Class 9 – Page 15 – Maharashtra Board Solutions

Introduction to Biotechnology Class 9 Science Chapter 17 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Prasanna

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 17 Introduction to Biotechnology Question Answer Maharashtra Board

Class 9 Science Chapter 17 Introduction to Biotechnology Question Answer Maharashtra Board

1. Each of the following statements is wrong. Rewrite them correctly by changing either one or two words.

a. Simple squamous epithelium is present in the respiratory tract.
b. Glandular epithelium is present in kidneys.
c. Chlorenchyma helps the plant to float in water.
d. Striated muscles are also called involuntary muscles.
e. Chloroplast is present in permanent tissue.
Answer:
a. False. Ciliated columnar epithelium is present in respiratory tract.
b. False. Cuboidal epithelium is present in kidneys.
c. False. Aerenchyma helps the plant to float in water.
d. False. Striated muscles are also called voluntary muscles.
e. False. Chlorenchyma is present in permanent tissue.

2. Identify the odd word and explain why it is odd.

a. Xylem, phloem, permanent tissue, meristematic tissue.
Answer:
Meristematic tissue. It has the ability to divide whereas the others have lost the ability to divide.

b. Epithelium, Muscle fibre, nerve fibre, the epidermis.
Answer:
Epidermis. It is a plant tissue whereas the others are animal tissues.

c. Cartilage, bone, tendon, cardiac muscle.
Answer:
Cardiac muscle. It is a muscular tissue whereas the others are connective tissues.

3. Write the names of the following tissues.

a. Tissue lining inner surface of mouth.
Answer:
Squamous epithelium.

b. Tissue joining muscles and bones.
Answer:
Tendon

c. Tissue responsible for increasing height of plants.
Answer:
Apical meristem.

d. Tissue responsible for increasing girth of stem.
Answer:
Lateral meristem.

4. Write the differences.
Simple tissue and complex tissues in plants.
Answer:

Simple Tissue in plants	Complex Tissues in plants
(i) They are made up of only one type of cells.	(i) They are made up of more than one type of cells.
(ii) They are found in all parts of the plant.	(ii) They are found in the vascular regions of the plant.
(iii) They perform different functions like storage	(iii) They mainly perform the function of
of food, support, giving strength etc.	conduction of water and food.
(iv) Examples – Parenchyma, Collenchyma, Sclerenchyma.	(iv) Example – Xylem and phloem.

5. Write short notes.

a. Meristematic tissue.
Answer:

Meristematic tissue is present in specific parts of a plant where growth takes place.
Cells of meristematic tissue contain thick cytoplasm, a conspicuous nucleus and a thin cell wall and are compactly packed together.
Vacuoles are usually absent in these cells and they are highly active.
The main function of meristematic tissue is to bring about plant growth.
According to the location, meristematic tissues are of three types: Apical meristem, intercalary meristem and lateral meristem.

b. Xylem
Answer:

Xylem is a complex permanent tissue in plants.
It consists of thick-walled dead cells.
The type of cells in xylem are trachieds, vessels, xylem fibres (dead cells) and xylem parenchyma (living cells).
Its structure is like interconnected tubes conduct water and minerals only in upward direction.

c. Striated muscles.
Answer:

The cells of striated muscles are long, cylindrical, multinucleate and have no branches.
These are alternate dark and light bands on these muscles.
As they are attached to bones, they are also called skeletal muscles.
They move as per our will, hence they are called voluntary muscles.
Striated muscles bring about movements of arms and legs, running, speaking etc.

d. Agro-complementary business.
Answer:
(i) The business that are complementary to agriculture and generate supplementary income for the farmers are called agro complementary business.

(ii) These include:

Animal Husbandry: It is practiced for milk production and for using the cattle as help in farming operations, e.g. cows and buffaloes are raised for milk whereas bulls and male buffaloes for pulling heavy loads.
Poultry farming: It is the rearing of egg and meat yielding chickens. Chickens raised for eggs are called layers while those raised for meat are called broilers.
Sericulture: It is the rearing of silkworms (moths) for silk production. The silk fibres obtained are processed, reeled and then woven into fabric.

e. Genetic engineering
Answer:

Genetic engineering is the deliberate modification of the characteristics of an organism by manipulating its genetic material.
An organism that is generated through genetic engineering is called a genetically modified organism (GMO).
Genetic engineering is applied in many fields like research, agriculture, industrial biotechnology and medicine.
In agriculture, genetic engineering is used in the production of cash crops, improvement in varieties of cash crops, increase in ability of plants to withstand environmental stresses.
In medicine, genetic engineering is used for vaccine production, early diagnosis of congenital disease, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

f. Sericulture
Answer:

Sericulture is the rearing of silkworms (moths) for production of silk.
Bombyx mori is the most commonly used variety of silkworm for this purpose.
The life cycle of silkworm consist of four stages namely egg, larva, pupa and adult.
Thousands of eggs deposited by female moths are incubated artificially to shorten the incubation period.
Larvae hatching out of eggs are released on mulberry plants.
Larvae are nourished by feeding on mulberry leaves.
After feeding for 3-4 days, larvae move to branches of mulberry plant.
The silk thread is formed from the secretion of their salivary glands.
Larvae spin this thread around themselves to form a cocoon. The cocoon may be spherical in nature.
Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
These fibres are unwound, processed and reeled. Various kinds of fabric is woven from silk threads.

6. Explain the meaning of biotechnology and its impact on agricultural management with suitable examples.
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called biotechnology. Impact of biotechnology on agricultural management:

Genetically modified crops (GM crops) are being produced by introducing changes in DNA of natural crops.
Normally such varieties are not found in nature.
Thus, new varieties are produced artificially Different useful characters are introduced in such varieties.
Some naturally occurring varieties cannot withstand environmental stress like frequently changing temperature.
Wet and dry famines, changing climates etc. However, GM crops can grow in any of such adverse conditions.
As GM crops are resistant to insect pests, pathogens, chemical weedicides, etc. the use of harmful chemicals like pesticides can be avoided.
Due to use of seeds of GM crops, there is improvement in nutritive value and decrease in loss of crops.

Draw neat and labelled diagrams of the following:

7. Which two main techniques are used in biotechnology? Why?
Answer:

The two main techniques used in biotechnology are – Genetic engineering and tissue culture.
These techniques are used to bring about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings.
Its uses are in the production of cash crops, improvement in varieties of cash crops, early diagnosis of congenital diseases, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

8. Discuss ‘Agritourism’ in the class and write a project on an agrotourism centre nearby. Present it in the class in groups.
Answer:

In agritourism, plantlets of flowering, medicinal, ornamental, vegetable plants and fruit trees are produced on a large scale by tissue culture technique.
By growing some of the plants fully, an agritourism centre can be developed.
If sufficient land is available, the emerging field of agritoursim would be a good business.
An agritourism centre consists of following:
(a) Mango, chikoo (sapota), guava, coconut, custard apple and some other regional fruit trees.
(b) Shade giving local or exotic attractive plants.
(c) Ornamental and flowering plants.
(d) Butterfly garden.
(e) Medicinal plant garden.
(f) Organic vegetables and fruits.
People visit places with such attraction in large numbers.
Selling plantlets/seedlings, fruits, vegetables at such places can be quite profitable.

9. Define the term tissue and explain the concept of tissue culture.
Answer:
A group of cells having the same origin, same structure and same function is called tissue. Concept of tissue culture:

Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium is called tissue culture.
Nowadays, a complete organism can be developed from a single cell or from tissue with the help of the tissue culture technique.
A liquid, solid or gel-like medium prepared from agar, which supplies nutrients and energy necessary for tissue culture is used in this technique.
Tissue culture can be used to grow plants on a large scale, which bear flowers, fruits of excellent quality, in shorter durations and are disease free.
The various processes involved in tissue culture are:

10. ‘Rearing of sheep is a livestock’. Justify this statement.
Answer:

The term livestock refers to animals reared for profit or for use.
Sheep provides us with wool, skin, meat and milk.
Therefore, rearing of sheep is a livestock.

Class 9 Science Chapter 17 Introduction to Biotechnology Intext Questions and Answers

Question 1.
Which components bring about important processes in the living organisms?
Answer:
Cells, tissues, organs, organ systems bring out important processes in living organisms.

Question 2.
Which is the smallest structural and functional unit of the body of living organisms?
Answer:
Cell is the smallest structural and functional unit of the body of living organisms.

Question 3.
Which type of muscle is the diaphragm of the respiratory system?
Answer:
Diaphragm is a skeletal muscle. It is an involuntary muscle that regulates breathing, although some voluntary control can be achieved.

Question 4.
What is the main difference between the growth of animals and plants?
Answer:

Growth in animals is uniform whereas the growth in plants occurs in specific parts of the body.
Growth in animals occurs for a certain period of life whereas the growth in plants occur throughout their life.

Question 5.
Why does the growth of a plant occur only at specific parts of the plant body?
Answer:
Growth of a plant occurs only at specific parts of the body due to the presence of meristematic tissues which contain dividing cells.

Question 6.
Suppose you want to grow a garden like the one shown in the picture, around your home or school. What would you do to achieve that? By which methods will you cultivate the seedlings?

Answer:

To grow a garden around the school or home, one must cultivate seedlings of different plants and those seedlings must be watered regularly and nurtured properly.
Seedlings can be cultivated by the following methods:
(a) Seed sowing: Seeds are sown, watered and allowed to germinate.
(b) Transplantation: Seeds are sown somewhere else and then the seedlings are transplanted in the garden.

Question 7.
You must have seen flowers of same variety but of 2 or 3 different colours borne by same plant. How is this possible?
Answer:

The different coloured flowers borne by the same plant are due to the pigments like anthocyanins, carotenoids etc. according to the genetic makeup of the plant.
Also, by using the latest techniques of biotechnology and tissue culture, it is possible to manipulate the genes for flower colour and get the desired flower colour.

Question 8.
What keeps the various organs and organ systems separate from each other? Why?
Answer:
The epithelial tissue keeps the various organs and organ systems separate from each other by forming a layer on the organs.

Question 9.
Why are epithelial tissues said to be simple tissues?
Answer:
Epithelial tissues are said to be simple tissues as they are made up of only one type of cells.

Question 10.
Why do slim persons feel more cold in winter than those who are obese?
Answer:

Below the skin, there is a tissue called adipose tissue.
The cells of these tissue are filled with fat droplets.
Due to storage of fat, it acts as an insulator and helps to retain heat in the body.
Obese people have more fat deposited in the adipose tissue as compared to slim persons. Therefore, slim persons feel more cold in winter than those who are obese.

Question 11.
Why can bones not be folded?
Answer:

Bone cells called osteocytes are embedded in solid ground substance made up of calcium phosphate.
This makes the bone hard, rigid and non- flexible. Therefore, bones cannot be folded.

Question 12.
Which other industries can be started as an extension of the plant nursery business?
Answer:
Businesses like Agritourism, Ecotourism, forest resorts and organic fruit gardens, yoga and meditation centres can be started as an extension of the plant nursery business.

Question 13.
To which places do people choose to go on vacation in order to relax when they are tired of crowds and stressful life?
Answer:
People choose to go to hill stations, beaches, forest resorts and places where there is lot of greenery to relax when they are tired of crowds and stressful life.

Question 14.
What is the inter-relationship between the two questions (7 and 8) above?
Answer:
The inter-relationship between the above two questions is that businesses like agritourism, ecotourism, forest resorts etc. can fetch a huge profit as people look for these kind of places to relax, to be away from the hustle and bustle of city life and feel close to nature.

Question 15.
Why are the cocoons transferred to boiling water before the pupa develops into an adult?
Answer:

Once the pupa develops into an adult, it will secrete a fluid to dissolve the silk and emerge out.
Hence the cocoons are transferred to boiling water before the pupa develops in to an adult to kill the silkworms before transformation is complete.

Question 16.
Why we cannot see our organs like the heart, blood vessels and intestines?
Answer:
We cannot see our organs like the heart, blood vessels and intestines as they are located inside the body.

Question 17.
What is meant by white revolution? Who was its pioneer? What benefits did it bring?
Answer:

White revolution was the programme launched by the National Dairy Development Board to increase the milk production in India.
Dr. Verghese Kurien was the pioneer of white revolution.
It transformed India from a milk-deficient nation to the largest producer of milk. It also helped the dairy farmers in directing their own development and empowering them.

It also helped to reduce the malpractices carried out by milk traders and merchants.

Answer the following questions:

Question 1.
Collect information about GM varieties of crops in your area and make a note of them. Also find out if there are adverse effects of GM crops on human beings and environment.
Answer:

Some of the GM crops are:
(a) Maize: MON 810, MON 863
(b) Rice: Golden rice
(c) Brinjal: BT brinjal
(d) Cotton: BT cotton
Effects of GM Crops on human beings: GM crops may cause toxicity and allergic reactions in humAnswer:
Effects of GM Crops on environment:
(a) GM crops may be toxic to pollinators and non-target species like butterflies.
(b) Many GM crops may be a threat to soil ecosystem as they secrete their toxins into the soil.

Question 2.
As shown in the figure, place an onion on each gas jar in such a way that its base (roots) will remain dipped in water. Measure and record the length of the roots of both onions on the first, second and third day. On the fourth day, cut off 1 cm of the roots of the onion in flask B Measure the length of the roots of both onions for the next five days and record your observations in the following table.

Length (cm)	Day 1	Day 2	Day 3	Day 4	Day 5
Flask A
Flask B

Answer:
(Students are expected to record their observation in the given table).

(a) Which onion has longer roots? Why?
Answer:
The onion in jar Ahas longer roots as it continues to grow due to the presence of meristematic tissue at root tip which had dividing cells.

(b) Why did the roots of the onion in jar B stop growing?
Answer:
The roots of onion in jar B stopped growing as the meristematic tissue present in the root tips were cut off.

Question 3.
Take a fresh and fleshy leaf of Rhoeo, lily or any other plant. Pull and press it, tearing it obliquely in such a way that its transparent epidermis will be visible at the cut margins. Take the transparent epidermis with the forceps and keep it in dilute safranin solution for 1 minute. Spread it on a slide, cover it with a cover-slip and observe it under a compound microscope.
Answer:

The cells of the epidermis are flat and polygonal and there are no intercellular spaces between them.
A single continuous layer is formed.
The epidermis is covered by a waxy cuticle.

Question 4.
Establish your own plant nursery near your school or home. Prepare the seedlings of flowering plants, fruit plants, and ornamental plants being grown in your area. Can you start a business in the future with the help of this activity? Think it over.
Answer:
(Students are expected to do this activity on their own.)

Yes, we can start a business of plant nursey in future. If sufficient land is available, the emerging field of agritourism would be a good business.

Question 5.
Observe the skin of the back of your hand with the help of a magnifying lens. Do you see the closely attached squarish and pentagonal shapes?
Answer:
Yes, we can see the squarish and pentagonal shapes. This is the stratified squamous epithelium present on the outer layer of skin.

Question 6.
Observe a permanent slide of blood smear under a compound microscope. What did you see?
Answer:

A permanent slide of blood smear under a compound microscope shows different types of cells like RBCs, WBCs and platelets.
The different types of WBCs that can be seen are eosin mphocytes.

Question 7.
Visit a modern cowshed nearby and record the following points – The number of cattle, their variety, total milk production, cleanliness in cattle-shed, arrangements for health care of cattle.
Answer:
Students are expected to do this activity on their own.

Question 8.
Collect more information about animal husbandry.
Answer:

The branch of agriculture which deals with the feeding, shelter, health and breeding of domestic animals is called animal husbandry.
It is a scientific and systematic management of livestock.
Animal husbandry practices include:
(a) Proper food and clean drinking water.
(b) Proper shelter
(c) Proper methods of breeding
(d) Prevention and cure of disease.
Animal husbandary practices serve as an alternative income for the farmers and help to satisfy the need of food for man.

Question 9.
Find out from the internet the average daily milk yield from local and exotic varieties of cow.
Answer:

Milk yield from local cow : 12-15 litres/day.
Milk yield from exotic cow: 15-30 litres/day.

Question 10.
Are the structure and functions of the bodies of plants and animals the same?
No. The structure and functions of the bodies of plants and animals are not the same. Some differences between plants and animals:
Answer:

Plants	Animals
Plants continue to grow throughout their life.	Animals grow only for a specific period of life.
Growth in plants in not uniform.	Growth in animals is uniform.
Plants have dividing and non-dividing tissues.	Animals do not have different dividing and nondividing tissues.
Plants are sedentary.	Animals generally move from place to place in search of food, shelter and partners.
Energy needs of plants are less.	Energy needs of animals are greater.
Plants can prepare their own food by photosynthesis.	Animals cannot prepare their own food. They depend on plants and other animals for their food.

Class 9 Science Chapter 17 Introduction to Biotechnology Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
Protective coverings in the animal body are called tissues.
(a) meristematic
(b) muscular
(c) epithelial
(d) bone
Answer:
(c) epithelial

Question 2.
Epithelial tissue is present in
(a) skin
(b) inner surface of blood vessels
(c) walls of the alveoli
(d) all of these
Answer:
(d) all of these

Question 3.
helps in selective transport of substances.
(a) Squamous epithelium
(b) Stratified epithelium
(c) Cuboidal epithelium
(d) Columnar epithelium
Answer:
(a) Squamous epithelium

Question 4.
helps in secretion of digestive juice, absorption of nutrients.
(a) Glandular epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Ciliated epithelium
Answer:
(b) Columnar epithelium

Question 5.
supports internal organs
(a) Areolar tissue
(b) Adipose tissue
(c) Lymph
(d) Tendon
Answer:
(a) Areolar tissue

Question 6.
tissue helps in insulation, supply of
Answer:
(b) Adipose tissue

Question 7.
connect muscles to bones.
(a) Ligaments
(b) Tendons
(c) Cartilages
(d) Nerves
Answer:
(b) Tendons

Question 8.
join two bones to each other.
(a) Cartilages
(b) Tendons
(c) Ligaments
(d) Muscles
Answer:
(c) Ligaments

Question 9.
meristem increases the length of root and stem.
(a) Intercalary
(b) Apical
(c) Lateral
(d) Permanent
Answer:
(b) Apical

Question 10.
showed that cells and tissues can be grown ex vivo.
(a) Frederick Campion Steward
(b) Gregor Johann Mendel
(c) Frederick Miescher
(d) Robert Brown
Answer:
(a) Frederick Campion Steward

Question 11.
is an example of genetically modified maize.
(a) Vaishali
(b) Vistive Gold
(c) MON 810
(d) Amflora
Answer:
(c) MON 810

Question 12.
is an exotic variety of cow.
(a) Holstein
(b) Gir
(c) Sahiwal
(d) Dangi
Answer:
(a) Holstein

Question 13.
is a layer chicken.
(a) Brahma
(b) Leghorn
(c) Cochin
(d) Aseel
Answer:
(b) Leghorn

Question 14.
is a broiler chicken.
(a) Leghorn
(b) Lehman
(c) Aseel
(d) Minorca
Answer:
(c) Aseel

Question 15.
is a broiler chicken.
(a) Minroca
(b) Ancona
(c) Leghorn
(d) Cochin
Answer:
(d) Cochin

Question 16.
Cartilage is found
(a) all around the cells in body
(b) in nose, ear, larynx, trachea
(c) between the muscles and skin
(d) around the blood vessels
Answer:
(b) in nose, ear, larynx, trachea

Question 17.
Which of the following is an exotic variety of COW?
(a) Plymouth Rock
(b) New Hampshire
(c) Black Rock
(d) Brown Swiss
Answer:
(d) Brown Swiss

Question 18.
epithelium is present in the inner surface of respiratory tract.
(a) Stratified
(b) Columnar
(c) Ciliated
(d) Cuboidal
Answer:
(c) Ciliated

Question 19.
Bones consist of osteocytes embedded in solid ground substance made up of
(a) calcium bicarbonate
(b) calcium carbonate
(c) calcium phosphate
(d) calcium sulphate
Answer:
(c) calcium phosphate

Question 20.
helps in growth of branches, formation of leaves and flowers.
(a) Lateral meristem
(b) Apical meristem
(c) Intercalary meristem
(d) Xylem
Answer:
(c) Intercalary meristem

Question 21.
tissue is present in the hard coat of seeds, outer covering of coconut.
(a) Parenchyma
(b) Xylem
(c) Collenchyma
(d) Sclerenchyma
Answer:
(d) Sclerenchyma

Question 22.
The cells of tissue are dead.
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma
(d) All of these
Answer:
(c) Sclerenchyma

Question 23.
Cuboidal epithelium is found in
(a) inner surface of mouth, blood vessels
(b) inner layer of skin
(c) tubules of kidney, salivary gland
(d) inner surface of respiratory tract
Answer:
(c) tubules of kidney, salivary gland

Find the odd man out:

Question 1.
Cell body, muscle fibre, axon, dendrites.
Answer:
Muscle fibre. It is a muscle cell whereas the others are parts of nerve cell.

Question 2.
Tracheids, vessels, companion cells, xylem fibres.
Answer:
Companion cells. It is an element of phloem whereas the others are elements of xylem.

Question 3.
Sieve tubes, tracheids, phloem parenchyma, phloem fibres.
Answer:
Tracheids. It is an element of xylem whereas the others are elements of phloem.

Question 4.
Holstein, Brown swiss, Bombyx mori, Jersey.
Answer:
Bombyx mori. It is a variety of silkworm whereas the others are varieties of cows.

Complete the analogy:

(1) Inner surface of mouth : Squamous epithelium :: Inner surface of intestine :
(2) Glandular epithelium : Secretion of sweat, oil :: Cuboidal epithelium:
(3) Respiratory tract: Ciliated columnar epithelium :: Kidney tubules:
(4) Outer layer of skin : Stratified epithelium :: Inner layer of skin :
(5) Muscular tissue : Movement :: Nervous tissue :
(6) Tendons : Join muscles to bones :: Ligaments :
Answer:
(1) Columnar epithelium
(2) Secretion of saliva
(3) Cuboidal epithelium
(4) Glandular epithelium
(5) Conduction of excitation
(6) Join two bones to each other

Match the columns:

Question 1.

Column ‘A’	Column ‘B’
(1) Squamous epithelium	(a) Secretion of digestive juice
(2) Cuboidal epithelium	(b) Selective transport of substances
(3) Columnar epithelium	(c) Protection of organs
(4) Stratified epithelium	(d) Secretion of saliva

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

Question 2.

Column ‘A’	Column ‘B’
(1) Brahma	(a) Local variety of cow
(2) Jersey	(b) Layer chicken
(3) Devin	(c) Exotic variety of cow
(4) Lehman	(d) Broiler chicken

Answer:
(1 – d),
(2 – C),
(3 – a),
(4 – h)

Question 3.

Column ‘A’	Column ‘B’
(1) Potato	(a) Vistive Gold
(2) Maize	(b) Amflora
(3) Soybean	(c) Vaishali
(4) Tomato	(d) MON 863

Answer:
(1-b),
(2-d),
(3-a),
(4-c)

Answer the following:

Question 1.
Explain the different types of muscular tissues.
Answer:

Question 2.
Explain the types of complex permanent tissues.
Answer:

Name of tissue	Xylem	Phloem
Characteristics	Consists of thick-walled dead cells	Consists of cytoplasm containing living cells.
Types of cells	Tracheids, vessels and xylem fibres – dead cells. Xylem parenchyma – living cells.	Sieve tubes, companion cells, phloem parenchyma – living cells. Phloem fibres – dead cells.
Function	A structure like interconnected tubes, conduct water and minerals only in upward direction.	Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in upward and downward direction

State whether the following statements are true or false. Correct the false statement:

Question 1.
Lateral meristem increase girth(diameter) of the root and stem.
Answer:
True

Question 2.
Due to use of seeds of GM crops, there is decrease in nutritive value and increase in loss of crops.
Answer:
False. Due to use of seeds of GM crops there is improvement in nutritive value and decrease in loss of crops.

Question 3.
In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
Answer:
True

Question 4.
Rare and endangered plants can be grown by tissue culture and can be protected from extinction.
Answer:
True

Question 5.
The silk thread is formed from the secretion of the sweat glands of silkworm.
Answer:
False. The silk thread is formed from the secretion of the salivary glands of silkworm.

Question 6.
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
Answer:
True

Question 7.
Phloem consists of thick-walled dead cells.
Answer:
False. Phloem consists of living cells containing cytoplasm.

Question 8.
Stratified squamous epithelium is present in inner surface of mouth, oesophagus, blood vessels and alveoli.
Answer:
False. Stratified squamous epithelium is present in the outer layer of skin.

Question 9.
Broiler chickens are raised for eggs.
Answer:
False. Broiler chickens are raised for meat.

Question 10.
Non-striated muscles bring about contraction and relaxation of heart.
Answer:
False. Cardiac muscles bring about contraction and relaxation of heart.

Name the following:

Question 1.
Tissue responsible for growth of branches, formation of leaves and flowers.
Answer:
Intercalary meristem.

Question 2.
Tissue that helps in insulation, supply of energy, storage of fats.
Answer:
Adipose tissue.

Question 3.
Tissue present in outer layer of skin.
Answer:
Glandular epithelium.

Question 4.
Tissue present in nose, ear, larynx, trachea.
Answer:
Cartilage.

Question 5.
Dead cells in xylem.
Answer:
Tracheids, vessels and xylem fibres.

Question 6.
Living cells in xylem.
Answer:
Xylem parenchyma.

Question 7.
Living cells in phloem.
Answer:
Sieve tubes, companion cells, phloem parenchyma.

Question 8.
Dead cells in phloem.
Answer:
Phloem fibres.

Question 9.
GM crops.
Answer:
BT cotton, Amflora, Golden Rice.

Question 10.
Local Indian varieties of cow.
Answer:
Sahiwal, Sindhi, Lai kandhari, Devni, Khillari, Dangi.

Question 11.
Exotic varieties of cow.
Answer:
Jersey, Brown swiss, Holstein.

Question 12.
Layers.
Answer:
Leghorn, Minorca, Ancona, Lehman.

Question 13.
Broilers.
Answer:
Brahma, Long, Cochin, Aseel.

Question 14.
Chickens reared for both eggs and meat.
Answer:
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock.

Give scientific reason.

Question 1.
Blood is a complex tissue.
Answer:

A complex tissue is made up of more than one type of cells.
In blood, cells of different types, colour and shapes are mixed together.
Blood contains erythrocytes (RBCs), leucocytes (WBCs) and platelets in a liquid plasma.
Therefore, blood is a complex tissue.

Define the following:

Question 1.
Tissue
Answer:
A group of cells having the same origin, same structure and same function is called tissue.

Question 2.
Differentiation
Answer:
When cells acquire a specific structure, shape and location and perform a specific function, it is called differentiation.

Question 3.
Biotechnology
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called ‘Biotechnology’.

Question 4.
Tissue culture
Answer:
Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is called tissue culture.

Write short notes

Question 1.
Animal husbandry
Answer:

In India, animal husbandry is practised for milk production and for using the cattle as help in farming operations.
Example – Cows and buffaloes are raised for milk and bulls and male buffaloes for pulling the heavy loads.
Local Indian varieties of cows like Sahiwal, Sindhi, Gir, Lai kandhari, Devni, Khillari, Dangi, etc. and exotic varieties like Jersey, Brown swiss, Holstein, etc. are kept for their milk.
Proper care of cattle is necessary for a clean and high yield of milk which includes:
(a) A balanced diet, i.e. all constituents of food should be given to cattle. It must include fibre- rich coarse food, fodder, and sufficient water.
(b) The cattle-shed should be clean and dry with proper ventilation and a roof.
(c) Cattle should be regularly vaccinated.

Question 2.
Poultry farming.
Answer:

Rearing of egg and meat yielding chickens is called poultry farming.
Chickens raised for laying eggs are called layers . e.g. Leghorn, Minorca, Ancona, Lehman.
Chickens raised for meat are called broilers, e.g. Brahma, Long, Cochin, Aseel.
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
The objectives behind development of new hybrid varieties from a cross between Indian varieties like Aseel and exotic varieties like Leghorn are to produce good quality chickens in large numbers.
To develop the ability to withstand high temperature, to use by-products of agriculture as poultry feed, etc.

Distinguish between:

Question 1.
Xylem and Phloem
Answer:

Xylem	Phloem
(i) Consists of mostly thick-walled dead cells.	(i) Consists of mostly living cells containing cytoplasm.
(ii) The types of cells include dead cells- tracheids, vessels and xylem fibres and living cells-Xylem parenchyma.	(ii) The types of cells include living cells – Sieve tubes, companion cells, phloem parenchyma and dead cells – phloem fibres.
(iii) Structure like interconnected tubes, conduct water and minerals only in upward direction.	(iii) Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in both upward and downward direction.

Question 2.
Striated muscles and Non-striated muscles.
Answer:

Striated muscles	Non-striated muscles
(i) Muscle cells are long, cylindrical, multinucleate and have no branches.	(i) Muscle cells are short, spindle-shaped, uninucleate and have no branches.
(ii) There are alternate dark and light bands on these muscles and they are attached to bones.	(ii) Dark and light bands are absent. Not attached to bones.
(iii) They move as per our will, hence they are called voluntary muscles.	(iii) They are not under the control of our will, hence they are called involuntary muscles.
(iv) These muscles bring about movements of arms and legs, running, speaking etc.	(iv) These muscles bring about movement of eyelids, passage of food through alimentary canal, contraction and relaxation of blood vessels etc.

Question 3.
Cartilage and Bone
Answer:

Cartilage	Bone
(i) They are present in nose, ear, larynx, trachea.	(i) They form the skeleton of the body.
(ii) They contain cells supported by fibrous, flexible, jelly-like ground substance.	(ii) They contain osteocytes embedded in solid ground substance made up of calcium phosphate.
(iii) Lubricates the surface of bones, gives support and shape to organs.	(iii) Supports and protects different organs, helps in movement.

Answer the following questions:

Question 1.
Close your eyes and try to identify different objects by feeling them with your hand. Why is it possible for you to identify things like a note-book, text-book, bench, compass-box, etc. only by touching them?
Answer:

We can identify objects just by touching them because of the memory that we retain in our brain.
The nervous tissue enables us to respond to the stimuli of touch.

Answer in detail:

Question 1.
Write the location, structure and function of following epithelial tissues:
(a) Squamous epithelium
(b) Stratified epithelium
(c) Glandular epithelium
(d) Columnar epithelium
(e) Ciliated epithelium
(f) Cuboidal epithelium
Answer:

Question 2.
Write the structure and function of the following connective tissues:
(a) Blood
(b) Lymph
(c) Areolar tissue
(d) Adipose tissue
(e) Cartilage
(f) Bones
(g) Tendons and ligaments
Answer:

Question 2.
Describe the structure of nervous tissue with the help of a neat labelled diagram.
Answer:

Nervous tissue enables us to respond to stimuli like touch, sound, odour, colour, etc.
Cells of the nervous tissue are specifically made to become excited and conduct the excitation from one part of the body to another.
The main part of the nerve cell is the cell body which contains the cytoplasm and the nucleus.
Numerous, small, branched fibres called dendrites arise from the cell body.
One of the fibres, is extremely long and is called
the axon.
The length of the nerve cell may be up to one metre.
Many nerve cells are bound together with the help of connective tissue to form a nerve.
Nervous tissue is present in the brain, spinal cord and the network of nerves spread all throughout the body.
In most animals, action in response to a stimulus occurs due to the integrated functioning of the nervous tissue and muscular tissue.

Question 3.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Question 4.
Explain types, location and function of Meristematic tissue in tabular form:
Answer:

Types	Apical	Intercalary	Lateral
Location	At tip of the root and stem	At the base of the petiole of leaves and branches	At sides of root and stem
Function	Increase the length of the root and stem	Growth of branches, the formation of leaves and flowers	Increases diameter of the root and stem

Question 5.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Name of Tissue	Parenchyma	Collenchyma	Sclerenchyma
Structure of cells	Living cells with thin cell wall and intercellular spaces.	Elongated living cells with thickened cell wall at corners due to cellulose and pectin.	Dead and fibrous cells with tapering ends, cell wall contains lignin.
Location	All parts like roots, stem, leaves, flowers and seeds.	At the base of leaf petiole, branches and stem	Stem, veins of leaves, hard coats of seeds, outer covering of coconut.
Functions	Support, storage of food and filling vacant spaces.	Support and flexibility to various parts.	Give strength and rigidity to parts of the plants.
Sub types	Chlorenchyma: Leaves, perform photosynthesis. Aerenchyma: Helps aquatic plants, leaves and stem to float.

Question 6.
Write down the applications of biotechnology (tissue culture) in floriculture, nurseries and forestry.
Answer:
Applications of biotechnology (tissue culture) in floriculture, nurseries and foresty:

Tissue culture can be used to grow those plants on a large scale which bear flowers, fruits of excellent quality.
Fully grown plants can be produced in shorter durations.
Plants can be grown on a large scale even if means of pollination or germinating seeds are not available.
For example, orchids or pitcher plants do not germinate but these plants can easily be produced by means of tissue culture.
In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
Bioreactors are useful for producing plantlets on a very large scale.
A large number of seedlings/plantlets can be produced in a short time using minimum resources and materials.
Usually, plants produced by tissue culture and genetic modification techniques are disease-free.
Plantlets produced by tissue culture technique of the meristem are virus-free.
Embryos produced using conventional hybridization technique between two or more varieties may not grow fully for some reasons.
However, embryos produced by tissue culture technique always complete their growth.
Rare and endangered plants can be grown using tissue culture technique and can thus be protected from extinction.
Similarly, various parts and seeds of such plants can be preserved by tissue culture and those varieties can be protected.

9th Std Science Questions And Answers:

Classification of Plants Class 9 Science Chapter 6 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Prasanna

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 6 Classification of Plants Question Answer Maharashtra Board

Class 9 Science Chapter 6 Classification of Plants Question Answer Maharashtra Board

1. Match the proper terms from columns A and C with the description in column B.

2. Complete the sentences by filling in the blanks and explain those statements.
(angiosperms, gymnosperms, spore, Bryophyta, thallophyta, zygote)
a. ……………….. plants have soft and fiber-like body.

b. ……………….. is called the ‘amphibian’ of the plant kingdom.
Answer:
Bryophyta plant is called the ‘amphibian’ of the plant kingdom

c. In pteridophytes, asexual reproduction occurs by ……………….. formation and sexual reproduction occurs by ………………..formation.
Answer:
Spore, zygote: Pteridophyta plants show alteration of generation. One generation reproduces by spore-formation and the next generation reproduces sexually by zygote formation.

d. Male and female flowers of ……………….. are borne on different sporophylls of the same plant.
Answer:
Gymnosperms bear their male and female flowers on different sporophylls of the same plant

3. Answer the following questions in your own words.

a. Write the characteristics of subkingdom Phanerogams.
Answer:

Plants which have special structures for reproduction and produce seeds are called Phanerogams.
In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food.
During the germination of the seed, the stored food is used for the initial growth of the embryo.
Depending upon whether seeds are enclosed in a fruit or not phanerogams are classified into gymnosperms and angiosperms.

b. Distinguish between monocots and dicots.
Answer:

	Dicots	Monocots
Seed	Two cotyledons	Single cotyledon
Root	Well developed, primary root (Taproot)	Fibrous roots
Stem	Strong, hard. e.g. Banyan tree	Hollow, e.g. Bamboo False, e.g. Banana Disc-like, e.g. Onion.
Leaf	Reticulate venation	Parallel venation
Flower	Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)	Flowers with 3 parts or in multiples of three (trimerous).

c. Write a paragraph in your own words about the ornamental plants called ferns.
Answer:

Ferns belong to the group of plants called Pteridophyta.
They have well-developed roots, stem and leaves but do not bear flowers and fruits.
They have separate tissues for the conduction of food and water.
They reproduce with the help of spores formed along the back or posterior surface of their leaves.
They reproduce asexually by spore formation and sexually by zygote formation.

d. Sketch, label and describe the Spirogyra.

e. Write the characteristics of the plants belonging to division Bryophyta.
Answer:

Bryophyta group of plants are called the amphibians of the plant kingdom because they grow in moist soil but need water for reproduction.
These plants are thalloid, multicellular and autotrophic.
They reproduce by spore-formation.
Their plant body structure is flat, ribbon-like, long, without true roots, stem and leaves.
Instead, they have stem-like or leaf-like parts and root-like rhizoids.
They do not have specific tissues for the conduction of food and water.
Examples: Moss (Funaria), Anthoceros, Riccia etc.?

4. Sketch and label the figures of the following plants and explain them into brief.
Marchantia, Funaria, Fern, Spirogyra.

Question 1.
Spirogyra.
Answer:

Spirogyra belongs to the division thallophyta. They are called as algae.
It grows mainly in water.
It does not have specific parts like root-stem- leaves-flowers but are autotrophic due to the presence of chlorophyll.
The plant body of Spirogyra is soft and fibre-like.
It has spirally arranged chloroplasts in its cell.

Question 2.
Funaria and Marchantia (Bryophyta)
Answer:

These plants are called ‘amphibians’ of the plant kingdom because they grow mostly in soil and need water for reproduction.
They do not have specific tissues for the conduction of food and water.
The plant body is fiat, ribbon-like long, without true roots, stem and leaves
Instead, they have stem-like or leaf-like parts and root like rhizoids.

Question 3.
Fern (Pteridophyta):
Answer:

They have well-developed roots, stem and leaves for the conduction of food and water.
They do not bear flowers and fruits.
They reproduce with the help of spores present along the back or posterior surface of the leaves.

5. Collect a monocot and dicot plant available in your area. Observe the plants carefully and describe them in scientific language.

6. Which criteria are used for the classification of plants? Explain with reasons.
Answer:
Criteria for classification of plants:

If plants do not bear flowers, fruits and seeds, they are non-seed bearing plants. If they bear flowers, fruit and seeds, they are seed-bearing plants.
Presence or absence of conducting tissues- Plants such as pteridophytes, gymnosperms and angiosperms which possess conducting tissues are included in vascular plants whereas thallophytes and bryophytes which do not possess conducting tissues are included under non-vascular plants.
Depending upon whether the seeds are enclosed in fruit or not, plants are classified as gymnosperms (naked-seeds) and angiosperms (seeds covered by fruit)
Depending upon the number of cotyledons in seeds, plants are classified into dicotyledons and monocotyledons

Class 9 Science Chapter 6 Classification of Plants Intext Questions and Answers

Can you recall?

Question 1.
How have living organisms been classified?
Answer:
(i) Organisms have been classified based on the following:

Cell structure
Body Organisation
Mode of nutrition
Reproduction

(ii) Organisms are also classified at kingdom level and groups and subgroups.

Activity-based questions

Question 1.
You may have seen a lush green soft carpet on old walls, bricks and rocks in the rainy season. Scrape it gently with a small ruler, observe it under a magnifying lens and discuss.
Answer:

It shows considerable tissue complexity and is differentiated into two main parts: a root and a shoot.
They have a variety of specialized tissues within these two regions of the body.
Same kind of cells are seen throughout the whole body except reproductive cells.

Question 2.
You may have seen ferns among the ornamental plants in a garden. Take a leaf of a fully grown fern and observe it carefully.
Answer:

New leaves typically expand by the unrolling in a tight spiral manner.
The anatomy of fern leaves can either be simple or highly divided.
They show the presence of spores formed along the back or posterior surface of their leaves.

Question 3.
Observe all garden plants like Cycas, Christmas tree, Hibiscus, Lily, etc. and compare them. Note the similarities and differences among them. Which differences did you notice in gymnosperms and angiosperms?
Answer:
Cycas and Christmas tree are gymnosperms, whereas Hibiscus and lily are angiosperms.

Similarities: These plants have special structures for reproduction and produce seeds. During the germination of the seed, the stored food is used for the initial growth of the embryo.
Differences: In gymnosperms, reproductive organs have cones whereas in angiosperms reproductive organs have flowers.
In gymnosperms, seeds are without natural coverings whereas in angiosperms seeds are enclosed in natural coverings called fruits.

Question 4.
Soak the seeds of corns, beans, groundnut, tamarind, mango, wheat, etc. in water for 8 to 10 hrs. After they are soaked, check each seed to see whether it divides into two equal halves or not and categorize them accordingly.
Answer:
Monocots: com, wheat (it cannot be divided into equal halves)
Dicots: beans, groundnut, tamarind and mango (it can be divided into two equal halves)

Class 9 Science Chapter 6 Classification of Plants Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The five-kingdom classification was proposed b7
(a) Robert Whittaker
(b) Robert Hooke
(c) Eichler
(d) Louis Pasteur
Answer:
(a) Robert Whittaker

Question 2.
In 1883, classified plants into two sub-kingdoms.
(a) Robert Whittaker
(b) Alexander Fleming
(c) Eichler
(d) Robert Hooke
Answer:
(c) Eichler

Question 3.
Ulothrix, ulva, sargassum belong to
(a) Bryophyta
(b) Thallophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(b) Thallophyta

Question 4.
is a bryophyte.
(a) Ulva
(b) Nephrolepis
(c) Funaria
(d) Equisetum
Answer:
(c) Funaria

Question 5.
In the seeds are naked.
(a) Pteridophyta
(b) Angiosperms
(c) Gymnosperms
(d) Bryophyta
Answer:
(c) Gymnosperms

Question 6.
In the flowers are reproductive organs.
(a) Angiosperms
(b) Gymnosperms
(c) Pteridophyta
(d) Bryophyta
Answer:
(a) Angiosperms

Question 7.
In the flowers are tetramerous or pentamerous.
(a) Monocotyledons
(b) Dicotyledons
(c) Gymnosperms
(d) Pteridophyta
Answer:
(b) Dicotyledons

Question 8.
In monocotyledonous plants, the stem is
(a) hollow
(b) false
(c) disc-like
(d) all of these.
Answer:
(d) all of these

Question 9.
Lycopodium belongs to
(a) Thallophyta
(b) Bryophyta
(c) Gymnosperms
(d) Pteridophyta
Answer:
(d) Pteridophyta

Question 10.
Leaves of show reticulate venation.
(a) Bamboo
(b) Banana
(c) Onion
(d) Banyan
Answer:
(d) Banyan

Question 11.
Various types of fungi like yeasts and moulds are included in the group
(a) Thallophyta
(b) Halophyte
(c) Xenophyta
(d) Angiosperms
Answer:
(a) Thallophyta

Question 12.
Bryophytes have a root-like structure called
(a) Nodes
(b) Rhizoids
(c) Nodules
(d) Aerenchyma
Answer:
(b) Rhizoids

Question 13.
reproduce with the help of spores formed along the back or posterior surface of their leaves.
(a) Halophyta
(b) Pteridophyta
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Pteridophyta

Question 14.
In ……………………….., the reproductive organs cannot be seen.
(a) Pteridophyta
(b) Cryptogams
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Cryptogams

Question 15.
are mostly evergreen, perennial and woody.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 16.
Gymnosperms bear male and female flowers on different of the same plant.
(a) Branches
(b) Roots
(c) Sporophylls
(d) Flowers
Answer:
(c) Sporophylls

Question 17.
In the seeds are not enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 18.
In the seeds are enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(d) Angiosperms

Question 19.
The plants whose seeds cannot be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(d) Monocotyledons

Question 20
The plants whose seeds can be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(c) Dicotyledons

Find the odd one out:

Question 1.
Ulothrix, Ulva, Nephrolepis, Sargassum
Answer:
Nephrolepis: It belongs to division Pteridophyta whereas the others belong to division thallophyta.

Question 2.
Funaria, Marchantia, Anthoceros, Spirogyra
Answer:
Spirogyra:

Question 3.
Marsilea, Pteris, Lycopodium, Riccia
Answer:
Riccia:

Question 4.
Cycas, Mango, Apple, Banyan
Answer:
Cycas:

Question 5.
Onion, Papaya, Wheat, Green peas
Answer:
Green peas:

Complete the analogy:

(1) Spirogyra : Thallophyta : : Riccia :
(2) Moss : Bryophyta : : Selaginella :
(3) Nephrolepis : Pteridophyta :: Ulothrix :
(4) Pteridophyta : Roots :: Bryophyta :
(5) Gymnosperms : naked seeds : : Angiosperms :
(6) Dicotyledon : Reticulate venation : : Monocotyledon:
(7) Bamboo stem: Hollow:: Onion Stem:
(8) Monocotylendon : Tap root:: Dicotyledon :
Answer:
(1) Bryophyta
(2) Pteridophyta
(3) Thallophyta
(4) Rhizoids
(5) Covered seeds
(6) Parallel venation
(7) Disc like
(8) Fibrous roots

Difference between:

Question 1.
Thallophyta and Bryophyta
Answer:

Thallophyta	Bryophyta
These plants grow mainly in water	They grow in moist soil but need water for reproduction

Question 2.
Gymnosperms and Angiosperms
Answer:

Gymnosperms	Angiosperms
No natural covering on seeds	Seeds are formed in fruits

Question 3.
Algae and Moss
Answer:

Algae	Moss
These plants mainly grow in water.	These plants need water for reproduction.

State whether the following statements are true or false. Correct the false statements:

(1) Thallophyta are called as the amphibians of the plant kingdom.
(2) Fungi like yeasts and moulds are included in division bryophyta.
(3) Moss (Funaria) belongs to division bryophyta.
(4) Bryophyta have specific tissues for conduction of food and water.
(5) Plants belonging to Thallophyta group are only unicellular.
(6) Pteridophytes have well developed roots, stems and leaves.
(7) Pteridophytes reproduce with the help of spores formed along the back or posterior surface of their leaves.
(8) Nephrolepis belongs to division Pteridophyta.
(9) Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into monocots and dicots.
(10) Gymnosperms are mostly evergreen, perennial and woody.
(11) Gymnosperms bear male and female flowers on different sporophylls of different plants.
(12) In Angiosperms, the seeds are covered by fruits.
(13) Dicotyledonous plants show reticulate venation.
(14) Moncotyledonous plants have trimerous flowers.
(15) In dicotyledonous plants, the stem is strong and hard.
Answer:
(1) False. Thallophyta plants grow mainly in water.
(2) False. Fungi like yeasts and moulds are included in division thallophyta.
(3) True
(4) False. Bryophyta do not have specialised tissuesfor conduction of food and water.
(5) False. Plants belonging to thallophyta group may be unicellular or multicellular.
(6) True
(7) True
(8) True
(9) False. Depending whether seeds are enclosed in. a fruit or not, angiosperms are classified into monocots and dicots.
(10) True
(11) False. Gymnosperms bear male and female flowers on different sporophylls of the same plant.
(12) True
(13) False. Dicotyledonous plants show parallel venation.
(14) True
(15) True.

Give name

Question 1.
What are ornamental plants are called?
Answer:
Ferns

Question 2.
Plants with two cotyledons are called.
Answer:
Dicots

Question 3.
Plants with single cotyledon are called.
Answer:
Monocots

Question 4.
Type of venation showed by hibiscus plant leaves
Answer:
Reticulate venation

Question 5.
Type of venation showed by lily plant leaves
Answer:
Parallel venation

One line answers

Question 1.
Which plants are mostly evergreen, perennial and woody?
Answer:
Gymnosperms are mostly evergreen, perennial and woody.

Question 2.
Which type of venation showed by dicot plants?
Answer:
Leaves of dicot plants show reticulated venation.

Question 3.
Which type of venation showed by monocot plants?
Answer:
Leaves of monocot plants show parallel venation

Question 4.
How are angiosperms classified into monocot and dicot?
Answer:
Depending whether seeds and enclosed in fruit or not, angiosperms are classified into monocot and dicot

Question 5.
In which division are fungi like moulds and yeast classified?
Answer:
Fungi like moulds and yeast classified in division thallophyta.

Question 6.
Plants belonging to which group may be unicellular or multicellular?
Answer:
Plants belonging to thallophyta group may be unicellular or multicellular

Give scientific reason

Question 1.
Thallophyta plants have thin and fibre like body
Answer:
Thallophyta: These plants grow mainly in water i.e. fresh water as well as in saline water, therefore they usually have a soft and fibre-like (filamentous) body.

Question 2.
Bryophyta plants are called the amphibian plants.
Answer:
Bryophyta: They grow in moist soil but need water for reproduction. Therefore, they are called ‘amphibians of plant kingdom’.

Question 3.
Gymnosperms bear their male and female flowers on different sporophylls of the same plant
Answer:
Gymnosperms: As these plants do not take the assistance of pollinators i.e. vectors, the male and female flowers are present on the different sporophyll of the same plant for successful fertilisation.

Write note on

Question 1.
August W. Eichler
Answer:
In 1883, Eichler, a botanist, classified the Kingdom Plantae into two subkingdoms. As a result, two subkingdoms, cryptogams and phanerogams were considered for plant classification.

Question 2.
Thallophyta
Answer:
These plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 3.
Bryophyta
Answer:
This group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 4.
Pteridophyta
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 5.
Phanerogams
Answer:
Plants which have special structures for reproduction and produce seeds are called phanerogams. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Question 6.
Gymnosperms
Answer:
Gymnosperms are mostly evergreen, perennial and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 7.
Angiosperms
Answer:
The flowers these plants bear are their reproductive organs. Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Complete the flow chart.

Question 1.
Living Organisms

Answer:

Question 2.
Kingdom: Plantae

Answer:

Distinguish between:

Question 1.
Bryophyta and Pteridophyta:
Answer:

Bryophyta	Pteridophyta
Bryophytes grow in soil but need water for reproduction.	Pteridophytes grow in soil.
Plant body is without specific parts like true roots, stem and leaves.	Plant body is differentiated into true roots, stem and leaves.
Conducting tissues for food and water absent.	Conducting tissues for food and water present.
Examples: Moss (Funaria), Marchantia, Anthoceros, etc.	Examples: Nephrolepis, Marsilea, Pteris, Adiantum, Lycopodium etc.

Question 2.
Angiosperms and Gymnosperms.
Answer:

Angiosperms	Gymnosperms
(i) In Angiosperms, the stems have branches.	(i) In Gymnosperms, the stems are without branches.
(ii) Reproductive organs are flowers.	(ii) Reprodcutive organs are cones.
(iii) Seeds are enclosed in natural coverings, i.e., fruits.	(iii) Seeds are not enclosed in natural coverings.
(iv) Examples: Mango, Bamboo, etc.	(iv) Examples: Cycas, Picea etc.

Question 3.
Cryptogams and Phanerogams.
Answer:

Cryptogams	Phanerogams
(iii) Their reproductive organs are hidden.	(iii) Their reproductive organs are exposed.
(iii) They reproduce by forming spores.	(iii) They reproduce by forming seeds.
(iii) They are less evolved plants.	(iii) They are highly evolved plants.
(iv) They are divided into Thallophyta,	(iv) They are divided into Gymnosperms and
Bryophyta, Pteridophyta.	Angiosperms.

Distinguish between:

Question 1.
Thallophyta
Answer:
Spirogvra, Ulothrix, Ulva, Sargassum

Question 2.
Bryophyta
Answer:
Moss (Funaria), Marchantia, Anthoceros, Riccia

Question 3.
Pteridophyta
Answer:
Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium

Question 4.
Gymnosperms
Answer:
Cycas, Picca (Christmas tree), Thuja (Morpankhi), Pinus (Deodar)

Question 5.
Angiosperms
Answer:
Tamarind, Mango, Apple, Lemon

Question 6.
Monocot plants
Answer:
Bamboo, bananas, corn, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

Question 7.
Dicot plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

Observe the figure and answer the questions

1. Dicot Plants

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Well developed, primary root (Tap root)

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Reticulate Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Strong and hard

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Two cotyledons

Question 6.
Give example of the following types of plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

2. Monocot Plants

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Fibrous roots

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 3 parts or in multiples of three (trimerous).

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Parallel Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Hollow, False or Disc-like

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Single cotyledons

Question 6.
Give example of the following types of plants
Answer:
Bamboo, bananas, com, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

3. Spirogyra

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division I Thallophyta.

Question 2.
Where does this plant grow?
Answer:
These plants grow mainly in water.

Question 3.
Are these types of plants unicellular or multicellular?
Answer:
They may be unicellular or multicellular and microscopic or large.

Question 4.
Are these types of plant autotropic?
Answer:
They are autotrophic due to the presence of chlorophyll but types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 5.
Do these plants have a root-stem-leaves-flowers system?
Answer:
They do not have specific parts like root-stem- leaves-flowers.

Question 6.
How is the body of these types of plants?
Answer:
These plants usually have a soft and fibre-like body.

4. Funaria

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division II Bryophyta.

Question 2.
Where does this plant grow?
Answer:
They grow in moist soil but need water for reproduction.

Question 3.
What are these group of plants called in the plant kingdom?
Answer:
This group of plants is called the ‘amphibians’ of the plant kingdom.

Question 4.
Are these types of plant autotropic?
Answer:
They reproduce by spore formation.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves.

Question 6.
What do these plants have instead of roots?
Answer:
They have root like rhizoids.

5. Fern

Question 1.
Which division of plants does this plant come? under?
Answer:
This plant come under Division III Pteridophy ta.

Question 2.
Where does this plant grow?
Answer:
They grow in soil.

Question 3.
How do these plants reproduce?
Answer:
These plants reproduce asexually by spore- formation and sexually by zygote formation.

Question 4.
Do these plants produce flowers and fruits?
Answer:
They do not bear flowers and fruits.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water.

Question 6.
Where are the spores formed in the plants body?
Answer:
The spores formed along the back or posterior surface of their leaves.

6. Cycas

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division III Phanerogams Division I Gymnosperms.

Question 2.
Explain structure of these types of plants?
Answer:
Gymnosperms are mostly evergreen perennial and woody.

Question 3.
How is stem and leaves of these types of plants?
Answer:
Their stems are without branches and the leaves form a crown.

Question 4.
Where are the male and female flowers located?
Answer:
These plants bear male and female flowers on different sporophylls of the same plant.?

Question 6.
Give some examples of these types of plants?
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

7. Monocot and Dicot plants

Question 1.
Which division of plants does this plant come under?
Answer:
ThisplantcomeunderDivisionlllPhanerogams Division II Angiosperms

Question 2.
How are the seeds of these types of planis?
Answer:
The seeds are formed within fruits thus these seeds are covered

Question 3.
How can we classify the plants according to their seeds in this division?
Answer:
The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Question 4.
How the venations are present on the leaves of these types of plants?
Answer:
These plants bear parallel or reticulated venations on the leaves.

Question 5.
How is the root system of these types of plants?
Answer:
The root systems of these types of plant are tap roots or fibrouš roots.

Complete the paragraph

Question 1.
Thallophyta plants grow mainly in …………….. . This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of …………….., is called algae. Algae show great diversity. They may be unicellular or …………….., and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a …………….. and fibre-like body. Various types of …………….. like yeasts and moulds which do not have …………….. are also included in this group.
Answer:
Thallophyta plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 2.
…………….. group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need …………….. for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by …………….. formation. The structure of the plant body of bryophytes is flat, ribbon?like long, without true …………….., stem and leaves. Instead, they have stem-like or leaf?like parts and root-like ……………. . They do not have specific …………….. for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.
Answer:
Bryophyta group of plants is called the ‘amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 3.
Plants from Pteridophyta group have well developed roots, stem and leaves and separate …………….. for conduction of food and water. But,
they do not bear …………….. and ……………… They reproduce with the help of …………….. formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce …………….. by spore-formation and sexually by …………….. formation. They have a well-developed conducting system.
Answer:
Plants from Pteridophyta group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 4.
Phanerogams plants which have special structures for …………….. and produce …………….. In these plants, after the process of reproduction, seeds are formed which contain the …………….. and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in …………….. a or not, phanerogams are classified into …………….. and ……………. .
Answer:
Phanerogams plants which have special structures for reproduction and produce seeds. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Question 5.
Gymnosperms are mostly …………….., perennial and woody. Their stems are without …………….. The leaves form a …………….. These plants bear male and female flowers on different …………….. of the same plant …………….. of these plants do not have natural coverings, i.e. these plants do not form …………….. and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.
Answer:
Gymnosperms are mostly evergreen, perennial • and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 6.
The flowers of Angiosperms plants bear are their …………….. organs Flowers develop into …………….. and seeds are formed within …………….. . Thus, these seeds are ……………..; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called …………….. plants and those whose seeds cannot be divided into equal parts are called …………….. plants.
Answer:
The flowers of Angiosperms plants bear are their reproductive orgAnswer: Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Answer the questions in detail:

Question 1.
Write the characteristics of Thallophyta.
Answer:

Thallophyta plants grow mainly in water.
The group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll are called algae.
Algae show great diversity They may be unicellular or multicellular and microscopic or large.
Some of these are found in freshwater while some are found in saline water.
Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.
Examples: Spirogyra, Ulothrix, Ulva, etc.

Question 2.
Write the characteristics of Gymnosperms.
Answer:

Gymnosperms are mostly evergreen, perennial and woody.
Their stems are without branches.
The leaves form a crown.
These plants bear male and female flowers on different sporophylls of the same plant.
Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms (gmnos: naked, sperms: seeds)
Examples: Cycas, Picea (christmas tree), Thuja, Pinus (deodar), etc.

Make concept diagram

Question 1.
Plant classification
Answer:

Question 2.
Taxonomy of carnivorous 1ant
Answer:

Question 3.
Taxonomy of mango plant
Answer:

9th Std Science Questions And Answers:

Useful and Harmful Microbes Class 9 Science Chapter 8 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Prasanna

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 8 Useful and Harmful Microbes Question Answer Maharashtra Board

Class 9 Science Chapter 8 Useful and Harmful Microbes Question Answer Maharashtra Board

1. Complete the statements using the proper option from those given below. Explain the statements. (mycotoxins, budding, rhizobium)

a. Yeast reproduces asexually by the …………………….. method.
Answer:
The yeast cells develop small round bodies on the parent cell. These are called buds. New daughter cells develop from these buds.

b. Toxins of fungal origin are called …………………….. .
Answer:
Mycotoxins are poisonous chemicals released into the food by fungi. This makes the food poisonous.

c. Leguminous plants can produce more proteins due to …………………….. .
Answer:

(i) Nitrogenous compounds are required to produce proteins.
(ii) Rhizobia produce nitrogenous compounds by fixing atmospheric nitrogen and make it available for their host plants like leguminous plants.

2. Write the names of microbes found in the following food materials.
yogurt, bread, root nodules of leguminous plants, idli, dosa, spoiled potato curry.
Answer:

Food materials	Microbes
Yogurt	Lactobacilli
Bread	Yeast
Root nodules of leguminous plants	Rhizobium
Idli	Yeast, bacteria
Dosa	Yeast, bacteria
Spoiled potato curry	Clostridium

3. Identify the odd word out and say why it is the odd one?

a. Pneumonia, diphtheria, chicken pox, cholera.
Answer:
Chickenpox. It is caused by a virus, whereas others are caused by bacteria.

b. Lactobacilli, rhizobia, yeast, clostridia.
Answer:
Yeast. It is a fungus, whereas the rest are bacteria.

c. Root rot, rust (tambura), rubella, mozaic.
Answer:
Rubella. It is a disease of humans, whereas the rest are diseases of plants.

4. Give scientific reasons.

a. Foam accumulates on a the surface of ‘dal’ kept for a long time in summer.
Answer:

Dal is rich in proteins.
During summer, bacteria attack the dal and cause fermentation resulting in the production of carbon dioxide gas.
Therefore, foam accumulates on the surface of the ‘dal’ kept for long time in summer.

b. Why are naphthalene balls kept with clothes to be put away.
Answer:

Naphthalene balls are balls of chemical pesticide and deodorant.
They help to kill or repel insects such as moths, cockroaches, mice etc.
Therefore, naphthalene balls are kept with clothes to be put away to prevent clothes from getting damaged.

5. Write down the modes of infection and the preventive measures against fungal diseases.
Answer:

Mode of infection: Contact with infected person or his/her belongings like clothes.
Preventive measure: Personal hygiene and avoid contact with infected person.

6. Match the pairs.

‘A’ group ‘B’ group
1. Rhizobium a. Food poisoning
2. Clostridium b. Nitrogen fixation
3. Penicillium c. Bakery products
4. Yeast d. Production of antibiotics
Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

7. Answer the following questions.

a. Which vaccines are given to infants? Why?
Answer:

Hepatitis A and B, DTP (Diphtheria, Tetanus, Pertussis.) Polio, MMR (Measles, Mumps, Rubella), Chicken pox, Influenza, Tetanus, BCG, Rotavirus, etc.
Vaccines consist of dead or weakend microbes. When these are swallowed or injected, the body produces antibodies to fight them.
These antibodies remain in the body and protect it from any future attack of the disease causing microbes.
Therefore, vaccines are given to infants for preventing diseases.

b. How is a vaccine produced?
Answer:

Vaccines are made using the disease causing bacteria or virus but in a form that will not harm the human beings.
Vaccine is made from dead or weakened microbes or their toxins.
Vaccine stimulates the immune system to produce antibodies which give life-long protection against the disease.
There are specific vaccines for specific diseases.

c. How do antibiotics cure disease?
Answer:
Antibiotics cure diseases by destroying or preventing the growth of harmful micro-organisms.

d. Are the antibiotics given to humans and animals the same? Why?
Answer:

Generally, antibiotics work against any harmful bacteria, whether it is attacking humans or animals.
But some of them are better suited to humans while some are better for animals. This is due to the adverse effects they show in different species.
Also, the dosages of antibiotics for humans and animals differ.

e. Why is it necessary to safely store the pathogens of a disease against which vaccines are to be produced?
Answer:

Pathogens are microbes which can cause diseases in us.
For the preparation of a vaccine, a particular pathogen is cultured and grown in a laboratory.
If these pathogens are not safely stored, they many get modified due to environmental factors, resulting in decrease in the efficiency of the vaccine.
Also, the live pathogens may escape and cause diseases in us.

8. Answer the following questions in brief.

a. What are ‘broad-spectrum antibiotics’?
b. What is fermentation?
Answer:

Yeast uses sugar for food.
Yeast grows and multiplies rapidly due to the carbon compounds in the sugar solution.
In the process of obtaining nutrition, yeast cells convert the carbohydrates in the food into alcohol and carbon dioxide.
Also, the bacteria Lactobacilli convert lactose, the sugar in milk into lactic acid.
This process is called fermentation.

c. Define ‘Antibiotic’.
Answer:

Carbon compounds obtained from some bacteria and fungi for destroying or preventing the growth of harmful micro -organisms are called antibiotics.
Antibiotics, a discovery of the 20th century, have brought a revolution in the field of medicine.
Antibiotics mainly act against bacteria. Some antibiotics can destroy protozoa.
Some antibiotics are useful against a wide variety of bacteria they are called broad-spectrum antibiotics. Examples – Ampicillin, amoxicillin, tetracycline, etc.
When the pathogen cannot be identified even though the symptoms of the disease are visible, broad-spectrum antibiotics are used.
Whenever a pathogenic micro-organism is definitely known, then narrow-spectrum antibiotics are used. Examples: Penicillin, gentamycin, erythromycin, etc.

Class 9 Science Chapter 8 Useful and Harmful Microbes Intext Questions and Answers

Can you recall?

Question 1.
What is meant by microbes? What are their characteristics?
Answer:
Microbes are tiny microscopic organisms which cannot be seen with the unaided eye.

Characteristics of Microbes.

They are the smallest organisms on earth.
They are composed of prokaryotic or eukaryotic cells.
They can be seen only with the help of a microscope.
They are found in any kind of environment ranging from coolest polar regions to hottest of deserts. Also found in soil, water and air.
Some of them are useful, whereas some of them are harmful micro-organisms.

Question 2.
How do you observe microbes?
Answer:
Microbes are observed using a microscope.

Answer the following.

Question 1.
Why are wineries located near Nashik in Maharashtra?
Answer:

Nashik in Maharashtra is the leading grape producer in the country as it has the soil suitable for the production of grapes.
Glucose and fructose, the sugars present in grape juice are fermented with the help of yeast to produce wines. Therefore, wineries are located near Nashik in Maharashtra.

Question 2.
Find out the uses of fungi to plants and animals?
Answer:

Fungi decompose the bodies of dead animals and convert them into simple carbon compounds. These substances easily mix with air, water and soil from where they are again absorbed by plants and enter the food chain.
Some fungi living in symbiotic association with plants help to absorb water and inorganic compounds like nitrate and phosphate.
Fungi are also used to derive antibiotics like penicillin which are useful to animals.
Ants grow fungi in their anthill and obtain food from it.
Some species of wasps and insects lay their eggs in the fungal bodies growing on trees, thus ensuring a food supply for their larvae.

Question 3.
What is the structure of lichen, a condiment? Where else is it used?
Answer:

Lichen is a symbiotic association between a fungus and an algae (Cyanobacterium).
Lichens are sensitive to environmental disturbances and are used in assessing air pollution in an area.
Lichens are also used in making dyes, perfumes and in traditional medicines.
A few lichen species are eaten by insects or animals such as reindeer.

Open-ended questions

Answer the following questions:

Class 9 Science Chapter 8 Useful And Harmful Microbes Exercise Question 1.
How is yoghurt made from milk? What exactly happens in this process?
Answer:

Milk contains sugar called lactose which is broken down with help of Lactobacilli.
Lactobacilli converts lactose into Lactic Acid. This process is called fermentation. As a result, the pH of milk decreases causing coagulation of milk proteins.
Thus, milk proteins are separated from other constituents of milk and milk changes to yoghurt.

Useful And Harmful Microbes Class 9 Exercise Question 2.
Sometimes, you may notice a black powder or white discs floating on the pickle or murabba, when a jar is opened after a long time. What exactly is this? Why are such food items not good to eat?
Answer:

A black powder or white disc floating on the pickle or murabba are fungi.
Different fungal species depend on host (pickle and murabba) for their growth and reproduction.
During this process, fungi secretes mycotoxins which are poisonous chemicals; which ultimately spoil the food. Consuming such food can cause food poisoning.
Therefore, such food items are not good to eat.

8 Useful And Harmful Microbes Class 9 Question 3.
How many different industries depend upon the Lactobacilli bacteria?
Answer:
Industries like milk products, cider, cocoa, pickles, pharmaceuticals depend on Lactobacilli bacteria.

Useful And Harmful Microbes Class 9 Question 4.
Which types of cottage industries and factories can be started in areas with abundant milk production?
Answer:
Cottage industries like the manufacture of milk products like ghee, cheese, paneer, curd, shrikhand, etc. and chocolate making can be started in areas with abundant milk production.

9th Class Science Chapter 8 Useful And Harmful Microbes Exercise Question 5.
Which changes do you notice in leather articles and gunny (jute) bags during the rainy season?
Answer:
In rainy season we can notice whitish-green cotton-like growth or black powder or white discs on leather articles and gunny (jute) bags during the rainy season as these articles are infected by fungus.

Class 9th Science Chapter 8 Useful And Harmful Microbes Exercise Question 6.
For how long afterwards can you use those articles?
Answer:
Those articles cannot be used for long as they wear out and do not last long.

Question 7.
Why do these articles not get spoilt during the summer or winter?
Answer:

Spores of fungi can germinate when there is sufficient moisture.
During summer or winter the weather is hot and dry and so fungus cannot grow in such weather.
Also microbes cannot survive extreme hot or cold temperatures of summer or winter. Therefore, these articles do not get spoilt during summer or winter.

Question 8.
Why do doctors advise you to take yoghurt or buttermilk if you have indigestion or abdominal discomfort?
Answer:

The Lactobacilli present in yoghurt or buttermilk help to restore the natural microbial flora in the intestine, thus helping in digestion and absorption of nutrients.
Also buttermilk helps to cool down the stomach and works as a laxative to ease the congestion during abdominal discomfort.

Question 9.
Sometimes, yoghurt becomes bitter and froths up. Why does this happen?
Answer:

Sometimes yoghurt becomes bitter due to excess fermentation by bacteria.
Excess amount of lactic acid is produced making the curd bitter.

Question 10.
Which different milk products are obtained at home by fermentation of the cream from the milk?
Answer:
Yoghurt, buttermilk, ghee, cheese, shrikhand, sour cream, etc.

Question 11.
Recently, it has been made compulsory in India and some other countries to mix 10% ethanol with fuels like petrol and diesel. What is the reason for this?
Answer:

Ethanol is a smokeless and high quality fuel. So it helps to reduce pollution when mixed with petrol or diesel.
As petrol or diesel is a fossil fuel less consumption of it will lead to resourceful use of it and making the country self-efficient by moving towards sustainable fuel like ethanol.

Question 12.
Chapattis made from wheat only swell up but bread becomes spongy, soft and easy to digest. Why is it so?
Answer:

The chapatti dough has water, which on heating converts into steam and tries to escape.
While doing so, it lifts up the upper layer of the chapatti. Therefore, the chapatti swells up.
Bread is made by adding yeast to the flour.
In the process of obtaining nutrition, the yeast cells convert the carbohydrates into alcohol and carbon dioxide.
When this dough is baked, the carbon dioxide escapes out making the bread spongy, soft and easy to digest.

