C) 0°
(D) 90°
Answer:
(B) 45°

Question 4.
Inside a bar magnet, the magnetic field lines
(A) are not present
(B) are parallel to the cross sectional area of the magnet
(C) are in the direction from N pole to S pole
(D) are in the direction from S pole to N pole
Answer:
(D) are in the direction from S pole to N pole

Question 5.
A place where the vertical components of Earth’s magnetic field is zero has the angle of dip equal to
(A) 0°
(B) 45°
(C) 60°
(D) 90°
Answer:
(A) 0°

Question 6.
A place where the horizontal component of Earth’s magnetic field is zero lies at
(A) geographic equator
(B) geomagnetic equator
(C) one of the geographic poles
(D) one of the geomagnetic poles
Answer:
(D) one of the geomagnetic poles

Question 7.
A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences
(A) a force but not a torque
(B) a torque but not a force
(C) both a force and a torque
(D) neither force nor a torque
Answer:
(C) both a force and a torque

2. Answer the following questions in brief.

Question 1.
What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
Answer:
i. When a magnet is cut into two pieces, then each piece behaves like an independent magnet.

ii. When a bar magnet is cut transverse to its length, the two pieces generated will behave as independent magnets of reduced magnetic length. However, the pole strength of all the four poles formed will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 1

iii. When the bar magnet is cut along its length, the two pieces generated will behave like an independent magnet with reduced pole strength. However, the magnetic length of both the new magnets will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 2

Question 2.
What could be the equation for Gauss’ law of magnetism, if a monopole of pole strength p is enclosed by a surface?
Answer:
i. According to Gauss’ law of electrostatics, the net electric flux through any Gaussian surface is proportional to net charge enclosed in it. The equation is given as,
øE = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\)

ii. Similarly, if a monopole of a magnet of pole strength p exists, the Gauss’ law of magnetism in S.I. units will be given as,
øE = ∫\(\vec{B}\) . \(\vec{dS}\) = µ0P

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

3. Answer the following questions in detail.

Question 1.
Explain the Gauss’ law for magnetic fields.
Answer:
i. Analogous to the Gauss’ law for electric field, the Gauss’ law for magnetism states that, the net magnetic flux (øB) through a closed Gaussian surface is zero. øB = ∫\(\vec{B}\) . \(\vec{dS}\) = 0

ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 3

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.

vii. Thus, according to the Gauss’ law of electrostatics øE = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\), where q is the positive charge enclosed.

viii. The situation is entirely different from magnetic lines of force. Gauss’ law of magnetism can be written as øB = ∫\(\vec{B}\) . \(\vec{dS}\) = 0
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

Question 2.
What is a geographic meridian? How does the declination vary with latitude? Where is it minimum?
Answer:
A plane perpendicular to the surface of the Earth (vertical plane) and passing through geographic axis is geographic meridian.

i. Angle between the geographic and the magnetic meridian at a place is called magnetic declination (a).
ii. Magnetic declination varies with location and over time. As one moves away from the true north the declination changes depending on the latitude as well as longitude of the place. By convention, declination is positive when magnetic north is east of true north, and negative when it is to the west. The declination is small in India. It is 0° 58′ west at Mumbai and 0° 41′ east at Delhi.

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 3.
Define the angle of dip. What happens to angle of dip as we move towards magnetic pole from magnetic equator?
Answer:
Angle made by the direction of resultant magnetic field with the horizontal at a place is inclination or angle of dip (ø) at the place.
At the magnetic pole value of ø = 90° and it goes on decreasing when we move towards equator such that at equator value of (ø) = 0°.

4. Solve the following problems.

Question 1.
A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 Am and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.
Answer:
Given that, (qm)1 = 200 Am
and (2l) = 5 cm = 5 × 10-2 m
∴ m = 200 × 5 × 10-2 = 10 Am²
For a bar magnet, magnetic dipole moment is,
m = qm (21)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
Ba = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 m}{r^{3}}\)
= 10-7 × \(\frac{2 \times 10}{(0.2)^{3}}\)
= 0.25 × 10-3
= 2.5 × 10-4 Wb/m²
The force acting on the pole will be given by,
F = qm Ba = 100 × 2.5 × 10-4
= 2.5 × 10-2 N

Question 2.
A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.
Answer:
Let true value of dip be ø. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’H = BH cos 30° Whereas, vertical component remains unchanged.
∴ For apparent dip of 45°,
tan 45° = \(\frac{\mathrm{B}_{\mathrm{V}}^{\prime}}{\mathrm{B}_{\mathrm{H}}^{\prime}}=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}} \cos 30^{\circ}}=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}} \times \frac{1}{\cos 30^{\circ}}\)
But, real value of dip is,
tan ø = \(\frac {B_V}{B_H}\)
∴ tan 45° = \(\frac {tan ø}{cos 30°}\)
∴ tan ø = tan 45° × cos 30°
= 1 × \(\frac {√3}{2}\)
∴ ø = tan-1 (0.866)

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 3.
Two small and similar bar magnets have magnetic dipole moment of 1.0 Am² each. They are kept in a plane in such a way that their axes are perpendicular to each other. A line drawn through the axis of one magnet passes through the centre of other magnet. If the distance between their centres is 2 m, find the magnitude of magnetic field at the midpoint of the line joining their centres.
Answer:
Let P be the midpoint of the line joining the centres of two bar magnets. As shown in figure, P is at the axis of one bar magnet and at the equator of another bar magnet. Thus, the magnetic field on the axis of the first bar magnet at distance of 1 m from the centre will be,
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 4
Ba = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
= 10-7 × \(\frac {2×1.0}{(1)^3}\)
= 2 × 10-7 Wb/m²
Magnetic field on the equator of second bar magnet will be,
Beq = \(\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\)
= 10-7 × \(\frac {1.0}{(1)^3}\)
= 1 × 10-7 Wb/m²
The net magnetic field at P,
Bnet = \(\sqrt {B_a^2+B_{eq}^2}\)
= \(\sqrt {(2×10^{-7})^2+(1×10^{-7})^2}\)
= \(\sqrt {(10^{-7})^2×(4+1)}\)
= √5 × 10-7 Wb/m²

Question 4.
A circular magnet is made with its north pole at the centre, separated from the surrounding circular south pole by an air gap. Draw the magnetic field lines in the gap. Draw a diagram to illustrate the magnetic lines of force between the south poles of two such magnets.
Answer:
i. For a circular magnet:
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 5

Question 5.
Two bar magnets are placed on a horizontal surface. Draw magnetic lines around them. Mark the position of any neutral points (points where there is no resultant magnetic field) on your diagram.
Answer:
The magnetic lines of force between two magnets will depend on their relative positions. Considering the magnets to be placed one besides the other as shown in figure, the magnetic lines of force will be as shown.
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 6

11th Physics Digest Chapter 12 Magnetism Intext Questions and Answers

Can you recall? (Textbook page no. 221)

Question 1.
What are the magnetic lines of force?
Answer:
The magnetic field around a magnet is shown by lines going from one end of the magnet to the other. These lines are named as magnetic lines of force.

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 2.
What are the rules concerning the lines of force?
Answer:
i. Magnetic lines of force originate from the north pole and end at the south pole.
ii. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
iii. The direction of the net magnetic field \(\vec {B}\) at a point is given by the tangent to the magnetic line of force at that point.
iv. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec {B}\)
v. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Question 3.
What is a bar magnet?
Answer:
Bar magnet is a magnet in the shape of bar having two poles of equal and opposite pole strengths separated by certain distance (2l).

Question 4.
If you freely hang a bar magnet horizontally, in which direction will it become stable?
Answer:
A bar magnet suspended freely in air always aligns itself along geographic N-S direction.

Try this (Textbook page no. 221)

You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced.
Answer:
When a small compass needle is kept at any position near a bar magnet, the needle always aligns itself in the direction parallel to the direction of magnetic lines of force.
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 7
Hence, by placing it at different positions, A, B, C, D,… as shown in the figure, the direction of magnetic lines of force can be traced. The direction of magnetic field will be a tangent at that point.

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Internet my friend: (Text book page no. 227)

https://www.ngdc.noaa.gov
[Students are expected to visit above mentioned link and collect more information about Geomagnetism.]

11th Std Physics Questions And Answers:

11th Physics Chapter 13 Exercise Electromagnetic Waves and Communication System Solutions Maharashtra Board

Class 11 Physics Chapter 13

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 13 Electromagnetic Waves and Communication System Textbook Exercise Questions and Answers.

Electromagnetic Waves and Communication System Class 13 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 13 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 13 Exercise Solutions 

1. Choose the correct option.

Question 1.
The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to green house effect is
(A) Infra-red radiation
(B) X ray
(C) Microwave
(D) Visible light
Answer:
(A) Infra-red radiation

Question 2.
Earth’s atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer:
(B) IR

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 3.
How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(A) depends on the values of p and e
(B) increases
(C) decreases
(D) remains same
Answer:
(D) remains same

Question 4.
The direction of EM wave is given by
(A) \(\bar{E}\) × \(\bar{B}\)
(B) \(\bar{E}\).\(\bar{B}\)
(C) along \(\bar{E}\)
(D) along \(\bar{B}\)
Answer:
(A) \(\bar{E}\) × \(\bar{B}\)

Question 5.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) h½
(B) h
(C) h3/2
(D) h²
Answer:
(A) h½

Question 6.
The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer:
(A) Microwave

Question 7.
If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer:
(A) 32000 m

2. Answer briefly.

Question 1.
State two characteristics of an EM wave.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

Question 2.
Why are microwaves used in radar?
Answer:
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 3.
What are EM waves?
Answer:
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4.
How are EM waves produced?
Answer:

  1. According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
  2. It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
  3. EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
  4. An electric charge at rest has an electric field in the region around it but has no magnetic field.
  5. When the charge moves, it produces both electric and magnetic fields.
  6. If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
  7. But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
  8. Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

Question 5.
Can we produce a pure electric or magnetic wave in space? Why?
Answer:
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6.
Does an ordinary electric lamp emit EM waves?
Answer:
Yes, ordinary electric lamp emits EM waves.

Question 7.
Why light waves travel in vacuum whereas sound wave cannot?
Answer:
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8.
What are ultraviolet rays? Give two uses.
Answer:
Production:

  1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
  2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
  3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

  1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
  2. Used in burglar alarms and security systems.
  3. Used to distinguish real and fake gems.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 9.
What are radio waves? Give its two uses.
Answer:

  1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
  2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Uses:

  1. Radio waves are used for wireless communication purpose.
  2. They are used for radio broadcasting and transmission of TV signals.
  3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 10.
Name the most harmful radiation entering the Earth’s atmosphere from the outer space.
Answer:
Ultraviolet radiation.

Question 11.
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer:
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12.
Name the three basic units of any communication system.
Answer:
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13.
What is a carrier wave?
Answer:
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14.
Why high frequency carrier waves are used for transmission of audio signals?
Answer:
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15.
What is modulation?
Answer:
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16.
What is meant by amplitude modulation?
Answer:
When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17.
What is meant by noise?
Answer:

  1. A random unwanted signal is called noise.
  2. The source generating the noise may be located inside or outside the system.
  3. Efforts should be made to minimize the noise level in a communication system.

Question 18.
What is meant by bandwidth?
Answer:
The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 19.
What is demodulation?
Answer:
The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20.
What type of modulation is required for television broadcast?
Answer:
Amplitude modulation is required for television broadcast.

Question 21.
How does the effective power radiated by an antenna vary with wavelength?
Answer:

  1. To transmit a signal, an antenna or an aerial is needed.
  2. Power radiated from a linear antenna of length l is, P ∝ (\(\frac {l}{λ}\))²
    where, λ is the wavelength of the signal.

Question 22.
Why should broadcasting programs use different frequencies?
Answer:
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23.
Explain the necessity of a carrier wave in communication.
Answer:

  1. Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
  2. The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
  3. Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.

Question 24.
Why does amplitude modulation give noisy reception?
Answer:
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25.
Explain why is modulation needed.
Answer:
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1.
Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 108 m/s.
Answer:
Given: λ = 250 m, c = 3 × 108 m/s
To find: Frequency (v)
Formula: c = v8
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{250}\) = 1.2 × 106 Hz
= 1.2 MHz

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 2.
Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 1018 Hz.
Solution:
Given: c = 3 × 108, v = 2 × 1018 Hz
To find: Wavelength (λ)
Formula: c = vλ
Calculation. From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{2×10^{18}}\) = 1.5 × 10-10
= 0.15 nm

Question 3.
The speed of light is 3 × 108 m/s. Calculate the frequency of red light of wavelength of 6.5 × 10-7 m.
Answer:
Given: c = 3 × 108 m/s, λ = 6.5 × 10-7 m
To find: Frequency (v)
Formula: c = vλ
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{6.5×10^{-7}}\) = 4.6 × 1014 Hz

Question 4.
Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: v = 8 GHz = 8 × 109 Hz,
c = 3 × 108 m/s
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{8×10^9}\) = 3.75 × 10-2
= 3.75 cm

Question 5.
In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 1010 What is the wavelength of the wave?
Answer:
Given: v = 2 × 1010 Hz, c = 3 × 108 m
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{2×10^{10}}\) = 1.5 × 10-2

Question 6.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?
Answer:
Given: B0 = 5 × 10-7 T, c = 3 × 108
To find: Amplitude of electric field (E0)
Formula: c = \(\frac {E_0}{B_0}\)
Calculation /From formula,
E0 = c × B0
= 3 × 108 × 5 × 10-7
= 150 V/m

Question 7.
A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 106 m)
Answer:
Given: h = 200 m,
Population density (n)
= 1000/km² = 1000 × 10-6/m² = 10-3/m²
R = 6.4 ×106 m
To find: Population covered
Formulae: i. A = πd² = π(\(\sqrt{2Rh}\))² = 2πRh
ii. Population covered = nA
Calculation /From formula (i),
A = 2πRh
= 2 × 3.142 × 6.4 × 106 × 200
≈ 8 × 109
From formula (ii),
Population covered = nA
= 10-3 × 8 × 109
= 8 × 106

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 8.
Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: h = 600 m, R = 6.4 × 106 m
To find: Range (d)
Height to get the double coverage (h’)
Formula: d = \(\sqrt{2hR}\)
Calculation: From formula,
d = \(\sqrt{2×600×6.4×10^6}\) = 87.6 × 10³ = 87.6 km
Now, for A’ = 2A
π(d’)² = 2 (πd²)
∴ (d’)² = 2d²
From formula,
h’ = \(\frac{(d’)^2}{2R}\)
= \(\frac{2d^2}{2R}\)
= 2 × h ……….. (∵ h = \(\frac{d^2}{2R}\))
= 2 × 600
=1200 m

Question 9.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 106 m.
Answer:
Given: ht = 32 m, hr = 50 m, R = 6.4 × 106 m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
dt = \(\sqrt{2Rh_t}\) = \(\sqrt{2×6.4×10^6×32}\)
= 20.238 × 10³ m
= 20.238 km
dr = \(\sqrt{2Rh_t}\)
= \(\sqrt{2×6.4×10^6×50}\)
= 25.298 × 10³ m
= 25.298 km
Now, d = dt + dr
= 20.238 + 25.298
= 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1.
i. What is a wave?
Answer:
Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 2.
What are Lenz’s law, Ampere’s law and Faraday’s law?
Answer:
Lenz’s law:
Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]

11th Std Physics Questions And Answers: