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Std 7 Science Chapter 5 Food Safety Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 5 Food Safety Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 5 Food Safety Question Answer Maharashtra Board

1. Complete the following statements by using the correct option from those given below.
(Irradiation, dehydration, pasteurization, natural, chemical)

Question a.
Drying the food grains from farms under the hot sun is called …………. .
Answer:
dehydration

Question b.
Materials like milk are instantly cooled after heating up to a certain high temperature. This method of food preservation is called ……….. .
Answer:
pasteurization

Question c.
Salt is a ……….. type of food preservative.
Answer:
natural

Question d.
Vinegar is a …………. type of food preservative.
Answer:
chemical

2. Answer the following questions in your own words. 

Question a.
How is milk pasteurised?
Answer:
Boil the milk at 80°C for 15 minutes and cool it quickly. This destroys the microbes present in the milk and it can remain for a longer duration. This process is called pasteurization of milk.

Question b.
Why should we not consume adulterated food materials?
Answer:
Different types of adulterants affect our health in different ways. Some adulterants cause abdominal discomfort or poisoning while some may affect the functioning of some organs if consumed over a long period of time or even cause cancer.

Question c.
What precautions do your parents take to keep foodstuffs safe?
Answer:
Our parents take following care to keep foodstuffs safe

1. Drying of grains.
2. Boiling of milk, soups and curries from time to time.
3. Refrigeration of vegetables, fruits, milk and cooked food.
4. Candying of jams.
5. Use natural preservatives like oil, spices, neem leaves, salt, etc.
6. Use chemical preservatives in sauces, ketchups, pickles, jams and squashes.

Question d.
How does food spoilage occur? Which are the various factors spoiling the food?
Answer:
Food spoilage is the process in which food deteriorates to the point in which it is not edible to humans or its quality of edibility becomes reduced.
Following are the factors for spoiling of food:

1. Bacteria causes food to spoil
2. Incorrect storage may spoil the food.
3. Infestation by pests.
4. Chemical reaction takes place in food and it gets spoiled.

Question e.
Which methods of food preservation would you use?
Answer:
I use following methods to preserve the food:

1. Freezing
2. Boiling
3. Salting of pickles
4. drying of grains
5. candying of jams.

3. What shall we do?

Question a.
There are vendors selling uncovered sweet-meats in open places in the market.
Answer:
We should tell them to sell the covered sweets because uncovered sweets are harmful to eat because it contains dust, dirt and germs. And buyers also should not purchase these uncovered sweets.

Question b.
A ‘pani-puriwala’ is serving the panipuri with dirty hands.
Answer:
We should tell him to wear gloves before serving panipuri because dirty hands contain dirt and germs which are harmful to us.

Question c.
We have purchased a large quantity of fruits and vegetables.
Answer:
We should keep them in refrigerator because in refrigerator, due to low temperature, fruits and vegetables do not spoil and biological and chemical reactions in fruits and vegetables are slowed down at low temperature.

Question d.
We need to protect foodstuffs from pests like rats, cockroaches, wall-lizards etc.
Answer:

1. If we do not protect foodstuffs from pests like rats, cockroaches, wall-lizards etc. then the food get spoiled by them and germs carried by them enters into the food.
2. If we eat this food then we may get food poisoning and we fall sick so we need to protect foodstuffs from pests like rat, cockroaches, wall lizards etc.

4. Find the odd-man-out. 

Question a.
Salt, vinegar, citric acid, sodium benzoate.
Answer:
Salt

Question b.
Lakhi dal, brick dust, metanyl yellow, turmeric powder.
Answer:
Turmeric powder

Question c.
Banana, apple, guava, almond.
Answer:
Almond

Question d.
Storing, freezing, settling, drying
Answer:
Storing

5. Complete the chart below.

Question a.

Answer:

6. Explain why this happens and suggest possible remedies.

Question a.
Qualitative wastage of food.
Answer:
Qualitative wastage of food happens due to wrong methods of protecting food, excessive use of food preservatives, over-cooking, washing the vegetables after cutting them, mis¬handling of fruits like grapes and mangoes, miscalculation of the time required to transport food from producers to consumers are some of the reasons of qualitative wastage of food.
Possible remedies:

1. Avoid overcooking of food.
2. Store grains and other perishable foodstuffs like vegetables, fruits, milk etc. using proper methods.
3. Wash fruits and vegetables before cutting it.

Question b.
The cooked rice is underdone.
Answer:
Sometimes in a hurry if we cook the rice it is underdone.
Possible remedies: Use proper pressure cooker to cook the rice.

Question c.
The wheat that was bought is a bit moist.
Answer:
Sometimes due to sudden rain, wheat gets a bit moist. Possible remedies: Do not store the moist wheat, it gets spoiled due to fungus so first sundry it and then store in a clean and dry container to avoid microbial growth.

Question d.
The taste of yoghurt is too sour/slightly bitter.
Answer:
The taste of yoghurt is too sour/slightly bitter means it is spoiled. This happens if it is not kept in the refrigerator.
Possible remedies:- Always keep the yoghurt in refrigerator to avoid biological and chemical reactions in food materials.

Question e.
Cut fruits turned black.
Answer:
Fruit contains an enzyme called polyphenol oxidase or tyrosinase that reacts with oxygen. The oxidation reaction basically forms a sort of rust on the surface of fruits so it turns black.
Possible remedies:

1. Coat the fruits with sugar syrup
2. Add lemon juice on fruits.

7. Give reasons.

Question 1.
Food remains safe at 5° Celsius.
Answer:
Food remains safe at 5° Celsius because at 5°C, micro-organisms stop growing.

Question 2.
Nowadays, food is served in buffet style during large gatherings.
Answer:
Nowadays, food is served buffet style during large gatherings because due to buffet style quantitative wastage of food can be avoided, as people take only as much as they could eat.

Project:

Question 1.
Go to your kitchen und take notes about the food safety measures and the food wastage you see there.

Question 2.
In a science exhibition demonstrate the various methods of detecting food adulteration.

Class 7 Science Chapter 5 Food Safety Important Questions and Answers

Complete the following statements by using the correct option from those given below.
(Irradiation, dehydration, pasteurization, natural, chemical)

Question 1.
………………………. is celebrated as “World Food Day”.
Answer:
16th October

Question 2.
FSSAI means ………………………. .
Answer:
Food Safety and Standardization Authority of India

Question 3.
………………………. gas is filled in tight packets of potato wafers.
Answer:
Nitrogen

Question 4.
Common name of acetic acid is ……………………….
Answer:
Vinegar

Question 5.
………………………. is sprayed on the gunny bags containing food grains.
Answer:
Melathion

Question 6.
………………………. is used in smoking method.
Answer:
Aluminium phosphide

Question 7.
………………………. and ………………………. are emitted by radioactive isotopes in irradiation method.
Answer:
X – rays, gamma rays

Question 8.
In Maharashtra, irradiation plants have been installed at ………………………. for onions and potatoes and at for spices and condiments.
Answer:
Lasalgaon, Navi Mumbai

Question 9.
………………………. found the pasteurisation method.
Answer:
Louis Pasteur

Question 10.
………………………. adulterant is added to turmeric powder.
Answer:
Metanyl yellow

Question 11.
………………………. adulterant is added to red chilly powder.
Answer:
Brickdust

Question 12.
………………………. is used to make fruits more attractive.
Answer:
calcium carbide

Question 13.
………………………. and ………………………. harmful chemicals are mixed with cold drinks.
Answer:
carbonated soda, phosphoric acid

Question 14.
Shopkeepers change the of the ………………………. food packets fo avoid a financial loss.
Answer:
expiry date

Question 15.
Milk vendors add ………………………. to the milk to appear as higher fat content.
Answer:
urea

Question 16.
The shelf life of potatoes and onions ………………………. due to slowed-down of sprouting.
Answer:
increases

Question 17.
Serving too much food to guests at traditional feasts and banquets leads to ………………………. wastage of food.
Answer:
Quantitative

Question 18.
Miscalculation of the time required to transport food from producers to consumers leads to ……………………… . wastage of food.
Answer:
Qualitative

Question 19.
Prevention of food spoilage by microbial growth and infestation by pests is called ………………………. .
Answer:
food protection

Question 20.
………………………. is an example of a chemical preservative.
Answer:
Sodium benzoate.

Say whether True or False,, Correct and rewrite the false statement.

Question 1.
To prevent adulteration of food, it is inspected by the “Food and Drug Administration department of the government”.
Answer:
True

Question 2.
Overcooking of food increases the quality of food.
Answer:
False. Over cooking of food spoils it

Question 3.
Peanuts become rancid then, it is not good to eat.
Answer:
True

Question 4.
Oil and ghee contain fats.
Answer:
True

Question 5.
Prevention of food spoilage by microbial growth and infestation by pests is called food wastage.
Answer:
False. Prevention of food spoilage by microbial growth and infestation by pests is called food protection.

Question 6.
Serve yourself only as much as you can eat.
Answer:
True

Question 7.
Salt, sugar and oil are naturally available preservatives.
Answer:
True

Question 8.
Pickles can be preserved by salting.
Answer:
True

Question 9.
Milk vendors add urea to the milk so that it appears to have higher fat content.
Answer:
True

Question 10.
Shelf life of fruits and vegetables decreases by gamma rays emitted by radio-active isotopes.
Answer:
False. Shelf life of fruits and ‘ vegetables increases by gamma rays emitted by radioactive isotopes.

Give scientific reasons.

Question 1.
Refrigerators are used in the kitchen.
Answer:

1. Biological and chemical reactions in food materials are slowed down at low temperature.
2. As a result food remains in good condition for a longer period. Therefore refrigerators are used in the kitchen.

Question 2.
Potatoes and onions are treated with gamma rays.
Answer:
Potatoes and onions are treated with gamma rays because irradiation with gamma rays prevents their wastage due to sprouting and increases their shelf life.

Question 3.
Grains are sun-dried.
Answer:
Grains are sun-dried to preserve them because on sun drying their water content gets reduced and hence they last longer.

Question 4.
We boil milk from time to time.
Answer:
We boil milk from time to time to kill microorganisms in it and thus prevent it from getting spoilt.

Question 5.
Jams or pickles get spoilt if their jars are not sealed properly.
Answer:
If the jar is not sealed properly, micro¬organisms from the air enter in the jar and start growing on the food. Oxygen in the air helps the micro-organisms and fungi to grow faster and hence bring about the spoilage of the jams and pickles.

Question 6.
Some vendors add urea to the milk.
Answer:
Some vendors add urea to the milk so that it appears to have higher fat content.

Question 7.
Food wastage should be avoided.
Answer:
In countries like India, the food requirement is plenty and people do not get proper meals even once in a day. If food is not wasted then it could have met the need of many others. Therefore, food should not be wasted and proper measures should be implemented to stop quantitative and qualitative wastage of food.

Question 8.
Adulterated food should not be consumed.’
Answer:

1. The health of all people is endangered by food adulteration.
2. Different types of adulterants affect our health in different ways.
3. Some adulterants cause abdominal discomfort or poisoning, while some may affect the functioning of some organs if consumed over a long period of time or even cause cancer.

Can you tell?

Answer the following questions:

Question 1.
How and where food is wasted?
Answer:
Quantitative wastage of food:
1. Wrong methods of farming like hand sowing of, ts seeds, inadequate threshing, improper storage and wrong methods of distribution are some reasons for quantitative wastage of food.

2. Besides, much food is wasted as a result of the custom of offering and serving too much food to guests at traditional feasts or banquets.

3. Had it not been wasted, all this food could have met the need of many others Qualitative wastage of food: Using wrong methods of protecting food, excessive use of food preservatives, over-cooking, washing the vegetables after cutting them, mis-handling of fruits like grapes and mangoes, mis-calculation of the time required to transport food from producers to consumers, are some of the causes of quantitative wastage of food.

Question 2.
How is the food adulterated?
Answer:
Food is adulterated by the following ways:

1. Removal of some important components of food. e.g. removal of fat content of milk, essence of cloves, cardamoms, etc.
2. Mixing of a low quality inedible or cheaper material or harmful colour with food.
3. Mixing of some harmful materials like small stones, fine sand, iron filings, urea, dung of horse, sawdust etc.

How will you find out if food has been adulterated?

Maharashtra State Board Class 7 Science Textbook Solutions

• The Living World: Adaptations and Classification Class 7 Science Textbook Solutions
• Plants: Structure and Function Class 7 Science Textbook Solutions
• Properties of Natural Resources Class 7 Science Textbook Solutions
• Nutrition in Living Organisms Class 7 Science Textbook Solutions
• Food Safety Class 7 Science Textbook Solutions
• Measurement of Physical Quantities Class 7 Science Textbook Solutions
• Motion, Force and Work Class 7 Science Textbook Solutions
• Static Electricity Class 7 Science Textbook Solutions
• Heat Class 7 Science Textbook Solutions
• Disaster Management Class 7 Science Textbook Solutions

Std 6 Science Chapter 1 Natural Resources – Air, Water and Land Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 6 Science Solutions Chapter 1 Natural Resources – Air, Water and Land Notes, Textbook Exercise Important Questions and Answers.

Class 6 Science Chapter 1 Natural Resources – Air, Water and Land Question Answer Maharashtra Board

1. Fill in the blanks and rewrite the completed statements.

Question a.
The layer of ozone gas absorbs ……………. rays that come from the sun to the earth.
Answer:
Ultraviolet (U.V) rays

Question b.
Of the total water available on the earth, fresh water forms ……….. percent.
Answer:
0.3 %

Question c.
Both …………. and ………… constituents are present in the soil.
Answer:
biotic, abiotic

2. why is it said that?

Question a.
The ozone layer is a protective shell of earth.
Answer:

1. The ultra violet (UV) rays coming from the sun are very harmful for living things.
2. The ozone layer present in the lower stratosphere absorb this U.V. rays and prevent them from reaching the earth. As a result life on earth is protected.
3. Therefore, it is said that the ozone layer is a protective shell of the earth.

Question 2.
Water is life.
Answer:

1. Water is a good solvent and it dissolves many substances.
2. The human blood is made of 70% water and the sap of plant also contains a very high proportion of water.
3. All the life processes would not take place in the absence of water.
4. Hence, without water no living organism can survive.
5. Therefore, it is said that ‘water is life’.

Question c.
Sea water is useful even though it is not potable.
OR
In what way is sea water useful even though it is salty?
Answer:

1. Many fish and aquatic animals live in sea water.
2. The water from the sea evaporates to form clouds which brings rain.
3. The sea water also helps the land to cool due to breezes.
4. The salt and minerals are also obtained from sea water.
5. Corals and pearls are obtained from sea animals.
6. Thus, sea water is useful even though it is not potable.

3.  What will happen if

Question a.

Question a.
Microbes in soil get destroyed.
Answer:

1. Microbes in the soil decompose dead plants and animals and convert it into humus. This humus supplies nutrients to the soil.
2. Humus also aerates soil and holds water in it. It makes the soil more fertile.
3. If microbes are destroyed, humus will not be formed and the soil will not become fertile, making it unsuitable for growth of plants. Also dead and decaying matter will accumulate on land.

Question b.
The number of vehicles and factories in your surroundings increases.
Answer:

1. Vehicles and factories are the major cause of air pollution.
2. They release harmful gases like carbon dioxide, carbon monoxide and sulphur dioxide into the air.
3. These pollutants are harmful to the environment and to the people living in the surrounding area.
4. Hence, if the number of vehicles and factories in our surroundings increases, the air pollution will also increase.

Question c.
The total supply of potable water is finished.
Answer:

1. Water plays very important role in the survival of living organism.
2. All living things are dependent on water.
3. A very small quantity of water is potable and can be used for drinking.
4. All bodily functions are regulated by water.
5. Therefore, if total supply of potable water is finished, plants and animals will not survive and there will be no life on earth.

4. Match the following. 

Question a.

Answer:

5. Name the following.

Question a.
Constituents of biosphere.
Answer:
Atmosphere, hydrosphere, lithosphere and all living things on earth.

Question b.
Biotic constituents of soil.
Answer:
Microbes, worms, insects, burrowing rhodents like rats, mice, roots of trees and plants.

Question c.
Fossil fuel.
Answer:
Crude oil from which we get kerosene, petrol, diesel, paraffin wax and tar.

Question d.
Inert gases in air.
Answer:
Neon, argon, helium, krypton, xenon.

Question e.
Gases that are harmful to ozone layer.
Answer:
Chlorofluorocarbon and carbon tetrachloride.

6. True or False?

Question a.
Land and soil is the same thing.
Answer:
False – Land consists of stones, soil and big rocks.

Question b.
The water in a lake is called ground water.
Answer:
False – Water trapped below the ground over the bedrocks is called ground water.

Question c.
It takes about thousand years to form a 25 cm thick layer of soil.
Answer:
False – It almost takes around thousand years to form a 2.5 cm thick layer of soil.

Question d.
Radon is used in decorative lights.
Answer:
False – Neon is used in decorative lights.

7. Answer in your own words. 

Question a.
Explain with the help of a diagram how soil is formed.
Answer:

1. The soil on the land is formed by a natural process.
2. The abiotic components of soil are supplied through the weathering of the bedrock.
3. Due to heat, cold wind and rain the bedrock breaks down into pieces.
4. Stones, sand and soil are formed from these pieces.
5. Microbes, worms, insects, rodents and roots of trees growing on land help in weathering of rocks.
   
   This process is slow, continuous and it takes a thousand years to form 2.5 cm thick layer of mature soil.

Question b.
Why is there a shortage of water even though it occupies about 71% of the earth’s surface?
Answer:

1. 71% of earth’s surface is covered with water of, which 97% is salty water present in seas and oceans and 2.7% water is available as ground water, ice and in other forms.
2. Only 0.3% water is available as fresh water which can be used for drinking.
3. All the living organisms require water to drink. The water is used in the industry and also for farming.
4. Due to increasing population and uncontrolled usage, we experience shortage of water.

Question c.
What are the various constituents of air? Write their uses.
Answer:
Air contains gases like oxygen, carbon dioxide, nitrogen, inert gases, water vapour and dust particles. The uses of constituents of air are as follows.

Question d.
Why are air, water and land considered to be valuable natural resources?
Answer:

1. The various components of air help and support the growth of living organisms.
2. Nitrogen is used to make proteins, oxygen is used for respiration and carbon dioxide is used to make food by plants.
3. Similarly, land supports growth of terrestrial plants and animals. It provides important minerals to plants and also to human beings.?
4. It is the shelter for worms, insects and rodents and supports their growth. Plants also cannot survive without land.
5. Water is necessary for carrying out all life processes in the living organisms and without water there will be no life on earth.
6. Hence land, air and water are considered valuable natural resources.

Activity:

Natural Resources Air, Water And Land Class 6 Questions And Answers Question 1.
Obtain detailed information about the work of the India Meteorological Department.

Natural Resources Air Water And Land Class 6 Questions And Answers Question 2.
Find a remedy for water scarcity.

Class 6 Science Chapter 1 Natural Resources – Air, Water and Land Important Questions and Answers

Fill in the blanks and rewrite the completed statements.

Class 6 Science Chapter 1 Natural Resources Air, Water And Land Exercise Question 1.
……………. gas, used for refrigeration and air conditioning, destroys the ozone layer.
Answer:
Chlorofluorocarbon or carbon tetrachloride

Natural Resources Air Water And Land Class 6 Exercise Question 2.
Air becomes ……………. at higher altitudes.
Answer:
rarer

Natural Resources Air Water And Land Class 6 Question 3.
………….. of land is reduced if green trees and bushes are grown in it.
Answer:
Erosion

Natural Resources Air Water And Land Question 4.
16th September is celebrated as ………. Day all over the world.
Answer:
Ozone Protection

Natural Resources Air Water And Land Question Answer Question 5.
………….. is the layer of air that surrounds the earth.
Answer:
Atmosphere

Natural Resources Air, Water And Land Question 6.
……………. occupies the largest part of the earth’s surface.
Answer:
Hydrosphere

Natural Resources Air Water And Land Class 6 Question 7.
Gases are not found in the …………… and beyond.
Answer:
exosphere

Choose the correct alternative:

Question 1.
………….. percentage of the land is covered by water.
(a) 70%
(b) 81%
(c) 71%
(d)80%.
Answer:
71%

Question 2.
The gas used in fluorescent tubes is ………………… .
(a) Argon
(b) Helium
(c) Neon
(d) Krypton.
Answer:
Krypton

Question 3.
The ozone layer is found in the lower part of …………… .
(a) atmosphere
(b) stratosphere
(c) mesosphere
(d) trophosphere.
Answer:
stratosphere

Question 4.
Gas released in air on combustion of fuel is …………….. .
(a) Hydrogen sulphide
(b) Carbon tetrachloride
(c) Nitrogen dioxide
(d) Oxygen
Answer:
Nitrogen dioxide

Question 5.
The proportion of humus in the upper layer of good fertile soil is about ……………… .
(a) 23% to 45%
(b) 33% to 50%
(c) 30% to 53%
(d) 13% to 33%
Answer:
33% to 50%

Match the following:

Question 1.

Answer:

Name the following:

Question 1.
Substances formed when fuel burns.
Answer:
Carbon dioxide, carbon monoxide, nitrogen dioxide, sulphur dioxide and smoke.

Question 2.
Layers of the atmosphere.
Answer:
Troposphere, stratosphere, mesosphere, ionosphere and exosphere.

Question 3.
Layers of land.
Answer:
Humus, mature soil, immature soil, small rocks and stones and bedrock.

Question 4.
Gas necessary for building proteins.
Answer:
Nitrogen.

State whether True or False. Correct if False.

Question 1.
The amount of gases in the air is greatest near the surface and becomes rarer at higher altitudes.
Answer:
True.

Question 2.
Fogs, clouds, snow, and rain are produced in the exosphere.
Answer:
False – Fogs, clouds, snow and rain are formed in the troposphere and lower stratosphere of the atmosphere.

Question 3.
Fossil fuels are formed from the dead remains of animals and plants buried underground for a long period.
Answer:
True.

Explain what will happen if:

Question 1.
Forests are destroyed.
Answer:

1. Soil will get eroded due to rains as roots of trees hold the soil.
2. The land will become barren as trees helps to increase the level of ground water.
3. Amount of carbondioxide in the air will increase as trees use carbondioxide for photosynthesis and release oxygen.
4. Natural habitat of many animals will get completely destroyed.

Question 2.
What would have happened if there was no air on the earth?
Answer:

1. Air contains gases like nitrogen, oxygen, carbon dioxide, dust particles and water vapour, which are used in various piofeesses in living organisms and environment.
2. If there is no air then there will be no life as oxygen is essential for all living beings to survive. Also atmosphere is a very important filter. It prevents die harmful elements from reaching the earth.
3. Hence without air, our earth would become a cold, dark planet without any life.

Answer the following:

Question 1.
What is humus?
Answer:
Humus is the topmost layer of the soil formed d by decomposition of remains of plants and animals and it makes the soil fertile.

Question 2.
What is land made up of?
Answer:
Land is made up of stones, soil, sand and big rocks.

Question 3.
Is land flat everywhere?
Answer:
No, land is not flat everywhere. It is flat in some regions and hilly in some regions.

Question 4.
Does man produce soil/ land?
Answer:
No, man does not produce soil/land, it is produced naturally.

Question 5.
What do you see on land?
Answer:
We see mountains, rivers, valleys, ocean, also terrestrial animals and plants. We also see roads, bridges, buildings etc.

Question 6.
What has man created on land?
Answer:
Man has dug wells, borewells to lift ground water. He has also constructed bunds and dams. He has also built many industries, buildings, roads for transport.

Question 7.
If a deep pit is dug in the ground, what do you see there?
Answer:
We see different layers of land.

Answer in your own words.

Question 1.
Explain with the help of diagram various layers of land.
Answer:

• Humus – Topmost layer, fertile layer formed by decomposition of remains of plants and animals. Immature
• Soil – Sand, soil, small stones, worms and insects.
• Layer of soil and small rocks – less soil and more rocks.
• Bedrocks – main minerals are obtained from this layer, determines colour and texture of soil.

Observe the picture and answer the questions.

Question 1.
Where do you see the birds?
Answer:
The birds are flying in the sky.

Question 2.
Where is the cow grazing?
Answer:
The cow is grazing in the pasture (field).

Question 3.
Locate the trees.
Answer:
The trees are seen along the river bank.

Question 4.
Where does the river come from?
Answer:
The river flows from the mountains towards the plains.

Question 5.
Where is the aeroplane?
Answer:
The aeroplane is in the sky above the clouds.

Question 6.
Where are fishes seen?
Answer:
Fishes are seen swimming in the river water.

Question 7.
On what is the sail boat floating?
Answer:
Sail boat is floating on the water.

Observe and discuss:

Question 1.
What is the similarity in the three pictures given above?
Answer:
All the pictures given above show large scale emission of smoke through different agencies. This smoke directly mixes with the atmosphere, disturbing the balance between the constituents of air and causing air pollution.

Observe and discuss:

Observe the distribution of water on the earth surface and complete the table.

Question 1.

Answer:

Observe given figure carefully and answer the following.

Question 1.
For which purpose is water being used?
Answer:
Water is being used for washing clothes, for bathing, farming, drinking, and industries.

Question 2.
Do other living things use water like we do?
Answer:
Animals do not use water like us. They use water only for drinking. Some animals like buffaloes, rhinoceros, elephants use water for cooling themselves during summers.

Question 3.
What are the constituents of soil? Classify them as biotic and abiotic constituents.
Answer:
The constituents of soil are humus, soil, sand, gravel, stones, bedrock, insects, worms, microbes, roots of trees and dead leaves, burrowing rodents like mice and rats.

Answer the following:

Question 1.
How would you save water? Give some measures you will adopt.
Answer:
Water can be saved in the following ways:

1. Repair the leaking taps and pipes and prevent wastage of water.
2. Take water in a bucket to wash a car, rather than using a hose pipe.
3. Close the tap when not required.
4. Store rainwater in underground tanks so that it can be used all round the year.
5. Water leftover after washing vegetables etc. can be used for watering plants in the garden.
6. Use water sparingly and reuse water wherever possible.

Question 2.
What measures would you take to prevent soil erosion?
Answer:

1. The trees, bushes and grass shall be planted in open spaces.
2. Shrubs will be planted along the river banks to prevent floods.
3. Proper drainage system will be provided so that there is no flooding of water.

Let’s try this

1. Take a transparent plastic bottle, a handful of soil, big stones, small stones, sand, some dry leaves and water.
2. Cut off the upper tapering part of the bottle. Put the rest of the materials in the lower part and add water.
3. Stir the mixture thoroughly and put it aside.

Observe it the next day and answer.

Question 1.
How does the mixture in the bottle look now?
Answer:
The mixture gets segregated into various layers. Heaviest substances settle down and lighter forms topmost layer.

Question 2.
Do you see the layers in it?
Answer:
Yes, we see the layers of soil.

Question 3.
What is seen in the different layers from top to bottom?
Answer:
The dry leaves are floating above the water at the top. Then the layer of the soil, which forms a layer above the sand. Below the sand we see a layer of small stones, and the big ones have settled down at the bottom of the bottle.

Question 4.
Obtain specimens of soil from various places and note the differences in the specimen with respect to colour, feel, texture and size of the particles.
Answer:

Question 5.
Observe how much water is used and for what purposes it is used in your house for a whole day. Record it in a chart. Discuss this data and find out how much water each person needs in your house.
Answer:

Total number of persons = 3
Total water used per person = \(\frac{50}{3}\) = 50 litres.
Approximately 50 litres of water is required for 1 person.

Maharashtra State Board Class 6 Science Textbook Solutions

• Natural Resources – Air, Water and Land Class 6 Science Textbook Solutions
• The Living World Class 6 Science Textbook Solutions
• Diversity in Living Things and their Classification Class 6 Science Textbook Solutions
• Disaster Management Class 6 Science Textbook Solutions
• Substances in the Surroundings – Their States and Proper Class 6 Science Textbook Solutionsties
• Substances in Daily Use Class 6 Science Textbook Solutions
• Nutrition and Diet Class 6 Science Textbook Solutions
• Our Skeletal System and the Skin Class 6 Science Textbook Solutions
• Motion and Types of Motion Class 6 Science Textbook Solutions
• Force and Types of Force Class 6 Science Textbook Solutions

Std 8 Science Chapter 16 Reflection of Light Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 8 Science Solutions Chapter 16 Reflection of Light Notes, Textbook Exercise Important Questions and Answers.

Class 8 Science Chapter 16 Reflection of Light Question Answer Maharashtra Board

Class 8 Science Chapter 16 Reflection of Light Textbook Questions and Answers

1. Fill in the blanks:

Question i.
The perpendicular to the mirror at the point of incidence is called …………. .
Answer:
The perpendicular to the mirror at the point of incidence is called the normal.

Question ii.
The reflection of light from a wooden surface is ……….. reflection.
Answer:
The reflection of light from a wooden surface is irregular reflection.

Question iii.
The working of a kaleidoscope is based on the properties of …………… .
Answer:
The working of a kaleidoscope is based on the properties of reflection of light.

2. Draw a figure describing the following: The reflecting surfaces of two mirrors make an angle of 90° with each other. If a ray incident on one mirror has an angle of incidence of 30°, i draw the ray reflected from the second mirror. What will be its angle of reflection?

Question a.
Draw a figure describing the following: The reflecting surfaces of two mirrors make an angle of 90° with each other. If a ray incident on one mirror has an angle of incidence of 30°, draw the ray reflected from the second mirror. What will be its angle of reflection?
Answer:
For the ray C, the angle of reflection = 60°.

3. How will you explain the statement ‘We cannot see the objects in a dark room’?

Question a.
How will you explain the statement ‘We cannot see the objects in a dark room’?
Answer:
In a room that is completely dark, no light falls on objects. Hence, no light enters our eyes. Hence, there is no sensation of vision, i.e., we cannot see the objects.

4. Explain the difference between regular and irregular reflection of light.

Question a.
Explain the difference between regular and irregular reflection of light.
Answer:
For regular reflection of light, the angles of incidence as well as the angles of reflection are the same for all parallel rays of light incident on the plane and smooth surface. Hence, the reflected rays are also parallel to one another.

For irregular reflection of light, the angles of incidence for parallel rays of light incident on the rough surface are not equal, and hence the angles of reflection are also not equal. Here, the reflected rays are not parallel to one another and spread over a large surface.

5. Draw a figure showing the following:
(a) Incident ray, (b) Normal, (c) Angle of incidence, (d) Angle of reflection, (e) Point of incidence, (f) Reflected ray.

Question a.
Draw a figure showing the following:
(a) Incident ray
(b) Normal
(c) Angle of incidence
(d) Angle of reflection
(e) Point of incidence
(f) Reflected ray.
Answer:

6. Study the following incident.

Swara and Yash were looking in a water-filled vessel. They could see their images clearly in the still water. At that instant, Yash threw a stone in the water. Now their images were blurred. Swara could not understand the reason for the blurring of the images.
Explain the reason for blurring of the images to Swara by answering the following questions:

Question i.
Is there a relation between the reflection of light and the blurring of the images?
Answer:
Yes.

Question ii.
Which types of reflection of light can you notice from this?
Answer:
Regular reflection of light when light is incident on the still water and irregular reflection of light when light is incident on the water as ripples are produced on its surface when a stone is thrown in the water.
Still water behaves as a plane and smooth surface while oscillating water behaves as a rough surface.

Question iii.
Are the laws of reflection followed in these types of reflection?
Answer:
Yes.

7. Solve the following examples.

Question a.
If the angle between the plane mirror and the incident ray is 40°, what are the angles of incidence and reflection?
Solution:
The angle between the plane mirror and the incident ray is 40°. Therefore, the angle of incidence (i) = the angle made by the incident ray with the normal to the plane mirror = 90° – 40° = 50°. The angle of reflection, r – i – 50°.

Question b.
If the angle between the mirror and reflected ray is 23°, what is the angle of incidence of the incident ray?
Solution:
The angle between the mirror and the reflected ray is 23°. Therefore, the angle of reflection (r) = the angle made by the reflected ray with the normal to the plane mirror = 90° – 23° = 67°.
∴ The angle of incidence, i = r = 67°.

Project:

Question a.
Apollo astronauts who stepped on the moon have kept some large mirrors there. Collect information about how the distance to the moon is measured using these.

Class 8 Science Chapter 16 Reflection of Light Important Questions and Answers

Rewrite the following statements by selecting the correct option:

Question 1.
If the angle made by the incident ray with the surface of a plane mirror is 30°, the angle of reflection must be …….. .
(a) 30°
(b) 90°
(c) 60°
(d) 15°
Answer:
If the angle made by the incident ray with the surface of a plane mirror is 30°, the angle of reflection must be 60°.

Question 2.
If the angle of incidence is 40°, the angle made by the reflected ray with the surface of the plane mirror must be ……. .
(a) 40°
(b) 50°
(c) 20°
(d) 80°
Answer:
If the angle of incidence is 40°, the angle made by the reflected ray with the surface of the plane mirror must be 50°.

Question 3.
If the angle of incidence is 20°, the angle made by the reflected ray with the normal to the surface must be ……… .
(a) 20°
(b) 70°
(c) 10°
(d) 40°
Answer:
If the angle of incidence is 20°, the angle made by the reflected ray with the normal to the surface must be 20°.

Question 4.
In a kaleidoscope, the mirrors are inclined to each other at ……. .
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Answer:
In a kaleidoscope, the mirrors are inclined to each other at 60°.

Question 5.
In a periscope, the mirrors are ………….. .
(a) parallel to each other
(b) at right angles to each other
(c) inclined at 45° to each other
(d) inclined at 60° to each other
Answer:
In a periscope, the mirrors are parallel to each other.

Find the odd one out and give the reason:

Question 1.
Plane mirror, Plywood, Wood, Rough tile.
Answer:
Plane mirror. In this case, regular reflection of light takes place. In other cases, reflection of light is irregular.

State whether the following statements are True or False. (If a statement is false, correct it and rewrite it.)

Question 1.
The sense of vision is the most important among our five senses.
Answer:
True.

Question 2.
In a periscope, the angle between the incident ray and the normal to the mirror is 30°.
Answer:
False. (In a periscope, the angle between the incident ray and the normal to the mirror is 45°.)

Answer the following questions in one sentence each:

Question 1.
What is an incident ray?
OR
Define incident ray.
Answer:
A ray of light falling on a surface is called an incident ray.

Question 2.
What is the point of incidence?
OR
Define point of incidence.
Answer:
The point at which the incident ray strikes the surface is called the point of incidence.
[Note: It is also the point of reflection.]

Question 3.
What is the normal?
OR
Define normal.
Answer:
The perpendicular to a surface at the point of incidence is called the normal.

Question 4.
What is the reflected ray?
OR
Define reflected ray.
Answer:
The ray of light that leaves the surface at the point of reflection (the same as the point of incidence) is called the reflected ray.

Question 5.
What is the angle of incidence?
OR
Define angle of incidence.
Answer:
The angle between the incident ray and the normal is called the angle of incidence.

Question 6.
What is the angle of reflection?
OR
Define angle of reflection.
Answer:
The angle between the reflected ray and the normal is called the angle of reflection.

Try this:

Switch off the light in your room at night for some time and then turn it on again.

Question 1.
Could you see the objects in the room clearly when the light was switched off?
Ans.
No.

Question 2.
What did you feel when it was turned on again?
Answer:
We could see the objects clearly. From the above activity you can notice that there is some connection between the sense of vision and light. When we switch off the light at night, the objects in the room cannot be seen and they can be seen as before when the light is switched on again. Thus, we can see objects when the light coming from these objects enters our eyes.

Answer the following questions:

Question 1.
What is reflection of light?
Answer:
When light rays fall on an object, their direction changes and they turn back. This is called the reflection of light.

Try this:

Material:
Torch light, mirror, a stand for hanging the mirror, black paper, comb, white paper, drawing board.
Activity :
1. Fit a white paper tightly over a table or drawing board.
2. Leaving out some portion in the middle of the comb, cover the rest with black paper so that light can only pass through the open central portion.
3. Hold the comb perpendicular to the white paper and throw torch light on its central portion.
4. Adjust the comb and torch so as to get light rays on the white paper. Now keep a mirror in the path of this ray of light as shown in the figure.
5. What do you observe?

Answer:
Light rays which fall on the mirror get reflected and travel in a different direction.

Question 2.
State the laws of reflection of light.
Answer:

1. The angle of reflection is equal to the angle of incidence.
2. The incident ray, the reflected ray and the normal lie in the same plane.
3. The incident ray and the reflected ray are on the opposite sides of the normal.

Try this:

Verification of the laws of reflection of light.
Equipment: Mirror, drawing board, pins, white paper, protractor, scale, pencil.
Activity:

1. Fit a white paper on the drawing board tightly as possible.
2. On the paper draw a line PQ indicating the position of the mirror.
3. Draw a perpendicular ON to PQ at point O.
4. Draw a ray AO making an angle of 30° with ON.
5. Fix two pins S and R along AO.
6. Fix the mirror to a stand and place it along PQ perpendicular to the drawing board.
7. Fix pins at T and U along the line joining the bottom of the reflected images of the pins at S and R.
8. Remove the mirror and join the points T and U and extend it up to O.
9. Measure ZTON.
10. Repeat steps 4 to 9 for angle of incidence equal to 45° and 60° and write down the angles in the following table.

What relation do you find between the angle of incidence and the angle of reflection? If you have done the experiment carefully, you will find that the angle of incidence is equal to the angle of reflection in all three cases. This verifies the laws of reflection.

Question a.
What will happen when a light ray is incident perpendicular to the mirror?
Answer:

Here,
r = i = 90°.
Hence the light ray, on reflection, will retrace the path.

Question 3.
Figures (a) and (b) show three parallel rays, shown in grey, incident on smooth and rough surfaces. The reflected rays drawn using laws of reflection are shown in red.
1. Rays reflected from which surface are parallel to one another?
2. What conclusion can you draw from the figure?
Answer:
1. Rays reflected from the smooth surface are parallel to one another.

2.When the reflecting surface is plane and smooth, the angles of incidence (i) as well as of reflection (r) are the same for all parallel rays incident on the surface. If i₁, i₂, i₃, … are the angles of incidence for incident parallel rays, and r₁, r₂, r₃, …, are the corresponding angles of reflection, then, i₁ = i₂ = i₃ = ……. = r₁ = r₂ = r₃ = ….. This is called regular reflection. Here, the reflected rays are parallel to one another. If the reflecting surface is rough and parallel rays are incident on it, then the angles of incidence are not equal and hence the angles of reflection are also not equal. Here, i₁ ≠ i₂ ≠ i₃ … and r₁ ≠ r₂ ≠ r₃ …, but r₁ = i₁, r₂ = i₂, r₃ = i₃ … as laws of reflection are obeyed. This is called irregular reflection. [Fig.(b)]. Here, the reflected rays are not parallel to one another and spread over a large surface.

Question 4.
What is regular reflection of light?
Answer:
The reflection of light from a plane and smooth surface is called regular reflection of light.

Question 5.
What is irregular reflection of light?
Answer:
The reflection of light from a rough surface is called irregular reflection of light.

Always remember:

1. Laws of reflection are followed in both and regular and irregular reflection.
2. The reflection of light in irregular reflection has not been obtained because the laws of reflection are not followed. They are obtained because the surface is rough (irregular).
3. In irregular reflection, the angles of incidence at different points are different. But at any one point, the angles of incidence and reflection are equal, i.e. i₁ = r₁, i₂ = r₂ …..

Can you recall?

Reflection of reflected light :

Question 1.
How do you see if the barber m a saloon has cut the hair on your neck properly or not?
Answer:
In a saloon, there are mirrors in your front and at back. The image of the back of your head is formed in the mirror at the back. The image of this image is formed in the mirror in front of you. Thus you can see how the hair at the backside of your head is cut.

Question 2.
What type of image do we see in a mirror? What happens to the left and right sides?
Answer:
The image in a plane mirror is upright (erect) and of the same size as the object, but the left and right sides are interchanged. Our right hand appears to be the left hand in the image and the left hand appears to be the right hand in the image. (This is called lateral inversion.)

Question 3.
How do we see the image of the moon in water?
Answer:
The moon is not self luminous. The sunlight falling on the surface of the moon is reflected. This reflected light is again reflected by water to give us the image of the moon.

Try this:

Kaleidoscope:

Activity:

1. Take three rectangular mirrors of the same size.
2. Using sticking tape, stick the mirrors together making a triangle with the reflecting surface facing inwards (see Fig.).
3. Take a white paper of triangular shape and fix it with a tape at one end of the mirrors closing that end.
4. Insert 4 – 5 coloured glass pieces in the hollow of the mirrors.
5. Close the other end also with a paper and make a hole in it.
6. Look through the hole towards light. You will see innumerable images of the glass pieces. These are formed due to reflections by the three mirrors.

You can see different designs in the kaleidoscope. The speciality of a kaleidoscope is that the designs do not easily repeat themselves. Every time the design is different. People making wall papers which are used to decorate walls and cloth designers use a kaleidoscope for making new designs.

Periscope:
Activity:

1. Take a cardboard box. Make slits in the top and bottom sides of the box and place two mirrors so that they make an angle of 45° with the sides of the box and are parallel to each other. Fix them with a sticking tape.
2. Make two windows of 1 inch each near the two mirrors. Now see through the bottom window.
3. Make note of what you see.

From the bottom window, one can see what is in front of the top window. This device is called a periscope. This is used in submarines to see objects above the surface of water. It is also used to observe and keep a watch on the objects or persons on the ground from an underground bunker. The kaleidoscope and periscope both use the properties of reflection of light.

[Note: In a periscope, the angle of incidence is 45° and the two plane mirrors are parallel to each other. Hence, the emergent ray is parallel to the incident ray.]

Example questions for practice:

Question 1.
If the angle between the plane mirror and the incident ray is 20°, what is the angle between the reflected ray
and the plane mirror?
Answer:
20°.

Question 2.
See Fig. In terms of O, what are the angles (i) AON (ii) BON (iii) AOB (iv) BOQ?

Answer:
(i) 90° – θ (ii) 90° – θ (iii) 180° – 2θ

Balbharati Maharashtra State Board 8th Std Science Textbook Solutions

• Human Body and Organ System Class 8 Science Textbook Solutions
• Introduction to Acid and Base Class 8 Science Textbook Solutions
• Chemical Change and Chemical Bond Class 8 Science Textbook Solutions
• Measurement and Effects of Heat Class 8 Science Textbook Solutions
• Sound Class 8 Science Textbook Solutions
• Reflection of Light Class 8 Science Textbook Solutions
• Man-made Materials Class 8 Science Textbook Solutions
• Ecosystems Class 8 Science Textbook Solutions
• Life Cycle of Stars Class 8 Science Textbook Solutions

Std 6 Science Chapter 14 Light and the Formation of Shadows Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows Notes, Textbook Exercise Important Questions and Answers.

Class 6 Science Chapter 14 Light and the Formation of Shadows Question Answer Maharashtra Board

1. Choose an appropriate word and fill in the blanks.

Question a.
A ……………. is a natural source of light.
Answer:
star

Question b.
A ………….. is an artificial source of light.
Answer:
candle

Question c.
When light passes through a prism, it gets separated into ………….. colours.
Answer:
seven

Question d.
The image obtained in the pinhole camera is …………. .
Answer:
inverted

Question e.
A shadow is formed when an …………… object comes in the way of light.
Answer:
opaque

Question f.
When a ………….. object comes in the way of light, light passes …………… it. options : seven, star, through, transparent, opaque, colors, shape, erect, inverted, luminous, candle.
Answer:
transparent, through

2. Write whether the following objects are luminous or non-luminous.

Question a.

Answer:

3. Match the following.

Question a.

Answer:

4. Write the answers to the following.

Question a.
What things are necessary for the formation of a shadow?
Answer:
Things necessary for the formation of a shadow are:

1. A source of light
2. An object
3. A surface or screen on which the shadow is formed

Question b.
When can an object be seen?
Answer:
We can see an object when reflected rays reach our eyes.

Question c.
What is a shadow?
Answer:

1. If an opaque object comes in the way of a light source, light does not pass through it.
2. As a result the light does not reach a wall or any other surface on the other side of the object.
3. That part remains dark. This dark part is called the shadow of the object.

Project:

Question 1.
Prepare a Newton’s disc.

Question 2.
Find out how to save electricity with the help of the sunlight we receive during the day.

Question 3.
Read a biography of Sir C. V. Raman and find out about the discoveries he made.

Class 6 Science Chapter 14 Light and the Formation of Shadows Important Questions and Answers

Fill in the blanks.

Question 1.
The ……………. is the main natural source of light.
Answer:
Sun

Question 2.
The light emitted by an electric torch is more …………….. than that obtained from a candle.
Answer:
intense

Question 3.
The left and right sides of the original object appear to be …………….. in the mirror.
Answer:
exchanged

Question 4.
The image is as far behind the mirror as the object is in ……………… of it.
Answer:
front

Question 5.
The …………….. of the image is the same as that of the object.
Answer:
height

Question 6.
The materials through which light passes is said to be …………… .
Answer:
transparent

Question 7.
The materials through which light does not pass is said to be ………….. .
Answer:
opaque

Question 8.
The materials through which light passes partially is said to be ………….. .
Answer:
translucent

Question 9.
If an ………….. object comes in the way of a light source, light does not pass it.
Answer:
opaque, through

Question 10.
The shadow of an object formed due to sunlight is ………….. in the mornings and evenings and ……………. in the afternoon.
Answer:
long, short

Question 11.
The shadow of an object is formed only when ………….. does not pass through the object.
Answer:
light

Question 12.
Stars are ………….. .
Answer:
luminous

Question 13.
Planets, satellites are ………….. .
Answer:
non-luminous

Question 14.
The largest sundial is at ………….., New Delhi.
Answer:
Jantar Mantar

Question 15.
………….. is celebrated as National Science Day.
Answer:
28th February

Question 16.
Light travels in a straight line. This is called ………….. .
Answer:
linear propagation of light

Question 17.
The image formed on the diaphragm of the pinhole camera is ………….. .
Answer:
inverted

Question 18.
The kind of shadow an object forms depends upon the ………….. between the ………….., the object and the ………….. or the ………….. on which the shadow is formed.
Answer:
relative distance, source of light, surface, screen

State whether following statements are True or False.

Question 1.
Light travels in a straight line.
Answer:
True

Question 2.
Stars are luminous.
Answer:
True

Question 3.
Image in a pinhole camera is inverted.
Answer:
True

Question 4.
In the afternoon, shadows are long.
Answer:
False

Question 5.
Fireflies are a natural source of light.
Answer:
True

Question 6.
We see the candle clearly when we bend the tube.
Answer:
False

Question 7.
We can see our image clearly in running water.
Answer:
False

Question 8.
Tracing paper is transparent.
Answer:
False

Question 9.
The light obtained from an electric torch is more intense than that obtained from a candle.
Answer:
True

Question 10.
28tn February is celebrated as “National Science day” since 1987 in India.
Answer:
True

Question 11.
Classify the following into natural and man-made/artificial sources of light. (tubelight, light bulb, torch, burning candle, the sun, fireflies, anglerfish, honey mushroom, stars in the night sky, oil lamps, lanterns)
Answer:

Question 12.
Identify the transparent, opaque and translucent objects from among the following. (piece of glass, wax paper, tinted glass, oil paper, white plastic, a tea kettle, a notebook, cloth, water, a wooden cupboard, sheet of notebook.)
Answer:

Question 13.
Classify the following into the type of images they form: Clear image, faint image, no image. (still clear water, cemented wall, wooden surface, new steel dish, flower, glossy granite cladding of a wall, mirror, butter paper).
Answer:

Question 14.
Relate images formed with the surfaces.
Answer:

1. The clear images are formed on plane surfaces.
2. Faint or no images are formed on rough surfaces.

Answer in one sentence.

Question 1.
What are luminous objects?
Answer:
The objects which emit light i.e. which themselves are a source of light, are called luminous objects.

Question 2.
What determines the intensity of light?
Answer:
The intensity of light is determined by the extent to which the objects emit light.

Question 3.
What are non-luminous objects?
Answer:
The objects that are not sources of light themselves are called as non-luminous objects.

Question 4.
What are artificial sources of light?
Answer:
Man-made objects which emit light are artificial sources of light.

Question 5.
What are natural sources of light?
Answer:
Natural substances, materials which emit light are called natural sources of light.

Question 6.
What is linear propagation of light?
Answer:
Property of light travelling in a straight line is linear propagation of light.

Question 7.
What is reflection of light?
Answer:
The rays of light falling on an object from a source of light are thrown back from the substance of that object. This is reflection of light.

Question 8.
How do we see objects around us?
Answer:
The rays of light falling on an object from a source of light are thrown back from the surface of that object. This is called reflection of light. We see the object when the reflected rays reach our eyes.

Question 9.
What is moonlight?
Answer:
Sunlight reflected from the surface of the moon reaching us, in which we see the moon is called the moonlight.

Question 10.
What type of image is formed in the mirror?
Answer:
The image formed in the mirror is ‘laterally inverted’ i.e. right side appears as left side and left side appears as right side.

Question 11.
What change do you see in the image if you decrease or increase your distance from the mirror?
Answer:
When the distance between object and mirror is increased the size of image decreases where as, when the distance is decreased the image size increases.

Question 12.
What difference do you find in the height of the image in the mirror and yourself?
Answer:
The size of the image in the mirror is the same as that of the object.

Question 13.
What is the image on the diaphragm of the pinhole camera?
Answer:
An inverted or an upside down image of the candle is seen on the diaphragm of the pinhole camera.

Question 14.
What do you mean by transparent object?
Answer:
The objects / materials through which light passes are said to be transparent.

Question 15.
What do you mean by opaque materials?
Answer:
The materials through which light does not pass are said to be opaque.

Question 16.
What do you mean by translucent materials?
Answer:
The materials through which light passes partially are said to be translucent.

Question 17.
How is the shadow in the morning, afternoon and evening?
Answer:
The shadows are long in the mornings and evenings and short in the afternoon.

Question 18.
What is shade of a tree?
Answer:
The shade of a tree is its shadow.

Question 19.
How many colours is sunlight made up of?
Answer:
Sunlight is made up of seven colours.

Give reasons for the following.

Question 1.
When we see in the mirror, we see our image in the mirror.
Answer:
When we see our face in the mirror, the light reflected from our face falls on the mirror and gets reflected back again. Hence, we see our image in the mirror.

Question 2.
Opaque materials cast shadow.
Answer:
An opaque material does not allow light to pass. Hence, it casts a shadow.

Question 3.
Transparent and translucent object do not cast a shadow.
Answer:
Translucent objects cast a faint shadow whereas transparent objects do not cast a shadow at all because they allow light to pass through them.

Use your brain power!

Question 1.
Why is the image on the diaphragm of the pinhole camera inverted?
Answer:
1. The pinhole camera works on the principle of light travelling in a straight line.
2. The rays of light from the candle flame go in all directions.
3. We consider only two rays that pass through the hole and fall on the screen.

4. The rays intersect at the pinhole.
5. Since the rays cross over at that point, the top of the object appears at the bottom of the image and the bottom of the image appears at the top. Thus, we see an inverted image of the candle.

Question 2.
How will you light up a dark room using reflected light?
Answer:
Focusing on the wall with torch light. Mirrors or reflectors can be used to get light from outside.

Question 3.
Try to start the TV by operating the remote control from behind it.
Answer:
T.V will not start.

Question 4.
In which step is the flame of the candle seenclearly? Why?

Answer:
In step 1 the flame of the candle is seen clearly because light travels in straight line.

Can you tell?

Question 1.
Can we see anything in total darkness?
Answer:
No, we cannot see anything in total darkness.

Question 2.
What helps us to see the objects around us?
Answer:
Reflected light helps us to see objects around us.

Question 3.
What does the light in each one of the pictures originate form?
Answer:

1. Bulb
2. Firefly
3. Candle
4. Sun

Question 4.
Name the natural sources of light.
Answer:
Sun, Fireflies

Question 5.
In which objects do we see our reflection?
Answer:
All objects reflect light rays, but the best reflectors of light are mirrors, still water in a lake, new steel dish i.e. smooth shiny surfaces.

Question 6.
What difference do you notice on looking through the windows in the picture? What causes the difference? The picture shows transperant, opaque, translucent window panes. Spot them.

Answer:

1. Through the first window we can see a clear picture of things outside.
2. Second window gives a faint image.
3. Through the third window, we can’t see anything
4. The difference in the image is due to the material of the window panes.
5. The first window pane is transparent.
6. The second window pane is translucent.
7. The third window pane is opaque.

Answer the following questions in brief.

Question 1.
List factors on which shadow depends.
Answer:
Shadow depends on relative distance between the source of light, the object and the surface on which the shadow is formed.

Question 2.
How we can see that light travels in straight line.
Answer:

1. In the morning or in the afternoon, rays of light enter a slit in a door, window or a small hole in the roof.
2. As these rays of light from the slit or the hole move towards the floor, the dust particles in their way are clearly seen.
3. Due to these particles, the path of light becomes visible to us.
4. Thus we can see that their path is along straight lines.

Question 3.
What is the difference between an object and its reflection? What causes the difference?
Answer:

1. Object and its reflection result in formation of images.
2. Reflections taking place from highly polished metals, mirrors, still water etc, form clear images.
3. Reflections taking place from wooden surface, flower, book form dull, blurred images.
4. The difference in reflections is caused by the surface of the object.
5. Regular reflections have smooth, polished surfaces, hence, image is clear.
6. Diffused reflections have hard, rough surfaces, hence, image is dull.

Question 4.
List characteristics of images in a plane mirror.
Answer:

1. The left and right sides of the original object appear to be exchanged in the mirror image.
2. The image is as far behind the mirror as the object is in front of it.
3. The size of the image is the same as that of the object.

Question 8.
State the characteristics of image formed by a pinhole camera.
Answer:
Characteristics of an image formed by a pinhole camera are as follows:

1. It is inverted/upside down.
2. It can be obtained on a screen – real image.

Try this.

Question 1.
Make your friend stand in between the torch and the wall. What happens?
Answer:
Friend’s Shadow forms on the wall.

Question 2.
Place a glass filled with water on a sheet of paper in the window so it receives direct sunlight. What is seen on the paper?
Answer:
We see rainbow colours on the paper.

Question 3.
Can we do the same in a dark room with the help of a prism and a torch? What do we learn from this?
Answer:
yes, we can. Light gets seperated into seven colour. From this we learn that white light contains seven colours.

Question 4.
If you dip the wire loop in the soap water and then blow it, soap bubbles are formed. Are the beautiful colours of the rainbow seen in these bubbles?
Answer:
Yes, splitting of white light into different colours takes place.

Question 5.
What do you see on holding a CD in the sun?
Answer:
CD reflects rainbow colours, and interesting : patterns.

Question 6.
Raise your right hand. In mirror which hand of the mirror image is raised?
Answer:
Left hand of the mirror image is raised.

Question 7.
Is there any difference between your height and height of the mirror image?
Answer:
The height remains the same.

Maharashtra State Board Class 6 Science Textbook Solutions

• Work and Energy Class 6 Science Textbook Solutions
• Simple Machines Class 6 Science Textbook Solutions
• Sound Class 6 Science Textbook Solutions
• Light and the Formation of Shadows Class 6 Science Textbook Solutions
• Fun with Magnets Class 6 Science Textbook Solutions
• The Universe Class 6 Science Textbook Solutions

Std 6 Science Chapter 8 Our Skeletal System and the Skin Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 6 Science Solutions Chapter 8 Our Skeletal System and the Skin Notes, Textbook Exercise Important Questions and Answers.

Class 6 Science Chapter 8 Our Skeletal System and the Skin Question Answer Maharashtra Board

1. Fill in the blanks with the proper word. 

Question a.
The place where two or more bones are connected is called a …………… .
Answer:
joint

Question b.
Cells of epidermis contain a pigment called ………… .
Answer:
melanin

Question c.
…………… and ……………. are the two layers of the human skin.
Answer:
epidermis, dermis

Question d.
The human skeletal, system is divided into ………….. parts.
Answer:
two

2. Match the pairs.

Question a.

Answer:

3. Right or wrong? If wrong, write the correct sentence. 

Question a.
Bones are soft.
Answer:
Wrong: Bones are hard.

Question b.
The human skeleton protects the internal organs.
Answer:
Right

4. Put a [✓] mark at the proper places.

Question a.
The system which gives our body. a definite shape to
(a) Excretory system [ ]
(b) Respiratory system [ ]
(c) Skeletal system [ ]
(d) Circulatory system [ ]
Answer:
(c) Skeletal system [✓]

Question b.
The ………… joint is seen in fingers and toes.
(a) Hinge joint [ ]
(b) Ball and socket joint [ ]
(c) Immovable joint [ ]
(d) Gliding joint [ ]
Answer:
(a) Hinge joint [✓]

5. Answer the following questions in your words.

Question a.
What are the functions of your skin?
Answer:
The functions of the skin are :

1. Protects the internal parts of the body like muscles, bones, organ systems etc.
2. Help to preserve the moisture in the body.
3. Synthesizing Vitamin D.
4. Regulates body temperature by releasing sweat.
5. Gives protection from heat and cold.
6. Functions as the sensory organ of touch

Question b.
What should you do to keep the bones strong and healthy?
Answer:
To keep the bones strong and healthy:

1. We should include calcium and phosphrous rich food in our diet.
2. We also include vitamin D rich food in our diet.
3. We get these from milk, milk products, leafy vegetables, meat and exposure to sunlight.
4. We should exercise regularly.

Question c.
What are the functions of human skeletal system?
Answer:
The functions of human skeletal system are:

1. Gives a definite shape to the body.
2. Provides support to the body.
3. Protects the delicate organs inside the body.

Question d.
Which are the various reasons due to which our bones might break?
Answer:
The bones in our body might break due to:

1. Lack of calcium and phosphorous.
2. Due to fracture because of an accident or a fall.
3. Lack of proper nutrition.
4. Due to deficiency of Vitamin D.

Question e.
What are the different types of bones? How many types are there? Give example of each.
Answer:
There are four types of bones in our body:

Types of Bone Example

1. Flat bones Sternum in the chest
2. Small bones Stirrup in each ear
3. Irregular bones Vertebra posterior (back side) of the body
4. Long bones Femur or thigh bone

6. What will happen if?

Question a.
There are no joints in our body.
Answer:
If there are no joints in our body, we will be standing like a tree without any movement, we can move only because of joints.

Question b.
There is no melanin pigment in our body.
Answer:
We will not have protection from ultraviolet rays. Our skin will become whitish.

Question c.
Instead of 33 vertebras in our body, we had one single and straight bone.
Answer:
We cannot bend down at our will.

7. Draw diagrams.

Question a.
Types of joints.
Answer:

Question b.
Structure of skin
Answer:

Activity:

Question 1.
Collect pictures of the different parts of the human skeletal system and paste them on chart paper. Write the functions of
each, too.

Question 2.
Collect the pictures, newspaper cuttings, etc. which show the skeletal systems of various animals and observe the differences between them.

Class 6 Science Chapter 8 Our Skeletal System and the Skin Important Questions and Answers

Choose the correct alternative:

Question 1.
The ………………….. protects the brain.
(a) skull
(b) rib cage
(c) spine
(d) none of above
Answer:
(a) skull

Question 2.
X-rays were discovered by …………………..  .
(a) Sir C.V.Raman
(b) Galileo
(c) Sir Isaac Newton
(d) Roentgen
Answer:
(d) Roentgen

Question 3.
The bones of ………………….. are immovable.
(a) hand
(b) leg
(c) spine
(d) skull
Answer:
(d) skull

Question 4.
We can move the bones of ………………….. in a 360° angle.
(a) elbow
(b) knee
(c) shoulder
(d) wrist
Answer:
(c) shoulder

Question 5.
Our body temperature usually remains constant at ………………….. °C.
(a) 32
(b) 35
(c) 37
(d) 40
Answer:
(c) 37

Question 6.
The part of the skin which maintains body temperature is ………………….. .
(a) Epidermis
(b) Dermis
(c) Subcutaneous layer
(d) Network of blood vessels and nerve fibers
Answer:
(c) Subcutaneous layer

Fill in the blanks:

Question 1.
Except for the …………….., none of the bones of the skull can move.
Answer:
lower jaw

Question 2.
The spinal cord originates from the …………… .
Answer:
brain

Question 3.
The longest and the strongest bone in the human body is …………… .
Answer:
femur

Question 4.
…………….. is the smallest bone in our body.
Answer:
stirrup

Question 5.
The vertical, flat bone in the chest is called the ……………… .
Answer:
sternum

Question 6.
The …………… joint moves in a 180° angle.
Answer:
hinge

Match the columns:

Question 1.

Answer:

Question 2.

Answer:

Right or Wrong? If wrong, write the correct answers:

Question 1.
The spine is a part of the appendicular skeleton.
Answer:
Wrong: The spine is a part of the axial skeleton.

Question 2.
Ankle joints are gliding joints.
Answer:
Right

Question 3.
All the skull joints are immovable.
Answer:
Wrong: Except lower jaw, all the skull joints are immovable.

Question 4.
The skin maintains normal body temperature.
Answer:
Right

Answer in one word:

Question 1.
The part which protects the heart and lungs.
Answer:
Rib cage

Question 2.
Ali falls down and his elbow is broken.
Answer:
Fracture

Question 3.
The image which spots the broken bone.
Answer:
X-ray

Question 4.
The biotic component of our body.
Answer:
Bone cell

Question 5.
The part which protects the vertebral column.
Answer:
Spine

Question 6.
The part which connects the bone in our body.
Answer:
Ligament

Question 7.
The organ which helps us to sense whether something is hot or cold.
Answer:
Skin

Question 8.
The pigment which gives colour to the skin.
Answer:
Melanin

Can you tell?

Question 1.
What is a fracture? How will you help a friend who has met with an accident and fractured his leg?
Answer:
Fracture is a crack or break in a bone. Fracture may occur due to accident or fall from height or injury. If my friend’s leg is fractured, then. I would –

1. Ask him to prevent any movement of the fractured part.
2. Get immediate medical help.
3. Take the x-ray image of the fractured or swollen part.
4. An x-ray image shows the exact spot where the bone is broken. This will help in providing proper treatment.

Question 2.
What are the properties of bones?
Answer:
The properties of bones are:

1. Bones are hard and not flexible.
2. Bone cells are composed of two main constituents:
   • Bone cells are biotic components.
   • Calcium carbonate, calcium phosphate, minerals and salts are abiotic components.
3. Calcium imparts strength to bones.
4. As we grow the size and length of bones increases upto a certain limit.

Question 3.
What is human skeletal system? How is it divided?
Answer:

1. All the bones together form a framework or a skeleton.
2. All the bones of the body along with cartilage together form the skeletal system.
3. The human skeletal system can be divided into two parts – the axial skeleton and the appendicular skeleton.
4. The axial skeleton consists of the skull, the spine and the rib cage. These are situated symmetrically along the central axis.
5. The appendicular skeleton consists of the bones of arms and legs on either side of the central axis.

Question 4.
What is a joint? What are it’s types?
Answer:
Joints are places where two or more than two bones are connected to each other.

Question 5.
Describe the structure of skin.
Answer:

1. Human skin is made up of two main layers- outermost layer, epidermis and layer below it called dermis.
2. Below dermis there is a network of blood vessels and nerve fibers.
3. Under this layer there is a subcutaneous layer, which maintains body temperature.
4. The epidermis has various layers.
5. There are sweat glands in the skin which secrete sweat.

Question 6.
Write a short note on melanin.
Answer:

1. Melanin is a pigment present in the cells of epidermis.
2. Melanin is synthesized in certain glands in the skin.
3. The percentage of melanin decides the fairness or darkness of the skin.
4. Melanin protects our skin and the inner parts from ultraviolet sunrays.

Use your brain power!

Question 1.
Which colour of the skin will give greater protection from sun’s rays?
Answer:
Darker colour will give greater protection.

Give scientific reasons:

Question 1.
We are able to bend down at our will.
Answer:

1. 33 bones of the spine are placed straight one above the other.
2. They are arranged flexibly.
3. Their flexibility allows us to bend down at our will.

Question 2.
Calcium is an important mineral.
Answer:

1. Calcium imparts strength to our bones.
2. If we are calcium deficient, possibility of bone fracture during a fall or an accident increases.
3. Hence, calcium is an important mineral.

Question 3.
Sweating helps to lower the body temperature.
Answer:

1. In the hot sun, the temperature of the skin rises.
2. The sweat is released.
3. The heat required for the evaporation of sweat is drawn from the body itself.
4. Hence, sweating lowers the body temperature.

Question 4.
Some people have jet black hair, while others have brown or reddish hair.
Answer:

1. It is melanin that determines the colour of our hair.
2. Jet black hair is due to pure melanin.
3. Brown hair is due to sulphur in the melanin.
4. Reddish hair is due to iron in the melanin.

Question 5.
Observe and discuss:
Your grandmother has wrinkles on her skin.
Answer:

1. As we grow older, the proportion of fat beneath the skin reduces.
2. However, previously tout skin does not shrink.
3. This causes wrinkles on the face of older people.

What will happen if:

Question 1.
If skin had no sweat glands.
Answer:
Skin regulates body temperature by releasing sweat. If skin had no sweat glands then we will not be able to maintain our body temperature at a constant 37°C.

Can you recall?

Question 1.
Which organ help us to sense whether something is hot or cold, rough or smooth, etc?
Answer:
The skin functions as the sensory organ of touch.

Question 2.
What happens when we walk or play in the hot sun?
Answer:
When we walk or play in the sun, we get tired, our skin secrete sweat.

Observe the figure and label as directed.

Question 1.

In the above figure label the parts marked a, b, c, d, e and hence show axial and appendicular
skeleton.
Answer:
(a) Skull
(b) Rib cage
(c) Spine
(d) Arm
(e) Leg
Axial skeleton: skull, the spine and rib cage
Appendicular: legs, arms

Question 2.

In the given figure, name the type of bones marked a, b, c, d. State where they are seen in our body.
Answer:
(a) Flat bones – rib cage and gliding joint
(b) small bones – finger
(c) Irregular bones – vertebal column
(d) Long bones – legs, arms

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Maharashtra State Board Class 6 Science Textbook Solutions

• Natural Resources – Air, Water and Land Class 6 Science Textbook Solutions
• The Living World Class 6 Science Textbook Solutions
• Diversity in Living Things and their Classification Class 6 Science Textbook Solutions
• Disaster Management Class 6 Science Textbook Solutions
• Substances in the Surroundings – Their States and Proper Class 6 Science Textbook Solutionsties
• Substances in Daily Use Class 6 Science Textbook Solutions
• Nutrition and Diet Class 6 Science Textbook Solutions
• Our Skeletal System and the Skin Class 6 Science Textbook Solutions
• Motion and Types of Motion Class 6 Science Textbook Solutions
• Force and Types of Force Class 6 Science Textbook Solutions

Std 6 Science Chapter 6 Substances in Daily Use Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 6 Science Solutions Chapter 6 Substances in Daily Use Notes, Textbook Exercise Important Questions and Answers.

Class 6 Science Chapter 6 Substances in Daily Use Question Answer Maharashtra Board

1. Fill in the blanks using proper works:

Question a.
Rubber made by vulcanization is a …………… material.
Answer:
hard

Question b.
Man-made materials are made by …………… natural materials.
Answer:
processing

Question c.
…………… thread was developed simultaneously in New York and London.
Answer:
Nylon

Question d.
Rayon is also known as …………… .
Answer:
synthetic silk

2. Answer the following questions. 

Question a.
Why did the need for man-made materials arise?
Answer:
The need for man-made materials arose due to the following reasons:

1. To meet the needs of an increasing population.
2. Human nature to try to make life more comfortable.
3. They can be made available in plenty at a low cost.
4. The reserve of natural substances is decreasing.

c

Question b.
Which are the natural materials obtained from plants and animals?
Answer:
Leather, jute, wool, cotton, silk are the natural substances obtained from plants and animals.

Question c.
What is vulcanization?
Answer:

1. Vulcanization is the process in which rubber is heated with sulphur for three to four hours.
2. Sulphur is mixed to give hardness to rubber.
3. The proportion of sulphur depends on the purpose for which the rubber is to be used.

Question d.
Which natural materials are used to obtain fibres?
Answer:
Cotton, wood pulp and various hydrocarbons obtained from mineral oils are used to obtain fibres.

3. What are we used for?

Question a.
What are we used for?
Answer:

1. Soil: It supports plant life and hence indirectly supports all living things. It is used for making clay pot, utensils, bricks etc.
2. Wood: It is used in paper industry. It is also used to make furniture.
3. Nylon: It is used to manufacture clothes, fishing nets, ropes, etc.
4. Paper. It is used in our textbooks, note books, currency notes, etc.
5. Rubber: It is used in the manufacture of erasers, tyres, rubber toys, rubber bands, etc.

4. How is paper manufactured? Write in your own words.

Question a.
How is paper manufactured? Write in your own words.
Answer:
Coniferous trees like pine trees are used to make paper.

1. The bark of the logs of these trees is first removed and the wood is broken into small pieces.
2. The mixture of these pieces with some chemicals is kept soaked for a long time to form pulp.
3. On completion of chemical process, fibrous substances from wood pulp are separated and some dyes are added.
4. The pulp is then passed through rollers, dried to form paper and finally wound on reels.

5. Give scientific reasons.

Question a.
We must use cotton clothes during summer.
Answer:

1. During summer we sweat more due to high temperature.
2. Cotton clothes absorb sweat.
3. Synthetic clothes are water repellent. They do not absorb sweat and we feel uncomfortable. Hence we must use cotton clothes in summer.

Question b.
We must observe economy in the use of materials.
Answer:

1. Due to excessive use of natural substances by human beings to fulfil their needs; they are getting depleted at a faster rate.
2. At the same time, it takes a very long time for these substances to get naturally formed again.
3. Hence, we must observe economy in the use of materials so that they are available for the future generation also.

Question c.
Saving paper is the need of the hour.
Answer:

1. Saving paper means saving trees as wood is used as the raw material to manufacture paper.
2. Trees are natural habitat for many Living things.
3. Trees help in increasing rainfall and water availability. Hence, saving paper helps in saving trees which in turn maintains balance in nature.

Question d.
Man-made materials have more demand.
Answer:

1. Man-made substances are waterproof, lightweight and easy for transportation.
   Substances in Daily Use
2. They are easier to use and can be made available in plenty at a low cost.
   Hence, there is more demand for man-made materials.

Question e.
Humus is a natural material.
Answer:

1. Humus is obtained from plant and animal wastes.
2. Micro-organisms act on these wastes and convert them into humus.
3. Hence, humus is a natural material.

6. Find out.

Question 1.
How is lac obtained from nature?
Answer:

1. Lac is a resinous substance secreted from the glands present in the skin of female lac insect.
2. Lac insects live on the Palash trees. In India lac is mainly produced in the states of Rajasthan and Bihar.

Question 2.
How are pearls obtained?
Answer:

1. Pearls are formed when a foreign particle such as a grain of sand or a small particle of rock accidentally enters the space between the mantle and shell of an oyster’s body.
2. Oysters cannot reject the particle, and as a defence mechanism its produces a shining coating called nacre on the particle layer by layer.
3. As the shiny layers get added, a pearl is formed.
4. Cultured pearls are artificially formed by inserting a bead in oyster shell and allowed to coat it with nacre over several years.

Activity:

Question 1.
Visit a rubber, paper or textile industry in your area and collect information about it.

Question 2.
Collect various samples of paper and note their uses.

Question 3.
Use blank pages from old note-books and make a new one.

Class 6 Science Chapter 6 Substances in Daily Use Important Questions and Answers

Fill in the blanks using proper works.

Question 1.
Natural rubber is obtained from …………… of trees.
Answer:
latex

Question 2.
Changes where the original constituent substances cannot be obtained again from the new substances are called …………… changes.
Answer:
irreversible

Question 3.
A paper factory in Maharashtra is situated at …………… .
Answer:
Ballarpur

Question 4.
Glass can be made from …………… and …………… .
Answer:
sand, calcium

Question 5.
Botanical name of rubber plant is …………… .
Answer:
Hevea brasiliensis

Question 6.
…………… obtained from mineral oils are used to make polymer chains.
Answer:
Hydrocarbons

Question 7.
The maximum production of rubber in India is in …………… .
Answer:
Kerala

Question 8.
…………… invented the process of vulcanisation.
Answer:
Charles Goodyear

Match the columns:

Question 1.

Answer:

State whether the following statements are ‘true’ or ‘false’.

Question 1.
We can find plastic in nature.
Answer:
False

Question 2.
Soap is a man-made substance.
Answer:
True

Question 3.
We should reuse available resources.
Answer:
True

Question 4.
Nylon clothes are good summer wear.
Answer:
False

Question 5.
Glass is a man-made substance.
Answer:
True

Question 6.
In irreversible changes original substances can be obtained again.
Answer:
False

Question 7.
Rayon is made up of cotton and wood pulp.
Answer:
True

Give two examples of each of the following:

Question 1.
Natural fibres
Answer:
cotton, silk

Question 2.
Synthetic fibres
Answer:
terylene, rayon

Question 3.
Biotic natural substances
Answer:
wool, jute

Question 4.
Abiotic natural substances
Answer:
air, water

Question 5.
Man-made substances.
Answer:
paper, glass

Classify the following substances in the table given below.
(iron, wood, brick, paper, terylene, stone, jute, air, silk, utensils, plastic, rayon, water, wool, dacron, lac, nylon, pearl)
Answer:

Define:

Question 1.
Natural substances.
Answer:
Substances available in nature are called natural substances.

Question 2.
Man-made substances.
Answer:
Man-made substances are new substances produced by processing naturally available resources.

Question 3.
Biotic substances.
Answer:
Natural substances obtained from living things are called biotic substances.

Question 4.
Abiotic substances.
Answer:
Natural substances that are not obtained from living things are called abiotic substances.

Question 5.
Plant-originated substance.
Answer:
A substance obtained from a plant is called a plant-originated substance.

Question 6.
Animal-originated substance.
Answer:
A substances obtained from an animal is called an animal-originated substance.

Question 7.
Hydrocarbons.
Answer:
Substances obtained from mineral oil are called hydrocarbons.

Answer the following in one or two sentences.

Question 1.
Why was Rayon named so?
Answer:
The threads of Rayon have shine and strength. They appeared to be shining bright like the sun’s rays. Hence, they were named ‘Rayon.

Question 2.
How are TV sets, refrigerators, etc. packed? Why?
Answer:
To pack TV sets, refrigerators, etc. big cartons and thermocol are used. These man-made substances are water resistant, light weight and easy for transportation.

Question 3.
Give the properties and uses of nylon.
Answer:
Nylon threads have a shine and are strong, transparent and water resistant. They are used to manufacture clothes, fishing nets, ropes, etc.

Question 4.
What is latex?
Answer:
Latex is a milky white natural substance produced in the stems of rubber trees.

Question 5.
Name the basic material used to obtain paper.
Answer:
Wood is the basic material used in the manufacture of paper.

Question 6.
What kind of paper is used for currency notes manufacturing?
Answer:
Flax fibre is used in the manufacture of currency notes.

Question 7.
Where was the process of making paper invented?
Answer:
The process of making paper was invented in China.

Answer in brief:

Question 1.
What are the advantages of synthetic fibre?
Answer:
Advantages of synthetic fibre are:

1. These fibres can be manufactured on a large scale.
2. They cost less.
3. They are strong and durable.
4. They can be used for a long time.
5. They are water repellent. They dyy easily.
6. They are light weight and comfortable to wear.
7. Clothes made from these threads are wrinkle free and scratch free.

Question 2.
Give the shortcomings of synthetic fibre.
Answer:

1. They are water repellent. Hence, they do not absorb sweat from the skin.
2. Continuous use of these clothes keeps the skin moist which may cause skin diseases.
3. Synthetic clothes are uncomfortable to wear especially in summer.
4. They catch fire easily.
5. If they catch fire, they stick to the skin and cause skin injuries.
6. These fibres are not decomposed by micro-organisms.

Question 3.
Write a short note on natural rubber.
Answer:

1. Rubber is a natural substance obtained by collecting the latex of certain trees.
2. The botanical name of this tree is ‘Hevea brasiliensis’
3. In India, the maximum production of rubber is in Kerala.

Question 4.
What are dacron, terylene and terene?
Answer:

1. Dacron, terelyne and terene are synthetic fibres prepared from hydrocarbons.
2. Various hydrocarbons obtained from mineral oil are used to make polymer chains.
3. A solution of such polymer is pressed through a strainer with fine holes.
4. The fibre formed after cooling are long and unbroken threads.
5. These threads have been named as dacron, terylene and terene.

Give scientific reasons:

Question 1.
Natural substances are depleting.
Answer:

1. Due to increase in population there is an increase in demand. To meet this demand, natural substances are used to a greater extent.
2. Due to human nature to make his life more comfortable, he learnt to use natural resources and also began to process them to make new substances. Hence natural substances are depleting at an alarming rate.

Can you tell?

Question 1.
Difference between leather, jute, wool, cotton and soil, water, metals.
Answer:

1. Leather, jute, wool are biotic natural substances.
2. Soil, water, metals are abiotic natural substances.

Question 2.
How are leather and wool different from jute and cotton?
Answer:
Leather and wool are obtained from animals while jute and cotton are obtained from plants.

Question 3.
Do you find plastic, nylon, brass or cement in nature?
Answer:
No, they are all man-made materials.

Question 4.
Can red chillies become green chillies again?
Answer:
No, the change from green chillies to red chillies is irreversible.

Question 5.
From which substances in nature can we get threads or fibre?
Answer:
Cotton plant, jute, silkworm.

Question 6.
What are clothes made from?
Answer:
Clothes are made from yarn obtained from fibre.

Classify the following substances according to their uses:
sand, soap, wool, window glass, bamboo, cotton, bricks, silk, leafy vegetables, cement, fruits, water, sugar.
Answer:

• For construction: Sand, window glass, bamboo, bricks, cement.
• As food: Leafy vegetables, fruits, water, sugar.
• At home: Soap for cleaning.
• For clothes: Wool, cotton, silk.

Make a list of objects, each of which can be made from several substances.
Answer:

Use your brain power!

Question 1.
Complete the table below, showing how substance of daily use are classified.

Answer:

1. Natural
2. Biotic
3. Cement
4. Animal Origin
5. Cotton

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Maharashtra State Board Class 6 Science Textbook Solutions

• Natural Resources – Air, Water and Land Class 6 Science Textbook Solutions
• The Living World Class 6 Science Textbook Solutions
• Diversity in Living Things and their Classification Class 6 Science Textbook Solutions
• Disaster Management Class 6 Science Textbook Solutions
• Substances in the Surroundings – Their States and Proper Class 6 Science Textbook Solutionsties
• Substances in Daily Use Class 6 Science Textbook Solutions
• Nutrition and Diet Class 6 Science Textbook Solutions
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• Motion and Types of Motion Class 6 Science Textbook Solutions
• Force and Types of Force Class 6 Science Textbook Solutions

Std 7 Science Chapter 7 Motion, Force and Work Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 7 Motion, Force and Work Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 7 Motion, Force and Work Question Answer Maharashtra Board

1. Fill ¡n the blanks with the proper words from the brackets.
(stationary, zero, changing, constant, displacement, velocity, speed. acceleration, stationary but not zero. inc reuses)

Question a.
If a body traverses a distance in direct proportion to the time, the speed of the body is ……………… .
Answer:
constant

Question b.
If a body is moving with a constant velocity, its acceleration is ……………… .
Answer:
zero

Question c.
……………. is a scalar quantity.
Answer:
Speed

Question d.
…………….. is the distance traversed by a body in a particular direction in unit time.
Answer:
Velocity

2. Observe the figure and answer the questions.

Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?

Question a.
Observe the figure and answer the questions

Sachin and Sarneer started on a motorbike from place A, took the turn at 13, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE’? Can this velocity be called average velocity?
Answer:
1. Actual distance = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\) + \(\overline{\mathrm{CD}}\) + \(\overline{\mathrm{DE}}\) = 3 + 4 + 5 + 3
Actual distance = 15 km

2. Displacement = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BD}}\) + \(\overline{\mathrm{DE}}\)
= 3 + 3 + 3
Displacement = 9 km

3. Speed = \(\frac{\text { Distance travelled }}{\text { Total time }}\)
Distance = 15 km = 15 × 1000 = 15000 m
Time = 1 hr = 1 × 60 × 60 = 3600 sec.
s = \(\frac{15000}{3600}\) or s = \(\frac{15 \mathrm{~km}}{1 \text { hour }}\) = 15km/hour
= 4.16 m/sec. or 15 km/hour

4. Velocity = \(\frac{\text { Distance travelled }}{\text { Total time }}\)
Displacement = 9 km = 9 × 1000 = 9000 m
Time = 1 hr = 1 × 60 × 60 = 3600 sec
V = \(\frac{9000}{3600}\) or V = \(\frac{9 \mathrm{~km}}{1 \text { hour }}\) = 9 km/hour
= 2.5 m/sec. or 9 km/hour

5. Yes, this velocity can be called as average velocity.

3. From the groups B and C, choose the proper words, for each of the words in group A.

Question a.
From the groups B and C, choose the proper words, for each of the words in group A.

Answer:

4. A bird sitting on a wire, flies, circles around and comes back to its perch. Explain the total distance it traversed during its flight and its eventual displacement.

Question a.
Answer:
The total distance the bird has traversed is the length of the distance covered by circling, but the eventual displacement are the bird is zero as its initial and final position are one and the same.

5. Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.

Question a.
Explain the following concepts in your own words with everyday examples: force, work, displacement, velocity, acceleration, distance.
Answer:
1. Force: The interaction that brings about the acceleration is called force.
e.g: An ox is pulling a cart, applying brakes to a bicycle, lifting heavy iron object with a crane.

2. Work: When an object is displaced by applying a force on it, work is said to be done.
e.g: A bucketful of water is to be drawn from a well and taken to the home by walking from well to home.

3. Displacement: The minimum distance
traversed by a moving body in one direction from the original point to reach the final point is called displacement.
e.g: A rolling of a ball from point A to point B in the same direction.

4. Velocity: Velocity is the distance traversed by a body in a specific direction in unit time.
e.g: A truck is covering a distance of 40km from A to D in a straight line in 1 hour.

5. Acceleration: It is change in velocity per second. It can be deduced.
Acceleration = \(\frac{\text { Change in velocity }}{\text { Time taken for change }}\)
e.g:

(i) In the above example a truck covered the distance AB at velocity of 60 km/hr, BC at 30 km/hr and CD at 40 km/hr. (ii) It means that the velocity for the distance CD is greater than the velocity for the distance BC. (iii) From the number of seconds required for this change in velocity to take place, the change in velocity per second can be deduced. This is called acceleration (iv) Distance: The length of the route actually traversed by a moving body irrespective of the direction is called distance.
e.g: Ranjit travelled 1km. from his home to school.

6. A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.

Question a.
A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C.

Answer:
As its initial and final positions are one and the same.
Initial Velocity = 2 cm/s.
Final Velocity = 4 cm/s
Time taken for the change in velocity from B to
D = 4 cm/s – 2 cm/s = 2 cm/s

7. Solve the following problems.

Question a.
A force of 1000 N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 10m. How much is the work done?
Answer:
Force (F) = 1000 N
displacement (s) = 10m
work done (W) = ?
W = Fs
= 1000 × 10
W = 10,000 Joule

Question b.
A cart with mass 20 kg went 50 m in a straight line on a plain and smooth road when a force of 2 N was applied to it. How much work was done by the force?
Answer:
Force (F) = 2 N
Displacement (s) = 50 m
Work done (W) = ?
W = Fs
= 2 × 50
W = 100 Joule

Project:

Question a.
Collect information about the study made by Sir Isaac Newton regarding force and acceleration and discuss it with your teacher.

Class 7 Science Chapter 7 Motion, Force and Work Important Questions and Answers

Fill in blanks:

Question 1.
Displacement is a …………. quantity.
Answer:
vector

Question 2.
The …………. of an object can change even while it is moving along a straight line.
Answer:
velocity

Question 3.
The …………. velocity can be different at different times.
Answer:
instantaneous

Question 4.
Change in velocity per second is called …………. .
Answer:
acceleration

Question 5.
The interaction that brings about the acceleration is called …………. .
Answer:
force

Question 6.
The scientist …………. was the first to study force and the resulting acceleration.
Answer:
Sir Isaac Nezvton

Question 7.
Ability to do work is called …………. .
Answer:
Energy

Question 8.
W = …………. × S.
Answer:
F

Question 9.
Unit of work is …………. and …………. .
Answer:
Joule, erg

Question 10.
Unit of force is …………. and …………. .
Answer:
Newton, dyne

Question 11.
Force is a …………. quantity.
Answer:
vector

Question 12.
The velocity at a particular time is called …………. velocity.
Answer:
instantaneous

Question 13.
The …………. of a body is the distance traversed per unit time.
Answer:
speed

Question 14.
Unit of acceleration is …………. and …………. .
Answer:
m/s² and cm/s²

Question 15.
Force is measured by the …………. that it produces.
Answer:
acceleration

Question 16.
Work done by a body with no displacement will be …………. .
Answer:
zero

Say whether True or False, correct the false 1 statements:

Question 1.
Velocity is distance travelled per unit of time.
Answer:
False. Speed is distance travelled per unit of time

Question 2.
In displacement, both distance and direction are taken into account.
Answer:
True

Question 3.
Speed = Distance/time.
Answer:
True

Question 4.
Change in speed per second is acceleration.
Answer:
False. Change in velocity per second is acceleration

Question 5.
Work done depends on the force and the displacement.
Answer:
True

Question 6.
C.G.S. unit of acceleration is m/s².
Answer:
False. C.G.S. unit of acceleration is cm/s².

Question 7.
M.K.S. unit of force is dyne.
Answer:
False. M.K.S. unit of force is Newton

Question 8.
Force is measured by the acceleration that it produces.
Answer:
True

Write the difference between the following:

Question 1.
Speed and Velocity
Answer:

Question 2.
Distance and Displacement
Answer:

Solve the following problems!

Question 1.
A bus travelled 200 km in the first 3 hours and then 100 kms for the next one and a half hours and then 120 kms for the next one and a half hours. What is the average velocity of the bus if it has moved in a straight line for the whole journey.
Answer:

Question 2.
See the diagram and calculate the Distance and Displacement travelled by the body from A to I.

Answer:
Distance travelled =
A → B → C → D → E → F → G → H + I
= 5 + 7 + 6 + 3 + 5 + 4 + 6 + 5
= 41 m
Displacement = A → I in a straight line shortest distance
= 1m

Use your brainpower:

Question 1.
The unit of acceleration is m/s², verify this.
Answer:

Question 2.
Acceleration is a vector quantity. Is force a vector quantity too?
Answer:
Yes, acceleration and force both are vector quantities, because both can be expressed completely only when magnitude and direction are given and the quantity which needs direction and magnitude both is called a vector quantity.

Maharashtra State Board Class 7 Science Textbook Solutions

• The Living World: Adaptations and Classification Class 7 Science Textbook Solutions
• Plants: Structure and Function Class 7 Science Textbook Solutions
• Properties of Natural Resources Class 7 Science Textbook Solutions
• Nutrition in Living Organisms Class 7 Science Textbook Solutions
• Food Safety Class 7 Science Textbook Solutions
• Measurement of Physical Quantities Class 7 Science Textbook Solutions
• Motion, Force and Work Class 7 Science Textbook Solutions
• Static Electricity Class 7 Science Textbook Solutions
• Heat Class 7 Science Textbook Solutions
• Disaster Management Class 7 Science Textbook Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

11th Maths Part 2 Differentiation Exercise 9.2 Questions And Answers Maharashtra Board

(I) Differentiate the following w.r.t. x

Question 1.
y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)
Solution:

Question 2.
y = √x + tan x – x³
Solution:

Question 3.
y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)
Solution:

Question 4.
y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)
Solution:

Question 5.
y = 7^x + x⁷ – \(\frac{2}{3}\) x√x – log x + 7⁷
Solution:

Question 6.
y = 3 cot x – 5e^x + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)
Solution:

(II) Diffrentiate the following w.r.t. x

Question 1.
y = x⁵ tan x
Solution:

Question 2.
y = x³ log x
Solution:

Question 3.
y = (x² + 2)² sin x
Solution:

Question 4.
y = e^x log x
Solution:

Question 5.
y = \(x^{\frac{3}{2}} e^{x} \log x\)
Solution:

Question 6.
y = \(\log e^{x^{3}} \log x^{3}\)
Solution:

(III) Diffrentiate the following w.r.t. x

Question 1.
y = x²√x + x⁴ log x
Solution:

Question 2.
y = e^x sec x – \(x^{\frac{5}{3}}\) log x
Solution:

Question 3.
y = x⁴ + x√x cos x – x² e^x
Solution:

Question 4.
y = (x³ – 2) tan x – x cos x + 7^x . x⁷
Solution:

Question 5.
y = sin x log x + e^x cos x – e^x √x
Solution:

Question 6.
y = e^x tan x + cos x log x – √x 5^x
Solution:

(IV) Diffrentiate the following w.r.t.x.

Question 1.
y = \(\frac{x^{2}+3}{x^{2}-5}\)
Solution:

Question 2.
y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)
Solution:

Question 3.
y = \(\frac{x e^{x}}{x+e^{x}}\)
Solution:

Question 4.
y = \(\frac{x \log x}{x+\log x}\)
Solution:
y = \(\frac{x \log x}{x+\log x}\)

Question 5.
y = \(\frac{x^{2} \sin x}{x+\cos x}\)
Solution:

Question 6.
y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)
Solution:

(V).

Question 1.
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution:
Let f(x) = ax² + bx + c …..(i)
∴ f(0) = a(0)² + b(0) + c
∴ f(0) = c
But, f(0) = 3 …..(given)
∴ c = 3 …..(ii)
Differentiating (i) w.r.t. x, we get
f'(x) = 2ax + b
∴ f'(2) = 2a(2) + b
∴ f'(2) = 4a + b
But, f'(2) = 2 …..(given)
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x² – 18x + 3

Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x² – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)² – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.

Question 2.
If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).
Solution:
f(x) = a sin x – b cos x
Differentiating w.r.t. x, we get
f'(x) = a cos x – b (- sin x)
∴ f'(x) = a cos x + b sin x


Now, f(x) = a sin x – b cos x
∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.
y = e^x . tan x
Diff. w.r.t. x

Solution:

Question 2.
y = \(\frac{\sin x}{x^{2}+2}\)
diff. w.r.t. x

Solution:

Question 3.
y = (3x² + 5) cos x
Diff. w.r.t. x

Solution:

Question 4.
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.
Solution:

State Board Class 11 Maths Solutions

• Differentiation Ex 9.1 Class 11 Maths Solutions
• Differentiation Ex 9.2 Class 11 Maths Solutions
• Differentiation Miscellaneous Exercise 9 Class 11 Maths Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

11th Maths Part 2 Differentiation Exercise 9.1 Questions And Answers Maharashtra Board

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x² + 3x – 1
Solution:
Let f(x) = x² + 3x – 1
∴ f(x + h) = (x + h)² + 3(x + h) – 1
= x² + 2xh + h² + 3x + 3h – 1
By first principle, we get

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get

(c) e^2x+1
Solution:

(d) 3^x
Solution:

(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get

(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get

(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get

(h) x√x
Solution:

Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) \(\sqrt{2 x+5}\) at x = 2
Solution:

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:

(iii) 2^3x+1 at x = 2
Solution:

(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get

(v) e^3x-4 at x = 2
Solution:

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x² + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:


∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.

Question 4.
Show that f(x) = x² is continuous and differentiable at x = 0.
Solution:

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.

f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x² if x ≤ 2 at x = 2.
Solution:

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:

Question 9.
Examine the function
f(x) = x² cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:

State Board Class 11 Maths Solutions

• Differentiation Ex 9.1 Class 11 Maths Solutions
• Differentiation Ex 9.2 Class 11 Maths Solutions
• Differentiation Miscellaneous Exercise 9 Class 11 Maths Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Miscellaneous Exercise 8 Questions and Answers.

11th Maths Part 2 Continuity Miscellaneous Exercise 8 Questions And Answers Maharashtra Board

(I) Select the correct answer from the given alternatives.

Question 1.
f(x) = \(\frac{2^{\cot x}-1}{\pi-2 x}\), for x ≠ \(\frac{\pi}{2}\)
= log √2, for x = \(\frac{\pi}{2}\)
(A) f is continuous at x = \(\frac{\pi}{2}\)
(B) f has a jump discontinuity at x = \(\frac{\pi}{2}\)
(C) f has a removable discontinuity
(D) \(\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3\)
Answer:
(A) f is continuous at x = \(\frac{\pi}{2}\)
Hint:

Question 2.
If f(x) = \(\frac{1-\sqrt{2} \sin x}{\pi-4 x}\), for x ≠ \(\frac{\pi}{4}\) is continuous at x = \(\frac{\pi}{4}\), then f(\(\frac{\pi}{4}\)) =
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(-\frac{1}{\sqrt{2}}\)
(C) \(-\frac{1}{4}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)
Hint:

Question 3.
If f(x) = \(\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^{2}}\), for x ≠ 0 is continuous at x = 0, then f(0) is
(A) \(\frac{10}{e^{2}}\)
(B) \(\frac{10}{e^{4}}\)
(C) \(\frac{5}{4}\)
(D) \(\frac{5}{2}\)
Answer:
(D) \(\frac{5}{2}\)
Hint:

Question 4.
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
(A) f is discontinuous at x = 2
(B) f is discontinuous at x = -4
(C) f is discontinuous at x = 0
(D) f is discontinuous at x = 2 and x = -4
Answer:
(B) f is discontinuous at x = -4
Hint:
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
= \(\frac{x^{2}-7 x+10}{(x+4)(x-2)}\)
Here f(x) is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when x = -4 or x = 2
But x = 2 does not lie in the given interval.
∴ x = -4 is the point of discontinuity.

Question 5.
If f(x) = ax² + bx + 1, for |x – 1| ≥ 3 and
= 4x + 5, for -2 < x < 4
is continuous everywhere then,
(A) a = \(\frac{1}{2}\), b = 3
(B) a = \(-\frac{1}{2}\), b = -3
(C) a = \(-\frac{1}{2}\), b = 3
(D) a = \(\frac{1}{2}\), b = -3
Answer:
(A) a = \(\frac{1}{2}\), b = 3
Hint:
f(x) = ax² + bx + 1, |x – 1| ≥ 3
= 4x + 5; -2 < x < 4
The first interval is
|x – 1| ≥ 3
∴ x – 1 ≥ 3 or x – 1 ≤ -3
∴ x ≥ 4 or x ≤ -2
∴ f(x) is same for x ≤ -2 as well as x ≥ 4.
∴ f(x) is defined as:
f(x) = ax² + bx + 1; x ≤ -2
= 4x + 5; -2 < x < 4
= ax² + bx + 1; x ≥ 4
f(x) is continuous everywhere.
∴ f(x) is continuous at x = -2 and x = 4.
As f(x) is continuous at x = -2,
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2}\left(a x^{2}+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5)\)
∴ a(-2)² + b(-2) + 1 = 4(-2) + 5
∴ 4a – 2b + 1 = -3
∴ 4a – 2b = -4
∴ 2a – b = -2 …..(i)
∵ f(x) is continuous at x = 4,
\(\lim _{x \rightarrow 4^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 4^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^{2}+b x+1\right)\)
4(4) + 5 = a(4)² + b(4) + 1
16a + 4b + 1 = 21
16a + 4b = 20
4a + b = 5 …..(ii)
Adding (i) and (ii), we get
6a = 3
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (ii), we get
4(\(\frac{1}{2}\)) + b = 5
∴ 2 + b = 5
∴ b = 3
∴ a = \(\frac{1}{2}\), b = 3

Question 6.
f(x) = \(\frac{\left(16^{x}-1\right)\left(9^{x}-1\right)}{\left(27^{x}-1\right)\left(32^{x}-1\right)}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then ‘k’ =
(A) \(\frac{8}{3}\)
(B) \(\frac{8}{15}\)
(C) \(-\frac{8}{15}\)
(D) \(\frac{20}{3}\)
Answer:
(B) \(\frac{8}{15}\)
Hint:

Question 7.
f(x) = \(\frac{32^{x}-8^{x}-4^{x}+1}{4^{x}-2^{x+1}+1}\), for x ≠ 0
= k, for x = 0,
is continuous at x = 0, then value of ‘k’ is
(A) 6
(B) 4
(C) (log 2) (log 4)
(D) 3 log 4
Answer:
(A) 6
Hint:

Question 8.
If f(x) = \(\frac{12^{x}-4^{x}-3^{x}+1}{1-\cos 2 x}\), for x ≠ 0 is continuous at x = 0 then the value of f(0) is
(A) \(\frac{\log 12}{2}\)
(B) log 2 . log 3
(C) \(\frac{\log 2 \cdot \log 3}{2}\)
(D) None of these
Answer:
(B) log 2 . log 3
Hint:

Question 9.
If f(x) = \(\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}\), for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is
(A) e⁷
(B) e³
(C) e¹²
(D) \(e^{\frac{3}{4}}\)
Answer:
(C) e¹²
Hint:

Question 10.
If f(x) = \(\lfloor x\rfloor\) for x ∈ (-1, 2), then f is discontinuous at
(A) x = -1, 0, 1, 2
(B) x = -1, 0, 1
(C) x = 0, 1
(D) x = 2
Answer:
(C) x = 0, 1
Hint:
f(x) = \(\lfloor x\rfloor\), x ∈ (-1, 2)
This function is discontinuous at all integer values of x between -1 and 2.
∴ f(x) is discontinuous at x = 0 and x = 1.

II. Discuss the continuity of the following functions at the point(s) or on the interval indicated against them.

Question 1.
f(x) = \(\frac{x^{2}-3 x-10}{x-5}\), for 3 ≤ x ≤ 6, x ≠ 5
= 10, for x = 5
= \(\frac{x^{2}-3 x-10}{x-5}\), for 6 < x ≤ 9
Solution:

Question 2.
f(x) = 2x² – 2x + 5, for 0 ≤ x ≤ 2
= \(\frac{1-3 x-x^{2}}{1-x}\), for 2 < x < 4
= \(\frac{x^{2}-25}{x-5}\), for 4 ≤ x ≤ 7 and x ≠ 5
= 7, for x = 5
Solution:
The domain of f(x) is [0, 7].
(i) For 0 ≤ x ≤ 2
f(x) = 2x² – 2x + 5
It is a polynomial function and is Continuous at all point in [0, 2).

(ii) For 2 < x < 4
f(x) = \(\frac{1-3 x-x^{2}}{1-x}\)
It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4).

(iii) For 4 ≤ x ≤ 7, x ≤ 5
f(x) = \(\frac{x^{2}-25}{x-5}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 does not lie in the interval.
∴ f(x) is continuous at all points in (4, 7] – {5}.

(iv) For continuity at x = 2:

∴ f(x) is continuous at x = 2.

(v) For continuity at x = 4:

∴ f(x) is continuous at x = 4.

(vi) For continuity at x = 5.
f(5) = 7

∴ f(x) is discontinuous at x = 5.
Thus, f(x) is continuous at all points on its domain except at x = 5.

Question 3.
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\), for x ≠ 0
f(0) = \(\frac{68}{15}\), at x = 0 on \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
Solution:
The domain of f(x) is [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(i) For [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}:
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero when cos x = 1,
i.e., x = 0
But x = 0 does not lie in the interval.
∴ f(x) is continuous at all points in [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}

(ii) For continuity at x = 0:


∴ \(\lim _{x \rightarrow 0} f(x) \neq f(0)\)
∴ f(x) is discontinuous at x = 0.

Question 4.
f(x) = \(\frac{\sin ^{2} \pi x}{3(1-x)^{2}}\), for x ≠ 1
= \(\frac{\pi^{2} \sin ^{2}\left(\frac{\pi x}{2}\right)}{3+4 \cos ^{2}\left(\frac{\pi x}{2}\right)}\), for x = 1, at x = 1
Solution:

Question 5.
f(x) = \(\frac{|x+1|}{2 x^{2}+x-1}\), for x ≠ -1
= 0, for x = -1, at x = -1
Solution:

Question 6.
f(x) = [x + 1] for x ∈ [-2, 2)
Where [*] is greatest integer function.
Solution:
f(x) = [x + 1], x ∈ [-2, 2)
∴ f(x) = -1, x ∈ [-2, -1)
= 0, x ∈ [-1, 0)
= 1, x ∈ [0, 1)
= 2, x ∈ [1, 2)

∴ \(\lim _{x \rightarrow-1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow-1^{+}} \mathrm{f}(x)\)
∴ f(x) is discontinuous at x = -1.
Similarly, f(x) is discontinuous at the points x = 0 and x = 1.

Question 7.
f(x) = 2x² + x + 1, for |x – 3| ≥ 2
= x² + 3, for 1 < x < 5
Solution:
|x – 3| ≥ 2
∴ x – 3 ≥ 2 or x – 3 ≤ -2
∴ x ≥ 5 or x ≤ 1
∴ f(x) = 2x² + x + 1, x ≤ 1
= x² + 3, 1 < x < 5
= 2x² + x + 1, x ≥ 5
Consider the intervals
x < 1 , i.e., (-∞, 1)
1 < x < 5, i.e., (1, 5) x > 5, i.e., (5, ∞)
In all these intervals, f(x) is a polynomial function and hence is continuous at all points.
For continuity at x = 1:

∴ f(x) is discontinuous at x = 5.
∴ f(x) is continuous for all x ∈ R, except at x = 5.

III. Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.

Question 1.
f(x) = x² + x – 3, for x ∈ [-5, -2)
= x² – 5, for x ∈ (-2, 5]
Solution:
f(-2) has not been defined.
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{-}}\left(x^{2}+x-3\right)\)
= (-2)² + (-2) – 3
= 4 – 2 – 3
= -1
\(\lim _{x \rightarrow-2^{+}} f(x)=\lim _{x \rightarrow-2^{+}}\left(x^{2}-5\right)\)
= (-2)² – 5
= 4 – 5
= -1
∴ \(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2} f(x) \text { exists. }\)
But f(-2) has not been defined.
∴ f(x) has a removable discontinuity at x = -2.

Question 2.
f(x) = x² + 5x + 1, for 0 ≤ x ≤ 3
= x³ + x + 5, for 3 < x ≤ 6
Solution:
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}\left(x^{2}+5 x+1\right)\)
= (3)² + 5(3) + 1
= 9 + 15 + 1
= 25
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}\left(x^{3}+x+5\right)\)
= (3)³ + 3 + 5
= 27 + 3 + 5
= 35
∴ \(\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3} f(x)\) does not exist.
∴ f(x) is discontinuous at x = 3.
∴ f(x) has a jump discontinuity at x = 3.

Question 3.
f(x) = \(\frac{x^{2}+x+1}{x+1}\), for x ∈ [0, 3)
= \(\frac{3 x+4}{x^{2}-5}\), for x ∈ [3, 6]
Solution:


∴ f(x) is continuous at x = 3.

IV. Discuss the continuity of the following functions at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous.

Question 1.
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
Solution:
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
= \(\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)}\)
∴ f(x) is not defined at x = 4 and x = -3.
∴ The domain of function f = R – {-3, 4}.
For x ≠ -3 and 4,

f(x) is discontinuous at x = 4 and x = -3.
This discontinuity is removable.
∴ f(x) can be redefined as
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\), for x ≠ 4, x ≠ -3
= -5, for x ∈ R – {-3, 4}, x = -3
= 2, for x ∈ R – {-3, 4}, x = 4

Question 2.
f(x) = x² + 2x + 5, for x ≤ 3
= x³ – 2x² – 5, for x > 3
Solution:

∴ f(x) is discontinuous at x = 3.
This discontinuity is irremovable.

V. Find k if the following functions are continuous at the points indicated against them.

Question 1.
f(x) = \(\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}\), for x ≠ 2
= k, for x = 2 at x = 2.
Solution:
f(x) is continuous at x = 2. …..(given)

Question 2.
f(x) = \(\frac{45^{x}-9^{x}-5^{x}+1}{\left(k^{x}-1\right)\left(3^{x}-1\right)}\), for x ≠ 0
= \(\frac{2}{3}\), for x = 0, at x = 0
Solution:
f(x) is continuous at x = 0 …..(given)

VI. Find a and b if the following functions are continuous at the points or on the interval indicated against them.

Question 1.
f(x) = \(\frac{4 \tan x+5 \sin x}{a^{x}-1}\), for x < 0
= \(\frac{9}{\log 2}\), for x = 0
= \(\frac{11 x+7 x \cdot \cos x}{b^{x}-1}\), for x < 0
Solution:

Question 2.
f(x) = ax² + bx + 1, for |2x – 3| ≥ 2
= 3x + 2, for \(\frac{1}{2}\) < x < \(\frac{5}{2}\)
Solution:

VII. Find f(a), if f is continuous at x = a where,

Question 1.
f(x) = \(\frac{1+\cos (\pi x)}{\pi(1-x)^{2}}\), for x ≠ 1 and at a = 1.
Solution:

Question 2.
f(x) = \(\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^{2}}\), for x ≠ π and at a = π.
Solution:

VIII. Solve using intermediate value theorem.

Question 1.
Show that 5^x – 6x = 0 has a root in [1, 2].
Solution:
Let f(x) = 5^x – 6x
5^x and 6x are continuous functions for all x ∈ R.
∴ 5^x – 6x is also continuous for all x ∈ R.
i.e., f(x) is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = 5¹ – 6(1) = -1 < 0
f(2) = (5)² – 6(2) = 13 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 1 and 2 such that f(c) = 0.
∴ There is a root of the given equation in [1, 2].

Question 2.
Show that x³ – 5x² + 3x + 6 = 0 has at least two real roots between x = 1 and x = 5.
Solution:
Let f(x) = x³ – 5x² + 3x + 6
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
Here, we have been asked to show that f(x) has at least two roots between x = 1 and x = 5.
f(1) = (1)³ – 5(1)² + 3(1) + 6
= 5 > 0
f(2) = (2)³ – 5(2)² + 3(2) + 6
= 8 – 20 + 6 + 6
= 0
∴ x = 2 is a root of f(x).
Also, f(3) = (3)³ – 5(3)² + 3(3) + 6
= 27 – 45 + 9 + 6
= -3 < 0
f(4) = (4)³ – 5(4)² + 3(4) + 6
= 64 – 80 + 12 + 6
= 2 > 0
∴ f(3) < 0 and f(4) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 3 and 4 such that f(c) = 0.
∴ There are two roots, x = 2 and a root between x = 3 and x = 4.
Thus, there are at least two roots of the given equation between x = 1 and x = 5.

11th Standard State Board Maths Solutions

• Continuity Ex 8.1 Class 11 Maths Solutions
• Continuity Miscellaneous Exercise 8 Class 11 Maths Solutions