Question 13.
Salt is applied on the inner surface of pickle jars and the pickle is covered with oil. Why is this done?
Answer:

Salt acts as a preservative. It prevents the growth of bacteria by forcing the microbes to lose water by osmosis. Hence, salt is applied on the inner surface of pickle jars.
Pickle is covered with oil as oil acts as preservative. It seals off the air from the item that is being pickled and provides an environment in which microbes cannot grow.

Question 14.
Which preservatives are mixed with ready to eat foods to prevent them from spoiling?
Answer:
Common salt, sugar, sodium benzoate, citric acid, sodium meta-bi-sulfite etc. are some 1 of the preservatives mixed with ready-to-eat foods to prevent them from spoiling.

Question 15.
Which plant and animal diseases are caused by micro-organisms and what are the 1 measures to be taken against them? Answer:
Plant diseases:

Citrus canker is a bacterial disease that affects 1 trees of citrus fruits.
Rust of wheat is a fungal disease that affects wheat crops.
Yellow vein mosaic is a viral disease which affects vegetables like bhindi (okra).

Preventive Measures:

Seeds which are healthy and disease-free should be selected for sowing.
Infected plants should be removed.
Plants should be sprayed with fungicides and germicides to prevent diseases.

Animal diseases:

Anthrax is a disease that affects cattle. It is caused by a bacterium.
Foot and mouth is a dangerous disease in cattle caused by a virus.
Rabies is a viral disease that affects animals.

Preventive Measures:

The place where animals are kept should be washed with germicides.
Animals should be dewormed regularly.
The animals should be treated with necessary antibiotics for infectious diseases.
They should be regularly vaccinated.
Take the animals to a veterinary hospital for proper treatment and vaccination.

Answer the following questions:

Question 1.
Bring ‘active dry yeast’ from the market. Mix a spoonful of yeast, two spoonfuls sugar with a sufficient quantity of lukewarm water in a bottle. Fix a colourless, transparent balloon on the mouth of that bottle.

What changes do you observe after 10 minutes? Mix limewater with the gas accumulated in the balloon. Collect that limewater in a beaker and observe it. What do you notice?

Answer:

After 10 minutes, the balloon is filled with a gas and gets inflated.
Lime water turns milky thus proving that the gas accumulated is carbon dioxide.

Class 9 Science Chapter 8 Useful and Harmful Microbes Additional Important Questions and Answers

Select the correct option:

Question 1.
The rod-shaped bacteria found in milk or buttermilk are called ……………………. .
(a) Rhizobium
(b) Clostridium
(c) Lactobacilli
(d) Saccharomyces
Answer:
(c) Lactobacilli

Question 2.
Yoghurt has a specific sour taste due to ……………………. .
(a) lactic acid
(b) citric acid
(c) acetic acid
(d) alcohol
Answer:
(a) lactic acid

Question 3.
Bacteria found in the root nodules of leguminous plants are ……………………. .
(a) clostridium
(b) streptococcus
(c) Lactobacilli
(d) Rhizobium
Answer:
(d) Rhizobium

Question 4.
A mutually beneficial relationship is called ……………………. .
(a) symbiosis
(b) parasitism
(c) autotropism
(d) none of these
Answer:
(a) symbiosis

Question 5.
Carbon compounds obtained from bacteria and fungi for destroying or preventing the growth of harmful micro-organisms are called ……………………. .
(a) probiotics
(b) antibiotics
(c) antibodies
(d) antigens
Answer:
(b) antibiotics

Question 6.
Fungi release ……………………. into the food, making the food poisonous.
(a) cyanotoxins
(b) dinotoxins
(c) mycotoxins
(d) cytotoxins
Answer:
(c) mycotoxins

Question 7.
……………………. produce bottle-shaped endospores in adverse conditions.
(a) Lactobacilli
(b) Clostridium
(c) Yeast
(d) Rhizobium
Answer:
(b) Clostridium

Question 8.
……………………. conducted important research on the toxin responsible for gas gangrene and the antitoxin responsible for treating it.
(a) Ida Bengston
(b) Van Ermengem
(c) Louis Pasteur
(d) Alexander Fleming
Answer:
(a) Ida Bengston

Question 9.
……………………. is a smokeless and high quality fuel.
(a) Methanol
(b) Ethanol
(c) Petrol
(d) Diesel
Answer:
(b) Ethanol

Question 10.
Antibiotics mainly act against ……………………. .
(a) bacteria
(b) viruses
(c) algae
(d) fungi
Answer:
(a) Bacteria

Question 11.
……………………. is a broad-spectrum antibiotic.
(a) Gentamycin
(b) Penicillin
(c) Amoxicillin
(d) Erythromycin
Answer:
(c) Amoxicillin

Question 12.
……………………. is a narrow-spectrum antibiotic.
(a) Ampicillin
(b) Amoxicillin
(c) Tetracyclin
(d) Penicillin
Answer:
(d) Penicillin

Question 13.
Antibiotic penicillin was discovered by ……………………. .
(a) Louis Pasteur
(b) Alexander Fleming
(c) Ida Bengston
(d) Van Ermengem
Answer:
(b) Alexander Fleming

Question 14.
……………………. proved that the anaerobic bacterium Clostridium botulinum is responsible for food poisoning.
(a) Louis Pasteur
(b) Ida Bengston
(c) Alexander Fleming
(d) Van Ermengem
Answer:
(d) Van Ermengem

Question 15.
AIDS is caused by ……………………. .
(a) virus
(b) bacteria
(c) protozoa
(d) fungi
Answer:
(a) Virus

Question 16.
Dengue is caused by ……………………. .
(a) droplets spread in air
(b) contact with infected person
(c) mosquito bite
(d) contaminated water and food
Answer:
(c)mosquitobite

Question 17.
Pneumonia is caused by ……………………. .
(a) virus
(b) bacteria
(c) fungi
(d) protozoa
Answer:
(b) bacteria

Question 18.
……………………. can be prevented by vaccination.
(a) Malaria
(b) AIDS
(c) Leprosy
(d) Chicken pox
Answer:
(d) Chicken pox

Question 19.
Malaria is caused by ……………………. .
(a) protozoa
(b) bacteria
(c) fungi
(d) virus
Answer:
(a) protozoa

Question 20.
Bird flu (H₇N₉) and swine flu (H₁N₁) are caused by ……………………. .
(a) bacteria
(b) protozoa
(c) fungi
(d) virus
Answer:
(d) virus

Question 21.
The Lactobacilli convert lactose, the sugar in the milk, into ……………………. .
(a) lactic acid
(b) acetic acid
(c) alcohol
(d) citric acid
Answer:
(a) lactic acid

Question 22.
The ……………………. destroys harmful microbes present in the milk.
(a) high pH
(b) neutral pH
(c) low pH
(d) none of these
Answer:
(c) low pH

Question 23.
Lactobacilli kill the harmful bacteria like ……………………. present in the alimentary canal.
(a) Rhizobium
(b) Saccharomyces
(c) Clostridium
(d) Alcanivorax
Answer:
(c) Clostridium

Question 24.
During fermentation, yeast cells convert carbohydrates into ……………………. .
(a) glucose and fructose
(b) alcohol and carbon dioxide
(c) proteins and fats
(d) fatty acids and amino acids
Answer:
(b) alcohol and carbon dioxide

Question 25.
Molasses is fermented with the help of yeast called ……………………. .
(a) Yarrowia lipolytica
(b) Alcanivorax
(c) Rhizobia
(d) Saccharomyces
Answer:
(d) Saccharomyces

Question 26.
A yeast ……………………. is used to absorb the toxins released during the production of palm oil.
(a) Yarrowia lipolytica
(b) Alcanivorax
(c) Saccharomyces cerevisiae
(d) Penicillium
Answer:
(a) Yarrowia lipolytica

Question 27.
The bacteria which spoil cooked food are ……………………. .
(a) Saccharomyces
(b) Lactobacilli
(c) Clostridium
(d) Rhizobium
Answer:
(c) Clostridium

Question 28.
……………………. can grow and reproduce only in living cells.
(a) Bacteria
(b) Viruses
(c) Fungi
(d) Protozoa
Answer:
(b) Viruses

Find the odd man out:

Question 1.
AIDS, Hepatitis, Leprosy, Dengue.
Answer:
Leprosy. It is caused by bacteria, whereas the rest are caused by viruses.

Question 2.
Cholera, Leprosy, Pneumonia, Influenza.
Answer:
Influenza. It is caused by a virus, whereas the rest are caused by bacteria.

Question 3.
Ampicillin, Amoxycillin, Penicillin, Tetracycline.
Answer:
Penicillin. It is a narrow-spectrum antibiotic, whereas others are broad-spectrum antibiotics.

Question 4.
Tetracycline, Penicillin, Gentamycin, Erythromycin.
Answer:
Tetracycline. It is a broad-spectrum antibiotic, whereas others are narrow-spectrum antibiotics.

Complete the analogy:

Question 1.
(1) Dengue : Virus :: Malaria : …………………………. .
(2) Hepatitis : Virus :: Pneumonia : …………………………. .
(3) Cholera : Bacteria :: Swine flu : …………………………. .
(4) Swine flu : HJNJ : : Bird Flu : …………………………. .
(5) Measles : Virus :: Ringworm : …………………………. .
(6) Yoghurt: Lactobacilli : : Bread : …………………………. .
(7) Oil spills: Alcanivorax :: Absorption of arsenic : …………………………. .
(8) Rhizobium : Nitrogen fixation : : Clostridium : …………………………. .
Answer:
(1) Protozoa
(2) Bacteria
(3) Virus
(4) HyN9
(5) Fungi
(6) Yeast
(7) Saccharomyces cerevisiae
(8) Food poisoning.

Match the columns:

Question 1.

Column ‘A’	Column ‘B’
(1) Leprosy	(a) Virus
(2) Ringworm	(b) Fungi
(3) Influenza	(c) Protozoa
(4) Malaria	(d) Bacteria

Answer:
(1 – d),
(2 – b),
(3 – a),
(4 – c)

State whether the following statements are true or false. Correct the false statements:

(1) Lactobacilli are aerobic bacteria.
(2) Lactobacilli converts lactose sugar into alcohol.
(3) Yoghurt has a specific sour taste due to acetic acid.
(4) The bacteria Clostridium are present in the root . nodules of leguminous plants.
(5) Yeast cell is a prokaryotic cell.
(6) The use of Rhizobium has helped to reduce the use of chemical fertilizers.
(7) Ethanol is a smokeless and high quality fuel.
(8) A yeast, Saccharomyces cerevisiae is used for absorbing toxins released during palm oil production.
(9) Gentamycin is a narrow-spectrum antibiotic.
(10) Antibiotics mainly act against bacteria.
(11) Oil spills in oceans are cleared with the help of Clostridium bacteria.
(12) Tetracycline is a narrow-spectrum antibiotic.
(13) Amoxicillin is a broad-spectrum antibiotic.
(14) Penicillin is a group of antibiotics obtained from a fungus Saccharomyces.
(15) Antibiotic Penicillin was discovered by Alexander Fleming.
(16) The bacteria Lactobacilli cause food-poisoning.
(17) Clostridium bacteria grow in aerobic conditions.
(18) AIDS is caused by a virus.
(19) Influenza is caused by a bacteria.
(20) Antibiotics useful to one person can be suggested to others also.
(21) Dengue is caused by a bacteria.
(22) Dandruff and ringworm are caused by fungi.
(23) Pneumonia is spread through droplets spread in air by infected person.
(24) Chicken pox spread due to contaminated food and water.
(25) Ida Bengston was honoured with the Typhoid Medal’ in 1947.
Answer:
(1) False. Lactobacilli are anaerobic bacteria.
(2) False. The Lactobacilli converts lactose sugar into lactic acid.
(3) False. Yogurt has a specific sour taste due to lactic acid.
(4) False. The bacteria Rhizobium are present in the root nodules of leguminous plants.
(5) False. Yeast cell is a eukaryotic cell.
(6) True.
(7) True.
(8) False. A yeast, Yarrowia lipolytic is used to absorb the toxins released during the production of palm oil.
(9) True.
(10) True.
(11) False. Oil spills in oceans are cleared with the help of Alcanivorax bacteria.
(12) False. Tetracycline is a broad-spectrum antibiotic.
(13) True.
(14) False. Penicillin is a group of antibiotics obtained from a fungus Penicillium.
(15) True.
(16) False. The bacteria Clostridium cause food? poisoning.
(17) False. Clostridium bacteria grow in anaerobic conditions.
(18) True.
(19) False. Influenza is caused by a virus.
(20) False. Antibiotics useful to one person cannot be suggested to others as different diseases require different antibiotics.
(21) False. Dengue is caused by a virus.
(22) True.
(23) True.
(24) False. Chicken pox spread due to contact with infected person.
(25) False. Ida Bengston was honoured with the ‘Typhus Medal’ in 1947.

Complete the statements using the proper option from those given below. Explain the statements: (mycotoxins, budding, Rhizobium, molasses, endospores, broad-spectrum, Lactobacilli)

Question 1.
Lactobacilli bacteria are used for making yoghurt.
Answer:
The lactobacilli convert lactose, the sugar in the milk, into lactic acid. As a result, the pH of milk decreases causing a coagulation of milk proteins. Milk changes into yogurt.

Question 2.
The use of Rhizobium has helped to reduce the use of chemical fertilizers.
Answer:
Rhizobium bacteria are found in the root nodules of leguminous plants. They help to convert atmospheric nitrogen into nitrogen compounds and provide it to the plants. This helps to reduce the use of chemical fertilizers and their adverse effects.

Question 3.
Amoxicillin is a broad-spectrum antibiotic.
Answer:
This antibiotic is useful against a wide variety of bacteria. It is used against pathogens which cannot be identified during symptoms of a disease.

Question 4.
Ethanol is produced by the fermentation of molasses.
Answer:
Molasses is produced from sugarcane juice. It is rich in carbohydrates. When it is fermented with the help of the yeast called Saccharomyces, ethanol (C2H5OH) is produced.

Question 5.
Clostridium bacteria produce bottle-shaped endospores.
Answer:
These endospores help them to survive in adverse conditions.

Give scientific reasons:

Question 1.
Lactobacilli are used for making yoghurt from milk.
Answer:

Lactobacilli convert lactose, the sugar in the milk, into lactic acid. This process is called fermentation.
As a result, the pH of milk decreases causing coagulation of milk proteins.
Thus, milk proteins are separated from other constituents of milk and milk changes into yoghurt.
Yoghurt has a specific sour taste due to lactic acid. The low pH destroys harmful microbes present in the milk. Therefore, Lactobacilli are used for making yoghurt from milk.

Question 2.
Antibiotics should be taken only when prescribed by a doctor.
Answer:

Antibiotics are a group of medicines used to kill disease-causing bacteria and certain protozoa.
The doctor selects and prescribes the antibiotic best suited for our disease.
If taken in extra dose, they can kill the useful bacteria present in our body.
If the course of antibiotics is not completed, the bacteria develop resistance to that antibiotic making it ineffective.
Therefore, antibiotics should be taken only when prescribed by a doctor.

Question 3.
Nowadays, seeds are coated with Rhizobial solution or powder before sowing.
Answer:

When seeds coated with Rhizobial solution or powder are sown, Rhizobia enter the plantlets.
This is called Rhizobial inoculation.
Rhizobia can produce nitrogenous compounds from atmospheric nitrogen.
This experiment has helped in the supply of nitrogen to cereal and other crops, besides leguminous crops.
Therefore, nowadays seeds are coated with Rhizobial solution or powder before sowing.

Question 4.
Antibiotics are not effective against common cold or influenza.
Answer:

Antibiotics are a group of medicines used to control inflections caused by bacteria.
Common cold or influenza is caused by a virus.
Antibiotics are not effective against viruses.
Therefore, antibiotics are not effective against common cold or influenza.

Question 5.
Cotton fabrics, gunny bags, leather items and wooden items do not last long.
Answer:

Microscopic spores of fungi are present in the air.
If there is sufficient moisture, spores germinate on cotton fabric, gunny bags, leather, wooden items etc.
The fungal hyphae (fibres of the fungus) penetrate deep into the material to obtain nutrition and to reproduce.
This causes the materials to wear and become weak.
As a result, cotton fabric, gunny bags, leather and wooden items do not last long.

Question 6.
Food on which fungi has grown cannot be eaten.
Answer:

Various species of fungi grow on food items like pickles, murabba, jam, sauce, chutney etc.
They use the nutrients in these food items for growth and reproduction.
During this activity, fungi release mycotoxins, certain poisonous chemicals, into the food and thus food becomes poisonous.
Hence, the food on which fungi have grown cannot be eaten.

Write short notes:

Question 1.
Rhizobial inoculation.
Answer:

Nowadays, seeds are coated with rhizobial solution or powder before sowing.
After sowing, Rhizobia enter the plantlets.
This is called Rhizobial inoculation.
This experiment has helped in the supply of nitrogen to cereal and other crops, besides leguminous crops.

Question 2.
Bio-remediation.
Answer:

Bio-remediation is a technique that involves the use of organisms to break down environmental pollutants.
Generally, fungi like yeast and bacteria are used for bio-remediation.
A yeast, Yarrowia lipolytica is used to absorb the toxins released during the production of palm oil and the heavy metals and minerals released in some other industrial processes.
Saccharomyces cerevisiae is used for absorption of a pollutant, arsenic.
Oil spills in oceans are cleaned with the help of Alcanivorax bacteria.

Question 4.
Clostridium.
Answer:

Clostridium are the bacteria that spoil food.
Out of about 100 different species of this bacterium, some are free living in the soil whereas some live in the alimentary canals of humans and other animals.
These bacteria are rod-shaped and produce bottle-shaped endospores in adverse conditions.
One special characteristic of these bacteria is that they cannot withstand the normal oxygen level of the air because they grow in anaerobic conditions.

Write down the mode of infection and preventive measures for the following:

Question 1.
Write down the causative pathogen, mode of infection and preventive measures of AIDS.
Answer:

Causative Pathogen: Virus.
Mode of infection: Through blood and semen of infected person and milk of mother suffering from AIDS.
Preventive measure: Safe sexual contact, avoid resuse of needles and injections.

Question 2.
Write down the modes of infection and preventive measures against Bird Flu (H7N9) and Swine Flu (HjN.,).
Answer:

Mode of infection: Contact with infected birds and animals.
Preventive measure: Personal hygiene, properly cooked meat.

Question 3.
Write down the modes of infection and preventive measures against Malaria and dengue.
Answer:

Mode of infection: Mosquito bite, unclean surroundings.
Preventive measure: Cleanliness of surroundings, preventing stagnation of water, controlling mosquitoes.

Question 4.
Write down the modes of infection and preventive measures against Pneumonia.
Answer:

Mode of infection: Droplets spread in air by infected person.
Preventive measure: Vaccination, avoiding contact with infected person.

Question 5.
Write down the modes of infection and preventive measures for leprosy.
Answer:

Mode of infection: Long term contact with infected person.
Preventive measure: Avoiding contact with infected persons and their belongings.

Question 5.
What are the modes of infection and preventive measures for Hepatitis?
Answer:

Mode of infection: Contaminated water and food.
Preventive measure: Use clean and filtered water, proper storage of food.

Question 6.
What are the modes of infection and preventive measures for Influenza.
Answer:

Mode of infection: Contact with infected person.
Preventive measure: Personal hygiene and avoiding contact with infected person.

Answer the following questions:

Question 1.
How is bread made?
Answer:

Bread is made using flour, yeast, salt and water. The yeast uses sugar as food.
In the process of obtaining nutrition, yeast cells convert the carbohydrates into alcohol and carbon dioxide. This process is called fermentation.
The bubbles of carbon dioxide given off cause the dough to rise.
This dough can be used to make bread.
When this dough is baked, more bubbles of carbon dioxide reformed due to heat. As the gas escapes, the bread rises and becomes soft and fluffy.

Question 2.
What is the advantage of Rhizobium to farmers?
Answer:

The use of Rhizobium has helped to reduce the use of chemical fertilizers and thereby their adverse effects.
It has also helped to reduce expenses on fertilizers and thus benefited the farmers.

Question 3.
How can we observe Lactobacilli in buttermilk?
Answer:

Smear a drop of fresh buttermilk on a glass slide.
Stain it with methylene blue and put a coverslip over it.
Observe the smear under the 10X objective of a compound microscope and then with the more powerful 60X objective.
The blue rod-shaped organisms moving about are Lactobacilli.

Question 4.
What is symbiosis? Give example.
Answer:

Symbiosis is a mutually beneficial relationship.
Example: Rhizobium living in root nodules of leguminous plants supply nitrates, nitrites and amino acids to that plant and in exchange get energy in the form of carbohydrates.

Complete the following table:

Question 1.
The spread and prevention of disease
Answer:

Question 2.
Different species of Clostridium bacteria and the diseases caused by them.
Answer:

Species of Clostridium	Diseases
Clostridium tetani	Tetanus
Clostridium perfringens	Food poisoning
Clostridium botulinum	Botulism (Paralysis of muscles)
Clostridium difficile	Colitis (Infection of the intestine)

Answer in detail:

Question 1.
How is alcohol produced?
Answer:

Alcohol is often produced along with sugar in sugar factories.
Molasses is produced from sugarcane juice. It is rich in carbohydrates.
Molasses is fermented with the help of the yeast Saccharomyces.
In this process, ethanol (C₂H₅OH) alcohol is produced as a primary product and ester and other alcohols are produced as secondary products.
Besides molasses, maize, barley and other grains are also used for industrial production of alcohol.
Glucose and fructose, the sugars present in grape juice are also fermented with the help of yeast to produce alcohol which is used to make wines.

Question 2.
Give the uses of Lactobacilli.
Answer:
Uses of Lactobacilli:

Various milk products like yoghurt, buttermilk, ghee, cheese, shrikhand, etc. can be obtained by fermentation of milk.
Lactobacilli fermentation is useful for large scale production of cider, cocoa, pickles of vegetables etc.
Lactobacilli and some other useful microbes taken together are used to treat abdominal discomfort.
Leavened fodder offered to domestic cattle like cows and buffaloes is fodder fermented with the help of lactobacilli.
The Lactobacilli fermentation process is used to make wine and some types of bread.

Question 3.
What is Penicillin? What is it used for?
Answer:

Penicillin is a group of antibiotics obtained from a fungus, penicillium.
It is used for controlling the infections caused by bacteria like Staphylococci, Clostridia, Streptococci, etc.
Medicines containing Penicillin are useful to treat certain bacterial infections of the ear, nose, throat and skin as well as diseases like Pneumonia and scarlet fever.

Question 4.
How was the antibiotic penicillin discovered?
Answer:

Alexander Fleming, a professor of microbiology at St. Mary’s Hospital had cultured varieties of bacteria and fungi in petri dishes in his laboratory.
On 3rd September 1928, while observing Staphylococci cultures, he made an interesting observation in one petri dish.
In that petri dish, fungal colonies had grown but the area around those colonies was clean and clear, i.e. the bacteria had actually been destroyed.
After further studies, he confirmed that the fungus growing there was Penicillium and its secretion had destroyed the bacterial colonies.
Thus, the first antibiotic – penicillin had been discovered accidentlly and this formed the basis to find cures for incurable diseases.

Question 5.
What are the precautions to be followed while taking antibiotics?
Answer:

Antibiotics should be taken only when prescribed by a doctor.
Don’t purchase any antibiotic from medical stores without a prescription from a doctor.
Don’t consume antibiotics on your own to treat common diseases like a throat infection, common cold or influenza.
Even if you feel well before completing of the prescribed course of the antibiotics, you must continue and complete it.
Don’t suggest to others the antibiotics which were useful to you.

Question 6.
How can we observe Rhizobium bacteria in the roots of leguminous plant?
Answer:

Take a plantlet of fenugreek, groundnut or any other bean and sterilize it with a 3 to 5% solution of hydrogen peroxide.
Afterwards, keep it in a 70% solution of ethyl alcohol for 4 to 5 minutes.
Clean the roots with sterile water and take thin sections of the root nodules.
Select a good section and place it an a solution of saffranin for 2 to 3 minutes.
Place the stained section on a glass slide, cover it with a coverslip and observe it under the compound microscope. The pinkish rod-shaped organisms are the Rhizobium bacteria.

9th Std Science Questions And Answers:

Measurement of Matter Class 9 Science Chapter 4 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Prasanna

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 4 Measurement of Matter Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 4 Measurement of Matter Question Answer Maharashtra Board

Class 9 Science Chapter 4 Measurement of Matter Question Answer Maharashtra Board

1. Give examples.

a. Positive radicals
Answer:
Na⁺– Sodium ion, K⁺ – Potassium ion

b. Basic radicals
Answer:
Na⁺ – Sodium ion, K⁺ – Potassium ion, Ag⁺ – Silver ion

c. Composite radicals
Answer:
\(\mathrm{SO}_{4}^{2-}, \mathrm{NH}_{4}^{+}\)

d. Metals with variable valency
Answer:
(a) Iron (Ferrum)
(i) Fe²⁺ – Ferrous [Iron – II]
(ii) Fe³⁺ – Ferric [Iron – III]

(b) Copper (Cuprum)
(i) Cu⁺ – Cuprous [Copper -1]
(ii) Cu²⁺ – Cupric [Copper – II]

e. Bivalent acidic radicals
Answer:
O^2- – Oxide, S^2- – Sulphide, \(\mathrm{CO}_{3}^{2-}\) – Carbonate

f. Trivalent basic radicals
Answer:
Al³⁺ – Aluminium, Cr³⁺ – Chromium, Fe³⁺ – Ferric.

2. Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.
Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen
Answer:

3. Write the steps in deducing the chemical formulae of the following compounds.
Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide
Answer:
In order to write the chemical formulae of compounds, it is necessary to know the symbols and valency of various radicals.

1. Sodium Sulphate:
Step – 1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
\(\mathrm{Na} \quad \mathrm{SO}_{4}\)
Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
Na₂ SO₄
(Sodium sulphate)

2. Potassium Nitrate:
Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
K NO₃
Step – 2 : To write the valency below the respective radical.
\(\begin{array}{cc}
\mathrm{K} & \mathrm{NO}_{3} \\
1 & 1
\end{array}\)
Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
KNO₃
(Potassium nitrate)

3. Ferric phosphate:
Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
Fe PO₄
Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
FePO₄
(Ferric phosphate)

4. Calcium oxide:
Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radicals on the right)
Ca O
Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
CaO
(Calcium oxide)

5. Aluminium hydroxide:
Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radical on the right)
Al OH
Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
Al(OH)₃
(Aluminium hydroxide)

6. Calcium carbonate:
Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radicals on the right)
Ca CO₃
Step – 2 : To write the valency below the respective radical.

Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
CaCO₃
(Calcium Carbonate)

7. Sodium dichromate:
Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radical on the right)
Na Cr₂O₇
Step – 2 : To write the valency below the respective radical.
\(\begin{array}{cc}
\mathrm{Na} & \mathrm{Cr}_{2} \mathrm{O}_{7} \\
1 & 2
\end{array}\)
Step – 3: To cross-multiply as shown by arrows the number of radicals.

Step – 4 : To write down the chemical formula of the compound.
Na2Cr₂O₇
(Sodium dichromate)

4. Write answers to the following questions and explain your answers.

a. Explain how the element sodium is monovalent.

Answer:

The number of protons or electrons (atomic number) in Sodium (Na) atom is 11. Therefore the electronic configuration of sodium atom is (2, 8,1).
In chemical reaction, sodium atom has the capacity to give away le_ from its outermost orbit to form Na+ ion with stable electronic configuration (2, 8).
As sodium atom gives away le- and a cation of sodium is formed, hence the valency of sodium is 1 and therefore, the element sodium is monovalent.

b. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate.
Answer:
M is a bivalent metal. Following are the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate:

(i) Compound of metal ‘M’ with radical sulphate
Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
M SO₄
Step – 2: To write the valency below the respective radical.

Step – 3: To cross multiply as shown by arrows the number of radicals.

Step – 4: To write down the chemical formula of the compound.
M SO₄

(ii) Compound of metal ‘M’ with radical phosphate.
Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
M PO₄
Step – 2: To write the valency below the respective radical.

Step – 3: To cross multiply as shown by arrows the number of radicals.

Step – 4: To write down the chemical formula of the compound.
M₃ (PO₄)₂

c. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.
Answer:

The mass of an atom is concentrated in its nucleus and it is due to the protons (p) and neutrons (n) in it.
Since an atom is very very tiny, it was not possible to measure atomic mass accurately. Therefore, the concept of relative mass of an atom was formed.
To express relative mass of an atom, reference of atom is considered. The two reference atoms were as follows:

(a) Hydrogen (H) atom: The hydrogen atom is the lightest. The relative mass of a hydrogen atom is 1 which has only 1 proton in its nucleus. On this scale, the relative atomic mass of many elements comes out to be fractional. Therefore, carbon was selected as a reference atom.

(b) Carbon (C) atom: The carbon atom is selected as reference atom. In this scale, the relative mass of a carbon atom is accepted as 12.

The relative atomic mass of 1 hydrogen (H) atom compared to the carbon (C) atom becomes

d. What is meant by Unified Atomic Mass.
Answer:

During earlier time, relative mass of an atom was considered for measuring the mass of an atom directly. But since the founding of unified mass, relative mass is not accepted henceforth.
Unified atomic mass is the unit of atomic mass called as Dalton.
Its symbol is ‘u’. lu = 1.66053904 x 10^-27 kg.

e. Explain with examples what is meant by a ‘mole’ of a substance.
Answer:

A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.
For example: Atomic mass of oxygen atom (O) is 16u. Thus, the molecular mass of oxygen molecule (O₂) is 16 x 2 = 32u. Therefore, 32 g of oxygen is 1 mole of oxygen.

5. Write the names of the following compounds and deduce their molecular masses.
Na₂SO₄, K₂CO₃, CO₂, MgCl₂, NaOH, AlPO₄, NaHCO₃
Answer:

6. Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:
‘sample m’ mass : 7g
Mass of constituent oxygen : 2g
Mass of constituent calcium : 5g
‘sample n’ mass : 1.4g
Mass of constituent oxygen : 0.4g
Mass of constituent calcium : 1.0g

Which law of chemical combination does this prove? Explain.
Answer:
(i) The expected proportion by weight of the constituent elements of quick lime that is calcium oxide would be from its known molecular formula CaO. The atomic mass of Ca and O are 40 and 16 respectively. This means, the proportion by weight of the constituent elements Ca and O in the compound CaO is 40 :16 which is 5 : 2.

(ii) Now, for the given sample’m’ of CaO = 5 g
mass of given sample = 7 g
mass of constituent Ca in sample’m’ = 5 g
mass of constituent O in sample’m’ = 2 g

(iii) This means that 7 g of calcium oxide contairis 5 g of calcium (Ca) and 2 g of oxygen (O); apd the proportion by weight of calcium and oxygen in it is 5 : 2.

(iv) Now, for the given sample ‘n’ of CaO mass of given sample CaO = 1.4 g
Mass of constituent Ca in sample ‘n’ = 1.0 g
Mass of constituent O in sample ‘n’ = 0.4 g
This means that 1.4g of calcium oxide contains 1.0 g of calcium (Ca) and 0.4 g of oxygen (O); and the proportion by weight of calcium and oxygen in it is 5 : 2.

(v) Above samples’m’ and ‘n’ of calcium oxide (CaO) shows that the proportion by weight of the constituent elements in different samples of a compound is always constant that is the proportion by weight of calcium (Ca) and oxygen (O) in different samples of calcium oxide (CaO) is constant.

(vi) The experimental value of proportion by weight of the constituent elements matched with the expected proportion calculated by molecular mass. This proves and verifies the law of constant proportion.

The law states that ‘The proportion by weight of the constituent elements in the various samples of a compound is fixed’.

7. Deduce the number of molecules of the following compounds in the given quantities.
32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine.

Class 9 Science Chapter 4 Measurement Of Matter Notes Question 1.
32g oxygen
Answer:
Given : Mass of oxygen (O₂) m = 32g
To find : Number of molecules in 32g of oxygen.
Solution : Atomic mass of oxygen (O) = 16
∴ Molecular mass of oxygen (O₂) M = 16 x 2 = 32
According to the formula, Number of moles in the given O₂ (n)

1 mol of O₂ contains 6.022 x 10²³ molecules that is 32 g of 02 contains 6.022 * 10²³ molecules of O₂.
32g of oxygen contains 6.022 x 10²³ molecules of oxygen.

Class 9 Science Chapter 4 Measurement Of Matter Answers Question 2.
90g water
Answer:
Given : Mass of water (H₂O) m = 90g.
To find : Number of molecules in 90g of water.
Solution : Molecular mass of (H₂O) M = (Atomic mass of H) x 2 + (Atomic mass of O) x 1
∴ Molecular mass of (H₂O) M = 1 x 2 +16
∴ Molecular mass of (H₂O) M = 18
According to the formula,
Number of moles in the given H₂O (n)

1 mol of H₂O contains 6.022 x 10²³ molecules.
5 mol of H₂O contains 5 x 6.022 x 10²³ molecules. = 30.11 x 10²³ molecules, that is 90g of H₂O contains 30.11 x 10²³ molecules of H20.
90g of water contains 30.11 x 10²³ molecules of water.

4 Measurement Of Matter Exercise Question 3.
8.8g carbon dioxide
Answer:
Given : Mass of Carbon dioxide (CO₂)m = 8.8g.
To find : Number of molecules in 8.8g of carbon dioxide.
Solution : Molecular mass of (CO₂)M = (Atomic mass of C) x 1 + (Atomic mass of O) x 2
∴ Molecular mass of (CO₂)M = 12 x 1 + 16 x 2 = 12 + 32
Molecular mass of (CO₂)M = 44
According to the formula, Number of moles in the given CO₂ (n)

∴ 1 mol of CO₂ contains 6.022 x 10²³ molecules.
∴ 0.2 mol of CO₂ contains 0.2 x 6.022 x 10²³ molecules.
= 1.2044 x 10²³ molecules,
that is 8.8g of CO₂ contains 1.2044 x 10²³ molecules of CO₂.
8.8g of CO₂ contains 1.2044 x 10²³ molecules of CO₂.

Class 9 Science Solutions Maharashtra Board Question 4.
7.1g chlorine
Answer:
Given : Mass of Chlorine (Cl₂)m = 7.1g.
To find : Number of molecules in 7.1g of chlorine.
Solution : Atomic mass of (Cl) = 35.5
∴ Molecular mass of chlorine (Cl₂)M = 35.5 x 2 = 71
According to the formula, Number of moles in the given Cl₂ (n)

∴ 1 mol of Cl₂ contains 6.022 x 10²³ molecules.
∴ 0.1 mol of Cl₂ contains 0.1 x 6.022 x 10²³ molecules.
= 0.6022 x 10²³ molecules,
that is 7.1g of Cl₂ contains 0.6022 x 10²³molecules of Cl_2.
7.1g of Cl₂ contains 0.6022 x 10²³ molecules of chlorine.

8. If 0.2 mol of the following substances are required how many grams of those substances should be taken? Sodium chloride, magnesium oxide, calcium carbonate
Answer:
Given : Number of moles of sodium chloride (NaCl) n = 0.2 mol
To find : Mass in grams of 0.2 mol of NaCl
Solution:
Molecular mass of (NaCl)M = (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1
= 23 x 1 + 35.5 x 1
= 23 + 35.5
Molecular mass of (NaCl)M = 58.5
According to the formula,
Number of moles in the given NaCl (n)

Mass of NaCl in grams (m) = 0.2 x 58.5
Mass of NaCl in grams (m) = 11.7 g
Mass of 0.2 mole of NaCl is 11.7g

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Maharashtra State Board Class 9 Science Solutions Question 1.
What is the type of chemical bond in NaCl and MgCl₂?
Answer:

The type of chemical bond in NaCl and MgCl₂ is ionic bond.

9th Class Science Chapter 4 Measurement Of Matter Question 2.
Determine the valencies of H, Cl, O and Na from the molecular formulae H₂, HC1, H₂O and NaCl.
Answer:
(i) In the molecular formula HCl

∴ The valency of H is 1 and Cl is 1.

(ii) In the molecular formula H₂O

∴ The valency of H is 1 and O is 2.

(iii) In the molecular formula NaCl

∴ The valency of Na is 1 and Cl is 1.
∴ From all the above, the valencies of the given elements are as follows : H = 1, Cl = 1, O = 2 and Na = l.

Measurement Of Matter Class 9 Exercise Answers Question 3.
How is an element indicated in Chemistry?
Answer:
In chemistry an element is indicated by its symbol.

Question 4.
Write down the symbols of the elements you know.
Answer:
Symbols of some elements are

Hydrogen – H
Helium – He
Boron – B
Carbon – C
Aluminium – A1

Question 5.
Write down the symbols for the following elements.
Antimony, Iron, Gold, Silver, Mercury, Lead, Sodium
Answer:
The symbols of given elements are as follows:

Antimony – Sb
Iron – Fe
Gold – Au
Silver – Ag
Mercury – Hg
Lead – Pb
Sodium – Na

Following are atomic masses of a few elements in Daltons and the molecular formulae of some compounds. Deduce the molecular masses of those compounds:

Atomic masses – H(l), 0(16), N(14), C(12), K(39), S(32) Ca(40), Na(23), Cl(35.5), Mg(24), Al(27)

Question 1.
Molecular formula – NaCl
Answer:
Molecular mass of NaCl (M)
= (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1
= (23 x 1) + (35.5 x 1)
= 23 + 35.5
= 58.5
∴ Molecular mass of NaCl (M) = 58.5

Question 2.
Molecular formula – MgCl₂
Answer:
Molecular mass of MgCl₂ (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of Cl) x 2
= (24 x 1) + (35.5 x 2)
= 24 + 71
= 95
∴ Molecular mass of MgCl₂ (M) = 95?

Question 3.
Molecular formula – KNO₃
Answer:
Molecular mass of KNO₃ (M)
= (Atomic mass of K) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (39 x 1) + (14 x 1) + (16 x 3)
= 39 + 14 + 48
= 101
Molecular mass of KNO₃ (M) = 101

Question 4.
Molecular formula – H₂O₂
Answer:
Molecular mass of H₂O₂ (M)
= (Atomic mass of H) x 2 + (Atomic mass of O) x 2
= (1 x 2) + (16 x 2)
= 2 + 32
= 34
∴ Molecular mass of H₂O₂ (M) = 34.

Question 5.
Molecular formula – A1C1₃
Answer:
Molecular mass of A1C1₃ (M)
= (Atomic mass of Al) x 1 + (Atomic mass of Cl) x 3
= (27 x 1) + (35.5 x 3)
= 27 + 106.5
= 133.5
∴ Molecular mass of A1C1₃ (M) = 133.5

Question 6.
Molecular formula – Ca(OH)₂
Answer:
Molecular mass of Ca(OH)₂ (M)
= (Atomic mass of Ca) x 1 + (Atomic mass of O + Atomic Mass of H) x 2
= (40 x 1) + (16 + 1) x 2
= 40 + (17 x 2)
= 40 + 34
= 74
∴ Molecular mass of Ca(OH)₂ (M)
= 74

Question 7.
Molecular formula – MgO
Answer:
Molecular mass of MgO (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of 0)xl
= (24 x 1) + (16 x 1)
= 24 + 16
= 40
Molecular mass of MgO (M) = 40

Question 8.
Molecular formula – H2S04
Answer:
Molecular mass of H2S04 (M)
= (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
= (1 x 2) + (32xl) + (16×4)
= 2 + 32 + 64
= 98
Molecular mass of H2S04 (M) = 98

Question 9.
Molecular formula – HN03
Answer:
Molecular mass of HN03 (M)
= (Atomic mass of H) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (lxl)+ (14xl)+ (16×3)
= 1 + 14 + 48
= 63
Molecular mass of HNOs (M) = 63

Question 10.
Molecular formula – NaOH
Answer:
Molecular mass of NaOH (M)
= (Atomic mass of Na) x 1 + (Atomic mass of O) x 1 + (Atomic mass of H) x 1
= (23 x 1) + (16 x 1) + (l x l)
= 23 + 16 + 1
= 40
Molecular mass of NaOH (M) = 40

Question 11.
How many molecules of water are there in 36 g water?
Answer:
Given : Mass of water (H₂O) m = 36g
To find : Number of molecules in 36g of water
Solution :
Molecular mass of (H₂O) M = (Atomic mass of H) x 2 + (Atomic mass of O) x 1 Molecular mass of (H₂O) M
= (1 x 2) + 16 x 1
Molecular mass of (H₂O) M = 18
According to the formula,
Number of moles in the given H₂O (n)

1 mol of H₂O contains 6.022 x 10²³ molecules.
∴ 2 mol of H₂O contains 2 x 6.022 x 10²³ molecules.
= 12.044 x 10²³ molecules, that is 36g of H₂O contains 12.044 x 10²³ molecules of H₂O.
36 g of water contains 12.044 x 10²³ molecules of water.

Question 12.
How many molecules of H2S04 are there in a 49 g sample?
Answer:
Given : Mass of Sulphuric acid (H₂SO₄) m = 49g
To find : Number of molecules in 49g of H₂SO₄
Solution:
Molecular mass of (H₂SO₄) M = (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
Molecular mass of (H₂SO₄)M = (1 x 2) + (32 x 1) + (16 x 4)
= 2 + 32 + 64
= 98.
According to the formula,
Number of moles in the given H₂SO₄ (n)

∴ 1 mol of H₂SO₄ contains 6.022 x 10²³ molecules.
∴ 0.5 mol of H₂SO₄contains 0.5 x 6.022 x 10²³ molecules.
= 3.011 x 10²³ molecules,
that is 49g of H₂SO₄ contains 3.011 x 10²³ molecules of H₂SO₄.
49 g of Sulphuric acid contains 3.011 x 10²³ molecules of H₂SO₄.

Question 13.
Fill the following tables.
Answer:

Question 14.
Complete the following chart.
Answer:

Question 15.
The relative atomic masses of some elements in the chart below are given. You have to find the relative atomic masses of the others.
Answer:

Question 16.
Classify the following radicals into simple radicals and composite radicals: (Use your brain power;
\(\begin{array}{l}
\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cl}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{Fe}^{2+}, \mathrm{ClO}_{3}^{-}, \mathrm{NH}_{4}^{+}, \mathrm{Br}^{-} \\
\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Cu}^{+}
\end{array}\)
Answer:

Simple radicals	Composite radicals
Ag⁺	\(\mathrm{SO}_{4}^{2-}\)
Mg²⁺	\(\mathrm{ClO}_{3}^{-}\)
Cl	\(\mathrm{NH}_{4}^{+}\)
Fe²⁺	\(\mathrm{NO}_{3}^{-}\)
Br^–
Na⁺
Cu⁺

Question 17.
Which are the basic radicals and which are the acidic radicals among the following?
\(\begin{array}{l}
\mathrm{Ag}^{+}, \mathrm{Cu}^{2+}, \mathrm{Cl}^{+}, \mathrm{I}, \mathrm{SO}_{4}^{2-}, \mathrm{Fe}^{3+}, \mathrm{Ca}^{2+}, \mathrm{NO}_{3} ; \mathrm{S}^{2}, \mathrm{NH}_{4}^{+} \\
\mathrm{K}^{+}, \mathrm{MnO}_{4}, \mathrm{Na}^{+}
\end{array}\)
Answer:

Basic Radical	Acidic Radical
(i)Ag⁺	(i) Cl^–
(ii) Cu²⁺	(ii) I^–
(iii) Fe³⁺	\(\text { (iii) } \mathrm{SO}_{4}^{2-}\)
(iv) Ca²⁺	\(\text { (iv) } \mathrm{NO}_{3}^{-}\)
Wnh;	\(\text { (v) } \mathrm{S}^{2-}\)
(vi) K⁺	\(\text { (vi) } \mathrm{MnO}_{4}^{-}\)
(vii) Na⁺

Give examples:

Question 1.
Make a list of elements in the monoatomic and in the diatomic molecular state. (Make a list and discuss;
Answer:

Elements in the monoatomic molecular state are: Helium (He), Neon (Ne), Argon (Ar), Sodium (Na), Copper (Cu),
Elements in the diatomic molecular state are:
Oxygen (O₂), Nitrogen (N₂), Hydrogen (H₂), Chlorine (Cl₂), Fluorine (F₂).

Problem-based questions

Answer the following questions:

Question 1.
Is it possible to weigh one molecule using a weighing balance?
Answer:
No, it is not possible to weigh one molecule using a weighing balance.

Question 2.
Will the number of molecules be the same in equal weights of different substances?
Answer:
No, the number of molecules will not be the same in equal weights of different substances.

Question 3.
If we want equal number of molecules of different substances, will it work to take equal weights of those substances.
Answer:
No, if we want equal number of molecules of different substances, it will not work to take equal weights of those substances.

Answer the following:

Question 1.
What is the Dalton’s atomic theory?
Answer:
Dalton’s Atomic theory-

All matter is made of atoms. Atoms are indivisible and indestructible.
All atoms of a given element are identical in mass and properties.
Compounds are formed by a combination of two or more different kinds of atoms.
A chemical reaction is a rearrangement of atoms.

Question 2.
How are compounds formed?
Answer:
Compounds are formed by a chemical combination of two or more different kinds of atoms.

Question 3.
What are the molecular formulae of salt, slaked lime, water, lime, limestone?
Answer:
The molecular formulae for
Salt – Sodium chloride – NaCl
Slaked lime – Calcium hydroxide Ca(OH)₂
Water – H₂O
Lime – Calcium oxide – CaO
Lime stone – Calcium carbonate – CaCO₃

Question 4.
From which experiments was it discovered that atoms have an internal structure? When?
Answer:

In 1911, Earnest Rutherford conducted a well known experiment called as ‘Gold foil experiment’.
From this experiment it was discovered that atoms have internal structure.

Question 5.
What are the two parts of an atom? What are they made up of?
Answer:
The two parts of atoms are nucleus and extra nuclear part. Nucleus is made up of positively charged protons and electrically neutral neutrons and the extra nuclear part is made up of negatively charged electrons revolving around the nucleus in different orbits.

Open-ended questions

Q.3. 2. Answer the following questions:

Question 1.
How will the compounds, MgCl₂ and CaO be formed from their elements?
Answer:
(1) Magnesium Chloride (MgCl₂)
Magnesium atom (Mg): Electronic configuration
\((2,8,2) \stackrel{-2 e^{-}}{\longrightarrow}\) Magnesium ion Mg²⁺ (2,8).
Chlorine atom (Cl). Electronic configuration \((2,8,7) \stackrel{+1 e^{-}}{\longrightarrow}\) Chloride ion Ch (2,8,8).
∴ Mg²⁺+ 2CT → MgCl₂ (Magnesium Chloride)

A Magnesium atom gives away 2e^– and a cation of Magnesium (Mg²⁺) is formed, hence, the valency of magnesium is two.
Two chlorine atoms takes le^– each and forms two anions of chlorine (2Cl^–) (chloride), and thus, the valency of chlorine is one.
After the give and take of electrons is over, the electronic configuration of all the resulting ions has a complete octet.
Due to the attraction between the unit but opposite charges on all the ions, one chemical bond known as ionic bond is formed between Mg²⁺ and 2C1^– each and the compound MgCl₂ is formed.

(2) Calcium Oxide (CaO)
Calcium atom (Ca): Electronic configuration
\((2,8,8,2) \stackrel{-2 e^{-}}{\longrightarrow}\) Calcium ion Ca²⁺ (2,8,8).
Oxygen atom (O). Electronic configuration (2,6)
\(\stackrel{+2 e^{-}}{\longrightarrow}\) Oxygen ion O^2- (2,8).
∴ Ca²⁺+ O^2- → CaO

A calcium atom gives away 2er and a cation of calcium (Ca²⁺) is formed, hence, the valency of calcium is two.
An oxygen atom takes 2e^– and forms anions of oxygen (O^2-) (oxide), and thus, the valency of oxygen is two.
After the give and take of electrons is over, the electronic configuration of both the resulting ions has a complete octet.
Due to the attraction between the unit but opposite charges on the two ions, one chemical bond known as ionic bond is formed between Ca²⁺ and O^2- and the compound CaO is formed.

Question 2.

Take 56 g calcium oxide in a large conical flask and put 18 g water in it.
Observe what happens.
Measure the mass of the substance formed.
What similarity do you find? Write your inference.

Answer:
(i) When 18 g of water is added to 56 g of calcium oxide, calcium oxide combines with water to form calcium hydroxide Ca(OH)₂

(ii) The mass of calcium hydroxide formed is 74 g.?

(iii) In this activity the total mass of reactants, Calcium oxide + Water = 56 g +18 g = 74 g.
It is equal to the mass of the product formed. Ca(OH)₂ = 74g.

This activity verifies the Law of Conservation of Matter, i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

Question 3.

Take a solution of calcium chloride in a conical flask and a solution of sodium sulphate in a test tube.
Tie a thread to the test tube and insert it in the conical flask.
Seal the conical flask with an airtight rubber cork.
Weigh the conical flask using a balance.
Now tilt the conical flask so that the solution in the test tube gets poured in the conical flask.
Now weigh the conical flask again.

Answer:

In this activity, a white precipitate of CaS04 in NaOl is seen in the conical flask after the reaction.
There is no change in the weight of the flask before and after the reaction.
This activity verifies the Law of Conservation of Matter i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

Question 4.
Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds : Calcium carbonate, Sodium bicarbonate, Silver chloride, Calcium hydroxide, Magnesium oxide, Ammonium phosphate, Cuprous bromide, Copper sulphate, Potassium nitrate, Sodium dichromate.

Answer:
Calcium carbonate – CaCO₃ Sodium bicarbonate – NaHCO₃ Silver chloride – AgCl, Calcium hydroxide – Ca(OH)₂, Magnesium oxide – MgO, Ammonium phosphate – (NH₄)₃PO₄, Cuprous bromide – CuBr, Copper sulphate – CuSO₄, Potassium nitrate – KNO₃, Sodium dichromate – Na₂Cr₂O₇.

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

(A) Select the correct option:

Question 1.
The proportion by weight of hydrogen and oxygen in water is ……………………….. .
(a) 8 : 1
(b) 2 : 1
(c) 1 : 2
(d) 1 : 8
Answer:
(d) 1: 8

Question 2.
The proportion by weight of carbon and oxygen in carbon dioxide is ……………………….. .
(a) 8 : 3
(b) 3 : 8
(c) 3 : 2
(d) 2 : 3
Answer:
(b) 3 : 8

Question 3.
A nucleus of an atom is made up of positively charged ………………………… and electrically neutral ……………………….. .
(a) protons; neutrons
(b) electrons; neutrons
(c) neutrons; protons
(d) neutrons; electrons
Answer:
(a) protons; neutrons

Question 4.
The size of an atom is determined by its ……………………….. .
Answer:
radius

Question 5.
Atomic radius is expressed in ……………………….. .
(a) milimetres
(b) centimetres
(c) nanometres
(d) picometres
Answer:
(c) nanomet res

Question 6.
The atomic size depends on the number of ………………………… in the atom.
(a) protons
(b) nucleus
(c) neutrons
(d) electron orbits
Answer:
(d) electron orbits

Question 7.
The mass of an atom is concentrated in its ……………………….. .
(a) protons
(b) nucleus
(c) neutrons
(d) electrons
Answer:
(b) nucleus

Question 8.
The total number of protons and neutrons in the atomic nucleus is called the ……………………….. .
(a) atomic number
(b) electronic configuration
(c) atomic mass number
(d) valency
Answer:
(c) atomic mass number

Question 9.
A ………………………… is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.
(a) mole
(b) dalton
(c) dozen
(d) gross
Answer:
(a) Mole

Question 10.
Avogadro’s number is denoted by the symbol ……………………….. .
(a) N_G
(b) N_v
(c) N_A
(d) N_D
Answer:
(c) N_A

Question 11.
A mole of any substance stands for ………………………… molecules.
(a) 60.22 x 10²³
(b) 6.022 x 10²²
(c) 6.022 x 10²³
(d) 60.22 x 10²²
Answer:
(a) 60.22 x 10²³

Question 12.
The capacity of an element to combine is called its ……………………….. .
(a) valency
(b) electronic configuration
(c) atomic number
(d) volence electrons
Answer:
(a) valency

Question 13.
Electronic configuration of sodium atom is ……………………….. .
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8,1)
Answer:
(d) (2,8,1)

Question 14.
Electronic configuration of chlorine atom is ……………………….. .
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8, 1)
Answer:
(b) (2, 8, 7)

Question 15.
Positively charged ions are called as ……………………….. .
(a) cations
(b) anions
(c) nucleous
(d) protons
Answer:
(a) cations

Question 16.
Negatively charged ions are called as ……………………….. .
(a) cations
(b) anions
(c) nucleus
(d) electrons
Answer:
(b) anions

Question 17.
Iron (Fe) exhibits the variable valencies ……………………….. .
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 2 and 4
Answer:
(b) 2 and 3

Question 18.
Cationic radicals are called as ………………………… radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer:
(a) basic

Question 19.
Anionic radicals are called as ………………………… radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer:
(b) acidic

Question 20.
The unit Dalton is used to express …………………………
(a) atomic mass
(b) atomic radius
(c) atomic number
(d) mass number
Answer:
(a) atomic mass

Question 21.
The valency of element with electronic configuration ………………………… is 2.
(a) (2,5)
(b) (2, 4)
(c) (2, 6)
(d) (2, 7)
Answer:
(c) (2, 6)

Question 22.
The symbol of Avogadro’s number is ……………………….. .
(a) ND
(b) N0
(c) NB
(d) NA
Answer:
(d) NA

Question 23.
………………………. is bicarbonate radical.
\((a) \mathrm{HCO}_{3}^{2-} (b) \mathrm{CO}_{3}^{-}
(c) \mathrm{HCO}_{3}^{-}
(d) \mathrm{CO}_{3}^{2-}\)
Answer:
\(\text { (c) } \mathrm{HCO}_{3}^{-}\)

Question 24.
Molecular formula of sodium sulphate is ……………………….. .
(a) Na(SO₄)₂
(b) Na₂SO₄
(c)Na₂(SO₄)₂
(d)NaSO₄
Answer:
(b) Na₂SO₄

Question 25.
………………………… is a composite radical.
(a) Fe³⁺
(b) Ca²⁺
(c) NH₄⁺
(d) S^2-
Answer:
(c) NH

Question 26.
A mole of any substance stands for ………………………… molecules.
(a) 6.022 x 10²³
(b) 6.022 x 10²²
(c) 60.22 x 10²³
(d) 60.22 x 10²²
Answer:
(a) 6.022 x 10²³

Question 27.
The mass of an atom is concentrated in its ………………………… .
(a) nucleus
(b) electrons
(c) extranuclear part
(d) protons
Answer:
(a) nucleus

Question 28.
………………………… g of water make 1 mole of water.
(a) 32
(b) 33
(c) 16
(d) 18
Answer:
(d) 18

Complete the analogy:

(1) Electron : extra nuclear part:: Neutron ………………………… .
(2) Sodium: (2, 8, 1):: Chlorine:: ………………………… .
(3) K : basic radical :: Br^– : ………………………… .
(4) Cut: simple radical:: NH₄⁺: ………………………… .
(5) Sodium sulphate: Na₂SO₄:: Potassium Sulphate: ………………………… .
(6) Mercurous: Hg⁺:: Mercuric : ………………………… .
(7) Positively charged ion : cation:: Negatively charged ion : ………………………… .
(8) 12: 1 dozen :: 144 : ………………………… .
(9) Hydrogen : \(\odot\) :: Copper : ………………………… .
(10) Law of constant proportions : J. L. Proust::
Law of conservation of matter : ………………………… .
Answer:
(1) nucleus
(2) (2, 8, 7)
(3) acidic radical
(4) composite radical
(5) K₂SO₄
(6) Hg²⁺
(7) anion
(8) 1 gross
(9) ©
(10) Antoine Lavoisier.

Match the columns:

Column A’	Column ‘B’
Example	Atomic radius (in metres)
(1) Water molecule (2) Haemoglobin molecule (3) Hydrogen atom	(a) 10^-10(b) 10^-9(c) 10^-8

Answer:
(1-b),
(2- c),
(3 – a)

Column ‘A’	Column ‘B’
Element	Atomic mass
(1) Neon	(a) 35.5
(2) Silicon	(b) 32
(3) Chlorine	(c) 28
(4) Sulphur	(d) 20

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b)

Column ‘A’	Column ‘B’
Molecule	Molecular mass in grams
(1) h₂	(a) 32 g
(2) H₂O	(b) 34 g
(3) O₂	(c) 2 g
(4) H₂O₂	(d) 18 g

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Column ‘A’	Column B’
Radicals	Names
(1) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)	(a) Carbonate
(2) \(\mathrm{ClO}_{3}^{-}\)	(b) Chromate
(3) \(\mathrm{CO}_{3}^{2-}\)	(c) Dichromate
(4) \(\mathrm{CrO}_{4}^{2-}\)	(d) Chlorate

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Answer the following in one sentence:

Question 1.
What are valence electrons?
Answer:
The electrons present in the outermost orbit of an atom are called valence electrons.

Question 2.
Give the formula to determine the number of moles of a substance.
Answer:
The formula to determine the number of moles of a substance is as given below.

Question 3.
What are basic radicals? Give examples.
Answer:
The radicals which are formed by removal of electrons from the atoms of metals are called as basic radicals, e.g., Na+, Cu2+

Question 4.
What are acidic radicals? Give examples.
Answer:
The radicals which are formed by adding electrons to the atoms of non-metals are called as acidic radicals, e.g., CT, S2-

State whether the following statement is ‘True’ or ‘False’. Correct the false statement.

(1) Molecular state of oxygen is monoatomic.
(2) The capacity of an element to combine is called its valency.
(3) Anionic radicals are basic radicals.
(4) The magnitude of charge on any radical is its atomic number.
(5) In a chemical reaction, mass of original matter and mass of matter newly formed as a result of chemical change are equal.
(6) The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 5.
(7) Relative mass of hydrogen is 1.
(8) The number of molecules in a given quantity of a substance is determined by its atomic mass.
(9) Avogadro’s number is 6.022 x 10²³
(10) Valency of sodium is 2.
Answer:
(1) False. Molecular state of oxygen is diatomic:
(2) True
(3) False. Anionic radicals are acidic radicals.
(4) False. Magnitude of charge on any radical is its valency.
(5) True
(6) False. The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 8.
(7) True
(8) False. The number of molecules in a given quantity of a substance is determined by its molecular mass.
(9) True
(10) False. Valency of sodium is 1.

Name the following:

Question 1.
Scientist who gave Law of Conservation of Matter.
Answer:
Antoine Lavoisier

Question 2.
Scientist who gave Law of Constant Proportion.
Answer:
J. L. Proust

Question 3.
What are protons and neutrons present in nucleus together called as?
Answer:
Nucleons

Question 4.
Unit used to express atomic radius.
Answer:
Nanometre

Question 5.
The number (p + n) in the atomic nucleus is called as?
Answer:
Atomic mass number

Question 6.
Name the unit of atomic mass.
Answer:
Dalton (u)

Question 7.
Write molecular formula of two ionic compounds containing chlorine.
Answer:
NaCl, MgCl₂

Question 8.
Give two monoatomic radicals.
Answer:
Na⁺, Cl^–

Question 9.
Give two examples of simple radicals.
Answer:
Ag⁺, O^2-

Give scientific reasons:

Question 1.
An atom is electrically neutral though it contains charged particles.
Answer:

An atom is made up of a nucleus and an extranuclear part. Protons and neutrons are present in the nucleus.
The nucleus is positively charged. The extranuclear part is made up of negatively charged electrons.
Protons are positively charged, electrons are negatively charged and neutrons are without any charge.
The magnitude of their charges is the same when they are equal in number.
Hence, the negative charge on all the extra, nuclear electrons together balances the positive charge on the
nucleus.
Therefore, an atom is electrically neutral though it contains charged particles.

Question 2.
Neon is chemically inert element.
Answer:

Atomic number of neon is 10, so its electronic configuration is (2, 8). There are 8 electrons in its 2nd shell, fulfilling its capacity.
Thus, neon has a complete octet.
It has a stable orbit therefore, it does not indulge in chemical reactions. Hence, neon is a chemically inert element.

Question 3.
The valency of sodium (Na) is one.
Answer:

The electronic configuration of sodium (Na) is (2, 8,1). It has 1 electron in its 3rd orbit.
It tends to give up this electron so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.
The loss of one electron leads to the formation of sodium ion (Na+) which is positively charged as it has lost one electron.

Question 4.
The valency of chlorine (Cl) is one.
Answer:

The electronic configuration of chlorine (Cl) is (2, 8, 7). It has 7 electrons in its 3rd orbit.
It tends to take one electron from another atom so that it has 8 electrons in the outermost orbit with electronic configuration (2,8,8) with stable state.
The gaining of one electron leads to formation of chloride ion (Cl^–) which is negatively charged as it has gained one electron.

Question 5.
The valency of Magnesium (Mg) is two.
Answer:

The electronic configuration of Magnesium (Mg) is (2,8,2), it has 2 electrons in its 3rd orbit.
It tends to give these ‘2’ electrons so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.
The loss of two electrons leads to the formation of Magnesium ion (Mg2+) which is double positively charged as it has lost two electrons.

Question 6.
Valency is always a whole number.
Answer:

The number of electrons that an atom of an element gives away, takes up or shares forming a bond is called the valency of that element.
These electrons are always in whole numbers and not in fractions.
Therefore, valency is always a whole number.

Question 7.
Atomic size of potassium is bigger than atomic size of sodium.
Answer:

The atomic size of an element depends on the number of electron orbits in the atom of that element.
The greater the number of orbits, the larger the size.
Atomic number of potassium (K) is 19. Hence, its electronic configuration is (2, 8, 8,1). While atomic number of sodium (Na) is 11. Hence its electronic configuration is (2, 8,1)
Number of orbits in potassium atom is 4, while that in sodium atom is 3.
Hence, atomic size of potassium is bigger than atomic size of sodium.

Question 8.
The atomic size of sodium is bigger than atomic size of Magnesium.
Answer:

The atomic size of an element depends on the number of electron orbits in the atom of that element.
If 2 atoms have the same outermost orbit, then the atom having the larger number of electrons in the outermost orbit is smaller than the one having fewer electrons in the same outermost orbit.
Atomic number of sodium (Na) is 11. Hence, its electronic configuration is (2, 8, 1) while atomic number of magnesum (Mg) is 12 and hence its electronic configuration is (2, 8, 2).
As compared to sodium atom Magnesum atom has larger number of electrons n its electronic configuration.
Therefore, atomic size of sodium is bigger than atomic size of Magnesium.

Write the names of the following compounds and deduce their molecular masses:

Atomic masses : H(1), 0(16), N(14), C(12), K(39), S(32), Ca(40), Na(23), C1(35.5), Mg(24), A1(27), P(31)

Question 1.
Molecular mass of K₂CO₃Answer:

Question 2.
Molecular mass of CO₂Answer:

Question 3.
Molecular mass of MgCl₂Answer:

Question 4.
Molecular mass of NaOH
Answer:

Question 5.
Molecular mass of AIPO₄Answer:

Question 6.
Molecular mass of NaHCO₃Answer:

Numerical.

Question 1.
Magnesium Oxide:
Answer:
Given : Number of moles of Magnesium oxide (MgO)n = 0.2 mol
To find : Mass in grams of 0.2 mol of MgO
Solution:
Molecular mass of (MgO)M
= (Atomic mass of Mg) x 1 + (Atomic mass of O) x 1
= 24 x 1 + 16 x 1
= 24 + 16
Molecular mass of (MgO)M = 40
According to the formula Number of moles in the given MgO (n)

Mass of MgO in grams (m) = 0.2 x 40
Mass of MgO in grams (m) = 8 g.
Mass of 0.2 mole of MgO is 8 g

Question 2.
Calcium Carbonate:
Answer:
Given : Number of moles of Calcium carbonate (CaCO₃) n = 0.2 mol
To find : Mass in grams of 0.2 mol of CaCO₃
Solution:
Molecular mass of (CaCO₃) M
= (Atomic mass of Ca) x 1 + (Atomic mass of C) x 1 + (Atomic mass of O) x 3
= (40 x l) + (12 x 1) +(16 x 3)
= 40+ 12+ 48
Molecular mass of (CaCO₃) M = 100
According to the formula Number of moles in the given CaCO₃ (n)

∴ Mass of CaCO₃ in grams (m) = 0.2 x 100
∴ Mass of CaCO₃ in grams (m) = 20 g
Mass of 0.2 mole of CaCO₃ is 20 g

State laws/Define the following:

Question 1.
Law of Conservation of Matter.
Answer:
In a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to chemical reaction.

Question 2.
Law of Constant Proportion.
Answer:
The proportion by weight of the constituent elements in the various samples of a compound is fixed.

Question 3.
Molecular Mass:
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance.

Question 4.
Mole
Answer:
A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.

Question 5.
Valency
Answer:
The capacity of an element to combine is called its valency.

Question 6.
Electronic definition of Valency
Answer:
The number of electrons that an atom of an element gives away or takes up while forming an ionic bond is called valency of that element.

Question 7.
Radicals
Answer:
The positively or negatively charged ions that take part independently in chemical reactions are called radicals.

Question 8.
Atomic size determination
Answer:
The size of an atom is determined by its radius. The atomic radius of an isolated atom is the distance between the nucleus of an atom and its outermost orbit.

Question 9.
Atomic mass number
Answer:
The number of protons and neutrons in the atomic nucleus is called the atomic mass number.

Question 10.
Unified mass
Answer:
Unified mass is the standard unit of atomic mass that quantifies mass on an atomic or molecular scale. Its symbol is ‘u’.
1 u = 1.66053904 x 10^-27 kg.

Question 11.
Molecular mass of a substance
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance. Like atomic mass, molecular mass is also expressed in the unit Dalton (u).

Answer the following questions:

Question 1.
What is variable valency?
Answer:

Under different conditions, the atoms of some elements give away or take up a different number of electrons.
In such cases, those elements exhibit more than one valency.
This property of elements is called variable valency.

Complete the following table:

Question 1.
Write down the cations and anions obtained from the compounds in the following chart.
Answer:

Answer the following questions:

Question 1.
Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds:

(a) Calcium carbonate
Answer:

∴ Chemical formula of Calcium carbonate is CaCO₃

(b) Sodium bicarbonate
Answer:

∴ Chemical formula of Sodium bicarbonate is NaHCO₃

(d) Calcium hydroxide Answer: Symbol Ca OH
Answer:

∴ Chemical formula of Calcium hydroxide is Ca(OH)₂

(e) Magnesium oxide
Answer:

∴ Chemical formula of Magnesium oxide is MgO

(f) Ammonium phosphate
Answer:

∴ Chemical formula of Ammonium phosphate is (NH₄)₃PO₄

(g) Cuprous bromide
Answer:

∴ Chemical formula of Cuprous bromide is CuBr.

(h) Copper sulphate
Answer:

∴ Chemical formula of Copper sulphate is CuSO₄.

(i) Potassium nitrate
Answer:

∴ Chemical formula of Potassium nitrate is KNO₃.

(j) Sodium dichromate
Answer:

∴ Chemical formula of Sodium dichromate is Na₂Cr₂O₇.

9th Std Science Questions And Answers:

Reflection of Light Class 9 Science Chapter 11 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Prasanna

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 11 Reflection of Light Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 11 Reflection of Light Question Answer Maharashtra Board

Class 9 Science Chapter 11 Reflection of Light Question Answer Maharashtra Board

1. Answer the following questions.

a. Explain the difference between a plane mirror, a concave mirror and a convex mirror with respect to the type and size of the images produced.
Answer:

	Plane mirror	Concave mirror	Convex mirror
Type of image	Virtual and Erect	Virtual (erect) as well as Real (inverted)	Virtual and Erect
Size of image	Same size	Diminished, Same size and magnified	Diminished

b. Describe the positions of the source of light with respect to a concave mirror in
1. Torch light
2. Projector lamp
3. Floodlight
Answer:
(a) Torch light: The source of light is placed at the focus.
(b) Projector lamp : The source of light is placed at the centre of curvature.
(c) Flood light : The source of light is placed just beyond the centre of curvature.

c. Why are concave mirrors used in solar devices?
Answer:

Solar devices like solar cooker or solar water heater use solar energy to cook food or heat water.
When sun rays fall on the concave mirror, they converge and come together in the focal plane.
Due to convergence, the intensity of sun rays increases and the food or water is heated faster. Hence, concave mirrors are used in solar- devices.

d. Why are the mirrors fitted on the outside of cars convex?
Answer:

A convex mirror is used as rear view mirror because they form erect, virtual, and diminished images.
This, allows the driver to view a large area in a small mirror.

e. Why does obtaining the image of the sun on a paper with the help of a concave mirror burn the paper?
Answer:

When sunrays fall on the concave mirror, they converge and come together in the focal plane.
Due to convergence, the intensity of sunrays increases.
Hence, image of the sun on a paper with the help of concave mirror bums the paper.

f. If a spherical mirror breaks, what type of mirrors are the individual pieces?
Answer:

When a spherical mirror breaks into smaller pieces, the radius of curvature and focal length does not change.
Hence, it will continue to behave like a spherical mirror only.

2. What sign conventions are used for reflection from a spherical mirror?
Answer:
According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.

The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
The distance measured vertically upwards from the principal axis are taken to be positive.
The distance measured vertically downwards from the principal axis are taken to be negative.
The focal length of a concave mirror is negative while that of a convex mirror is positive.

3. Draw ray diagrams for the cases of images obtained in concave mirrors as described in the table on page 122.

Answer:
(a) A ray diagram for object at infinity for a concave mirror.

Image position	Nature of image
At focus	Real, inverted and point image

(b) A ray diagram for object beyond centre of curvature for a concave mirror.

An object beyond centre of curvature for a concave mirror

Image position	Nature of image
Between the centre of curvature and focus.	Real, inverted and diminished.

(c) A ray diagram for object at the centre of curvature for a concave mirror.

Object at centre of Curva fu re be a concave mirror.

Image position	Nature of image
At the centre of curvature.	Real, inverted and same size

(d) A ray diagram for object between F and C for a concave mirror.

Object between F & C for a concave mirror

Image position	Nature of image
Beyond the centre of curvature.	Real, inverted and magnified.

(e) A ray diagram for obj ect at focus for a concave mirror.

Object at focus for a concave mirror.

Image position	Nature of image
At infinity.	Real, inverted and highly magnified.

(f) A ray diagram for object between pole and focus for a concave mirror.

Image position	Nature of image
Behind the mirror.	Virtual, erect and magnified.

4. Which type of mirrors are used in the following?
Periscope, floodlights, shaving mirror, kaleidoscope, street lights, headlamps of a car.

Answer:

Objects	Type of Mirror
Periscope	Plane mirror
Floodlights	Concave mirror
Shaving mirror	Concave mirror
Kaleidoscope	Plane mirror
Street lights	Convex mirror
Head lamps of car	Concave mirror

5. Solve the following examples

a. An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?
Solution:
Given: Object size (h₁) = 7 cm
Object distance (u) = -25 cm
Focal length (f) = -15cm
To find: Image distance (u) = ?
Image size (h₂) = ?

The screen should be kept 373 cm in front of the mirror. The image is real.

The height of the image is 10.5 cm, it is an inverted and enlarged image.

b. A convex mirror has a focal length of 18 cm. The image of an object kept in front of the mirror is half the height of the object. What is the distance of the object from the mirror?
Solution:
Given: Image size (h₂) = ¹/₂h₁
Focal length (f) = 18 cm
To find: Object distance (u) ?

The object is placed in front of the convex mirror at a distance of 18 cm.

c. A 10 cm long stick is kept in front of a concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Solution:
Given: Object size (h₁) = 10 cm
Object distance (u) = -20 cm
Focal length (f) = -10 cm

The height of the image is 10 cm and it is a real and inverted image.

6. Three mirrors are created from a single sphere. Which of the following:
pole, centre of curvature, radius of curvature, principal axis – will be common to them and which will not be common?
Answer:

Centre of curvature and Radius of curvature will be common for all three pieces.
Pole and Principal axis will not be common.

Class 9 Science Chapter 11 Reflection of Light Intext Questions and Answers

Class 9 Science Chapter 11 Reflection Of Light Question 1.
What is light
Answer:
Light is a form of electromagnetic radiation that produces the sensation of vision.

9th Class Science Chapter 11 Reflection Of Light Exercise Question 2.
What is a mirror?
Answer:
A mirror is a reflecting surface which reflects light and creates clear images.

9th Class Science Chapter 11 Reflection Of Light Exercise Answer Question 3.
Principal Focus of Concave and Convex Mirror.
Answer:

Principal Focus of the Concave Mirror	Principal Focus of the Convex Mirror
(i) Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror. (ii) It is formed in front of the mirror. (iii) Focus of concave mirror is real.	(i) Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror. (ii) It is formed behind the mirror. (iii) Focus of convex mirror is virtual.

9th Class Science Chapter 11 Reflection Of Light Notes Question 4.
If we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror. Why does it happen?
Answer:

When we hold a page of a book in front of the mirror, the image of the words appear laterally inverted.
The image of every point of the word is formed behind the mirror at the same distance from the mirror
Because of this the left and right side of the image is interchanged.
Hence, if we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror.

11 Reflection Of Light Exercise Question 5.
Which letters of the English alphabet form images that look the same as the original letters?
Answer:
A, H, I, M, O, T, U, V, W, X, Y

9th Science Chapter 11 Reflection Of Light Exercise Question 6.
When a person stands in front of a plane mirror, how is the image formed? What is the nature of the image?
Answer:

The image of a person is formed from every point of the source, thereby forming an extended image of the whole source.
The image formed would be virtual, upright and left-right reversed.

Answer the following questions:

Reflection Of Light Class 9 Questions And Answers Question 1.
Place two plane mirrors at an angle of 90a to each other. Place a small object between them. Images will be formed in both mirrors. How many images do you see? Now change the angle between the mirrors as given in the following table and count the number of images each time. How is this number related to the measure of the angle?
Answer:
The Relation between the angle between the mirrors and the number of images formed is given by
\(n=\frac{360^{\circ}}{\mathrm{A}}-1\)
n = number of images,
A = angle between the mirrors

Angle	Number of images
120°	2
90^s	3
60®	5
45®	7
30®	11

Class 9 Science Chapter 11 Reflection of Light Additional Important Questions and Answers

Can you recall?

9th Class Science Chapter 11 Reflection Of Light Question 1.
What is meant by reflection of light and what are the types of reflection?
Answer:
The bouncing back of light when it hits an opaque surface is called reflection of light. The two types of reflection are regular and irregular reflection.

Reflection Of Light Class 9 Exercise Answers Question 2.
What are the laws of reflection.
Answer:

The incident ray, reflected ray and normal all lie in the same plane at the point of incidence.
The angle of incidence is equal to the angle of reflection.
The incident ray and the reflected ray lie on opposite sides of the normal.

Choose and the correct option:

Class 9th Science Chapter 11 Reflection Of Light Question Answer Question 1.
If the reflected rays do not actually meet, such an image is called as image.
(a) real
(b) virtual
(c) magnified
(d) inverted
Answer:
(b) virtual

Class 9 Science Chapter 11 Reflection Of Light Exercise Solutions Question 2.
In a plane mirror, the perpendicular distance of the image from the mirror is equal to
(a) the perpendicular distance of the source from the object.
(b) the perpendicular distance of the source from the mirror.
(c) the parallel distance of the source from the object.
(d) the parallel distance of the source from the mirror.
Answer:
(b) the perpendicular distance of the source from the mirror

Reflection Of Light Class 9 Notes Pdf Maharashtra Board Question 3.
The image formed in a convex mirror is always
(a) virtual, smaller and behind the mirror
(b) virtual, smaller and in front of the mirror
(c) real, smaller and behind the mirror
(d) real, smaller and in front of the mirror
Answer:
(a) virtual, smaller and behind the mirror

Reflection Of Light Class 9 Solutions Question 4.
images can be displayed on a screen.
(a) Virtual
(b) Real
(c) Virtual and erect
(d) Virtual and inverted
Answer:
(b) Real

9th Class Science Chapter 11 Reflection Of Light Exercise Pdf Question 5.
A concave mirror is also called as a mirror.
(a) converging
(b) diverging
(c) plane
(d) outward curved
Answer:
(a) converging

9th Science Chapter 11 Reflection Of Light Question 6.
The centre of the mirror surface is called its
(a) pole
(b) centre of curvature
(c) principal axis
(d) focus
Answer:
(a) pole

Class 9 Science Chapter 11 Question Answer Reflection Of Light Question 7.
According to the new sign convention, the of the mirror is taken as origin.
(a) focus
(b) pole
(c) optical centre
(d) centre of curvature
Answer:
(b) pole

Class 9 Science Chapter 11 Reflection Of Light Exercise Question 8.
A convex mirror is also called as a mirror.
(a) converging
(b) plane
(c) diverging
(d) inward curved
Answer:
(c) diverging

Reflection Of Light Class 9 Maharashtra Board Question 9.
In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be
(a) same height as that of the person
(b) double the height of the person
(c) half the height of the person
(d) quarter the height of the person
Answer:
(c) half the height of the person

Reflection Of Light Exercise 9th Class Question 10.
If the inner surface of the spherical mirror is reflecting, then it is a mirror, and if the outer surface is reflecting then it is mirror.
(a) convex, concave
(b) convex, plane
(c) concave, plane
(d) concave, convex
Answer:
(d) concave, convex

9th Std Science Chapter 11 Reflection Of Light Question 11.
The image formed by a concave mirror is
(a) always virtual and erect
(b) always virtual and inverted
(c) virtual if the object is placed between the pole and the focus
(d) virtual if the object is beyond the focus
Answer:
(c) virtual if the object is placed between the pole and the focus

Question 12.
No matter how far you stand from a spherical mirror, your image appears erect. The mirror may be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer:
(d) either plane or convex

Question 13.
In case of a concave mirror, an erect image is
(a) real and enlarged
(b) real and diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer:
(d) virtual and enlarged

Question 14.
A rear view mirror of a car is
(a) plane mirror
(b) concave mirror
(c) convex mirror
(d) cylindrical mirror
Answer:
(c) convex mirror

Question 15.
An image of an object placed at infinite distance from a concave mirror is formed at
(a) the focus of the mirror
(b) behind the mirror
(c) centre of curvature
(d) infinity
Answer:
(a) the focus of the mirror

Question 16.
A ray of light parallel to principal axis after reflection from concave mirror passes through
(a) centre of curvature
(b) focus
(c) pole
(d) optical centre
Answer:
(b) focus

Question 17.
The image made by a plane mirror is a image.
(a) real
(b) virtual
(c) inverted
(d) diminished
Answer:
(b) virtual

Question 18.
The size of the image of an object placed at the focus of a concave mirror is
(a) erect
(b) very large
(c) same size
(d) diminished
Answer:
(b) very large

Question 19.
For virtual images, the height is while for real images, it is
(a) positive, positive
(b) negative, positive
(c) negative, negative
(d) positive, negative
Answer:
(d) positive, negative

Find the odd man out:

Question 1.
Torches, flood lights, head lamps of vehicles, rear view mirror.
Answer:
Rear view mirror – In rear view mirrors, convex . mirror is used. Concave mirrors are used in the rest.

Question 2.
Side mirrors of cars, parking mirrors, flood lights, mirror fitted in shops.
Answer:
Flood lights – In flood lights concave mirror is used. Convex mirrors are used in the rest.

Question 3.
Virtual and enlarged, virtual and diminished, real and inverted, real and magnified
Answer:
Virtual and diminished type of image is not formed by a concave mirror. All the other types of images are formed by a concave mirror.

Question 4.
Image is laterally inverted, image is of same size, image is at same distance, image is diminished.
Answer:
Image is diminished is not a characteristic of image formed in a plane mirror. Rest of them are characteristics of plane mirror.

Answer the following in one sentence:

Question 1.
What kind of mirror will a doctor use to concentrate on teeth, eyes, ears etc.?
Answer:
The doctor will use a concave mirror to concentrate on teeth, eyes, ears etc.

Question 2.
What do the nature, position and size of the image depend on?
Answer:
The nature, position and size of the image depend upon the distance of the object from the reflecting surface.

Question 3.
Give the expression for mirror formula.
Answer:
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Question 4.
State any four uses of concave mirror.
Answer:
Concave mirrors are used in torches, headlights, shaving mirrors, dentists’ mirrors, solar devices etc.

Question 5.
What are the two types of spherical mirror?
Answer:
Convex mirror and concave mirror are the two types of spherical mirror.

Match the columns:

Question 1.

Column ‘A’	Column ‘B’
(1) Plane mirror	(a) Rear view mirror
(2) Concave mirror	(b) At laughing gallery
(3) Convex mirror	(c) At a hair dresser
(4) Irregular curved mirror	(d) At a dentist

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Question 2.

Column ‘A’	Column ‘B’
(1) Plane mirror	(a) Virtual and diminished image
(2) Concave mirror	(b) Virtual and same size image
(3) Convex mirror	(c) Real and inverted image

Answer:
(1 – b),
(2 – c),
(3 – a)

State whether the following statements are true or false. Correct the false statements:

(1) If the mirrors are kept at right angle to each other, then the number of images formed will be 4.
(2) A convex mirror is used in flood lights.
(3) A concave mirror always forms a magnified image.
(4) Images formed by convex mirrors are always virtual.
(5) The distance between the focus and the pole is called the radius of curvature.
(6) Reflection from a spherical mirror obeys laws of reflection.
(7) The reflecting surface of a concave mirror is curved.
(8) Distances measured in the direction of the incident light are taken as positive.
(9) If the image is erect, the height of the image is negative.
(10) A real image can be displayed on a screen.
(11) A concave mirror always forms a real and inverted image.
(12) Doctors use diverging beam of light to study teeth, ears and eyes.
Answer:
(1) False. if the mirrors are kept at right angle to each other then the number of images formed will be 3.
(2) False. a concave mirror is used in flood lights.
(3) False. a concave mirror can sometimes form a diminished image as well.
(4) True
(5) False. the distance between the focus and the pole is called the focal length.
(6) True
(7) True
(8) True
(9) False. if the image is erect, the height of the image is positive.
(10) True
(11) False. a concave mirror can also form a virtual and erect image.
(12) False. doctors use a converging beam of light to study teeth, ears and eyes.

Give scientific reasons:

Question 1.
A concave mirror is called a converging mirror.
Answer:

When rays of light parallel to the principal axis are incident on concave mirror, they converge.
After convergence, they meet at one point on the principal axis, hence concave mirror is called converging mirror.

Question 2.
Concave mirrors are used in torches and in car headlights.
Answer:

Concave mirrors are used in torches and car headlights because when a source of light is placed at the focus of a concave mirror, a parallel beam of light rays is obtained.
This helps us to see things upto a considerable distance in the darkness.

Question 3.
A dentist uses a concave mirror while examining teeth.
Answer:

A concave mirror produces an erect, virtual and magnified image of an object placed between its pole and focus.
A dentist uses this principle to get a clear and distinct image of teeth, hence, a dentist uses a concave mirror.

Solve the following numerlcals.

Tips for solving numerical:

Object distance (u) is always -ve
If Image distance (u) is +ve then image is behind the mirror and virtual. if u is -ve then image is in front of the mirror and real.
Object height (h₁) is always +ve since it is erect.
Image height (h₂) can be +ve for virtual and -ve for real.

Type – A

Question 1.
A bird is sitting in front of two plane mirrors inclined at an angle of 600 to each other. How many images does the bird see in the mirror?
Solution:
Given : Angle between mirror A = 60⁰
To find: Number of images formed n = ?

The brrd sees 5 images in the mirror.

Question 2.
A coin is kept in front of two plane mirrors inclined to each other. If 3 images of the coin are seen then what is the angle A between the mirrors?
Solution:
Given: no. of images formed n =3
To find: Angle between mirror A =?

The mirrors are inclined atan angle of 900 to each other.

Question 3.
An image is formed 5 cm behind a convex mirror of focal length 10 cm. At what distance is the object placed from the mirror?
Solution:
Given: Image distance (u) = 5 cm
Focal length (f) = 10 an
To find: Object distance (u) = ?

The object is placed at a distance of 10 cm in front of the mirror.

Question 4.
An object placed 20 cm in front of a convex mirror is found to have an image 15cm behind the mirror. Find the focal length of the mirror.
Solution:
Given: Object distance (u) = -20 cm
Image distance (u) = 15 cm
To find: focal length (f) = ?

The focal length of the convex mirror is 60 cm.

Numerical For Practice

Question 5.
An object is placed at a distance of 36 cm from a concave mirror of focal length 12 cm. Find the image distance.
Answer:
-18 cm

Question 6.
An arrow is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the image distance.
Answer:
11.1 cm

Type – B

Question 1.
An object 4cm in height is placed at a distance of 36 cm from a concave mirror. The image is formed 18 cm in the front of the mirror. Find the height of the image.
Solution:
Given: Object height (h₁) = 4 cm
Image distance (u) = -18 cm
Object distance (u) = -36 cm
To find: Height of image (h₂) = ?

The height of the image is 2 cm and it is inverted.

Question 2.
An object 2 cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3 cm high. Find the image distance.
Solution:
Given: Object height (h₁) = 2 cm
Object distance (u) = -16 cm
Image height (h₂) = -3 cm
To find: Image distance (u) = ?

The image is formed at a distance of 24 cm in front of the mirror.

Numericals For Practice

Question 3.
An object 10cm in height is placed at a distance of 36 cm from a concave mirror. 1f the image is formed at a distance of 18 cm in front of the mirror, find the height of the image.
Answer:
-5cm

Question 4.
A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror. Find the image distance.
Answer:
-80cm

Type – C

Question 1.
Rajashree wants to get an inverted image of height 5 cm of an object kept at a distance of 30 cm from a concave mirror. The focal length of the mirror is 10 cm. At what distance from the mirror should she place the screen? What will be the type of the image, and what is the height of the object?
Solution:
Given:
Focal length = f = -10 cm,
Object distance = u = -30 cm
Height of the image = h₂ = 7 cm
To find: Height of the object = h₁ = ?
Image distance = u =?

Rajashree has to place the screen 15 cm to the left of the mirror.
Magnification formula

The height of the object is 10 cm. Thus, the image will be real and diminished.

Question 2.
A 10 cm long stick is kept horizontally in front of the concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Solution:
The stick is kept parallel to the Principal axis. Distance between A and P is 20 cm. Say u₁ = 20 cm.
Hence, the other end of the stick is at distance, u₂ = (u₁ + 10) = 30 cm from pole of the mirror.

Using mirror formula for concave mirror,
Solution:

Here, negative signs indicate that images are formed on the left of the mirror.

The length of the image formed ?s given by, u = u₂ – u₁ = 15 – (-20) = 5cm.
The length of the image is 5 cm.

Numerical For Practice

Question 3.
An object 2 cm in height is placed at a distance of 16 cm from a concave mirror. If the focal length of the mirror is 9.6 cm., find the image distance, nature and size of the image.
Answer:
u = -24 an, h₂ = -3 cm; real, inverted and enlarged.

Question 4.
An arrow of 2.5cm height is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.
Answer:
v = 11.1cm, h₁= 1.1cm; virtual and in dimirrished form.

Define the following:

Question 1.
Centre of curvature of mirror (C)
Ans.
The centre of the sphere of which the mirror is a parte is called the centre of curvature of the mirror.

Question 2.
The radius of curvature (R)
Answer:
The radius of the sphere of which the mirror is a part, is called the radius of curvature of the mirror.

Question 3.
Pole (P)
Answer:
The centre of the mirror surface is called its pole.

Question 4.
The principal axis of a mirror
Answer:
The straight line passing through the pole and centre of curvature of the mirror is called its principal axis.

Question 5.
The focus of a concave mirror (F)
Answer:
Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror.

Question 6.
Focus of a convex mirror (F)
Answer:
Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror.

Question 7.
Focal length of a mirror (f)
Answer:
The distance (f) between the pole and the principal focus of the mirror is called the focal length. This distance is half of the radius of curvature of the mirror. \(f=\frac{R}{2}\)

Answer the following in short:

Question 1.
What are the rules for drawing ray diagrams for the formation of image by spherical mirror?
Answer:
The rules are as follows :

If an incident ray is parallel to the principal axis, then the reflected ray passes through the principal focus.
If an incident ray passes through the principal focus of the mirror, the reflected ray is parallel to the principal axis.
If an incident ray passes through the centre of curvature of the mirror, the reflected ray traces the same path back.

Distinguish between:

Question 1.
Convex mirror and Concave mirror
Answer:

Convex mirror	Concave mirror
(i) In a convex mirror, the reflecting surface is on the outer side. (ii) It is called a diverging mirror. (iii) The focus of a convex mirror is virtual. (iv) It can form only a virtual image. (v) It can form only a diminished image.	(i) In a concave mirror, the reflecting surface is on the inner side. (ii) It is called a converging mirror. (iii) The focus of a concave mirror is real. (iv) It can form a real as well as a virtual image. (v) It can form an enlarged, diminished as well as the same size image.

Question 2.
Real image and Virtual image
Answer:

Real image	Virtual image
(i) A real image is formed only when the reflected rays actually meet at a point. (ii) Real images can be obtained on a screen. (iii) All real images are inverted.	(i) A virtual image is formed only when the reflected rays appear to meet at a point. (ii) Virtual images cannot be obtained on a screen. (iii) All virtual images are erect.

Answer the following questions:

Question 1.
If we keep the mirrors parallel to each other, how many images will we see ?
Answer:
When two mirrors are kept parallel to each other infinite images are formed, this is because light gets reflected infinite times.

Answer in detail:

Question 1.
What sign conventions are used for reflection from a spherical mirror?
Answer:

According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.

The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
Distance measured vertically upwards from the principal axis are taken to be positive.
Distance measured vertically downwards from the principal axis are taken to be negative.
The focal length of a concave mirror is negative while that of a convex mirror is positive.

Question 2.
Draw ray diagrams for the image obtained in convex mirrors.
Answer:

Image position	Nature of image
Behind the mirror.	(A) Virtual, (B) Erect (C) Diminished

Question 3.
In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be half the height of the person. Explain.
Answer:

Proof:

In the figure, the point at the top of the head, the eyes and a point at the feet of a person are indicated by H, E and F respectively.
R and S are midpoints of HE and EF respectively.
The mirror PQ is at a height of NQ from the ground and is perpendicular to it. PQ is the minimum height of the mirror in order to obtain the full image of the person.

For this, RP and QS must be perpendicular to the mirror.

Minimum height of the mirror
PQ = RS
= RE + ES
\(=\frac{\mathrm{HE}}{2}+\frac{\mathrm{EF}}{2}=\frac{\mathrm{HF}}{2}\)
= Half of the person’s height.

Question 4.
Determine the sign of magnification in each of the 6 cases in the table and verify that they are same using formulae
\(\mathbf{M}=\frac{h_{2}}{h_{1}}\) and \(\mathbf{M}=\frac{-v}{u}\)
Answer:

Question 5.
Explain the images formed by concave mirrors with respect to position of the image and object and also the Nature and size of image
Answer:

9th Std Science Questions And Answers:

Study of Sound Class 9 Science Chapter 12 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Prasanna

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 12 Study of Sound Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 12 Study of Sound Question Answer Maharashtra Board

Class 9 Science Chapter 12 Study of Sound Question Answer Maharashtra Board

1. Fill in the blanks and explain.

a. Sound does not travel through ……………………….……….. .
b The velocity of sound in steel is ……………………….………… than the velocity of sand in water.
c. The incidence of ……………………….………… in daily life shows that the velocity of sound is less than the velocity of light.
d. To discover a sunken ship or objects deep inside the sea, ……………………….………… technology is used.

2. Explain giving scientific reasons.

a. The roof of a movie theatre and a conference hall is curved.
Answer:

Sound waves get reflected from the walls and roof of a room multiple times. This causes a single sound to be heard not once but continuously. This is called reverberation.
Due to reverberation, some auditoriums or some particular seats in an auditorium have inferior sound reception. This can be compensated with curtains.
Ceilings of these halls are made curved so that sound after reflecting from the ceiling, reaches all parts of the hall and the quality of sound improves.

b. The intensity of reverberation is higher in a closed and empty house.
Answer:

Reverberation occurs due to multiple reflections of sound.
The furniture in the house acts as a sound-absorbing material.
So if the house is closed and empty, a reflection of sound will be maximum and hence, intensity of reverberation is higher.

c. We cannot hear the echo produced in a classroom.
Answer:

For distinct echoes, the minimum distance of the reflecting surface from the source of sound must be 17.2 m.
Benches in the classroom are sound absorbing materials which prevent echo of sound.
Because of these two reasons echo is not heard in a classroom.

3. Answer the following questions in your own words.

a. What is an echo? What factors are important to get a distinct echo?
Answer:

An echo is the repetition of the original sound because of reflection by some surface.
At 22°C, the velocity of sound in air is 344 m/s.
Our brain retains a sound for 0.1 seconds Thus, for us to be able to hear a distinct echo, the sound should take more than 0.1 seconds after starting from the source to get reflected and. come back to us.
We know that,
Distance = speed x time
= 344 m/s x 0.1 s
= 34.4 m
Thus, to be able to hear a distinct echo, the reflecting surface should be at a minimum distance of half of the above, i.e. 17.2 m.
As the velocity of sound depends on the temperature of air, this distance depends on the temperature.

b. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.
Answer:

Goighumat with a height of 51 metres and diameter of 37 metres with 3 metres thick walls is spread over approximately 1700 square metres.
This meets the conditions for echo i.e. : 17.2 metres minimum.
The dome of the golghumat is curved and hence, sound reflects multiple times before reaching the observer.
This is the reason for multiple echoes being produced.

c. What should be the dimensions and the shape of classrooms so that no echo can be produced there?
Answer:

Dimensions: The distance between opposite walls in a classroom must be less than 17.2 m so that the reflected sound returns to the observer within 0.1 s.
Shape: The classrooms should have curved ceilings and walls so that the reflected sound is directed towards the observer instantly within 0.1 s

4. Where and why are sound-absorbing materials used?
Answer:
The sound absorbing materials are used in :

School, cinema hall, concert hall, houses or places where quality of sound is important.
In the absence of sound absorbing material the sound will undergo multiple reflection causing reverberation of sound.

5. Solve the following examples.

a. The speed of sound in air at O °C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?
Answer:
Given:
Initial speed of sound at 0°C 332 m/s.
Final speed of sound -344 m/s.
Rate of increase per degree rise in temp. = 0.6m/s
To find:
Temperature when speed is 344m/s
Formulae:
Increase in temperature

Temperature when the speed of sound is 344 m/sis 20°C

b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her? (The velocity of sound in air is 340 m/s?)
Answer:
Given : Speed of sound (v) = 340 m/s
Time taken (f) = 4 sec
To find : Distance (s) = ?

The lightning has struck at a distance of 1360 m from the observer.

c. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.
1. What is the velocity of sound in air?
2. What is the distance between the two walls? (Ans: 330 m/s; 1650 m)
Answer:
Given:
Distance of the closer wall (S₁) = 660 m
Time of echo from closer wall = 4 sec
∴ Time taken (t₁) = 4/2 sec = 2 sec
Time of echo from distant wall = 6 sec
∴ Time taken (t₂) = 6/2 sec = 3 sec
To find :
Velocity of sound in air (y) =?
Distance between two walls (S₁ + S₂) = ?

The velocity of sound in air is 330 mIs and the distance between two walls is 1650 m.

d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How may times as fast as the other? (Ans: In A; Twice)
Answer:
In A; Thrice

e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw? (Ans: Temperature of B is 4 times the temperature of A.)
Given:
Mass of Helium in bottle A = (mA) = 10gm
Mass of Helium in bottle B = (mB) = 40gm

The temperature of B is 4 times the temperature of A

Class 9 Science Chapter 12 Study of Sound Intext Questions and Answers

Study Of Sound Class 9 Notes Maharashtra Board Question 1.
How does the velocity of sound depend on its frequency?
Answer:
The velocity of sound is directly proportional to its frequency
ν = υ λ
when ν = velocity
υ = frequency
λ = wavelength

9th Class Science Chapter 12 Study Of Sound Exercise Question 2.
The molecular weight of oxygen gas (O₂) is 32 while that of hydrogen gas (H₂) is 2. Prove that under the same physical conditions, the velocity of sound in hydrogen is four times that in oxygen.
Answer:
Given:
Molecular wt of Oxygen (M_o) =32
Molecular wt of hydrogen (M_H) = 2
To Find:
V_H = 4 v_o

Hence, proved that velocity of sound in hydrogen is four times that in oxygen.

Answer the following:

Study Of Sound Class 9 Maharashtra Board Question 1.
How will you reduce reverberation in public halls or buildings?
Answer:
(i) Reverberation in public halls or buildings will be reduced by using sound absorbing materials like curtains on wall, carpets on the floor.
(ii) By keeping the windows open, as sound will not get reflected.

12 Study Of Sound 9th Class Exercise Question 2.
How is ultrasound used in medical science?
Answer:

Sonography: Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
Echocardiography: Echocardiography is a test that uses ultrasonic sound waves to produce live images of your heart.

9th Science Chapter 12 Study Of Sound Exercise Question 3.
To hear the echo distinctly, will the distance from the source of sound to the reflecting surface be same at all temperatures? Explain your answer.
Answer:

No,the distance from the source of sound to the reflecting surface will not be the same at all temperatures.
Velocity of sound is directly proportional to the square root of temperature.
One of the conditions of echo is that the time interval between the original and reflected sound should be more than 0.1 sec.
So if the temperature increases, the velocity of sound increases and the reflected sound reaches in less than 0.1 sec.
So for echo to be heard the distance between the observer and the reflecting surface has to increase.

9th Science Chapter 12 Study Of Sound Question 4.
When is the reflection of sound harmful?
Answer:

Reflected sound of high intensity called as noise is disturbing and harmful to the ears.
When sound reverberates i.e it undergoes multiple reflections, poor quality of sound is produced.

9th Class Science Chapter 12 Study Of Sound Question Answer Question 5.
What kind of waves are created when a stone is dropped in water ?
Answer:

When a stone is dropped in water, the particles of water oscillate up and down.
These oscillations are perpendicular to the direction of propagation of the wave, such waves are called transverse waves.

Answer the following question:

12 Study Of Sound 9th Class Question 1.
Observe the graph/ diagram and discuss your observation.
Answer:

Fig. A shows changes in density. The region where particles are crowded is called compression and where they are far apart are rarefaction.
Fig. B show change in pressure. The lines represent layers of air. The regions when lines are crowded are high pressure regions while when they are far apart are of low pressure.
Fig. C shows changes in density or pressure. The crest represents high pressure region while trough represents low pressure region.

Answer the following question:

Study Of Sound Class 9 Question Answer Question 1.
How are the frequencies of notes sa, re, ga, ma, pa, dha, ni related to each other?
Answer:
The frequencies of notes sa, re, ga, ma, pa, dha, ni are related in the ratio.

i.e if first Sa is 240Hz then the next Sa will be 240 x 2 = 480Hz

Class 9 Science Chapter 12 Study Of Sound Exercise Question 2.
What is the main difference between the frequencies of the voice of a man and that of a woman?
Answer:

Voice of a woman is high pitch i.e shorter wavelength and higher frequency
Voice of man is low pitch i.e larger wavelength and smaller frequency.

Question 3.
Try this;

(a) In the above activity, what will happen if you lift one of the tubes to some height?
Answer:
If one of the tubes is lifted, angle of incidence will not be equal to angle of reflection, hence, the sound will not be clearly audible.

(b) Measure the angle of incidence 01 and the angle of reflection 02. Try to see if they are related in any way.
Answer:
Angle of incidence is same as the angle of reflection.

Class 9 Science Chapter 12 Study of Sound Additional Important Questions and Answers

Can you recall?

12.Study Of Sound Question 1.
How is the direction of the oscillation of the particles of the medium related to the direction of propagation if the sound wave?
Answer:

Sound travels as a longitudinal wave.
In a longitudinal wave, the particle of the medium oscillate parallel to the direction of propagation of the wave.

Choose and write the correct option:

Class 9 Science Chapter 12 Study Of Sound Question 1.
The unit of frequency is ……………………………… .
(a) Hertz
(b) m/s²
(c) Decibels
(d) m/s
Answer:
(a) Hertz

Study Of Sound Class 9 Exercise Question 2.
The normal hearing range for humans is ……………………………… .
(a) 0 Hz to 20 Hz
(b) greater than 20,000 Hz
(c) 20 Hz to 20,000 Hz
(d) none of these
Answer:
(c) 20 Hz to 20,000 Hz

Class 9th Science Chapter 12 Study Of Sound Question Answer Question 3.
Sound will not travel through ……………………………… .
(a) Vacuum
(b) Liquid
(c) Solid
(d) Gases
Answer:
(a) vacuum

Class 9 Science Chapter 12 Study Of Sound Question Answer Question 4.
SI unit of ………………………………. is Hertz (Hz).
(a) Wavelength
(b) Frequency
(c) Speed of wave
(d) Velocity
Answer:
(b) frequency

Reflection Of Sound Class 9 Question 5.
The velocity of sound is inversely proportional to the ……………………………… .
(a) Pressure
(b) Square root of temperature
(c) Square root of density
(d) Humidity
Answer:
(c) square root of density

Question 6.
Sound waves with frequency greater than 20 kHz are called ……………………………… .
(a) Infrasound
(b) Ultrasound
(c) Sonic
(d) Damped sound
Answer:
(b) ultrasound

Question 7.
The loudness of a sound depends upon ……………………………… .
(a) Amplitude
(b) Speed
(c) Density
(d) Wavelength
Answer:
(a) Amplitude

Question 8.
……………………………… are used in sonography.
(a) High frequency ultrasound
(b) Stationary waves
(c) High frequency infrasound
(d) High frequency micro waves
Answer:
(a) High frequency ultrasound

Question 9.
The ……………………………… receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve.
(a) Cochlea
(b) Tympanic cavity
(c) Stapes
(d) Pinna
Answer:
(a) Cochlea

Find the odd one out:

Question 1.
Bats, rats, cats, dolphins
Answer:
Cats: cannot produce ultrasonic sound.

Question 2.
Clothes, paper, curtains, mirror
Answer:
Mirror: is a good reflector of sound, while others are poor reflectors.

Question 3.
Submarines, icebergs, internal organ, sunken ships.
Answer:
Internal organ: sonography is used , while for others sonar system is used.

Question 4.
Temperature, density, molecular weight, pressure
Answer:
Pressure: for a fixed temperature, the speed of sound does not depend on the pressure of the gas, all other factors affect speed of sound.

Answer in one sentence:

Question 1.
How can one produce sound?
Answer:
Vibration set up in an object produces sound (or) sound is produced when an object is disturbed and starts vibrating.

Question 2.
What is velocity of sound wave ?
Answer:
The distance covered by a point on the wave in unit time is the velocity of the sound wave.

Question 3.
What is the minimum distance of the reflecting surface to hear an echo ?
Answer:
To be able to hear a distinct echo, the reflecting surface should be at a minimum distance of 17.2 m.

Match the columns:

Question 1.

Column ‘A’	Column B’	Column C
(1) Transverse wave	(a) Particles oscillate parallel to direction of propagation	(i) Wave produced in a slinky
(2) Longitudinal wave	(b) Particles oscillate perpendicular to direction of propagation	(ii) Frequency less than 20 Hz
(3) Ultrasound	(c) Echo formation is heard under particular conditions	(iii) Wave produced in string
(4) Infrasound	(d) High frequency waves	(iv) Frequency between 20 Hz to 20000 Hz
(5) Audible frequency	(e) Low frequency waves	(v) Frequency greater than 20000 Hz

Answer:
(1-b- iii),
(2– a- i),
(3 – d – v),
(4 – e – ii),
(5 -c- iv)

Question 2.

Column A’	Column ‘B’	Column C
(1) Amplitude	(a) T	(i) Pitch of sound
(2) Frequency	(b) A	(ii) Loudness of sound
(3) Wavelength	(c) υ	(iii) Reciprocal of frequency
(4) Time period	(d) λ	(iv) v/υ

Answer:
(1 -b – ii),
(2 -c – i),
(3-d – iv),
(4 – a – iii)

Name the following:

Question 1.
A form of energy which produces sensation of hearing in our ears.
Answer:
Sound energy

Question 2.
Repetitions of sound due to reflection .
Answer:
Echo

Question 3.
The audible range of sound for human being.
Answer:
20 Hz to 20,000 Hz

Question 4.
A method to obtain images of internal organs of the human body.
Answer:
Sonography

Question 5.
The matter or substance through which sound gets transmitted.
Answer:
Solid, liquid, gases

Question 6.
Three major parts of the ear.
Answer:
External ear, the middle ear and the inner ear.

Question 7.
Any two examples in which infrasound is produced.
Answer:
Pendulum, earthquake.

Question 8.
Name the living beings that can produce ultrasound.
Answer:
Bats, dolphins, mice.

Give scientific reasons:

Question 1.
Bats can navigate in dark.
Answer:

The ultrasonic sound produced by bats, gets reflected on hitting an obstacle.
This reflected sound is received by their ears and they can locate the obstacle and estimate its distance even in the dark.
Hence, bats can navigate in dark.

Question 2.
A SONAR system is installed in a ship.
Answer:

A SONAR system determines the depth of the sea.
It locates underwater hills, valleys, icebergs, submarines and sunken ships. It also locates the positions of other ships or submarines.
Hence a SONAR system is installed in a ship.

Question 3.
Sound travels faster in iron than in air.
Answer:

Sound requires a material medium for its propagation and travels in the form of a longitudinal wave.
The denser the medium, faster is the propagation of sound.
Hence, sound travels faster in iron than in air.

Solve the following:

Type – A

Formula:
\(\text { (i) Velocity }=\frac{\text { distance }}{\text { time }}\)

Question 1.
Ultrasonic waves are transmitted downwards into the sea with the help of a SONAR. The reflected sound is received after 4 s. What is the depth of the sea at that place? (Velocity of sound in seawater = 1550 m/s)
Answer:
Given:
Time to hear echo = 4 sec
Time taken by sound waves to reach the bottom 4 of sea (t) = 4/2 sec = 2 sec
Velocity of sound in sea water (v) = 1550 m/s
To find:
Depth of sea(s) = ?

The depth of the sea at that place is 3100 m.

Question 2.
A person standing near a hill fires a gun and hears the echo after 1 second. If speed of sound in air is 340 m/s. Find the distance between the hill and the person.
Answer:
Given:
Time to hear echo = 1 sec 1
Time taken (t) = 1/2 sec
Velocity of sound (v) = 340 m/s
To find:
Distance (s) = ?

Distance between the person and hill is 170 m.

Numerical For Practice

Question 3.
If you hear the thunder 20 seconds after you see the flash of lightning, how far from you has the lightning occurred? (Speed of sound in air = 340 m/s)
Answer:
6800m

Question 4.
Aboy observes smoke from a cannon 3 seconds before he hears the bang. If the cannon is 1020 m from the observer, find the velocity of sound.
Answer:
340 rn/s

Question 5.
A soldier standing between the two buildings fires a gun. He heard the echo of the sounds from the first building after 2 seconds and echo from the second building after 3 seconds. Find the distance between two buildings. (Speed of sound in air = 340 m/s)
Answer:
850m

Type – B

\(Formula:
(i) Velocity = Frequency \times Wavelength
(ii) Velocity =\frac{\text { distance }}{\text { time }}\)

Question 1.
Sound waves of wavelength 1 cm have a velocity of 340 mIs in air. What is their frequency? Can this sound be heard by the human ear?
Answer:
Given:
wave length (λ) = 1cm = 1/100
Velocity of sound (v) = 340 m/s
To fInd :
frequency (u) = ?
Formulae:
ν = υ λ
Solution:

The frequency of the sound waves is 34000 Hz. The frequency is higher than 20000 Hz and therefore, this sound cannot be heard by the human ear.

Question 2.
How long will it take for a sound wave of 25 cm wavelength and 1.5 kHz frequency, to travel a distance of 1.5 km?
Answer:
Given:
frequency (u) = 1.5 kHz = 1500 Hz

\(\begin{array}{l}
=\frac{1500}{375} \\
=4 \mathrm{sec}
\end{array}\)
The sound wave takes 4 sec to travel the distance of 1.5 km.

Question 3.
Calculate distance travelled by a sound wave having frequency 1000 Hz and wavelength 0.25 m, if it travels for 5 seconds in a certain medium.
Answer:
Given:
frequency (u) = 1000 Hz
wavelength (λ) = 0.25 m
time (t) = 5 seconds
To find :
Distance (d) =?
Formulae:
ν = υ λ
Solution:

The distance travelled by the sound wave is 1250 m.

Question 4.
The audible range of sound is 20 Hz to 20000 Hz. At 22°C in air speed of sound is 344 mIs. Express the range of sound in terms of wavelength by calculating the respective values.
Answer:
Given:
frequency (u₁) 20 Hz
frequency ( u₂) = 20,000 Hz
velocity (v) = 344 rn/s
To find :
Wavelengths λ₁ and λ₂= ?
Formulae:
ν = υ λ
Solution:

Audible range of wavelength of sound is from 17.2 x 10^-3 m to 17.2 m.

Numerical For Practice

Question 5.
A sound wave has frequency 320 Hz and wavelength 0.25 m. How much distance will it travel in 10 second?
Answer:
The distance travelled is 800 m.

Type – C

Question 1.
Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gin respectively. In which bottle will sound travel faster? How many timés as fast as the other?
Answer:
Given:
Mass of hydrogen in bottle A (m_A) = 12gm
Mass of hydrogen in bottle B(m_B) = 48gm
To find:
In which bottle sound travels faster.
Formulae:

Since both bottles are identical hence, the volume is the same, i.e. v
Dividing (j) and (ii),

(i) Vivacity of sound will be more in bottle A.
(ii) Velocity of sound in bottle A (V_A) is twice of that in bottle B (v_B)

Numerical For Practice

Question 2.
Argon gas is filled in two identical bottles X and Y. The mass of the gas in the two bottles is 5 gm and 25gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw?
Answer:
(Temperature of Y is 5 times the temperature of X.)

Type – D

Numerical For Practice

Question 1.
Velocity of sound in air at 0°C is 332nVs. It increases by 0.6ni/s for each °Celsius rise in temperature. At what temperature of ait the velocity will be 359m1s?
Answer:
45°C

Question 2.
Velocity of sound In air at 0°C is 332m/s It increases by 0.6mIs for each degree Celsius rise In temperature. What will be the velocity of sound at 60°C?
Answer:
368 rn/s

Define the following:

Question 1.
Wave length (λ)
Answer:
The distance between two consecutive compressions (or crests) or two consecutive rarefactions (or troughs) is called the wavelength.

Question 2.
Amplitude (A)
Answer:
The maximum value of pressure or density is called amplitude.

Question 3.
Frequency (υ)
Answer:
The frequency of a sound wave is defined as the number of complete oscillations of density (or pressure of the medium) per second.

Question 4.
Time Period (T)
Answer:
The time taken for one complete oscillation of pressure or density at a point in the medium is called the time period.

Question 5.
Echo
Answer:
An echo is the repetition of the original sound because of reflection by some surface.

Question 6.
Transverse waves
Answer:
Oscillations of the particles of the medium vibrate at right angles to the direction of propagation of the wave are called transverse waves.

Question 7.
longitudinal waves
Answer:
The particles of the medium oscillate about their central or mean position in a direction parallel to the propagation of wave is called as longitudinal waves.

Question 8.
Velocity of wave
Answer:
The distance covered by a point on the wave (for example the point of highest density or lowest density) in unit time is the velocity of the sound wave.’

Distinguish between:

Question 1.
Infrasound and Ultrasound
Answer:

	Infrasound		Ultrasound
(i)	Longitudinal waves whose are below 20 Hz are called Infrasound waves. frequencies Infrasonic or	(i)	Longitudinal waves whose frequencies lie- above 20,000 Hz are called Ultrasonic or ultrasound waves.
(ii)	Whales, elephants produce sound in the infrasound range.	(ii)	Bats produce (30 kHz to 50 kHz) frequency and dolphins produce ultrasound (100 kHz).

Question 2.
Transverse waves and Longitudinal waves
Answer:

Transverse waves	Longitudinal waves
(i) Particles of the medium vibrate at right angles to the direction of propagation of the wave. (ii) They produce crests and troughs. (iii) For transverse waves, a wavelength is made up of one crest and one trough.	(i) Particles of the medium vibrate parallel to the direction of propagation of the wave. (ii) They produce compression and rarefaction. (iii) For longitudinal waves, a wavelength is made up of one compression and one rarefaction.

Question 3.
Consider two cases
(A) whistle of train (B) roar of a lion

(I) In which case the sound is high pitch?
Answer:
Whistle of a train is high pitch as compared to roar of a lion, as the frequency is higher.

(II) What is the real cause of sound production? Explain with examples.
Answer:

Vibrations in the object are responsible to produce a sound.
Vibration is a rapid to and fro motion of an object.
Sometimes the vibrations may be strong enough to be seen by eyes, e.g. string vibrations in string instruments, vibration on mobile phone, blowing air in the cap of your pen by holding it near the lips.

(III) Three sounds 5 Hz, 500 Hz and 50,000 Hz are produced by different sources.
(a) Which sound will be heard by humans?
(b) Which sounds may be produced by bats?
(c) Which sounds may be produced by elephants?
Answer:
(a) 500 Hz – Humans can hear sounds in the range of 20 Hz-20,000 Hz
(b) 50,000 Hz – Bats produce ultrasonic sounds above 20,000 Hz
(c) 5 Hz – Elephants can produce infrasonic sounds below 20 Hz

Question 6.
Suppose you and your friend are on the moon. Will you be able to hear any sound
Answer:
Sound waves need a material medium for their propagation. Since there is no atmosphere on the moon, we cannot hear any sound on the moon.

Answer in detail:

Question 1.
What are the factors on which velocity of sound in gaseous medium depend?
Answer:
The velocity of sound in a gaseous medium depends on the physical conditions i.e. the temperature, density of the gas and its molecular weight.

Temperature (T): The velocity of sound is directly proportional to the square root of the temperature of the medium. This means that increasing the temperature four times doubles the velocity.
\(\text { v } \alpha \sqrt{\mathrm{T}}\)
Density(p): The velocity of sound is inversely proportional to the square root of density. Thus, increasing the density four times, reduces the velocity to half its value.
\(\mathrm{v} \alpha \frac{1}{\sqrt{\rho}}\)
Molecular weight (M): The velocity sound is inversely proportional to the square root of molecular weight of the gas. Thus, increasing the molecular weight four times, reduces the velocity to haff its value.
\(\mathrm{v} \alpha \frac{1}{\sqrt{\mathrm{M}}}\)

Question 2.
What are the uses of ultrasonic sound?
Answer:
Uses of ultrasonic sound are as follows:

For communication between ships at sea.
To join plastic surfaces together.
To sterilize liquids like milk by killing the bacteria in it so that the milk keeps for a longer duration.
Echocardiography which studies heartbeats, is based on ultrasonic waves (Sonography technology).
To obtain images of internal organs in a human body.
In industry to clean intricate parts of machines where hands cannot reach.
To locate the cracks and faults in metal blocks.

Question 3.
Explain with the help of a neat labelled diagram the working of human ear.
Answer:

The ear is an important organ of the human body.
When sound waves fall on the eardrum, it vibrates and these vibrations are converted into electrical signals which travel to the brain through nerves.
The ear can be divided into three parts:
(a) Outer ear
(b) Middle ear
(c) Inner ear.

(a) Outer ear or Pinna
The outer ear collects the sound waves and passes them through a tube to a cavity in the middle ear. Its peculiar funnel like shape helps to collect and pass sounds into the middle ear.

(b) Middle ear
There’ is a thin membrane in the cavity of the middle ear called the eardrum. When a compression in a sound wave reaches the eardrum, the pressure outside it increases and it gets pushed inwards. The opposite happens when a rarefaction reaches there. The pressure outside decreases and the membrane gets pulled outwards. Thus, sound waves cause vibrations of the membrane.

(c) Inner ear
The auditory nerve connects the inner ear to the brain. The inner ear has a structure resembling the shell of a snail. It is called the cochlea. The cochlea receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve. The brain analyses these signals.

Question 4.
Write a short note on SONAR
Answer:

(i) SONAR is the short form for Sound Navigation and Ranging. It is used to determine the direction, distance and speed of an underwater object with the help of ultrasonic sound waves. SONAR has a transmitter and a receiver, which are fitted on ships or boats.

(ii) The transmitter produces and transmits ultrasonic sound waves. These waves travel through water, strike underwater objects and get reflected by them. The reflected waves are received by the receiver on the ship.

(iii) The receiver converts the ultrasonic sound into electrical signals and these signals are properly interpreted. The time difference between transmission and reception is noted. This time and the velocity of sound in water give the distance from the ship, of the object which reflects the waves.

(iv) SONAR is used to determine the depth of the sea. SONAR is also used to search underwater hills, valleys, submarines, icebergs, sunken ships etc.

Question 5.
Write a short note on Sonography. How is it misused?
Answer:

Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
This is useful in finding out the cause of swelling, infection, pain, condition of the heart, the state of the heart after a heart attack as well as the growth of foetus inside the womb of a pregnant woman.
This technique makes use of a probe and a gel.
The gel is used to make proper contact between the skin and the probe so that the full capacity of the ultrasound can be utilized.
High-frequency ultrasound is transmitted inside the body with the help of the probe.
The sound reflected from the internal organ is again collected by the probe and fed to a computer which generates the images of the internal organ.
As this method is painless, it is increasingly used in medical practice for correct diagnosis.
This technique is used by many people to find out gender of an unborn baby and this often leads to the incidence of female foeticide.

9th Std Science Questions And Answers:

Std 9 History Chapter 5 Question Answer Maharashtra Board

February 6, 2025January 4, 2025 by Bhagya

Balbharti Maharashtra State Board Class 9 History Solutions Chapter 5 Education Notes, Textbook Exercise Important Questions and Answers.

Education Class 9 Questions And Answers Maharashtra Board

Class 9 History Chapter 5 Education Questions And Answers Maharashtra Board

History Class 9 Chapter 5 Question Answer Maharashtra Board

1. Choose the right option and rewrite the sentence:

Question 1.
The Scientist who developed the Param – 8000 supercomputer _________.
(a) Dr Vijay Bhatkar
(b) Dr R. H. Dave
(c) P. Parthasarathy
(d) None of the above
Answer:
(a) Dr. Vijay Bhatkar

Question 2.
The magazine Jeevan Shikshan is published by _______.
(a) Balbharati
(b) University Education Commission
(c) MSCERT
(d) NCERT
Answer:
(c) MSCERT

Question 3.
The educational institution called IIT is famous for educational in the area of _______.
(a) Agriculture
(b) Medicine
(c) Skilled managers
(d) Engineering
Answer:
(d) Engineering

2. Do as directed.

Question 1.
Complete the table below with details about individuals in the education field and their work.

Individual	Work
First education minister of India	Maulana Abul Kalam Azad.
—————–	Chairman of the University Education Commission
Prof Sayyad Rauf	…………..
……………………..	Kosbad Project

Answer:

Individual	Work
First education minister of India	Maulana Abul Kalam Azad.
Dr. Sarvapalli Radhakrishna	Chairman of the University Education Commission
Prof Sayyad Rauf	Preparation of common state-wide curriculum for Std. I to VII for Maharashtra.
Anutai Wagh	Kosbad Project

Question 2.
Obtain information about the National Council of Educational Research and Training, from the internet, and present it in the form of a timeline.
Answer:

1st September 1961	The National Council of Educational Research and Training (NCERT) was established in Delhi.
27th January 1967	The Maharashtra State Bureau of Textbook Production and Curriculum Research (Balbharati) was established in Pune.
1986	The National Policy on Education was adopted.
1988	The ‘Primary Education Curriculum was prepared.
1995	The Competency-based primary education curriculum.
1975	India met with success in its attempt to use a satellite for the purpose of education.
1970	The United Nations declared the year 1970 as the International Education Year. A seminar was organised in New Delhi and led to the establishment of an open university.
1974	The government appointed a committee under the chairmanship of P. Parthasarathy

3. Explain the following statements giving reasons for your answer:

Question 1.
The District Primary Education Programme was undertaken.
Answer:
(i) In 1994, the District Primary Education Programme (DPEP) was started with the objective of universalisation of primary education. It was implemented in seven States including Maharashtra.

(ii) The plan envisaged 100% attendance in primary schools, arresting student drop-out, education for girls and for the physically handicapped.

(iii) It included programmes such as research on and evaluation of primary education, alternative education, creating societal awareness, etc.

Question 2.
The NCERT was established.
Answer:
(i) National Council of Educational Research and Training (NCERT) was established in Delhi on the 1st of September 1961.

(ii) Its main objective is to help the Central Government in matters of school education in the context of a comprehensive policy and in implementing educational schemes. The NCERT was given the responsibility of educational research and development, training, extension, educational programmes, restructuring of school curriculum and textbooks.

(iii) It has played a central role in designing school curricula and textbooks with the cooperation of the Central Board of Secondary Education (CBSE).

(iv) It provides guidance and cooperation in the area of primary and secondary education to the state governments.

(v) It has been involved in designing workbooks and handbooks for teachers, in teacher training, in developing teaching-learning techniques and conducting talent search examinations at the national level.

Question 3.
The farmers were benefited by the Indian Agricultural Research Institute.
Answer:
(i) The Indian Agricultural Research Institute was given the status of a University in 1958 and work began in departments such as the development of the agriculture sector, research, well-equipped laboratories, soil science, agricultural sciences, economic botany and other departments.

(ii) Research also began on wheat, pulses, oilseeds, vegetables and many other problems.

(iii) Its most significant achievement is the fundamental research it has conducted on the methods of taking multiple crops in a year, which has been of great benefit to farmers.

4. Write notes:

Question 1.
Indira Gandhi National Open University
Answer:
(i) Indira Gandhi National Open University (IGNOU) was established with the objective that the stream of education should make
its way into every average household of the country.

(ii) The United Nations had declared 1970 as the International Education Year. In the same year, a seminar was organised in New Delhi on the subject of Open University by the Government of India Departments of Education and Social Welfare, Information and Broadcasting, University Grants Commission along with UNESCO. The idea of establishing an open university evolved in this seminar.

(iii) In 1974, the government appointed a committee under the chairmanship of P. Parthasarathy and the Open University took shape on 20 September, 1985 in accordance with its recommendations and suggestions. It was named after Prime Minister Indira Gandhi. Those who are unable to get a college education in the formal way are given concessions in eligibility criteria, age and other conditions for admission to this university.

(iv) In 1990, the IGNOU started an audio-visual distance education programme through Akashvani and Doordarshan. It conducted more than one thousand curricula of various branches.
It provides facilities for education through 58 training centres in the country and 41 centres in foreign countries.

Question 2.
Kothari Commission
Answer:
(i)In 1964 a Commission was appointed under the Chairmanship of Dr D. S. Kothari. Dr J. P. Naik made valuable contributions to the work of this Commission.

(ii)The Commission also recommended the 10+2+3 pattern for secondary, higher secondary and university education. This system was implemented from 1972.

(iii)The Commission also suggested a uniform national system of education, the inclusion of the mother tongue, Hindi and English in education, as well as continuing education, adult education, education by correspondence and open universities to make education trickle down to the lowest rungs of society.

(iv)It also recommended increased provision in the government’s budget for expenditure on education of neglected sections like the scheduled castes and tribes. Maharashtra government adopted the 10+2+3 pattern in 1972 and conducted the first Std X Board exam in 1975.

Question 3.
Bhabha Atomic Research Centre
Answer:
(i) Bhabha Atomic Research Centre (BARC) has conducted valuable research in the fields of nuclear physics, solid state physics, spectroscopy, chemical and life sciences.

(ii) It also started a school to train scientists for setting up nuclear reactors.

Question 4.
Balbharati
Answer:
(i) The Maharashtra State Bureau of Textbook Production and Curriculum Research (Balbharati) was established in Pune on 27 January 1967.

(ii) Balbharati prepares textbooks for school children. Textbooks are made in eight languages, namely, Marathi, Hindi, English, Urdu, Kannad, Sindhi, Gujarati and Telugu. ‘Kishor’, a monthly magazine for children, is also published by Balbharati.

5. Answer the following questions in detail:

Question 1.
Which programmes were included in the Operation Blackboard scheme?
Answer:
(i) The government made funds available to help improve the standard of the schools and to fulfil minimum educational needs such as at least two proper classrooms, toilets, one of the two teachers to be female, a blackboard, maps, laboratory apparatus, a small library, a playground, sports equipment, etc. This scheme helped the primary education system to gain some momentum.

(ii) In 1994, this scheme was expanded and provision was made for one additional classroom and for appointing one more teacher in schools with an enrolment of more than 100 students, with priority given to girls’ schools, schools with a majority of scheduled caste and tribe students and schools in rural areas.

(iii) It was also made binding upon the State government to appoint female teachers to fifty per cent of the posts in schools.

Question 2.
What role do agriculture schools/colleges play in the development of agriculture?
Answer:
(i) The Indian Agricultural Research Institute was given the status of a University in 1958 and work began in departments such as the development of the agriculture sector, research, well-equipped laboratories, soil science, agricultural sciences, economic botany and other departments. Research also began on wheat, pulses, oilseeds, vegetables and many other problems.

(ii) Its most significant achievement is the fundamental research it has conducted on the methods of taking multiple crops in a year, which has been of great benefit to farmers.

Question 3.
Describe with examples of the progress that India has made in the field of medicine.
Answer:
(i) The Indian Council for Medical Research (ICMR) was established for conducting research in the medical field. It was given the responsibility of cooperating with universities, medical colleges, government and non-government research institutes and giving them guidance and financial support for research activities.

(ii) Twenty-six centres were started in different parts of the country for research on various diseases. Their research has made it possible to control tuberculosis and leprosy.

(iii) The All India Institute of Medical Sciences (AIIMS) was established to give further impetus to advanced education and research in medicine.

(iv) It was given the responsibility of undergraduate and postgraduate courses in medicine. Colleges for undergraduate and postgraduate education in most branches of medicine, good research facilities and well- equipped hospitals are the significant features of this institute.

(v) The institute also provides medical treatment to the common people at nominal rates. It has established special colleges for training in nursing, and super-speciality centres for treatment of disorders of the heart, brain and eyes.

(vi) For further development of the medical field, the Medical Council of India was restructured in 1958 and was entrusted with the task of determining criteria for quality of medical education, its supervision and inspection.

(vii) These institutes have the responsibility of conducting research on various diseases, developing tests and standardisation of medicines.

Question 4.
Write a note on the curricular and co-curricular activities conducted in your school.
Answer:
(i) Languages – Our school motivates us to participate in theatres or plays organised on works of Shakespeare and Munshi Premchand. We also take part in elocution and poetry recitation competitions to enhance public speaking. Our school makes every child read the newspaper headlines from Marathi and English dailies in the auditorium turn wise.

(ii) Maths and Science – Scientific experiments mentioned in our textbook are practically performed by each of us under the supervision of our teachers. On Sundays, our Math teachers take special classes on Vedic Maths. Our school has also organised trips to Nehru Planetarium and ‘Jantar Mantar’ to give us an understanding of the astronomical aspects of science. Every year our school conducts Homi Bhabha scholarship exam. A display of the best 25 science projects is done in the school hall.

(iii) Social Studies – Field studies are taken up to explain to us the topography of Indian landscapes. The school library holds a large number of book and documentaries over world history trips to museums and historical places.

Class 9 History Chapter 5 Education Additional Important Questions and Answers

Choose the correct option from the given options and rewrite the statements:

Question 1.
In 1995, the _____ scheme was started to provide proper nourishment for the students.
(a) free ration
(b) foodgrain subsidy
(c) mid-day meal
(d) medical-check up
Answer:
(c) mid-day meal

Question 2.
_______was the first education minister of Independent India.
(a) Vallabhbhai Patel
(b) Maulana Abdul Kalam Azad
(c) Sachchidanand Sinha
(d) J.B.Kriplani
Answer:
(b) Maulana Abdul Kalam Azad

Question 3.
_____ recommended increased provision in government’s budget for expenditure on education of neglected sections like the SC/ST.
(a) University Education Commission
(b) Mudaliar Commission
(c) National Council of Educational Research and Training
(d) Kothari Commission
Answer:
(d) Kothari Commission

Question 4.
_______, a scientist with ISRO played an important role in providing satellite education.
(a) K. Kasturirangan
(b) Eknath Chitnis
(c) A. S. Kirankumar
(d) G. Madhavan Nair
Answer:
(b) Eknath Chitnis

Question 5.
In 1990 _____ started an audio-visual distance education programme through Akashwani and Doordarshan.
(a) ISRO
(b) NCERT
(c) IGNOU
(d) CSIR
Answer:
(c) IGNOU

Question 6.
The National Institute for Research in Mathematical and Physical Sciences’ in ______ was established in 1962.
(a) Chandigarh
(b) Kerala
(c) Karnataka
(d) Tamil Nadu
Answer:
(d) Tamil Nadu

Question 7.
India’s first IIT was setup at _______ at West Bengal in 1951.
(a) Kharagpur
(b) Midnapore
(c) Hooghly
(d) Howrah
Answer:
(a) Kharagpur

Identify and write the wrong pair in the following sets:

Question 1.
(1) TCS – obtained an American contract for software production.
(2) FTII – Systematic training in all aspects of film-making.
(3) C-DAC – developed Param – 8000
(4) IIM – imparting training in industrial design.
Answer:
Wrong Pair : IIM – imparting training in industrial design.

Question 2.
(1) The Advanced Centre for Treatment Research and Education in Cancer – Branch of Tata Memorial Centre
(2) AIIMS – advanced education and research in medicine.
(3) DPEP – Charts showing learning competencies upto Std.V
(4) National Policy on Education 1986 – Common core curriculum was framed for all states.
Answer:
Wrong Pair : DPEP – Charts showing learning competencies upto Std.

Question 3.
(1) National Institute of Design – Ahmedabad
(2) Maharashtra State Board of Secondary and Higher Secondary Education – Pune
(3) The National Institute for Research in the Mathematical and Physical Science – Delhi
(4) Indian Institute of Technology – West Bengal
Answer:
Wrong Pair: The National Institute for Research in the Mathematical and Physical Science – Delhi

Question 4.
(1) J.P. Naik – Kothari Commission
(2) P. Parthasarthy – Committee on Open University
(3) Sayyad Rauf – Drafted common state wide curriculum
(4) Dr. Sarvapalli Radhakrishnan – Director of NCERT.
Answer:
Wrong Pair : Dr. Sarvapalli Radhakrishnan Director of NCERT.

Question 5.
(1) Dave Committee – minimum levels of learning.
(2) The Art Department – designing workbooks and hand books.
(3) Kosbad project – Education of Adivasis
(4) Council of Scientific and Industrial Research – Promote scientific research.
Answer:
Wrong Pair : The Art Department – designing workbooks and hand books.

Question 6.
(1) MSCERT – Jeevan Shikshan
(2) Balbharati – Kishor
(3) The Maharashtra State Board of Secondary and Higher Secondary Education – Countrywide classroom.
(4) SITE – Satellite Education.
Answer:
Wrong Pair : The Maharashtra State Board of Secondary and Higher Secondary Education – Countrywide classroom

Question 7.
(1) Tarabai Modak – Kuranshalas
(2) Anutai Wagh – Kosbad Projects
(3) Eknath Chitnis – Education Minister
(4) R. H. Dave – Competency based Primary Education Curriculum.
Answer:
Wrong Pair: Eknath Chitnis – Education Minister

Complete the concept maps:

Question 1.
Objectives of Education:
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 1

Question 2.
Central Council for Research in Indian Medicine and Homeopathy:
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 2

Question 3.
Departments of the Indian Agricultural Research Institute:
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 3

Question 4.
Role of Dave Committee:
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 4

Question 5.
Features of the IIT
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 5

Question 6.
IIMs in India
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 6

Question 7.
Concept map of basic responsibilities of NCERT
Answer:
Maharashtra Board Class 9 History Solutions Chapter 5 Education 11

Write short notes:

Question 1.
AIIMS
Answer:
(i) The All India Institute of Medical Sciences (AIIMS) was established to give further impetus to advanced education and research in medicine.

(ii) It was given the responsibility of undergraduate and post-graduate courses in medicine.

(iii) Colleges for undergraduate and postgraduate education in most branches of medicine, good research facilities and well-equipped hospitals are the significant features of this institute.

(iv) The institute also provides medical treatment to the common people at nominal rates.

(v) It has established special colleges for training in nursing, and super-speciality centres for treatment of disorders of the heart, brain and eyes.

Question 2.
Achievements of CSIR
Answer:
(i) In the post-independence period in 1950, the Council of Scientific and Industrial Research (CSIR) was established with the objective to promote scientific research in the country and to take the benefits of the research to all the people.

(ii) The achievements of the CSIR include making the ink used for marking voters’ fingers during elections, medicines for malaria, elephantiasis and tuberculosis, water purification technology, reduction in the time required for bamboo production.

(iii) It also used DNA fingerprinting for the first time in India, conducted a genetic study of the Adivasis of the Andamans and proved that those tribes are 60,000 years old and developed the earthquake early warning system.

(iv) It has also played an important role in the use of neem as a pesticide, use of turmeric for healing wounds and in the case of the patents for varieties of rice.

(v) The CSIR has prepared a digital encyclopedia of Indian traditional knowledge and made it available in eight international languages.

Question 3.
NCERT
Answer:
(i) National Council of Educational Research and Training (NCERT) was established in Delhi on the 1st of September 1961.

(iii) It has played a central role in designing school curricula and textbooks with the cooperation of the Central Board of Secondary Education (CBSE). ’

(iv) It provides guidance and cooperation in the area of primary and secondary education to the state governments.

Read the passage and answer the questions and answer the questions based on it.

Passage I

National Council of Educational Research and Training. This body was established in Delhi on the 1st of September 1961. Its main objective is to help the Central Government in matters of school education in the context of a comprehensive policy and in implementing educational schemes. The NCERT was given the responsibility of educational research and development, training, extension, educational programmes, restructuring of school curriculum and textbooks. It has played a central role in designing school curricula and textbooks with the cooperation of the Central Board of Secondary Education (CBSE). It provides guidance and cooperation in the area of primary and secondary education to the state governments. It has been involved in designing workbooks and handbooks for teachers, in teacher training, in developing teaching-learning techniques and conducting talent search examinations at the national level.
SCERTs were established in all States on the lines of the NCERT. The Maharashtra State Council of Educational ‘Research and Training was established in Pune, This institute performs various functions such as , improving the standard of primary education, in service training of teachers, training in the areas of syllabii and evaluation, vocational guidance for students after Std X and Std XII exams and other , educational functions This institute is known as the , academic authority. It brings out a periodical called Jeevan Sikshan. ‘

Question 1.
What was the main objective behind establishment of NCERT?
Answer:
The main objective behind establishment of NCERT is to help the central government in matters of school education in the context of a comprehensive policy and in implementing educational schemes.

Question 2.
SCERT was established only in Maharashtra State on the lines of the NCERT. Identify True or False.
Answer:
False.

Question 3.
State 4 research institutes in India.
Answer:
Following are the 4 research institutes in India:

Council of Scientific & Industrial Research.
Bhabha Atomic Research Centre.
Indian Council of Medical Research.
Indian Agricultural Research Institute.

Passage II

National Policy on Education 1986: According to this policy, certain changes of a fundamental nature were brought about in primary, secondary and higher , secondary education in keeping with the changing needs of society. Under this policy, a common core curriculum was framed for all States. The expectation , is that, by this means, all students in India will get equal educational opportunity. There is scope in the national curriculum for individual States to bring in flexibility in accordance with their cultural, geographical and historical needs.
The ‘Primary Education Curriculum 1988’ was prepared on the basis of the plan of action designed at the national level for the effective implementation of the National Policy on Education 1986. The competency based primary Education curriculum 1995: Even as the Primary Education Curriculum , 1988 was being implemented, a Committee was formed at the national level with Dr. R. H. Dave as its Chairman, to determine minimum levels of learning, The Dave as Committee developed charts of minimum , levels of learning in language, mathematics and environmental studies up to Std V. The charts showed the sequence of learning of the competencies meant for each class.
Use of a satellite: In 1975, India met with success in its attempt to use a satellite for the purpose of education, Eknath Chitnis, a scientist with ISRO, played an , important role in it.
SITE (Satellite Instructional Television Experiment) , was undertaken for educational purposes under the leadership of the Space Applications Centre at Ahmedabad. The concept of satellite education came , out of this experiment. America had helped India in , this programme. The programme helped to make provisionsforagoodstandardofeducationinruralareas.

Question 1.
What was done to implement the National Policy on Education 1986?
Answer:
For the effective implementation of National Policy on Education 1986, the Primary Education Curriculum 1988 was prepared on the basis of the action plan formulated at the National level.

Question 2.
How did the Dave committee determine the minimum levels of learning?
Answer:
To determine the minimum levels of learning the Dave committee developed charts in language, mathematics and environmental studies upto Std. V. Sequential learning competencies were depicted on these charts to set standards through levels of learning.

Question 3.
Who is your favourite educationist? State his/ her educational ideas which impacted the society.
Answer:
(i) My favourite educationist are Rabindranath Tagore, Dr. Annie Besant and Dr. Sarvapalli Radha Krishnan.

(ii) They all enlightened the society with their educational ideas and deeds. Dr. Annie Besant in particular was iconic.

(iii) Her ideas of education were imbibed with political freedom, women’s liberation and universal brotherhood. Her teachings of ‘Home Rule’ and ‘theosophy’ impacted our society during British Rule. She taught these ideas through political lectures to motivate the youth.

Answer the following questions in detail:

Question 1.
What does the District Primary Education Programme envisage?
Answer:
(i) In 1994, the District Primary Education Programme (DPEP) was started with the objective of universalisation of primary education. It was implemented in seven States including Maharashtra.

(ii) The plan envisaged 100% attendance in primary schools, arresting student drop-out, education for girls and for the physically handicapped.

(iii) It included programmes such as research on and evaluation of primary education, alternative education, creating societal awareness, etc.

Question 2.
State the contribution of Tarabai Modak in educating adivasis.
Answer:
(i) Tarabai Modak began work in the field of education in Bordi and Kosbad.
(ii) She started Anganwadis for Adivasi children.
(iii) She worked hard to bring in the Teaming by doing’ method, to start kuranshalas (meadow schools) and to spread vocational technical education.

Question 3.
What purpose did the Kosbad Project serve?
Answer:
(i) Anutai Wagh established an institution at Kosbad for the progress of Adivasis in Thane district.
(ii) It is known as the Kosbad Project.
(iii) She started creches, nurseries, primary schools, adult education classes, balsevika training schools, etc. for the education of Adivasis.

Question 4.
What was proposed by the Mudaliar Commission?
Answer:
(i) In 1952-53, the Mudaliar Commission was appointed.
(ii) At that time, the pattern of education consisted of 11 + 4 years for the first degree or of 11 + 1 + 3 years.
(iii) The Commission studied secondary education, the nature of the curriculum, medium of instruction, teaching methodology and made certain recommendations.
(iv) This Commission proposed the concept of Higher Secondary Education. However, it was found difficult to implement it all over the country.

Question 5.
State the recommendation of Kothari Commission.
Answer:
(i) The Kothari Commission recommended the 10+2+3 pattern for secondary, higher secondary and university education. This system was implemented from 1972.

(ii) The Commission also suggested a uniform national system of education, the inclusion of the mother tongue, Hindi and English in education, as well as continuing education, adult education, education by correspondence and open universities to make education trickle down to the lowest rungs of society.

(iii) It also recommended increased provision in the government’s budget for expenditure on education of neglected sections like the scheduled castes and tribes.

Question 6.
What do you know about the Education Board of your state?
Answer:
(i) The Maharashtra State Board of Secondary and Higher Secondary Education was established on the 1st of January 1966 at Pune.
(ii) This Board conducts the school leaving examinations of Std. X and Std. XII.
(iii) It also publishes a periodical called ‘Shikshan Sankraman’.

Question 7.
State the functions of MSCERT.
Answer:
(i) The Maharashtra State Council of Educational Research and Training was established in Pune.
(ii) This institute performs various functions such as improving the standard of primary education,
in-service training of teachers, training in the areas of syllabi and evaluation, vocational guidance for students after Std X and Std XII exams and other educational functions.

Question 8.
How did the concept of satellite education evolve in India?
Answer:
(i) SITE (Satellite Instructional Television Experiment) was undertaken for educational purposes under the leadership of the Space Applications Centre at Ahmedabad.
(ii) The concept of satellite education came out of this experiment. America had helped India in this programme.
(iii) The programme helped to make provisions for a good standard of education in rural areas.

Question 9.
Discuss the responsibilities of Indian Council for Medical Research.
Answer:
(i) Indian Council for Medical Research (ICMR) was established for conducting research in the medical field.
(ii) It was given the responsibility of cooperating with universities, medical colleges, government and non-government research institutes and giving them guidance and financial support for research activities.

Question 10.
Which institute works for advanced research in cancer?
Answer:
The Advanced Centre for Treatment, Research and Education in Cancer is a branch of the Tata Memorial Centre. It functions as the national centre for treatment, research and education in relation to cancer.

Observe the pictures/personality and write relevant information:

Question 1.
Maharashtra Board Class 9 History Solutions Chapter 5 Education 8
Answer:
(i) Kothari Commission recommended the 10+2+3 pattern for secondary, higher secondary and university education.
(ii) The objectives of Education were :

Modernisation of Education.
Social and national integration.
Nurture of social, moral, spiritual values.
Education and Productivity.
Securing and strengthening of democracy.

(iii) The following are the other recommendations made by the Kothari commission:

A uniform national system of education
inclusion of mother tongue (along with Hindi and English)
Continuing education
Adult education
Education by correspondence
Open Universities.

Question 2.
Maharashtra Board Class 9 History Solutions Chapter 5 Education 9
Answer:
(1) The open university was established with the objective that the stream of education should make its way into every average household of the country.

(2) In 1974, the government appointed a committee under the chairmanship of P. Parthasarthy and open university took shape on 20th September 1985 in accordance with its recommendations and suggestions.
It was named after Prime Minister – Indira Gandhi.

(3) In 1990, IGNOU started an audio-visual distance education programme through Akashwani and Doordarshan.

(4) It provides facilities for education through 58 centres in the country and 41 centres in foreign countries.

Question 3.
Maharashtra Board Class 9 History Solutions Chapter 5 Education 10
Answer:
(1) CSIR was established with the objective to promote scientific research in the country and to take the benefits of the research to all the people.

(2) CSIR conducted research in fields like:

Physics, chemistry, pharmaceuticals, food processing and mining

(3) The achievements of CSIR include:

Making the ink used for marking voter’s fingers during elections
Medicines for malaria, elephantiasis and tuberculosis
water purifying technology
DNA fingerprinting for the first time in India.

(4) CSIR has prepared a digital encylopoedia of Indian traditional knowledge and made it available in eight international languages.

9th Std History Questions And Answers:

Std 9 Political Science Chapter 5 Question Answer Maharashtra Board

February 6, 2025January 4, 2025 by Bhagya

Balbharti Maharashtra State Board Class 9 Political Science Solutions Chapter 5 India and Other Countries Notes, Textbook Exercise Important Questions and Answers.

India and Other Countries Class 9 Questions And Answers Maharashtra Board

Class 9 Political Science Chapter 5 India and Other Countries Questions And Answers Maharashtra Board

Political Science Class 9 Chapter 5 Question Answer Maharashtra Board

1. Choose the right option and rewrite the sentence:

Question 1.
The country that has a free international border with India _______.
(a) Pakistan
(b) Bangladesh
(c) Nepal
(d) Myanmar
Answer:
(c) Nepal

Question 2.
The countries that have tense relations with India _______.
(a) Pakistan and China
(b) Nepal and Bhutan
(c) Myanmar and Maldives
(d) Afghanistan and America
Answer:
(a) Pakistan and China

Question 3.
Factors that have an influence on the relations between India and Pakistan _____.
(a) Difference in the world view of both the countries
(b) Kashmir issue
(c) Nuclear Rivalry
(d) All of the above
Answer:
(d) All of the above

2. Explain with reasons whether the following statements are true or false:

Question 1.
India has an important position among the South Asian countries.
Answer:
True.

India’s position in Asia is geographically and politically important.
Afghanistan, Pakistan, Sri Lanka, Bangladesh, Myanmar, Nepal, Bhutan, China and Maldives are India’s neighbours.
The values of equality and mutual respect have a great importance in India’s foreign policy. India has established relations with neighbouring countries on the basis of these values.
India is the biggest country in the Indian subcontinent. Similarly, India is economically and technologically more advanced. It is thus natural that India has a great influence among the South Asian countries.

Question 2.
Sino-Indian relations are friendly.
Answer:
False.

The conflict between India and China is connected to two issues : (a) Border issue and (b) The status of Tibet.
The border dispute between India and China is related to the Aksai Chin area and the Macmahon line.
China claims that the area, South of Aksai Chin and Macmahon line (Arunachal Pradesh) is Chinese Territory. China is not willing to agree that Macmahon Line is an international border.
India made several efforts to sort this border dispute by means of dialogue, but it has not met with much success. In 1962, China Attacked India.
When China was stepping up its military control in Tibet, Dalai Lama took asylum in India. This issue has been responsible for the conflict between the two countries.

Question 3.
India sent a peacekeeping force to help the Sri Lankan government.
Answer:
True.

India has friendly relations with the southern neighbour Sri Lanka.
Political instability in Sri Lanka after 1985 was a result of the differences between the Tamil and the Sri Lankan Government. At that time,
India had sent a peace keeping force to help the Sri Lankan government.
Friendly relations with Sri Lanka are important from the point of view of security in the Indian Ocean.

3. Complete the following chart.

Question 1.

Treaties/Exchanges	Countries
…………..	India and Pakistan
Macmahon Line	……………..
…………………	India and Bangladesh
Import of natural gas	…………..
…………..	India and Africa

Answer:

Treaties/Exchanges	Countries
Shimla Agreement, Tashkent Agreement	India and Pakistan
Macmahon Line	India and China
Treaties about sharing river waters and borders	India and Bangladesh
Import of natural gas	Myanmar
Civil Nuclear Agreement	India and Africa

4. Write short notes :

Question 1.
Shimla Agreement
Answer:
(i) India tried to solve Indo-Pak conflict bi-laterally. The Shimla Agreement of 1972 was based on these principles.

(ii) Pakistan considers Kashmir as the biggest hurdle in establishing relations with India.

(iii) The first Indo-Pak war took place in 1965, over Kashmir. The Tashkent Agreement was signed in 1966, but nothing much came out of it.

(iv) Even though the 1971 war was about the creation of Bangladesh, it also had the dimension of the Kashmir problem to it.

(v) The Shimla Agreement in 1972 provided a new format to the interaction between India and Pakistan.

Question 2.
Indo-Nepalese Friendship Treaty
Answer:
(i) The foundation of friendship between India and Nepal was laid with Indo-Nepalese Friendship Treaty in 1950.

(ii) According to this treaty, Nepalese citizens can not only enter India, but they are also permitted to take up government jobs and carry on trade in India.

(iii) The year 1990 marked the beginning of Nepal’s transition to democracy.

(iv) Nepal is dependent upon India for economic progress, basic facilities, food related needs, trade and energy needs.

(v) India sent a lot of help to Nepal at the time of the 2015 earthquake.

Question 3.
MacMahonLine
Answer:
(i) The conflict between India and China is connected to two issues :

Border issue
The status of Tibet

(ii) The border dispute between India and China is related to Aksai Chin area and the Macmahon line.
(iii) China claims that the area South of Aksai Chin and Macmahon line (Arunachal Pradesh) is Chinese Territory.
(iv) China is not willing to agree that Macmahon Line is an international border.
(v) India made several efforts to sort this border dispute by means of dialogue, but it has not met with much success.
(vi) In 1962, China attacked India.

Question 4.
Indo-Afghanistan Relation
Answer:
(i) There is a lot of political instability in Afghanistan.

(ii) The dominance of the terrorist organisation, Taliban is responsible for this.

(iii) India has extended help to bring peace, security and stability, curb violence and establish a democratic government.

(iv) Similarly, India is also helping Afghanistan in re-establishing communication facilities that have got destroyed due to war, build roads, cooperate in the fields of science and technology, build schools, health facilities and irrigation projects.

5. Answer the following questions in brief:

Question 1.
Explain the background of cooperative relations between India and America.
Answer:
India and America are two powerful nations who have democratic political systems. Right from the beginning, America was India’s most important trading partner.

(i) Many Indians have gone to America for education or work. Due to the presence of these non-resident Indians, the cultural, social and economic relations between America and India have been increasing.

(ii) After the end of Cold War, the military relations between India and America have increased on a large scale.

(iii) After India accepted free market economy, the speed of economic progress slowly increased. It resulted in further strengthening the trade relations between India and America.

(iv) When India conducted nuclear tests in 1998, there was tension in the relations between the two countries.

(v) Many rounds of discussions took place after that to improve the relations. Through these rounds of talks, America came to believe that India will use its nuclear weapons with responsibility. This completely transformed the Indo-American relations.

(vi) The Civil Nuclear Agreement signed in 2005 that was approved by the U.S. Congress in 2008 was an important landmark in Indo-American relations.

Question 2.
Give some information about the efforts taken by India for the establishment of democracy in neighbouring countries with examples.
Answer:
India upheld sovereignty and democratic values in Nepal, Afghanistan, Sri Lanka and Pakistan.

(i) Nepal was a Hindu Kingdom which adopted democratic government in 2006. India backed up its decision of adopting popular rule by helping Nepal with constitutional guidelines.

(ii) India also extended friendly relations to Ashraf Ghani through SAARC and other international platforms after the collapse of Taliban there.

(iii) India sent a peacekeeping force to Sri Lanka to bring about a cease fire between Sinhalese and Tamils and enforce peace and democracy.

(iv) The regime of Pervez Musharraf had adopted aggressive foreign policies and oppressed public opinion in Pakistan which led to violence and chaos. India had also criticised military rule in Pakistan.

Question 3.
What is the role of South Asian Association for Regional Cooperation?
Answer:
SAARC is a regional organisation established by the South Asian countries in 1985.

(i) The aim of this organisation is to encourage economic co-operation between South Asian nations and thereby achieve the development of the South Asian region.

(ii) The organisation of SAARC is a platform for South Asian countries to come together and discuss common questions and interests.

(iii) Some common interests of South Asian countries are alleviation of poverty, development of agriculture and technological revolution.

(iv) Some treaties were signed on the SAARC platform to make it easy for South Asian countries to trade with each other.

(v) As a part of this, it has been decided that South Asia should be made into a free trade area. For an equitable development of South Asian nations, a treaty about forming South Asian Free Trade Area was signed and South Asian University was established.

6. Give your own opinion about :

Question 1.
What remedies would you suggest in order to reduce the tensions in the relations between India and Pakistan?
Answer:
(i) To reduce tensions in the relations between India and Pakistan, I suggest that

(ii) Cultural exchange between the two nations must be enhanced by allowing and inviting artists of these nations to perform in each other’s countries. Films and sports would also work wonders to improve relations between them.

(iii) Since India is the largest democracy in the world, it should assist and uphold establishment of democratic systems in Pakistan, where government is either overpowered or practically governed by military.

(iv) Pakistan should be convinced through world opinion at international summits, conferences and organisations such as the UN.

(v) India should ascertain that the media does not mislead people in both the nations, by telecasting exaggerated content which might negatively influence the relations.

(vi) India should appeal the Pakistani Government for preservation of Hindu and Sikh pilgrimage places and cultural sites. India must also promote tourism in both the countries, so that civilians across borders have cordial relations.

(vii) Incidents of cross-border terrorism, violence and infiltration must be taken to international forums. Government of India should frame suitable policies for retrieval of Indian subjects, Prisoners of War and fishermen held in captivity of Pakistani Government.

Question 2.
Do you agree with the statement: ‘Hurdles are created in the way of internal development due to tense relations of India with neighbouring countries/ Explain with reasons.
Answer:
(i) Yes, I agree that hurdles are created in the way of internal development due to tense relations of India with neighbouring countries. Due to tense relations with the neighbours, India is unable to focus on the internal law and order situation and social welfare.

(ii) Non-cordial relations with the neighbours will force India to embark on military and nuclear mobilisation to retaliate the pressure of foreign aggression. This might culminate into a war.

(iii) War with neighbours will disturb the peace in the Indian subcontinent and compel India to divert its resources and efforts towards defence and manufacturing weapons.

(iv) Money and resources which should be invested for constructive purposes like health, education and infrastructure would be diverted towards destruction.

(v) A war with her neighbours can put tremendous pressure on India’s economy giving rise to problems like scarcity and inflation.

Question 3.
Do you think that strong healthy relations of India with America are conducive to India’s economic development?
Answer:
(i) Yes, I believe that strong healthy relations of India with America are conducive to India’s economic development.

(ii) America is an undisputable super power of the world and India is on its way to become a super power. Co-operation between these two countries can thwart terrorism in the world, which is a big challenge in the path of economic development.

(iii) Both countries follow democratic systems of government and free market economies. Co-operation from America is a must to enhance India’s economic prospects.

(iv) The non-resident Indians in America have helped to improve on Indian Foreign Reserves. With Indians taking up more employment in America, India can boost its foreign exchange.

(v) Call centres of American MNCs and other ways of outsourcing have provided employment opportunities to the Indian youth. American companies such as Coca Cola, Colgate and MacDonald established their franchises in India, whereas Indian IITians have dominated the Silicon Valley. An Indian – Sundar Pichai, today is the CEO of ‘Google’ – an American company.

(vi) As American companies find Indian economy very promising, FDI has also increased. The ’Make in India’ drive is now seeking empowerment of Indian economy and employment of Indians through such FDI’s.

Class 9 Political Science Chapter 5 India and Other Countries Additional Important Questions and Answers

Choose the correct option from the given options and rewrite the statements:

Question 1.
in l972,______ provided a new format to the interaction between India and Pakistan.
(a) Tashkent Agreement
(b) Shimla Agreement
(c) Civil Nuclear Agreement
(d) Nuclear Non-Proliferation Treaty
Answer:
(b) Shimla Agreement

Question 2.
Aung Sang Su Kyi is credited with having established democracy in ______.
(a) Bhutan
(b) Myanmar
(c) Cambodia
(d) the Maldives
Answer:
(b) Myanmar

Question 3.
The border dispute between India and China is related to ______ area and the Macmahon line.
(a) Sir Creek Area
(b) Siachien
(c) Aksai Chin
(d) Baltistan
Answer:
(c) Aksai Chin

Question 4.
Due to a series of tripartite talks between ____, China and India, Sino-Indian relations are improving slowly.
(a) Pakistan
(b) America
(c) Russia
(d) Japan
Answer:
(b) America

Question 5.
When China was stepping up its military control in Tibet,_____ took asylum in India.
(a) Dalai Lama
(b) Avalokiteshwara
(c) Bodhisattvas
(d) Vajarayana followers
Answer:
(a) Dalai Lama

Question 6.
The ______ signed in 2005 that was approved by the U.S. Congress in 2008 was an important landmark in Indo-American relations.
(a) Comprehensive Test Ban Treaty
(b) Nuclear Non-Proliferation Treaty
(c) Civil Nuclear Agreement
(d) Treaty of Friendship and Co-operation
Answer:
(c) Civil Nuclear Agreement

Question 7.
The policy of increasing trade with South-East Asian nations is known as the ______ policy.
(a) Act East
(b) Pact East
(c) Look East
(d) Seek East
Answer:
(c) Look East

Question 8.
The Summit Conference of India and Africa was held in 2015 with representatives from all _____ African nations attending it.
(a) 54
(b) 56
(c) 58
(d) 53
Answer:
(a) 54

Question 9.
India has co-operated in the project to produced ______ on a large scale, using the water source in Bhutan.
(a) Irrigation facilities
(b) Hydel power
(c) Tube wells
(d) Water reservoirs
Answer:
(b) Hydel power

Question 10.
_______has agreed to extend cooperation and technological help to India in the development and security of coastal areas.
(a) Malaysia
(b) Indonesia
(c) Brunei
(d) Japan
Answer:
(d) Japan

Question 11.
European Nations, especially ______ invest on a large scale in the field of technology in India.
(a) Greece and Italy
(b) Germany and France
(c) Belgium and Netherlands
(d) Spain and Portugal
Answer:
(b) Germany and France

Question 12.
The dominance of the terrorist organisation _______ is responsible for political instability in Afghanistan.
(a) Lashkar-e-Taiba
(b) Jamaat-ud-Dawa
(c) Taliban
(d) Al-Qaeda
Answer:
(c) Taliban

State whether the following statements are true or false with reasons:

Question 1.
Write about India’s relation with Maldives.
Answer:
(i) India’s relations with Maldives have been friendly right from the beginning. This small nation to the South of India is dependent on India for a lot of reasons.

(ii) Trade relations have been established between the two since 1981.

(iii) India has helped Maldives in the development of basic facilities or infrastructure, health and communication.

(iv) From 2006 onwards, co-operation in the military field also started between the two countries.

(v) They have entered into treaties with respect to cooperation in space research, conservation of historical objects and tourism. Similarly, they have decided to co-operate in the area of fighting against terrorism.

Write short notes:

Question 1.
SAARC
Answer:
(i) SAARC is a regional organisation established by the South Asian countries in 1985.

(ii) The aim of this organisation is to encourage economic co-operation between South Asian nations and there by, achieve the development of the South Asian region.

(iii) The organisation of SAARC is a platform for South Asian countries to come together and discuss common questions and interests.

(iv) Some common interests of South Asian countries are alleviation of poverty, development of agriculture and technological revolution.

(v) Some treaties were signed on the SAARC platform to make it easy for South Asian countries to trade with each other.

Do as directed:

Complete the following charts:

Question 1.

S.No	Column ’A’	Column ’B’
(1)	Joint Naval Exercises	……….
(2)	Oil imports	………..
(3)	Security in the Indian Ocean	……….
(4)	Hydel Power Plant	………….

Answer:

S.No	Column ’A’	Column ’B’
(1)	Joint Naval Exercises	Japan
(2)	Oil imports	United Arab Emirates
(3)	Security in the Indian Ocean	Sri Lanka
(4)	Hydel Power Plant	Bhutan

Question 2.

S.No.	Help extended by India	Beneficiary Nations
(1)	Sent Peacekeeping Force to restore peace	Sri Lanka
(2)	Supplies during 2015 earthquake	Nepal
(3)	Communication facilities, roads, science and technology, schools, health and irrigation projects	Afghanistan
(4)	Helped in the freedom struggle	Bangladesh
(5)	Cooperation in the project to produce Hydel Power	Bhutan

Answer:

S.No.	Help extended by India	Beneficiary Nations
(1)	Sent Peacekeeping Force to restore peace	Sri Lanka
(2)	Supplies during 2015 earthquake	Nepal
(3)	Communication facilities, roads, science and technology, schools, health and irrigation projects	Afghanistan
(4)	Helped in the freedom struggle	Bangladesh
(5)	Cooperation in the project to produce Hydel Power	Bhutan

Prepare the following flow charts:

Question 1.
Maharashtra Board Class 9 Political Science Solutions Chapter 5 India and Other Countries 4
Answer:

Question 2.
Maharashtra Board Class 9 Political Science Solutions Chapter 5 India and Other Countries 5
Answer:

Question 3.
Maharashtra Board Class 9 Political Science Solutions Chapter 5 India and Other Countries 6
Answer:

Answer the following questions in brief.

Question 1.
Illustrate the differences in world views between India and Pakistan.
Answer:
(i) India and Pakistan have totally different world views.

(ii) In line with the Indian world view, India opposed the military alliances of the Cold War and tried to solve the Indo-Pak conflict bilaterally.

(iii) On the other hand, Pakistan tried to maintain relations with the Islamic world and China entered into a military agreement with America.

Question 2.
Describe the Kashmir issue between India and Pakistan.
Answer:
(i) Pakistan considers Kashmir as the biggest hurdle in establishing relations with India.

(ii) The first Indo-Pak war took place in 1965 over Kashmir.

(iii) The Tashkent Agreement was signed in 1966, but nothing much came out of it.

(iv) Even though the 1971 war was about the creation of Bangladesh, it also had the dimension of the Kashmir problem to it.

(v) The Shimla Agreement in 1972 provided a new format to the interaction between India and Pakistan.

(vi) In 1999, there was a conflict between India and Pakistan over the infiltration by Pakistan in the Kargil region.

(vii) Even today, the Kashmir question is the basic cause of conflict between the two countries.

Question 3.
Sino-Pak Friendship has affected the security of India. Explain.
Answer:
(i) The threat to India’s security is increasing due to friendship between China and Pakistan, supply of weapons from China to Pakistan and also transfer of missile and nuclear weapons technology.

(ii) The increasing friendship between China and Pakistan and increasing influence of China over India’s other neighbours is a cause of concern for India.

Question 4.
How is India trying to resolve its differences with China?
Answer:
(i) India has always made efforts to improve relations with China.

(ii) A joint executive group has been set up to solve the Sino-Indian border dispute.

(iii) When the rate of economic growth of both India and China increased, the trade relations between both the countries began to get strengthened.

(iv) Because of improvement in economic and trade relations between India and China and a series of tripartite talks between Russia, China and India, Sino-Indian relations are improving slowly.

(v) Even though the border dispute is not completely resolved, it has taken a back seat and relations in other areas have become more important.

Question 5.
How was Bangladesh created?
Answer:
(i) Today’s Bangladesh is the erstwhile East Pakistan. When Pakistan was created, it had two parts: West and East Pakistan.

(ii) There was linguistic difference between them. There were other political conflicts as well.

(iii) The movement to liberate Bangladesh emerged out of this.

(iv) This movement tried to free Bangladesh from the dominance of West Pakistan.

(v) India helped Bangladesh in its freedom struggle. Bangladesh was formed in 1971.

Question 6.
How could India and African nations benefits from each other?
Answer:
(i) Having close relations with Africa will be beneficial from the point of view of both.

(ii) Many countries in Africa are developing rapidly.

(iii) MaharashtraBoardSoLutions.com diness to help in the all-round development of African youth. India has also agreed to give economic help and even grants to bring about development in different areas like technology, agriculture, tourism, etc.

(iv) India’s energy needs can be fulfilled by the energy rich countries like Egypt, Nigeria, Angola and Sudan.

(v) India is making efforts to increase trade with Africa.

9th Std Political Science Questions And Answers:

Std 9 History Chapter 9 Question Answer Maharashtra Board

February 6, 2025January 4, 2025 by Bhagya

Balbharti Maharashtra State Board Class 9 History Solutions Chapter 9 Changing Life 1 Notes, Textbook Exercise Important Questions and Answers.

Changing Life 1 Class 9 Questions And Answers Maharashtra Board

Class 9 History Chapter 9 Changing Life 1 Questions And Answers Maharashtra Board

History Class 9 Chapter 9 Question Answer Maharashtra Board

1. Choose the proper option and rewrite the completed statements:

Question 1.
The first open-heart surgery under the leadership of Dr N Gopinath was successfully performed in the city of _____.
(a) Chennai
(b) Vellore
(c) Hyderabad
(d) Mumbai
Answer:
(b) Vellore

Question 2.
_______is known as the father of the ‘Jaipur foot’.
(a) Dr N. Gopinathan
(b) Dr Pramod Sethi
(c) Dr Mohan Rao
(d) None of the above
Answer:
(b) Dr Pramod Sethi

2. Identify the wrong pair and rewrite the corrected one:

Question 1.
(1) Dr N. Gopinath – open heart surgery
(2) Ramchandra Sharma – a skilled craftsman
(3) Dr Subhash Mukhopadhyaya – test tube baby
(4) Dr Mohan Rao – polio
Answer:
Wrong Pair: Dr Mohan Rao – polio

3. Explain the concepts:

Question 1.
The institution of the family.
Answer:
The institution of family can be explained as follows:

During the pre-independence period, the institution of the family was an important identifying feature of Indian society.
India was known all over the world as the country of joint families.
However, the wave of globalisation has now given an impetus to the system of nuclear families.

Question 2.
Jaipur Foot technology
Answer:
(i) Dr Pramod Sethi designed and manufactured artificial limbs, noses and ears, with the help of the skilled craftsman Ramchandra Sharma.

(ii) These artificial body parts manufactured with the help of the Jaipur foot technology have made it easy for the differently-abled to walk bare-foot on rough surfaces, run, go cycling, work in the fields, climb trees, and even climb mountains, etc.

Question 3.
Urbanization
Answer:
(i) Urbanisation is a process of the concentration of population in a city or urban area. Increasing population is one major reason for urbanisation.

(ii) A few other factors affecting urbanisation are air, water and the economic as well as social organisations necessary for community life.

(iii) In the context of post-independence India, the increase in urban population is also due to factors such as a reduction in the mortality rate, industrialisation, unavailability of means of livelihood in rural areas, job opportunities in cities and the resulting migration.

Question 4.
Changing economic life
Answer:
(i) Earlier, every village was economically self-sufficient. A majority of the villagers were dependent on farming. Farm produce used to be distributed among artisans as payment for their work.
(ii) Now this situation has changed. Rural areas are engaged in agriculture and occupations ancillary to farming while urban society is engaged in non-agricultural production and the service sector.

4. Give reasons:

Question 1.
The campaign for pulse polio immunisation was initiated.
Answer:
The campaign of pulse polio immunisation was initiated because:

Before 1978, every year, six Indian infants out of every 10 who were bom faced fatal health problems in the very first year of their birth.
The immunisation programme was taken up to overcome the threat of polio, measles, tetanus, TB, diphtheria, and whooping cough.
The ‘pulse polio’ immunisation programme started in 1995 has controlled polio.

Question 2.
The Rural Water Supply Scheme was started.
Answer:
The State started the Rural Water Supply Scheme for sinking wells and providing piped water because:

(i) It is necessary to give priority to facilities for public hygiene and health-care. However even today, the rural population faces problems related to year-round provision of drinking water, toilets, open drainage, narrow roads, inadequate electrification and lack of facilities for medical treatment.

(ii) There is a lack of good quality educational facilities from primary level to higher education, centres for recreation and libraries.

(iii) The collective development scheme had an important place in all the first four five year plans of the government of India. The State of Maharashtra registered remarkable achievements under this scheme. By 1971,1677 small dams had also been built.

5. Answer the following questions in 25-30 words:

Question 1.
Which kinds of discrimination does the Constitution prohibit?
Answer:
(i) According to our Constitution, all Indians are equal before the law, and they cannot be discriminated against on the basis of religion, race, caste, gender or place of birth.

(ii) All citizens have the right to freedom of speech and expression, freedom to assemble peacefully without arms, freedom to form associations, freedom to move, to live and settle down anywhere on Indian territory and to practice any occupation.

(iii) Indian citizens living anywhere in India enjoy the right to nurture their language, script and culture.

Question 2.
What is the aim of the social welfare programme?
Answer:
(i) The objective to establish the welfare state has been incorporated in the Constitution itself.

(ii) The social welfare programmes aim to make opportunities for full-time employment, health care, education and development available to all Indian citizens. Indian society is riddled with large scale economic, social, educational and cultural inequalities.

(iii) However, it is necessary that opportunities for development reach women, children, the differently-abled, the scheduled castes and tribes as well as other minorities.

(iv) Hence on 14th June 1964, the government of India constituted the Ministry of Social Welfare. Under this ministry, various programmes are implemented for nutrition and child- development, social security and social protection, women’s welfare and development.

Question 3.
What are the challenges facing rural development?
Answer:

(i) In 1961, 82% of India’s population resided in villages. In 1971, the percentage was 80.01%.
(ii) So far, the rural area has shouldered the responsibility of providing food grains and other raw material to urban areas, supplying labour to the industrial sectors of the city, and taking care of natural resources.
(iii) However there are three major challenges with reference to rural development, namely, bringing about economic development, developing facilities to meet social needs and finally bringing about a change of attitude in matters concerning society, culture and ways of thinking.
(iv) It is also equally necessary to expedite irrigation projects and bring about land reforms.

6. Write a brief review of the significant events in the field of public health in India.
Answer:
The Constitution of India states that the primary duty of the government is to raise the people’s standard of living, to ensure proper nutrition and to improve public health. The Ministry of Health and Social Welfare at the Centre helps the state governments in this regard.

(i) To make primary health services as well as medical care available to rural people, the tribals and the poor was an objective of the Sixth Five Year plan. Efforts were made towards helping people maintain good health by giving recognition to Unani, Homoeopathy, Ayurveda and Naturopathy in addition to Allopathy.

(ii) In 1962, the first successful open heart surgery was performed under the leadership of Dr N. Gopinath at Christian Medical College Hospital at Vellore in Tamil Nadu. Hence, it is no longer necessary to go abroad for such treatment.

(iii) The invention of the ‘Jaipur Foot’ has transformed the lives of the disabled in India.

(iv) Before 1968, if a person lost a leg in an accident, he had to suffer for the rest of his life. In order to remedy the situation, Dr Pramod Sethi designed and manufactured artificial limbs, noses and ears, with the help of the skilled craftsman Ramchandra Sharma.

(v) These artificial body parts manufactured with the help of the Jaipur foot technology have made it easy for the differently-abled to walk bare-foot on rough surfaces, run, go cycling, work in the fields, climb trees, and even climb mountains, etc.

(vi) They save on the expenses on shoes as there is no need for footwear on these artificial limbs. They can bend their legs at the knees and sit cross-legged.

(vii) These feet are also convenient while working in water or under wet conditions.

Class 9 History Chapter 9 Changing Life 1 Additional Important Questions and Answers

Choose the correct option from the given options and rewrite the statements:

Question 1.
On 14th June 1964, the government of India constituted the Ministry of ________.
(a) Home Affairs
(b) Social Justice
(c) Social Welfare
(d) Education
Answer:
(c) Social Welfare

Question 2.
According to the 1971 Census, _____ people in India belonged to scheduled castes.
(a) 21%
(b) 22%
(c) 23%
(d) 24%
Answer:
(b) 22%

Question 3.
UNESCO awarded its international prize for the spread of literacy to ______ in 1972 and tribes.
(a) Gujarat
(b) Bengal
(c) Maharashtra
(d) Karnataka
Answer:
(c) Maharashtra

Question 4.
‘Sinhagad Express’ on the Pune-Mumbai route and ‘ _____’ on the Mumbai-Kolkata route were started which had no class divisions.
(a) Karmabhoomi Express
(b) Howrah Mail
(c) Shalimar Express
(d) Gitanjali Express
Answer:
(d) Gitanjali Express

Question 5.
A _______ was formed to help the development of industry in rural areas.
(a) Village Industries Planning Committee
(b) Co-operative Credit Societies
(c) Industrial Development Association
(d) Rural Industrial Council
Answer:
(a) Village Industries Planning Committee

Question 6.
A settlement smaller than a village is called a _____.
(a) Town
(b) Taluka
(c) Hamlet
(d) Locality
Answer:
(c) Hamlet

Identify and write the wrong pair in the following sets:

Question 1.
(1) Kothari Commission – Agricultural Universities
(2) Durga – Open heart surgery
(3) Pramod Sethi – Artifical limbs
(4) Dr Mohan Rao – Kidney transplant
Answer:
Wrong Pair : Durga – Open heart surgery

Question 2.
(1) Christian Medical College – Tamil Nadu
(2) Vidya Niketan – Chikhaldara
(3) Gitanjali Express – Pune-Mumbai route
(4) First successful artificial conception – Kolkata
Answer:
Wrong Pair : Gitanjali Express – Pune-Mumbai route

Question 3.
(1) Ministry of Social welfare established in -1962
(2) Pulse polio immunisation started – 1995
(3) Maharashtra received a prize for literacy in Maharashtra -1972
(4) Rural Electrification Corporation – 1969
Answer:
Wrong Pair : Ministry of Social welfare established in -1962

Question 4.
(1) Village Industries Planning Committee – Development of Industries in rural areas
(2) Vidya Niketan – residential highschools
(3) Rural water supply schemes – sinking wells
(4) Objective of Sixth Five Year Plan – Advanced technology
Answer:
Wrong Pair : Objective of Sixth five year plan – Advanced technology

Do as Directed:

Complete the table

Question 1.
Answer:

Institutes	Location
Rural Electrification Co-operative societies	Andhra Pradesh, Gujarat, Karnataka, Maharashtra and Uttar Pradesh
Agriculture Universities	Rahuri, Akola, Parbhani and Dapoli in Maharashtra
Vidya Niketan	Satara, Aurangabad, Nasik and Chikhaldara

Question 2.
Answer:

Rural Community in Pre-Globalisation period	Urban Community in Pre-Globalisation period
Priority to farming and ancillary vocations (jobs)	Priority to nonagricultural production and services.
Small in size, homogeneous in terms of language, culture and tradition.	Large in size, heterogeneous in terms of languages, cultures and traditions.
Elementary kinds of vocations; trend of sending villagers to cities rather than accommodating outsiders in rural occupations.	Large industries, production for global consumption, accommodating outsiders.
Large proportion of vocations running in the family	Less proportion of vocations running in the family.
Priority to the head of the family and to the family as an institution. Joint family system.	Family secondary, priority to individuals. Trend towards – nuclear families.

Complete the web diagram

Question 1.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 9 Changing Life 1 2

Question 2.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 9 Changing Life 1 3

Write short notes:

Question 1.
Rural Development
Answer:
(i) In 1961, 82% of India’s population resided in villages.

(ii) In 1971, the percentage was 80.01%. So far, the rural area has shouldered the responsibility of providing food grains and other raw material to urban areas, supplying labour to the industrial sectors of the city, and taking care of natural resources.

(iii) There are three major challenges with reference to rural development, namely, bringing about economic development, developing facilities to meet social needs and finally bringing about a change of attitude in matters concerning society, culture and ways of thinking.

(iv) It is equally necessary to expedite irrigation projects and bring about land reforms.

Question 2.
Social Welfare
Answer:
(i) The objective to establish the welfare state has been incorporated in the Constitution itself. India is the first country in the world to do so.

(ii) The social welfare programmes aim to make opportunities for full-time employment, health care, education and development available to all Indian citizens.

(iii) It is necessary that opportunities for development reach women, children, the differently abled, the scheduled castes and tribes as well as other minorities.

(iv) This was the greatest challenge facing the government during the post-independence period

(v) That is why, on 14 June 1964, the government of India constituted the Ministry of Social Welfare.

(vi) Under this ministry various programmes are implemented for nutrition and child development, social security and social protection, women’s welfare and development.

Question 3.
Kidney transplant
Answer:
(i) Since kidney transplant can now be performed in India, doctors have succeeded in saving patients’ lives. Such surgeries were not so common in India till 1971.

(ii) But in 1971, it was successfully performed at the Christian Medical College hospital at Vellore in Tamil Nadu.

(iii) Dr. Johny and Dr. Mohan Rao transplanted the kidney donated by a living person into the patient’s body.

(iv) Nowadays, patients from underdeveloped countries come to India for such surgeries.

Question 4.
Test-tube baby
Answer:
(i) Right from early times, having a baby is considered to be of utmost importance in the Indian family system.
(ii) The support of ‘test-tube baby’ technology has been available to childless couples since 1978.
(iii) It was under the supervision of Dr. Subhash Mukhopadhyaya that the experiment of the test-tube baby was successfully carried out at Kolkata.
(iv) This was the first successful attempt at artificial conception. A girl named Durga was born.
(v) This technology helped to solve the problem of childless couples who wanted children.

Explain statements with reason:

Question 1.
Rural electrification is essential.
Answer:
Rural electrification is essential because:

(i) Electricity is essential for the development of rural areas. Automatic pumps are needed to water the fields.

(ii) Electricity is also needed to preserve perishable food items such as milk, eggs, fruits, and vegetables, to run fertilizer projects and for fans and TV, and for lights for studying at night.

(iii) During the first five-year plan, electricity had been provided to three thousand villages in India. By 1973, it reached 1,38,646 villages. From 1966, it was decided to supply more electricity for pumps and borewells.

(iv) ‘Rural Electrification Corporation’ was established in 1969.

Question 2.
It is necessary to give special attention to rural areas.
Answer:
(i) It is necessary to give priority to facilities for public hygiene and health-care.

(ii) Even today the rural population faces problems related to year round provision of drinking water, toilets, open drainage, narrow roads, inadequate electrification and lack of facilities for medical treatment.

(iii) There is a lack of good quality educational facilities from primary level to higher education, centres for recreation and libraries. That is why it is necessary to give special attention to rural areas.

Question 3.
The Rural Local Self-Government was empowered with more authority.
Answer:
(i) Keeping in mind the goal of rural development, the collective development scheme has come into force.

(ii) Plans were made for improving farming technology, increasing the area of irrigated land, extending educational facilities, enacting the laws for land reform under this scheme.

(iii) It also aimed at increasing agricultural produce as well as expanding the means of transport, health services and scope for education in the rural areas.

(iv) The government undertook this work through the medium of Gram Panchayats which included the people from all castes and tribes.

(v) To achieve this, the Gram Panchayats, Panchayat Samitis and Zilla Parishads were empowered with more authority.

Read the following passage and answer the questions below:

Public health: The Constitution of India states that ‘ the primary duty of the government is to raise the people’s standard of living, to ensure proper nutrition i and to improve public health. The ministry of health i and Social Welfare at the Centre helps the state governments in this regard. To make primary health services as well as medical care available to rural people, the tribals and the poor was an objective of the Sixth Five Year plan. Efforts were made towards helping people maintain good health by giving recognition to Unani, Homoeopathy, Ayurveda and Naturopathy in addition to Allopathy, Progress in the medical field has made the life of Indians relatively free of health concerns. In 1962, the first successful open heart surgery was performed under the leadership of Dr. N. Gopinath at Christian Medical College Hospital at Vellore in Tamil Nadu, Hence, it is no longer necessary to go abroad for such treatment.

Question 1.
Prepare a concept map of recognised health care and medical treatments in India.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 9 Changing Life 1 1

Question 2.
Who was Dr. N. Gopinath? State his achievement?
Answer:
Dr. N. Gopinath was an experienced surgeon at Christian Medical College Hospital of Vellore under whose leadership an Open Heart Surgery was performed for the first time in India in 1962.

Question 3.
Which disease among children do you consider most threatful? Why?
Answer:
I would consider Polio as one of the most threatening diseases. It renders an infant physically impaired or incapacitated for life. Without physical development, it is not possible for a child to lead a normal life. He/she becomes dependent on others for daily chores and cannot become self-reliant.

Answer the following questions in detail:

Question 1.
State the achievements of Christian Medical College Hospital.
Answer:
(i) In 1962, the first successful open heart surgery was performed under the leadership of Dr. N. Gopinath at Christian Medical College Hospital at Vellore in Tamil Nadu.

(ii) In 1971, kidney transplant was successfully performed at the Christian Medical College hospital at Vellore in Tamil Nadu. Dr Johny and Dr Mohan Rao transplanted the kidney donated by a living person into the patient’s body.

Question 2.
Explain, how ‘Jaipur Foot’ has transformed the lives of the disabled in India?
Answer:
(i) The invention of the ‘Jaipur Foot’ has transformed the lives of the disabled in India. Before 1968, if a person lost a leg in an accident, he had to suffer for the rest of his life.

(ii) In order to remedy the situation, Dr Pramod Sethi designed and manufactured artificial limbs, noses and ears, with the help of the skilled craftsman Ramchandra Sharma.

(iii) These artificial body parts manufactured with the help of the Jaipur foot technology have made it easy for the differently abled to walk bare-foot on rough surfaces, rim, go cycling, work in the fields, climb trees, and even climb mountains, etc.

(iv) They save on the expenses on shoes as there is no need for footwear on these artificial limbs. They can bend their legs at the knees and sit cross-legged.

(v) These feet are also convenient while working in water or wet conditions.

Question 3.
What measures were taken up to uplift the scheduled castes and tribes?
Answer:
(i) According to the 1971 Census, 22% people in India belonged to scheduled castes and tribes.

(ii) Laws were enacted so that they got educational scholarships, and representation in Parliament as well State Assemblies. Some seats are also reserved for them in Government services.

Question 4.
Give an account of the collective development scheme that was featured in the first four Five Year Plans.
Answer:
(i) The collective development scheme had an important place in all the first four five year plans of the government of India.

(ii) The State of Maharashtra registered remarkable achievements under this scheme. For example, ‘Zilla Parishads’ was established in 1962.

(iii) Nutritious diet scheme was launched in 1970-1971. The State also started the ‘Rural Water Supply Scheme’ for sinking wells and providing piped water.

(iv) By 1971,1677 small dams had also been built.

9th Std History Questions And Answers:

Std 9 History Chapter 4 Question Answer Maharashtra Board

February 6, 2025January 4, 2025 by Bhagya

Balbharti Maharashtra State Board Class 9 History Solutions Chapter 4 Economic Development Notes, Textbook Exercise Important Questions and Answers.

Economic Development Class 9 Questions And Answers Maharashtra Board

Class 9 History Chapter 4 Economic Development Questions And Answers Maharashtra Board

History Class 9 Chapter 4 Question Answer Maharashtra Board

1.A Choose right option and write the sentence:

Question 1.
On 19th July, 1969 ______ major banks in India were nationalised.
(a) 12
(b) 14
(c) 16
(d) 18
Answer:
(b) 14

Question 2.
_________ declared a 20 – point programme.
(a) Pandit Nehru
(b) Lai Bahadur Shastri
(c) Indira Gandhi
(d) P.V. Narasimha Rao
Answer:
(c) Indira Gandhi

B. Identify the wrong pair and rewrite the corrected one:

Question 1.
(1) Kawasajee Dawar – Iron and Steel factory.
(2) Dr. Datta Samant – Leadership of mill workers.
(3) N. M. Lokhande – Holiday for mill workers.
(4) Narayan Surve – Depiction of lives of workers through poems.
Answer:
Wrong Pair : Kawasajee Dawar – Iron and Steel factory.

2.A Complete the following chart:

Question 1.

Five Year Plan	Duration	Purposes
First	……………….	Agriculture and Social Development
Second	1956 -1961	Industrialisation
Third	………………	Elimination of inequality, increasing opportunities for employment, increase in national income.
………….	1969 -1974	Scientific research, health and family planning.
Fifth	…………….	…………………..

Answer:

Five Year Plan	Duration	Purposes
First	1951 -1956	Agriculture and Social Development
Second	1956 -1961	Industrialisation
Third	1961 -1966	Elimination of inequality, increasing opportunities for employment, increase in national income.
Fourth	1969 -1974	Scientific research, health and family planning.
Fifth	1974 -1979	Making India economically self-sufficient by alleviating poverty.

B. Explain the concepts:

Question 1.
Mixed Economy
Answer:
(i) In a capitalist system, the means of production are privately owned. In a socialist economy, the means of production are owned by the State in the name of the society.
(ii) A mixed economy works both in the private and the public sector.
(iii) In order to achieve economic development of modem India, we gave preference to a model of ‘mixed economy’. We can see three parts in this kind of economy:

(a) Public Sector: The industries in this sector are completely under the control and management of the government. For example, production of defence equipment.

(b) Private Sector: The industries in this sector are owned by private industrialists. Of course, the government supervises and controls them too. For example, consumer goods.

(c) Joint Sector: In this sector, some industries are owned by private industrialists, while some are run under government management.

(iv) For mixed economy to run smoothly there is a need for co-ordination between the private sector and the public sector.
(v) This system aims at maximisation of production and popular participation on a large scale.
(vi) An effort has been made in this system to bring together the good aspects of both the capitalist and the socialist systems.
(vii) A mixed economy cannot ignore the profit motive, entrepreneurship, discipline, time bound planning, etc.

Question 2.
20 – Point Programme
Answer:
20-Point Programme: Prime Minister Indira Gandhi announced the 20-Point Programme on 1st July 1975 and resolved to make efforts towards rapidly becoming a developed nation. The main provisions of the 20 point programme are as follows:

Land ceiling for cities and agricultural land, equal division of wealth, minimum wages for workers, increase in water conservation schemes.
Workers’ participation in industry, national training scheme, freeing bonded labour.
Prevention of tax evasion, economic crimes and smuggling.
Regulation of prices of basic necessities, improvement in the public distribution system.
Improvement in the textile industry by developing handloom sector, waiving loans for weaker sections, housing, communication facilities, making educational equipment available to schools.

3.A Give reasons.

Question 1.
Independent India opted for mixed economy.
Answer:
(i) Our Prime Minister Pandit Nehru adopted the middle path rather than taking recourse to any extremes. Some countries had adopted Capitalism, while some had adopted Socialism. Each type of economy had its own advantages.

(ii) In a capitalist system, the means of production are privately owned. In a socialist economy, the means of production are owned by the State in the name of the society.

(iii) A mixed economy works both in the private and the public sector. In order to achieve economic development of modem India, we gave preference to a model of ‘mixed economy’.

Question 2.
Banks were nationalised in 1969.
Answer:
(i) During the tenure of Prime Ministers Pandit Nehru and Lai Bahadur Shastri, banking was a monopoly of the private sector.

(ii) These banks represented different industrial groups. The Directors of these banks were working towards developing industrial sector and increasing its profit.

(iii) In order to stop this, the government nationalised the ‘Imperial Bank’ in 1955 and it got converted into State Bank of India.

(iv) This Bank opened several branches all over the country in a short while and played a major role in development.

(v) India had adopted a mixed economy after independence. Nationalisation of banks was essential to cover the deficits if they occurred while implementing different schemes.

(vi) Also the profit of these banks would come into the government treasury once they were nationalised. Prime Minister Indira Gandhi nationalised 14 banks on 19th July, 1969.

Question 3.
Mill workers went on strike.
Answer:
(i) During the Diwali of 1981, the workers expected to get a bonus of 20%.
(ii) The Rashtriya Mill Mazdoor Sangh, which was negotiating with the employers, agreed upon 8 to 17% bonus without taking the workers into confidence.
(iii) The cut in the bonus proved to be the cause of unrest. Some workers went to Dr. Datta Samant.
(iv) They asked him to accept their leadership. Workers of 65 mills came together and Dr. Datta Samant led the strike. On 18th January 1982, two and a half lakh workers went on strike.

B. Answer the following questions in 25 to 30 words.

Question 1.
Which programmes were started in the Eighth Five Year Plan?
Answer:
Following programmes were started in the Eighth Five Year Plan:

Pradhanmantri Rozgar Yojana
Mahila Samriddhi Yojana
Rashtriya Samajik, Arthik Sahayya Yojana
Midday Meal Scheme
Indira Mahila Scheme
Ganga Kalyan Scheme

Question 2.
Which heavy industries were set up under the Second Five Year Plan?
OR
Which projects were started in the Second Five Year Plan?
Answer:
(i) Iron and steel industries at Durgapur, Bhilai and Rourkela; Chemical fertilisers plant at Sindri; rail engine factory at Chittaranjan; factory of railway bogies at Perambur, Ship building factory at Vishakhapattanam and other heavy industries were set up in the Public Sector.

(ii) Huge dams like Bhakra-Nangal, Damodar, etc. were built to make water available for agriculture.

Class 9 History Chapter 4 Economic Development Additional Important Questions and Answers

Choose the correct option from the given options and and rewrite the statements:

Question 1.
India removed the import restriction on several commodities for _____ countries.
(a) BRICS
(b) SAARC
(c) Third – World
(d) Middle – East countries
Answer:
SAARC

Question 2.
The efforts of _____ resulted in the weekly Sunday holiday for mill workers from 1st January 1882.
(a) Mahatma Jyotirao Phule
(b) Sane Guruji
(c) Babu Jagjeevan Ram
(d) Narayan Meghaji Lokhande
Answer:
Narayan Meghaji Lokhande

Question 3.
During the tenure of Prime Minister Indira Gandhi ,________ a group in the Congress Party, made a demand for nationalisation of commercial banks.
(a) Congress Socialist Party
(b) Congress Forum for Socialist Action
(c) Congress wing of Socialist Propaganda
(d) Agency of Socialist Congress
Answer:
Congress Forum for Socialist Action

Question 4.
In 1995, India became a member of _______.
(a) World Trade Organisation
(b) International Monetary Fund
(c) GATT
(d) SAARC
Answer:
World Trade Organisation

Question 5.
In order to overcome drought and food shortages ______ undertook the experiment of Green Revolution.
(a) Chaudhary Charansingh
(b) Gulzarilal Nanda
(c) H D Deve Gowda
(d) Lal Bahadur Shastri
Answer:
Lal Bahadur Shastri

Question 6.
During the tenure of Prime Minister P.V. Narasimha Rao, _______ was our Finance Minister.
(a) John Mathai
(b) Yashwant Sinha
(c) P. Chidambaram
(d) Dr. Manmohan Singh
Answer:
Dr. Manmohan Singh

Question 7.
The first textile mill was started in Mumbai on 11th July, 1851 by _______.
(a) Pirojsha Burjorji Godrej
(b) Jamsedji Nusserwanji Tata
(c) Kawasjee Dawar
(d) Azim Hashim Premji
Answer:
Kawasjee Dawar

Question 8.
The new government of Janata Party ended the fifth Five Year plan by March 1978 and started the _______ from April 1978.
(a) Development Plan
(b) Enrolling Plan
(c) Rolling Plan
(d) Redemption Plan
Answer:
Rolling Plan

Question 9.
Manmohan Singh signed the agreement with the WTO and launched the policy of privatisation, ______ and globalisation.
(a) Liberalisation
(b) Industrialisation
(c) Maximisation
(d) Profitisation
Answer:
Liberalisation

Question 10.
On 18th January 1982, two and a half lakh workers went on a strike under the leadership of _________.
(a) Dhondo Keshav Karve
(b) Baba Amte
(c) Dr. Datta Samant
(d) Anna Hazare
Answer:
Dr. Datta Samant

Question 11.
The _______ plan is a reflection of the liberalisation and free market policy adopted in 1991.
(a) Seventh
(b) Eighth
(c) Fifth
(d) Sixth
Answer:
Eighth

Question 12.
The government nationalised the ‘Imperial Bank’ in 1955 and it got converted into ______
(a) Syndicate Bank
(b) Bank of India
(c) State Bank of India
(d) Reserve Bank of India
Answer:
State Bank of India

Question 13.
The term ‘Girangaon’ means ________.
(a) Town of hills
(b) Town of mills
(c) Group of industries
(d) Cottage industries
Answer:
Town of mills

Identify and write the wrong pair in the following sets:

Question 1.
(1) Public Sector – Controlled and managed by government
(2) Mixed Economy – Coordination in private and public sector
(3) 20 point programme – Nationalisation of 13 textile mills.
(4) Increased oil prices – Invasion of Kuwait by Iraq.
Answer:
Wrong Pair : 20 point programme Nationalisation of 13 textile mills.

Question 2.
(1) Congress Forum for Socialist Action – demanded nationalisation of banks.
(2) Rashtriya Mill Mazdoor Sangh – negotiations with the employer.
(3) Imperial Bank – converted in State Bank of India.
(4) National Stock Exchange – existed before WTO.
Answer:
Wrong Pair : National Stock Exchange – existed before WTO.

Question 3.
(1) SAPTA – Preferential trade
(2) SAARC – removal of import restrictions by India
(3) SEBI – established in 1998
(4) WTO – regulate global trade
Answer:
Wrong Pair : SEBI – established in 1998

Question 4.
(1) Barrister A.R.Antulay – leader of Rashtriya Mill Mazdoor Sangh
(2) Dr. Manmohan Singh – Finance Minister
(3) Indira Gandhi – nationalised 14 banks
(4) V.P. Singh – waived off loans of the farmers
Answer:
Wrong Pair : Barrister A.R.Antulay – leader of Rashtriya Mill Mazdoor Sangh

Question 5.
(1) Babasaheb Bhonsale – Chief Minster of Maharashtra
(2) Namdev Dhasal – Marathi Poet
(3) Shahir Sable – Programmes of public education
(4) Chandrashekhar – Associate of Mahatma Jyotirao Phule.
Answer:
Wrong Pair : Chandrashekhar – Associate of Mahatma Jyotirao Phule.

Complete the concept maps and flow charts of various programmes and schemes introduced:

Question 1.
Answer:

SECOND FIVE YEAR PLAN (1956-1961)
Iron and Steel industries	Durgapur, Bhilai, Rourkela
Chemical Fertilisers Plant	Sindri
Rail Engine Factory	Chittaranjan
Factory of railway bogies	Perambur
Ship building Factory	Vishakhapattanam

Question 2.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 1

Question 3.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 2

Question 4.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 3

Question 5.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 4

Question 6.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 5

Question 7.
Answer:
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 6

Give Reasons:

Question 1.
India established the Planning Commission to deal with economic repercussions of British colonialism,
Answer:
(i) The colonial power had exploited India economically. The country faced severe problems like poverty, unemployment, population growth, low standards of living, low productivity of agriculture and industries and backwardness in the fields of knowledge, science and technology. Planning was essential to solve these problems.

(ii) India established the Planning Commission in 1950 with Prime Minister Pandit Jawaharlal Nehru as its Chairman.

(iii) It formulated India’s five year plans that included rural and agricultural development, : balanced industrialisation, provision for a minimum standard of living and economic development consistent with democratic ideals.

(iv) It focused on people’s participation and individual development in the formulation and implementation of the five year plans.

Read the passage and answer the questions.

Passage I

Issues of workers: The first textile mill was started in Mumbai on 11th July, 1851 by Kawasjee Dawar. Eventually, mills started in Dadar, Parel, Bhaykhala, Shivdi, Prabhadevi and Worli. This part came to be i known as Girangaon or ‘town of mills’. In the 1980s, the increasing unrest among workers was due to the economic conditions in other sectors. In some industries the wages of the workers were increasing. They were also getting more amounts as bonus. They were getting more facilities than the textile mill workers.
In the Diwali of 1981, the workers expected to get a bonus of 20%. The Rashtriya Mill Mazdoor Sangh, which was negotiating with the employers, agreed upon 8 to 17% bonus without taking the workers into confidence. The cut in the bonus proved to be the cause of unrest. Some workers went to Dr. Datta Samant. They asked him to accept their leadership, Workers of 65 mills came together and Dr. Datta Samant led the strike. On 18th January 1982, two and a half lakh workers went on strike. Girangaon mills stopped running making it seem like Mumbai’s heart topped throbbing.
The Chief Minister of Maharashtra, Barrister A. R. , Antulay set up a committee to solve this issue. Later Babasaheb Bhosale became the Chief Minister of Maharashtra State. He insisted that as per law, he would talk only with the Rashtriya Mill Mazdoor Sangha. Dr. Datta Samant demanded that the law be revoked. In the beginning, the striking workers received help from their native villages. It was also , not very difficult for them to help each other. They set up departmental committees and distributed food grains, assistance in the form of funds, etc. The left parties had supported the strike. As the strike , dragged on, efforts were made to split the striking workers’ ranks. Even as the strike completed 6 months, the central government completely ignored it. The workers started a Jail Bharo Agitation’. In , September 1982, one and a half lakh workers took a march on the Legislative Assembly of Maharashtra State. It didn’t help at all. The strike completed a year. This was the first strike to have gone on for a year. In this period, about one and a half lakh workers became unemployed. As polyester had come into greater demand than cotton cloth, the sale of mill , cloth had already got affected. The mills moved from Mumbai to Surat in Gujarat. The Central government nationalised 13 textile mills. Appointment of i arbitrators did not help to resolve the issue.

Question 1.
Name the worker’s union mentioned in the passage. Why were the workers agitated with it?
Answer:
(i) The worker’s organisation mentioned in the passage is Rashtriya Mill Mazdoor Sangh.
(ii) The workers were agitated because they received less bonus and facilities in their textile mills, whereas economic conditions in the other sectors of the economy were far better.
(iii) The Rashtriya Mill Mazdoor Sangh that negotiated with the employers, settled at 8 to 17% of bonus as against the expectation of 20% by the workers.
(iv) The cut in the bonus led to the workers agitation.

Question 2.
State the contribution of Dr. Datta Samant in raising workers issues in Maharashtra?
Answer:
(i) Dr. Datta Samant was the leader of two and a half lakh workers who went on strike.
(ii) He demanded that the law be revoked which had empowered Rashtriya Mill Mazdoor Sangh alone to talk on behalf of the workers.

Question 3.
What is mixed economy?
Answer:
(i) In a capitalist system, the means of production are privately owned. In a socialist economy, the means of production are owned by the State in the name of the society.
(ii) A mixed economy works both in the private and the public sector.
(iii) In order to achieve economic development of modem India, we gave preference to a model of ‘mixed economy’.

Passage II

General Agreement on Tariffs and Trade (GATT) existed at the international level before the World Trade Organisation came into being. It regulated commerce. In India there were opposed, extreme views about the World Trade Organisation. Yet India decided to take its membership. The provisions of the World Trade Organisation are regarding grants, import-export, foreign investment, agriculture, technology and services. The sectors of electricity, water transportation, education and health rapidly commercialised since India became a member of the World Trade Organisation. As per the various reports of the World Trade Organisation, India has made a considerable improvement in different areas like reduction in the below poverty line (BPL) population, decline in infant mortality, availability of facilities regarding drinking water and waste water management. India signed the South Asian Preferential Trade Agreement (SAPTA) along the lines of the World Trade Organisation. India removed the import restrictions on several commodities for SAARC countries. India also gave discounts on import duties. India opened up the insurance sector to private and foreign investment.

Question 1.
Study the passage and give the full forms of GATT and SAPTA.
Answer:
GATT – General Agreement on Tariff and Trade.
SAPTA – South Asian Preferential Trade Agreement.

Question 2.
Explain, in your own words, the economic policy of “Make in India”.
Answer:
’Make in India’ is an economic policy as well as an organised progamme of the Government of India to encourage and promote various companies and enterprises to start manufacturing and producing goods in India. The aim of this initiative is to generate employment, enhance the skill of Indians and also to attract foreign investment.

Question 3.
What do the provisions of the WTO cater to?
Answer:
The provisions of the WTO cater to grants, import -export, foreign investment, agriculture, technology and services.

Answer the following questions in detail:

Question 1.
What is the fundamental principle of the Five Year Plans? Which economic goals are set forth by the Government?
Answer:
Fundamental principle of planning: A general principle of planning is the proportionate distribution of the resources of a country and the appropriate use of the human resources to fulfill the needs of the people.
Goals of Plans:
The general goals of the economic planning of India are as follows:

Increase in the national income.
Bringing about rapid industrialisation by focusing on the basic industries.
Bringing about an increase in agricultural production so that the country becomes self-sufficient in foodgrain production.
Use the human resources in the country optimally by availing increasing employment opportunities.
Remove inequality in earnings and wealth.
Maintain stable prices of commodities.
Controlling the growth of population through family planning.
Improve the standard of living by eradicating poverty.
Develop social services.
Make the economy self-sufficient.

Question 2.
Describe the prolonged agitation of the mill workers under Dr. Datta Samant.
Answer:
(i) In the beginning, the striking workers received help from their native villages. It was also not very difficult for them to help each other.

(ii) They set up departmental committees and distributed food grains, assistance in the form of funds, etc

(iii) The left parties had supported the strike. As the strike dragged on, efforts were made to split the striking workers’ ranks.

(iv) Even as the strike completed 6 months, the central government completely ignored it. The workers started a ‘Jail Bharo Agitation’.

(v) In September 1982, one and a half lakh workers took a march on the Legislative Assembly of Maharashtra State.

(vi) It didn’t help at all. The strike completed a year. This was the first strike to have gone on for a year.

(vii) In this period, about one and a half lakh workers became unemployed.

(viii) As polyester had come into greater demand than cotton cloth, the sale of mill cloth got affected. The mills moved from Mumbai to Surat in Gujarat.

(ix) The Central government nationalised 13 textile mills. Appointment of arbitrators did not help to resolve the issue.

Question 3.
Which enterprises came under government due to the lack of capital investment?
Answer:
(i) The industries like defence, scientific research, education, roads, railways, waterways, sea port and airport development require huge capital investment but the returns in these areas are delayed.
(ii) Not many private industrialists are keen to invest in these areas. In such a situation, the government has to take the initiative.

Question 4.
State the Industrial policy of 1973.
Answer:
(i) The industrial policy of 1973 increased the speed of development.
(ii) Priority was given in this policy to control the influence of heavy industries, industrial families and foreign industries and remove the imbalance in regional development.
(iii) The government focused on the development of small scale industries and cottage industries.

Question 5.
What were the aims of the ‘Third Five Year Plan’?
Answer:
(i) This plan was aimed at bringing about a balance in industries and agriculture.
(ii) The other goals of the plan included increase in national income, heavy industries, development in transport and mineral industry, alleviation of poverty and to expand the opportunities for employment.

Question 6.
What was the background for . the nationalisation of banks?
Answer:
(i) India had adopted a mixed economy after independence.
(ii) Nationalisation of banks was essential to cover the deficits if they occurred while implementing different schemes.
(iii) Also the profits of these banks would come into the government treasury once they are nationalised.

Question 7.
State the cultural contributions of the working class.
Answer:
(i) The working class has contributed culturally as well through folk theatre, folk art and literature. Anna Bhau Sathe, Shahir Amar Sheikh, Shahir Sable were popular for their programmes aimed at public education.
(ii) Poets like Narayan Surve, Namdev Dhasal, etc., portrayed the real life of the workers through their poems.

Question 8.
Who took responsibility to bring the Indian economy in sync with the global mainstream?
Answer:
(i) The year 1991 is very important in the history of modem India. After the 10th General elections, P. V. Narasimha Rao became the Prime Minister of India.

(ii) With Dr. Manmohan Singh as Finance Minister, he adopted the new economic policy of linking India’s economy with the global economy.

(iii) For this, fundamental changes were brought about in the Indian economy. Indian economy was brought in tune with the global mainstream.

Question 9.
Which corrective measures were taken by Manmohan Singh to relieve the crisis of the Indian economy?
Answer:
(i) Dr. Singh undertook many corrective measures. The situation began to change.
(ii) He removed the restrictions on foreign investments.
(iii) He restricted the licence system to 18 industries. In view of the increasing losses in the public sector industries, he opened up the public sector for investment by private industries.
(iv) In order to bring the share market under control, he established the Securities and Exchange Board of India (SEBI) in 1992.
(v) National Stock Exchange was computerised. He gave priority to remove the spectre of recession.

Question 10.
State the impact of corrective measures taken by Dr. Manmohan Singh on the economy of our country.
Answer:
(i) Foreign Investment in India grew during the first tenure of Dr. Manmohan Singh as Finance Minister.
(ii) India could recover the gold mortgaged with the Bank of England.
(iii) The government got the support of the capitalist class as well as the middle class.
(iv) As the government opened up the telecom sector, mobile phone services started all over the country.
(v) Dr. Manmohan Singh signed the agreement with the World Trade Organisation and launched the policy of “privatisation, liberalisation and globalisation.”

Question 11.
What were the objectives of the WTO?
Answer:
(i) In 1995, India became a member of the World Trade Organisation (WTO).
(ii) The organisation had the following objectives: to free trade between countries, to put an end to all those discriminatory laws, restrictions, rules and policies that are hurdles in the way of international free trade and to regulate global trade with the help of a formal multi – party mechanism.

Question 12.
Do you agree that membership with the WTO has ushered in changes in the economic sector of our country? State the reason for your answer.
Answer:
(i) Yes, of course. The provisions of the World Trade Organisation have included grants, import-export, foreign investment, agriculture, technology and services.

(ii) The sectors of electricity, water transportation, education and health have been rapidly commercialised since India became a member of the World Trade Organisation.

Question 13.
State the improvements which have occurred in our economy as per the reports of WTO.
Answer:
As per the various reports of the World Trade Organisation, India has made considerable improvement in different areas like reduction in the below poverty line (BPL) population, decline in infant mortality, availability of facilities regarding drinking water and waste water management.

Question 14.
Which economic challenges were faced by the Government of Prime Minister Chandra Shekhar?
Answer:
(i) During the tenure of Chandra Shekhar the rate of inflation was 17%.
(ii) Economic growth rate had decreased by 1.1%. India had foreign exchange reserves only enough to last for imports for a week.
(iii) It had become difficult to pay back its loan and the interest on it.
(iv) In May 1991, the government had tried to control the situation by selling some of its gold reserves and by mortgaging some.

Question 15.
Which economic liabilities were incurred by V. P. Singh government?
Answer:
(i) The V. P. Singh government had incurred a liability of over 10 thousand crore rupees on the economy by waiving off the loans of all the farmers.
(ii) The proportion of internal loans of central and state government together to the Gross Domestic Product was 55%.
(iii) In 1980-81 foreign loan was 2350 crore dollars. It increased to 8380 crore dollars in 1990-91.
(iv) At this time, India had foreign exchange reserves of only 100 crore dollars.
(v) This also had the background of the increased oil prices due to the invasion of Kuwait by Iraq.
(vi) It became difficult for India to raise a loan. Even the non-resident Indians started withdrawing their deposits in foreign currency from India.

Identify the picture and give relevant information.

Question 1.
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 7
Answer:
(i) Nationalisation of banks was essential to cover the deficits if they occurred while implementing different schemes and profit of these banks came under the government treasury.

(ii) Also the profit of these banks would come into the government treasury once they were nationalised. Prime Minister Indira Gandhi nationalised 14 banks on 19th July, 1969.

Question 2.
Maharashtra Board Class 9 History Solutions Chapter 4 Economic Development 8
Answer:
(i) P. V. Narasimha Rao and Manmohan Singh.
(ii) They removed the restrictions on foreign investments.
(iii) Restricted the license system to 18 industries.
(iv) Opened up the public sector for investment by private industries.
(v) Established the SEBI (Securities and Exchange Board of India) in 1992 to regulate and bring share market under control.
(vi) Foreign investment in India grew.
(vii) India could recover the gold mortgaged with the Bank of England.

9th Std History Questions And Answers: