Month: October 2024

Jharkhand Board JAC Class 10 Science Important Questions Chapter 15 Our Environment Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 15 Our Environment

Additional Questions and Answers

Question 1.
Distinguish between:
(1) Biodegradable wastes and Non-biodegradable wastes
Answer:

Biodegradable wastes	Non-biodegradable wastes
1. The waste materials which can be broken down by biological processes are called biodegradable wastes.	1. The waste materials which cannot broken down by biological processes are called non-biodegradable wastes.
2. Disposal of these can be done by the activity of decomposers.	2. Disposal of these cannot be done by the activity of decomposers.
3. Biodegradable wastes are organic substances which are the products of living organisms.	3. Synthetic man-made substances are there in non- biodegradable wastes.
4. Decaying vegetables and fruits, peels, scales, dead bodies, etc. are biodegradable wastes.	4. Glass, plastics, polythenes, chemical pesticides, etc. are non-biodegradable wastes.

(2) Producer organisms and Consumer organisms
Answer:

Producer organisms	Consumer organisms
1. These are autotrophic organisms.	1. These are heterotrophic organisms.
2. They transform solar energy into chemical energy and prepare food through photosynthesis for themselves as well as for other organisms.	2. They consume other organisms or their products for their food.
3. Green plants in terrestrial ecosystem and various types	3. Animals are consumers in any ecosystem.
4. Producers are not divided in any specific category.	4. Consumers are divided as herbivores, carnivores, omnivores and parasites categories.

Question 2.
Give scientific reasons for the following statements:
(1) Consumers are directly or indirectly dependent on green plants for energy.
Answer:
In ecosystem, green plants absorb solar energy and transform it in chemical energy. Thus at first trophic level green plants produce food by transforming chemical energy from available solar energy.

Consumers consume other organisms or their products as food for their energy requirement. Flow of energy occurs in form of food from green plants to herbivores and from it to carnivores. Consumers are heterotrophic so their energy requirement is satisfied from lower trophic level.

Thus, consumers are directly or indirectly dependent on green plants for energy.

(2) Energy flow is always unidirectional in an ecosystem.
Answer:
The solar energy enters in an ecosystem through the process of photosynthesis performed by green plants. The energy that is captured by green plants does not revert back to the sun.

From green plants energy passes to upper trophic levels of consumers. Energy obtained by herbivores does not come back to the green plants. As it moves progressively through the various trophic levels it is no longer available to s the previous level. Thus, energy flow is always unidirectional in s an ecosystem.

(3) The trophic relationships of organisms occur in form of food webs.
Answer:
In nature, organisms are interlinked for food to form food chain. Among various ecosystems, each one is having definite food chain. But single organism is never dependent only on one organism for food. So trophic relationships for food can never be explained as straight chain. The individuals involved in food chain of one ecosystem are also linked with food chains of other ecosystems.

So, trophic relations for food do not operate in isolation but interlinked to form complex net called food web.
Thus, the trophic relationships of organisms occur in form of food web.

(4) It is necessary to prevent the uncontrolled use of pesticides and other chemicals.
Answer:
Pesticides and other chemicals are used in order to protect the crops from diseases or pests. Such chemicals are non-biodegradable. The amount of these chemicals increases in the soil or water bodies due to uncontrolled use of it. From soil or water they enter in the body of plants and then it passes in trophic levels of food chain.

They get accumulated progressively at each trophic level as these chemicals are not degradable. Due b to biomagnification, high concentration of such chemical causes lethal effects on higher level carnivores animals. Some such chemicals prove lethal for organisms of lower trophic level too. These chemicals reach upto human beings too through food.

Thus, it is necessary to prevent the uncontrolled use of pesticides and other chemicals.

(5) The main responsible factor for ‘ depletion of ozone layer is CFC.
Answer:
The ozone layer in stratosphere absorbs s harmful ultraviolet radiation from the sunlight and b prevents it from reaching the earth.

CFC (Chlorofluorocarbon) adds chlorine (Cl^–) in atmosphere. The chlorine atom reacts with ozone and removes an atom of oxygen one by one. One atom of chlorine can decompose many molecules of ozone in this fashion. CFC thus causes about 80% of the total depletion of ozone.

Thus, the main responsible factor for depletion of ozone layer is CFC.

(6) The production of household waste can be reduced by few common steps.
Answer:

• Buy loose fruits and vegetables in order to avoid packaging.
• Limit the use of disposable items such as razors, pens, nappies, paper napkins, cups, plates, etc.
• Buy reusable carry bags from super-markets. Stop using plastics.
• Use rechargeable batteries.
• Do not throw old clothes and shoes but donate them.
• Reuse the back of papers as scrap paper.
• Biodegradable wastes can be decomposed by composting at home. This will reduce the amount of wastes which go out of the house.

Thus, follow the principle of reuse, recycle and reduce which can substantially reduce the amount of household waste.

Question 3.
Carefully observe the given diagram and answer the questions related it:
(1) Only 10% energy is transferred from one trophic level to its upper trophic level. From following diagram how much energy available to second and third trophic level?

Answer:
At second trophic level 1 joule energy available.
At third trophic level 0.1 joule energy available.

Questions :
(1) Give the name of ’a’ and mention its category.
(2) Give the name of ‘b’ and mention its function.
Answer:
(1) a – Carnivores
Categories : Primary carnivores and higher carnivores

(2) b – Decomposers (Bacteria and fungi)
Function: Decompose dead bodies and excretory wastes

Questions:
(1) Which trophic level has maximum number of organisms? Why?
(2) Why number of secondary consumers are less than herbivores?
(3) Why lion is placed at the top of pyramid?
Answer:
(1) Maximum number of organisms are at first trophic level because trees and shrubs are producers supplies food and energy to giraffe.

(2) Because secondary consumers obtained only 10 % energy from the tropic level of harbivores.

(3) Because lion is higher carnivores having less population as shown in the pyramid.

(4) Observe satellite picture showing ozone layer over Antarctica. What is shown in picture? Give the name of compound responsible for it.

Answer:
Picture shows ozone hole which is caused by CFC (Chlorofluorocarbon) compound

(5) Fill the blank in following given table for terrestrial food chain :

Answer:

• a – Third order consumer
• b – Primary consumer
• c – Producer
• d – Heterotrophic
• e – Heterotrophic
• f – Autotrophic
• g – Snake
• h – Rat

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) Who affects the environment the most?
Answer:
Human activities affect the environment most.

(2) Which is main functional unit of an environment?
Answer:
Ecosystem is main functional unit of an environment.

(3) What happens to the dead plants and dead animals?
Answer:
Decomposers decompose complex organic matter into inorganic constituents that are present in dead plants and dead animals.

(4) What are various orders of food chain named as?
Answer:
Various levels of food chain are named as trophic levels.

(5) How many trophic levels occur in food chain?
Answer:
There are three or four trophic levels in a food chain.

(6) Which trophic level has maximum and which trophic level has minimum number of organisms?
Answer:
In a food chain, maximum number of organisms are at first trophic level (at producers level) and minimum number of organisms are at fourth trophic level (higher carnivores).

(7) Can any single organism occupy two different trophic levels? Yes or no. Explain.
Answer:
Yes. Rat occupies herbivores trophic level as primary consumer and also occupies carnivores trophic level as secondary consumer.

(8) Construct food chain from following organisms:
(a) Frog, insects, birds, grass.
(b) Fish, algae, small animals, large fish
Answer:
(a) Grass → insects → frog → birds
(b) Algae → small animals → fish → large fish

(9) Why all food chains begin with organisms that contain chlorophyll?
Answer:
Chlorophyll containing organisms are able to capture solar energy and covert this into chemical energy. Because of it energy flow is continued in an ecosystem.

(10) Among snake, insects, frog, plants, hawk which one is an organism of fourth trophic level?
Answer:
Snake

(11) Among grass, deer, lion by whom maximum energy is transported to its upper trophic level?
Answer:
Grass

(12) Mention the amount of solar energy captured by the green plants out of the total energy that reaches on the earth from the sun and mention the percentage of energy, which is transferred from one trophic level to next higher one.
Answer:
The green plants in a terrestrial ecosystem capture about 1 % of the energy of sunlight falling on their leaves. An average 10% energy is transferred from one trophic level to the next higher one.

(13) List two methods of safe disposal of the non-biodegradable waste.
Answer:

• By recycling
• By underground dumping

(14) Which chemical is used in fire extinguishers? How it is harmful?
Answer:
CFCs (chlorofluorocarbons) is used in fire extinguishers. It is harmful as it leads to depletion of ozone layer.

(15) Why is straight-line food chain is not common in the natural ecosystem?
Answer:
In the natural ecosystem each organism is generally eat two or more other kinds of organisms which is in turn are eaten by several other organisms. So, straight-line food chain is not common in the natural ecosystem.

(16) What is the function of ozone in the upper atmosphere?
Answer:
Ozone shields the surface of the earth from harmful ultraviolet radiation from the sun.

(17) Name two natural and two artificial ecosystems.
Answer:

(18) Name the natural cleansing agent in an ecosystem.
Answer:
Decomposers (e.g., bacteria, fungi) and Scavengers (e.g., kite, crow).

(19) Why are autotrophs called producers and heterotrophs called consumers?
Answer:
Autotrophs are green plants which perform photosynthesis and produce organic compounds (carbohydrates). So, they are called producers. Heterotrophs consume food from other organisms. So they are called consumers.

Question 2.
Define : OR Explain the terms :
(1) Ecosystem
Answer:
A system formed by an interaction between biotic components and the physical environment is called an ecosystem.

(2) Decomposers
Answer:
The microorganisms that breakdown the complex organic substances into simple inorganic substances are known as decomposers.

(3) Food chain
Answer:
Living organisms of an ecosystem depend on each other for their food requirement and form a chain turmed as food chain.

(4) Food web
Answer:
The network of overlapping food chains operating in an ecosystem is called a food web.

(5) Artificial ecosystem
Answer:
An ecosystem which is created and maintained by humans is called artificial ecosystem.

(6) Biological magnification
Answer:
A phenomenon of progressive increase in the concentration of non-biodegradable substances in organisms at each successive trophic level is known as biological magnification.

(7) Biodegradable substances
Answer:
The substances which can be disposed off naturally by the action of micro¬organisms / decomposers are called biodegradable substances.

(8) Non-biodegradable substances
Answer:
The substances that cannot be converted into harmless simpler forms by the action of microorganisms are called non-biodegradable substances.

(9) Producers
Answer:
All green plants and certain bacteria which can produce food by the process of photosynthesis are called producers.

(10) Consumers
Answer:
Those organisms which are dependent on producers for their nutritional requirement are called consumers.

Question 3.
Fill in the blanks:

1. Waste substances that are broken down by microbes are called ………………..
2. ……………….. is formed by interaction between biotic community and abiotic factors.
3. Use of ……………….. chemical is hazardous for ozone layer.
4. Farm and garden are ……………….. ecosystem.
5. A parasite is in the class of ……………….. category.
6. All animals are ……………….. in food chain.
7. Food chain ends with ……………….. organisms from where simple inorganic substances return back to the soil.
8. Solar energy is transformed into ……………….. by chlorophyllous organisms.
9. Main energy source in an ecosystem is ………………..
10. A problem of ……………….. is due to entry of non-biodegradable substance.
11. UV radiation can cause ……………….. cancer in human beings.
12. ……………….. is flow in form of food in an ecosystem.

Answer:

1. biodegradable
2. Ecosystem
3. CFC
4. artificial (man-made)
5. consumer
6. consumers
7. decomposers
8. chemical energy
9. Sun
10. biological magnification
11. skin
12. Energy

Question 4.
State whether the following statements true or false:

1. Microbes decomposing dead bodies are located at trophic level of producers.
2. Sun is ultimate energy source for ecosystems on this earth.
3. Wastes which we generated are biodegradable or non-biodegradable.
4. Oxygen gets released under the effect of ultraviolet rays on ozone molecules.
5. Ecosystem is formed only with biotic components.
6. Level of energy decreases as one goes to upper trophic level in an ecosystem.
7. Only 10% energy is used at each trophic level from the available energy and remaining is saved for next level of organisms.
8. Number of consumers are always more than producers.
9. UV radiation absorbs ozone that is deadly poison at ground level.
10. In ecosystem food relationships among organisms are in form of food web.
11. Water is abiotic factor in pond ecosystem.
12. In food chain energy flow is bidirectional.
13. All organisms can use only chemical energy.
14. Food web is formed due to alternative path for food availability.
15. Decomposers are natural scavengers.

Answer:

1. False
2. True
3. True
4. False
5. False
6. True
7. False
8. False
9. False
10. True
11. True
12. False
13. True
14. True
15. True

Question 5.
Chart – diagram based questions :
1. A food chain with four trophic levels is shown in following diagram:

(1) Which trophic level indicates carnivores?
(2) Arrange trophic levels in correct manner on the basis of the amount of available energy.
Answer:
(1) T₃ and T₄
(2) T₁ > T₂ >T₃ > T₄

2. In the chart of energy flow given below identify the decomposers and explain your choice:

Answer:
In the chart B indicates decomposers. They decompose dead materials available from A, C and D.

In a given food web,
(1) Identify which one is secondary as well as tertiary consumers.
(2) Identify which one has minimum conc. of pesticides.
Answer:
(1) Hawk
(2) Producers, i.e., food grains, small plants, grass.

Question 6.
Match the following:
(1)

Column I	Column II
1. Animals feed only on plants	p. Carnivores
2. Animals feed on other animals	q. Decomposers
3. Animals feed on plants as well as animals	r. Herbivores
4. Organisms obtain nutrition from decaying organic matter	s. Omnivores

Answer:
(1 – r), (2 – p), (3 – s), (4 – q).

(2)

Column I	Column II	Column III
1. Omnivores	a. Scavenger	p. Grass
2. Producer	b. Secondary consumer	q. Crow
3. Detritus feeding	c. Biconsumer	r. Deer
4. Carnivores	d. Primary consumer	s. Tiger
5. Herbivores	e. Transformer of solar energy	t. Fox

Answer:
(1 – c – q), (2 – e – p), (3 – a – t), (4 – b – s), (5 – d – r).

(3)

Column I	Column II
1. Autotroph	p. Convert complex organic matter into inorganic
2. Primary consumer	q. Convert inorganic constituent into organic with the help of sunlight
3. Secondary consumer	r. Plant feeder
4. Decomposer	s. Animal feeder

Answer:
(1 – q), (2 – r), (3 – s), (4 – p).

Question 7.
Choose the correct option from those given below each question:
1. Which process is significant in a primary producers for to sustain an ecosystem?
A. Biological magnification
B. Photosynthesis
C. Respiration
D. Water absorption
Answer:
B. Photosynthesis

2. Animals that fed on both producer as well as consumer trophic level are called …………….
A. herbivores
B. carnivores
C. omnivores
D. decomposers
Answer:
C. omnivores

3. There are more herbivores than carnivores in a food chain because …………….
I. carnivores have bigger appetites.
II. a large number of herbivores are needed to support a small number of carnivores.
III. there are more plants for herbivores to eat.
IV much energy is lost in the transfer of energy from herbivores to carnivores.
A. I and III
B. II and IV
C. I and II
D. All of given
Answer:
B. II and IV

4. In a food chain deer feeds on plants. Deer is eaten by a lion. Which of the following statement is correct for it?
I. Lion is secondary consumer and deer is primary consumer.
II. Lion is predator and deer is prey.
III. Lion is primary consumer and deer is secondary consumer.
IV. Lion is tertiary consumer and deer is secondary consumer.
A. I and II only
B. III and IV only
C. II and III only
D. I and IV only
Answer:
A. I and II only

5. Which is possible food chain in pond?

Producer	Primary consumer	Secondary consumer
A. Algae	Fish	Larva of mosquito
B. Fish	Algae	Larva of mosquito
C. Larva or mosquito	Algae	Fish
D. Algae	Larva of mosquito	Fish

Answer:
Alage, Larva of mosquito, Fish

6. Which of the following option is correct for decomposers feeding on dead organism?

Bacteria	Fungi	Virus
A. No	Yes	Yes
B. Yes	No	Yes
C. Yes	Yes	No
D. Yes	Yes	Yes

Answer:
Yes, Yes, No

7. In an ecosystem ……………
A. unidirectional flow of both energy and nutrients.
B. cyclic flow of both energy and nutrients.
C. cyclic flow of energy and unidirectional flow of nutrients.
D. unidirectional flow of energy and cyclic flow of nutrients.
Answer:
A. unidirectional flow of both energy and nutrients.

8. A rat is in which category of consumer animals?
A. Herbivores
B. Carnivores
C. Omnivores
D. Decomposer
Answer:
C. Omnivores

9. Which of the following problem is a result of man-made activities?
A. Breaking food chain due to deforestation
B. Global warming
C. Depletion of ozone layer
D. All of the given
Answer:
D. All of the given

10. In an ecosystem third trophic level is always occupied by which animals?
A. Herbivores
B. Carnivores
C. Producer
D. Decomposers
Answer:
B. Carnivores

11. In the agricultural farm with wheat crop, snake, peacock, kite, rat, etc. occur in food chain. Pesticides are spray regular time interval. Which of the following animal has lowest concentration of pesticides?
A. Snake
B. Rat
C. Kite
D. Peacock
Answer:
B. Rat

12. Which of the following statement is wrong in case of food chain?
A. Decomposers can derive their food from all other nutritional substances except producers.
B. The one which are not producers are consumers.
C. More energy is available at lower trophic levels.
D. One organism may be associated with more than one food chain.
Answer:
A. Decomposers can derive their food from all other nutritional substances except producers.

13. Which of the following is an example of biodegradable waste?
A. Vegetables or fruits
B. Glass
C. Plastic
D. Metal
Answer:
A. Vegetables or fruits

14. Which of the following is an example of non-biodegradable waste?
A. Fruits
B. Vegetables
C. Paper
D. Polythene
Answer:
D. Polythene

15. Ecosystem is an interacting system made up of ……………
A. organisms and their physical surroundings.
B. producers and consumers.
C. producers and their physical surroundings.
D. consumers and their physical surroundings.
Answer:
A. organisms and their physical surroundings.

16. What is being form due to entry of harmful chemical in food chain?
A. Bioequilibrium
B. Eutrophication
C. Biological magnification
D. Biological regulation
Answer:
C. Biological magnification

17. What is formed by interdependent organisms to each other for their nutrition?
A. Biosphere
B. Food chain
C. Ecosystem
D. Environment
Answer:
B. Food chain

18. In food chain, organisms which synthesise carbohydrates by the conversion of radiant energy are called ……………
A. primary producers
B. primary consumers
C. primary carnivores
D. primary decomposers
Answer:
A. primary producers

19. In the food chain comprising of a frog, snake, grass, grasshopper, which organisms is at third trophic level in it?
A. Snake
B. Frog
C. Grass
D. Grasshopper
Answer:
B. Frog

20. An ecosystem has producers, herbivores and carnivores but without which of following it is not complete?
A. Secondary carnivores
B. Higher carnivores
C. Omnivores
D. Decomposers
Answer:
D. Decomposers

21. Which trophic level has highest energy in an ecosystem?
A. Producers
B. Herbivores
C. Primary carnivores
D. Higher carnivores
Answer:
A. Producers

22. Which trophic level, lowest energy available in an ecosystem?
A. Producer
B. Herbivores
C. Carnivores
D. Decomposers
Answer:
C. Carnivores

23. Which of following is wrong statement?
A. All green plants and blue-green algae are producers.
B. Green plants derived their food from organic compounds.
C. Producers synthesise their food by using inorganic elements.
D. Green plants transformed solar energy into chemical energy.
Answer:
B. Green plants derived their food from organic compounds.

24. What is the role of decomposers in an ecosystem?
A. Converts inorganic elements in simple form
B. Converts organic compound in inorganic elements
C. Converts inorganic elements in organic compounds
D. Both B and C
Answer:
B. Converts organic compound in inorganic elements

25. Which of the following constitute a food chain?
A. Grass, wheat and mango
B. Grass, goat and human
C. Goat, cow and elephant
D. Grass, fish and goat
Answer:
B. Grass, goat and human

26. Which of the following are environment friendly practices?
A. Carrying cloth-bags to purchases in while shopping.
B. Switching off unnecessary lights and fAnswer:
C. Use reusuable razors, nappies and pens.
D. All of the given.
Answer:
D. All of the given.

27. When frog uses grasshopper as food, then energy flow occurs from ……………
A. producer to decomposer.
B. producer to primary consumer.
C. primary consumer to secondary consumer.
D. primary carnivores to higher carnivores.
Answer:
C. primary consumer to secondary consumer.

28. Statement X: Decomposers satisfy their energy requirement directly from producers.
Statement Y: The amount of available energy in an ecosystem depends on the rate of photosynthesis carried out by the producers.
Which option is correct for statements X and Y?
A. X and Y both correct.
B. X and Y both incorrect.
C. X correct and Y incorrect.
D. X incorrect and Y correct.
Answer:
D. X incorrect and Y correct.

29. Statement X: Reduction in number of prey altimately results in reduction of the number of predator.
Statement Y: Ozone layer absorbs harmful ultraviolet radiations.
Which option is correct for statements X and Y?
A. X and Y both correct.
B. X and Y both incorrect.
C. X correct and Y incorrect.
D. X incorrect and Y correct.
Answer:
A. X and Y both correct.

30. Usually green plants are at first level in an ecosystem because ……………
A. they are widely spread on this earth.
B. they are fixed at one place in soil.
C. they synthesise food with the help of sunlight.
D. herbivores are more than carnivores.
Answer:
C. they synthesise food with the help of sunlight.

31. Which is the correct statement regarding food chain?
A. All consumers are not producers.
B. More energy is available at lower trophic levels.
C. One organism may be involved in more than one food chain.
D. All of the given
Answer:
D. All of the given

32. Which organism is more affected by biological magnification problem arising due to entry of DDT in pond ecosystem?
A. Man
B. Birds habitat near the pond
C. Fishes in pond
D. Aquatic plants in pond
Answer:
A. Man

33. Which of the following groups have only non- biodegradable components?
1. wood, paper, leather
2. polythene, detergent, PVC
3. plastic, detergent, glass
4. plastic, grass, animal dung
A. 3
B. 4
C. 1 and 3
D. 2 and 3
Answer:
D. 2 and 3

34. By whom the number of organisms are limited at trophic level of food chain?
A. Less availability of energy at higher tropic level
B. Sufficient food supply
C. Polluted air
D. Water
Answer:
A. Less availability of energy at higher tropic level

35. In which category does the human beings fall in an ecosystem?
A. Herbivores
B. Carnivores
C. Decomposers
D. Omnivores
Answer:
D. Omnivores

Question 8.
Answer as directed : (Miscellaneous)
(1) Give the full form of : CFCs, UNEP
Answer:

• CFCs – Chloroflurocarbons
• UNEP – United Nations Environment Programme

(2) Use of disposable plastic plates should be prevented because … (Complete the sentence.)
Answer:
It is made-up of non-biodegradable substance.

(3) Find correct sequence for food chain …
I. producers → decomposers → herbivores
II. producers → herbivores → carnivores
III. decomposers → producers → herbivores
IV. producers herbivores decomposers → carnivores → omnivores
Answer:
II. producers herbivores → carnivores

(4) Who am I?
I am a product of oxygen under the influence of UV radiation and I absorb UV rays.
Answer:
Ozone (O₃)

(5) Select the mismatched pair in the following and correct it.
I. Parasite – Consumers which obtain liquid food nutrients from host organisms.
II. Agro-ecosystem – A man-made ecosystem
III. UV radiation – Causes skin cancer in human beings.
IV Ecosystem – Physical components of environment.
Answer:
IV. is a mismatched pair. Ecosystem consists of an interaction of physical components of environment and biotic components.

(6) When a population of animal ‘X’ is introduced into the biotic community of an ecosystem, the number of lions increased while the number of deer decreased. Which is correct option for animal ‘X’?
Answer:
X animal is a predator of deer and a prey of lion.

(7) Among the following, which have posed a threat to the environment?
Body spray, bacteria, aerosols, CFCs, wild grass
Answer:
Body spray, aerosols, CFCs

(8) A food chain comprising hawk, snake, paddy and mice. If 15,000 J of energy available to mice then how much energy will be in paddy and how much energy will be available to hawk.
Answer:
15,000 J of energy will be in paddy and 150 J of energy will be available to hawk.

(9) Smallest energy transfer : secondary consumers to tertiary consumers::
largest energy transfer : …………….
Answer:
producers level

(10) In which unit is amount of atmospheric ozone expressed?
Answer:
Dobson Unit (DU)

Value Based Questions With Answers

Question 1.
Improvement in our lifestyle have resulted in generation of greater amount of waste material. We use more and more disposable things due to our changed lifestyle and wrong attitude.

Questions:

1. Name the two factors that have contributed to the deterioration of our environment.
2. Why do we generate lot of waste now-a-days?
3. Why most of the wastes from our homes contain more and more non-biodegradable substances?

Answer:

1. Advancement in the technology, excessive non-biodegradable waste generation that have contributed to the deterioration of our environment.
2. We are generating lot of waste now-a-day because of improvement in our lifestyles due to advanced technology and our changed attitude. We do not have environment conscience and thus we generate more waste.
3. Changes in packaging have resulted in our waste becoming non-biodegradable. We have tendency of “use and throw” which is causing more non-biodegradable substances in our wastes.

Question 2.
Advanced agricultural practice leads to excessive use of fertilisers, pesticides and irrigation. Use of such agricultural products can cause some health issues in human beings. Uncontroled use of chemicals can also cause adverse effect on our environment.

Questions:
1. Name the ecological process involved in the above case.
2. How is soil affected by this phenomena?
3. Why is pesticide persist for long time in an environment?
Answer:
1. Biological magnification.

2. Soil is affected in following concern :

• Pesticides are toxic that accumulate in soil.
• Salinity of soil increases.
• Soil becomes non-fertile.

3. Pesticides are non-biodegradable wastes. They are not converted into simple substances by decomposers.

Practical Skill Based Questions With Answers

Question 1.
The diagram shows the organisms in a habitat.

Which of the following indicates the correct feeding relationships of these organisms shown by four students?

Answer:

Question 2.
(1) A man consumed curd or buttermilk after meal which help in the process of digestion. Milk is fermented into curd by Lactobacillus bacteria. Imagine a food chain and state the category of Lactobacillus and human.
(2) Instead of curd, if we use milk than which trophic level is occupied by us?
Answer:
(1) Lactobacillus – Third trophic level → consumer / decomposer
Human – Fourth trophic level → consumer

(2) Third trophic level / second level of consumer

Question 3.
Observe the picture carefully and answer the following questions :

1. State the category according to the food eaten by encircled organisms.
2. State the trophic level observed in the picture.
3. State two abiotic components of an environment.
4. Which is an important process of an ecosystem observed in a picture?

Answer:

1. Squirrel and deer are both herbivores.
2. Producer (Trophic level – I) and herbivores (Trophic level – II)
3. Soil, light
4. Flow of energy

Memory Map:

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.3

Question 1.
State which pairs of triangles in the given figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Answer:
1.

In ΔABC and ΔPQR,
∠A = ∠P = 60°, ∠B = ∠Q = 80° and
∠C = ∠R = 40°
∴ By AAA criterion, ΔABC – ΔPQR.

2.

In ΔABC and ΔQRP,
\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}\), \(\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{2.5}{5}=\frac{1}{2}\) and \(\frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\)
Thus, \(\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}\)
∴ By SSS criterion, ΔABC ~ ΔQRP.

3.

No, the given triangles are not similar as
\(\frac{MP}{DE}=\frac{1}{2}\), \(\frac{LP}{DF}=\frac{1}{2}\), but \(\frac{LM}{EF}\) = \(\frac{2.7}{5}≠\frac{1}{2}\)

4.

In ΔMNL and ΔQPR,
\(\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{2.5}{5}=\frac{1}{2}\), \(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}\)
∠M = ∠Q = 70°
∴ By SAS criterion, ΔMNL ~ ΔQPR.

5.

No, the given triangles are not similar as two given corresponding sides are proportionate but the corresponding included angles are not equal.

6.

In ΔDEF, ∠D = 70°, ∠E = 80°
∠F = 180° – 70° – 80° = 30°
In ΔPQR, ∠Q = 80°, ∠R = 30°
∴ ∠P = 180° – 80° – 30° = 70°
Thus, in ΔDEF and ΔPQR.
∠D = ∠P, ∠E = ∠Q and ∠F = ∠R
∴ By AAA criterion, ΔDEF ~ ΔPQR.

Question 2.
In the given figure ΔODC – ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Answer:

In ΔDOC, ∠COB is an exterior angle.
∴ ∠COB + ∠DOC = 180°
∴ 125° + ∠DOC = 180°
∴ ∠DOC = 55°
Again, ∠COB = ∠ODC + ∠DCO
∴ 125° = 70° + ∠DCO
∴ ∠DCO = 55°
Now, ΔODC ~ ΔOBA
∴ ∠OAB = ∠OCD
∴ ∠OAB = 55°
Thus, ∠DOC = 55°, ∠DCO = 55° and
∴ ∠OAB = 55°.

Question 3.
Diagonal AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{OA}{OC}=\frac{OB}{OD}\)
Answer:
Given:
In trapezium ABCD, AB || DC and diagonals AC and BD intersect at O.
To prove : \(\frac{OA}{OC}=\frac{OB}{OD}\)

Proof: In trapezium ABCD, AB || CD.
∴ ∠CAB = ∠ACD and ∠DBA = ∠BDC
(Alternate angles) ……(1)
Then, in ΔOAB and ΔOCD.
∠OAB = ∠OCD and ∠OBA = ∠ODC [By (1)]
∴ By AA criterion, ΔOAB ~ ΔOCD.
∴ \(\frac{OA}{OC}=\frac{OB}{OD}\)

Question 4.
In the given figure \(\frac{QR}{QS}=\frac{QT}{PR}\) and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Answer:

In ΔPQR, ∠1 = ∠2, i.e., ∠PQR = ∠PRQ
∴ PR = QP
Now, \(\frac{QR}{QS}=\frac{QT}{PR}\)
∴ \(\frac{QR}{QS}=\frac{QT}{QP}\)
In ΔTQR, P and S are points on QT and QR respectively and \(\frac{QR}{QS}=\frac{QT}{QP}\)
∴ By theorem 6.2, SP || RT.
∴ ∠QPS = ∠QTR and ∠QSP = ∠QRT (Corresponding angles)
Now, in ΔPQS and ΔTQR.
∠QPS = ∠QTR,
∠QSP = ∠QRT and
∠PQS = ∠TQR (Same angle)
∴ By AAA criterion, ΔPQS ~ ΔTQR.

Question 5.
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS
Answer:
Given:
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS.
To prove ΔRPQ ~ ΔRTS

Proof : ∠P = ∠RTS
∴∠RPQ = ∠RTS.
In ΔRPQ and ΔRTS,
∠RPQ = ∠RTS and
∠PRQ = ∠TRS (Same angle)
∴ By AA criterion, ΔRPQ ~ ΔRTS.

Question 6.
In the given figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Answer:

ΔABE ≅ ΔACD (Given)
∴ AB = AC and AE = AD (CPCT)
∴ \(\frac{AE}{AC}=\frac{AD}{AB}\)
Now, in ΔADE and ΔABC,
\(\frac{AE}{AC}=\frac{AD}{AB}\)
and ∠DAE = ∠BAC (Same angle)
∴ By SAS criterion, ΔADE ~ ΔABC

Question 7.
In the given figure, altitudes AD and CE of ΔABC intersect each other at the point P.
Show that:
1. ΔAEP ~ ΔCDP
2. ΔABD ~ ΔCBE
3. ΔAEP ~ ΔADB
4. ΔPDC ~ ΔBEC
Αnswer:

1. In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Right angles)
∠EPA = ∠DPC (Vertically opposite angles)
∴ By AA criterion, ΔAEP ~ ΔCDP.

2. In ΔABD and ΔCBE,
∠ABD = ∠CBE (Same angle)
∠ADB = ∠CEB (Right angles)
∴ By AA criterion, ΔABD ~ ΔCBE.

3. In ΔAEP and ΔADB,
∠AEP = ∠ADB (Right angles)
∠EAP = ∠DAB (Same angle)
∴ By AA criterion, ΔAEP ~ ΔADB.

4. In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Right angles)
∠PCD = ∠BCE (Same angle)
∴ By AA criterion, ΔPDC ~ ΔBEC.

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Answer:
Given: E is a point on the side AD produced of parallelogram ABCD and BE intersects CD at F
To prove : ΔABE ~ ΔCFB

Proof: In parallelogram ABCD,
∠A = ∠C (Opposite angles)
∴ ∠BAE = ∠FCB ………..(1)
E lies on AD extended in parallelogram ABCD.
∴ AE || BC
∴ ∠AEB = ∠CBE (Alternate angles)
∴ ∠AEB = ∠CBF …………..(2)
Now, in ΔABE and ΔCFB,
∠BAE = ∠FCB [By (1)]
∠AEB = ∠CBF [By (2)]
∴ By AA criterion, ΔABE ~ ΔCFB.

Question 9.
In the given figure, ABC and AMP are two right triangles, right-angled at B and M respectively. Prove that :
1. ΔABC ~ ΔAMP
2. \(\frac{CA}{PA}=\frac{BC}{MP}\)
Answer:

In ΔABC and ΔAMP
∠ABC = ∠AMP (Right angles)
∠BAC = ∠MAP (Same angle)
∴ By AA criterion, ΔABC ~ ΔAMP [Result (1)]
Since ΔABC ~ ΔAMP, \(\frac{CA}{PA}=\frac{BC}{MP}\) [Result (2)]

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC – ΔFEG, show that:
1. \(\frac{CD}{GH}=\frac{AC}{FG}\)
2. ΔDCB ~ ΔHGE
3. ΔDCA ~ ΔHGF
Answer:

ΔABC ~ ΔFEG
∴ ∠A = ∠F, ∠B = ∠E and ∠ACB = ∠FGE …………….(1)
CD is the bisector of ∠ACB and GH is the bisector of ∠FGE.
∴ ∠ACD = ∠BCD = \(\frac{1}{2}\)∠ACB ……(2)
and ∠FGH = ∠EGH = \(\frac{1}{2}\)∠FGE ………..(3)
So, from (1), (2) and (3).
∠ACD = ∠FGH and ∠BCD = ∠EGH ……(4)
Now, in ΔDCB and ΔHGE,
∠B = ∠E [By (1)]
∠BCD = ∠EGH [By (4)]
Hence, by AA criterion,
ΔDCB ~ ΔHGE [Result (2)]
Again, in ΔDCA and ΔHGF
∠A = ∠F [By (1)]
∠ACD = ∠FGH [By (4)]
Hence, by AA criterion,
ΔDCA – ΔHGF [Result (3)]
Now, ΔDCA – ΔHGF
∴ \(\frac{CD}{GH}=\frac{AC}{FG}\) [Result (1)]

Question 11.
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Answer:

In ΔABC, AB = AC
∴ ∠ABC = ∠ACB
∴ ∠ABD = ∠ECF (∵ E lies on CB extended and F lies on AC.)
AD ⊥ BC
∴ ∠ADB = 90°
EF ⊥ AC
∴ ∠EFC = 90°
Now, in ΔABD and ΔECF
∠ABD = ∠ECF
∠ADB = ∠EFC (Both right angles)
∴ By AA criterion, ΔABD ~ ΔECF

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ΔABC ~ ΔPQR.
Answer:

Hence, by SSS criterion, ΔABD ~ ΔPQM.
∴ ∠ABD = ∠PQM
∴ ∠ABC = ∠PQR
Now, in ΔABC and ΔPQR,
\(\frac{AB}{PQ}=\frac{BC}{QR}\) and ∠ABC = ∠PQR
Hence, by SAS criterion, ΔABC ~ ΔPQR.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB. CD.
Answer:

In ΔCDA and ΔCAB.
∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Same angle)
∴ By AA criterion, ΔCDA ~ ΔCAB
∴ \(\frac{CD}{CA}=\frac{CA}{CB}\)
∴ CB . CD = CA . CA
∴ CA² = CB.CD

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides Pg and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Answer:

In ΔABC, AD is a median.
∴ BD = DC
Take point E on AD extended such that AD = DE and draw BE and CE.
∴ AE = 2AD.
In quadrilateral ABEC, diagonals AE and BC bisect each other.
∴ ABEC is a parallelogram.
∴ BE = AC (Opposite sides) ……(1)
Similarly, in ΔPQR, PM is a median.
∴ QM = MR
Take point N on PM extended such that PM MN and draw QN and RN.
∴ PN = 2PM
In quadrilateral PQNR, diagonals PN and QR bisect each other.
∴ PQNR is a parallelogram.
∴ QN = PR (Opposite sides) …………..(2)
Now,
∴\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (Given)
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\) [By (1) and (2)]
∴\(\frac{AB}{PQ}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ By SSS criterion ΔABE ~ ΔPQN.
∴ ∠BAE = ∠QPN
∴ ∠BAD = ∠QPM …………..(3)
In the same manner, it can be proved that ΔACE ~ ΔPRN
∴ ∠CAE = ∠RPN
∴ ∠CAD = ∠RPM …………..(4)
Adding (3) and (4).
∠BAD + ∠CAD = ∠QPM + ∠RPM
∴ ∠BAC = ∠QPR
Now, in ΔABC and ΔPQR,
\(\frac{AB}{PQ}=\frac{AC}{PR}\) and ∠BAC = ∠QPR
Hence, by SAS criterion, ΔABC ~ ΔPQR.

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same. time a tower casts a shadow 28 m long. Find the height of the tower.
Answer:

Here, AB is the vertical pole and AC is its shadow while PQ is the tower and QR is its shadow.
As both the shadows are measured at the same time, ∠C and ∠R both represent the elevation of the sun.
∴ ∠C = ∠R
In ΔABC and ΔPQR,
∠C = ∠R
∠B = ∠Q (Right angles)
∴ By AA criterion, ΔABC ~ ΔPQR.
∴ \(\frac{AB}{PQ}=\frac{BC}{QR}\)
∴ \(\frac{6}{PQ}=\frac{4}{28}\)
∴ PQ = \(\frac{6 \times 28}{4}\)
∴ PQ = 42 m
Thus, the height of the tower is 42 m.

Question 16.
If AD and PM are medians of triangles ABC and PQR respectively; where ΔABC ~ ΔPQR prove that \(\frac{AB}{PQ}=\frac{AD}{PM}\)
Answer:

(∵ AD and PM are medians of ΔABC and ΔPQR.)
Also, ∠ABC = ∠PQR
∴ ∠ABD = ∠PQM
Now, in ΔABD and ΔPQM,
\(\frac{AB}{PQ}=\frac{BD}{QM}\) and ∠ABD = ∠PQM
∴ By SAS criterion, ΔABD ~ ΔPQM.
∴ \(\frac{AB}{PQ}=\frac{AD}{PM}\)

Jharkhand Board JAC Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Additional Questions and Answers

Question 1.
Solve the following examples :
1. An object is placed at a distance of 20 cm in front of a convex mirror having focal length 0.3 m. Find the position, nature and magni-fication of the image formed by the mirror.
Answer:
a = 12 cm, virtual, erect and diminished, m = 0.6

2. The radius of curvature of a concave mirror is 12 cm. The image obtained by it is real and has magnification \(\frac { 3 }{ 2 }\). Find the object distance and image distance.
Answer:
u = – 10cm, v = – 15cm

3. A convex mirror having radius of curvature 3 m is fitted as a rear view mirror in a bus. Find the position, nature and magnification of the image of a car standing behind the bus at 6 m from the mirror.
Answer:
u = 1.2m, virtual, erect and diminished, m = 0.2

4. An object of height 2 cm is placed at 32 cm from a concave mirror. If the image formed by the mirror is real, inverted and 1.5 times the object in size, find the focal length of the mirror and the image distance.
Answer:
f = 19.2 cm, u = – 48cm

5. The refractive index of diamond relative to glass is 1.61. If the refractive index of glass is 1.5, find the absolute refractive index of diamond.
Answer:
n_da = 2.415

6. The absolute refractive index of ice is 1.31. The speed of light in vacuum is 3 x 10⁸ms^-1. Then find the speed of light in ice.
Answer:
v = 2.29 x 10⁸ ms^-1

7. The speed of light is 2 x 10⁸ms^-1 in a glass having refractive index 1.5. In a liquid the ; speed of light is 2.5 x 10⁸ms^-1. Find the absolute refractive index of the liquid.
Answer:
n_1a = 1.2

8. A ray of light incident at an angle of 45° on a rectangular glass slab having refractive index \(\sqrt{2}\) from air. Find the angle of refraction, Take the refractive index of air as 1.
Answer:
r = 30°

9. The image of a candle is formed at 10 cm from a concave lens having focal length 15 cm. At what distance is the candle placed from the mirror?
Answer:
u = – 30 cm

10. A virtual Image is obtained, at 20 cm in front of a lens, when an object is placed at 60 cm in front of the lens. Calculate the focal length s of the lens. State the type of the lens.
Answer:
f = – 30 cm, concave lens

11. The focal length of a lens is – 25 cm. Calculate the power of the lens. Mention the type of the lens.
Answer:
P = – 4 D, concave lens

12. The power of a lens is +5 D. An object is placed at a distance equal to twice the focal length of the lens. (1) Calculate the focal length of the lens. (2) Find the position of the image.
Answer:

1. f = 20cm
2. on the opposite side of the object at a distance of 40 cm from the lens

13. The focal length of a convex lens is 25 cm. The image is obtained at a distance of 75 cm on the other side of the lens. Find the position of the object. What is the nature of the image?
Answer:
u = – 37.5cm; real, inverted and magnified

14. The power of a lens is -4D. A virtual and erect image is obtained at 5 cm from it. Find the position of the object.
Answer:
u = – 6.25 cm

15. A ray of light covers 100 m in vacuum in a certain time. How much distance will it cover in glass of refractive index 1.5 in the same time?
Answer:
\(\frac { 100 m }{ 1.5 }\) = 66.67 m

16. An object is placed at a distance of 25 cm from a concave mirror of focal length 20 cm. Find the image distance and nature of the image.
Answer:
– 100 cm; real, inverted and enlarged

17. An object, 4 cm in height, is placed at 25 cm from a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be kept, so that a sharp image of the object can be obtained on the screen? Find the nature and height of the image.
Answer:
v = – 37.5 cm; size/height = 6 cm; Image is real, inverted and enlarged.

18. A doctor has prescribed a corrective lens of power +1.5 D to a patient. Find the focal length of the lens in centimetre. Is the prescribed lens diverging or converging?
Answer:
f = 67 cm, converging

19. The Sun rays are incident on a concave mirror, parallel to its principal axis. If the image of the Sun is formed at 12 cm distance from the pole of the mirror, find its radius of curvature.
Answer:
24 cm

20. An object is placed at 12 cm in front of a concave mirror. It forms a real and inverted image. Where is the image formed, if the image height is double the object height?
Answer:
In front of the concave mirror at a distance of 24 cm from it.

21. What is the position of the image when an object is placed at 10 cm in front of a concave mirror? The image is virtual and double in size as
compared to the object.
Answer:
Behind the mirror at a distance of 20 cm from it.

22. Find the size of the image formed by a concave lens of focal length 20 cm, when an object having 4 cm height is placed at 60 cm from it.
Answer:
Image size/height = 1.0 cm

23. Determine the nature and size of the image, and the type of mirror for the image formed by mirrors corresponding to magnifications + 1, -1, +0.5, -0.5, +5.0 and -5.0.
Answer:

No.	Magni-fication (m)	Nature and size of Image	Type of Mirror
1.	+ 1	Virtual, erect and of the same size as the object	Plane
2.	– 1	Real, inverted and of the same size as the object	Concave
3.	+ 0.5	Virtual, erect and diminished	Convex
4.	– 0.5	Real, inverted and diminished	Concave
5.	+ 5.0	Virtual, erect and magnified	Concave
6.	– 5.0	Real, inverted and magnified	Concave

24. An object of height 4 cm is placed at 18 cm from a concave mirror having focal length 12cm. Find the position, nature and height of the image.
Answer:
v = – 36 cm, h’ = – 8 cm. Real, inverted and enlarged

25. A convex lens forms a real and inverted image of an object at a distance of 40 cm from it. What is the distance of the object from the lens if the size of the image is the same as that of the object? Determine the power of the lens.
Answer:
u = – 40 cm, P = + 5.0 D

26. An object of height 5 cm is placed at 10 cm from a convex mirror of focal length 15 cm. Find the position, nature and size of the image.
Answer:
v = 6 cm, h’ = 3 cm. Virtual, erect and diminished

27. The refractive index of glass with respect to water is 1.12. Find the absolute refractive index of water if the absolute refractive index of glass is 1.5.
Answer:
1.34

28. Light propagating in a rectangular glass slab enters into water. The refractive index of water with respect to glass is 0.9. The angle of incidence at the surface separating the two media is 26° 48′. Find the angle of refraction. Take sin 26° 48′ = 0.45 approximately.
Answer:
30°

29. An object is placed perpendicular to the principal axis of a concave lens of focal length 30 cm. Find the position of the image when the object is at a distance of 20 cm from the lens.
Answer:
v = – 12 cm

30. The power of a convex lens is +4.0 D. At what distance from the lens should the object be placed to obtain its real and inverted image of the same size as the object on the screen?
Answer:
u = – 50 cm

Question 2.
Give four points of difference between the following terms/quantities:
(1) Concave mirror and Convex mirror
Answer:

Concave mirror	Convex mirror
1. Its inner surface is reflecting.	1. Its outer surface is reflecting.
2. It forms a real or virtual image of an object depending on the position of the object.	2. It always forms a virtual image.
3. The real image can be smaller or of the same size as the size of the object or bigger than the object. The virtual image is always bigger than the object.	3. The image is always smaller than the object.
4. This mirror is used in a torch, headlight of vehicles, search-light, solar cooker, etc.	4. This mirror is used as a rear-view (wing) mirror in vehicles.

(2) Real Image and Virtual Image
Answer:

Real Image	Virtual Image
1. In this case, the rays of light emanating from an object meet after reflection or refraction and hence, an image is formed.	1. In this case, the rays of light emanating from an object (when extended backwards) appear to meet after reflection or refraction.
2. It can be obtained on a screen.	2. It cannot be obtained on a screen.
3. It is inverted compared to an object.	3. It is erect compared to an object.
4. In case of a spherical mirror it is formed towards the reflecting surface in front of the mirror.	4. In case of a spherical mirror it is formed behind the reflecting surface of the mirror i.e., in the mirror itself.

(3) Convex lens and Concave lens
Answer:

Convex lens	Concave lens
1. It is thick at the centre and thin at the edges.	1. It is thin at the centre and thick at the edges.
2. It converges parallel rays incident on it.	2. It diverges parallel rays incident on it.
3. It forms a real or virtual image of an object depending on the position of the object.	3. It always forms a virtual image.
4. The virtual image obtained with it is always enlarged.	4. The virtual image obtained with it is always diminished.

Question 3.
Give scientific reasons for the following statements:
(1) In a headlight, torch and search-light, a concave mirror (of small aperture) is used.
Answer:
In a headlight, torch and search-light, the light source is placed at the principal focus of a concave mirror. As a result, an intense parallel beam of light comes out from a headlight, torch, search-light.

[Note: Actually a parabolic mirror = reflector is used in a headlight, torch and search-light.]

(2) The rear-view mirror (back-view mirror) of the vehicles is a convex mirror.
Answer:
The field view of a convex mirror is very large, i.e., it gives over-all view of vast area of the backside of a vehicle and it forms a virtual, erect and diminished image of a backside object. Hence, the vehicle can be driven safely by viewing backside fully.

(3) A big concave mirror is used in a solar furnace and solar cooker.
Answer:
Parallel rays of light coming from the Sun fall on the concave mirror fitted in a solar s furnace and solar cooker and after reflection, these rays meet at the principal focus F of the concave i mirror. The greater the aperture of the mirror, the greater is its power to collect light. As a result, abundant amount of heat energy is produced.

(4) For shaving as well as for make-up a concave mirror is used.
Answer:
If a concave mirror is held such that the face is between the pole and principal focus of the concave mirror, a virtual, erect and magnified image of the face is obtained behind the mirror. This is of great help in shaving as well as make-up.

[A dentist also uses a concave mirror to obtain an enlarged (magnified) image of teeth.]

(5) It is difficult to pierce a fish swimming in a river.
Answer:
The rays of light from a swimming fish emerge from river – water and enter into air. Thus, they travel from a denser medium to a rarer medium. Therefore they bend away from the normal at the interface.

Hence, the correct/true position of the fish cannot be determined by the observer from air. That is why it is difficult to pierce a fish swimming in a river.

(6) A watch repairer uses a convex lens.
Answer:
If an object is kept between optical centre O and principal focus F of a convex lens, we get a virtual, erect and enlarged image of the object on the same side of the object, beyond the distance 2f from the lens. Thus, by keeping a watch between the lens and its principal focus, the watch repairer can see minute parts of the watch clearly, so that repairing becomes easy.

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence :
(1) Can a magnified image be formed by a convex mirror? State Yes or No.
Answer:
No

(2) What is the angle of reflection when a ray c of light falls normally on a mirror?
Answer:
0°

(3) In a ray diagram, how do you draw the normal at point A on a spherical mirror?
Answer:
By drawing a line joining point A on the spherical mirror with the centre of curvature of the mirror.

(4) State the mirror formula to find the focal length of a spherical mirror.
Answer:
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\)

(5) The focal length of a convex mirror is 15cm. What is its radius of curvature?
Answer:
R = 2f = 30 cm

(6) Which two quantities change when a ray of light
Answer:
wavelength and velocity

(7) The magnification produced by concave lens is always less than 1. Why?
Answer:
The image formed by a concave lens (for any position of an object) is always diminished.

(8) Does the refractive index of a medium depend on the frequency of light passing S through it?
Answer:
No. [From n = \(\frac{c}{v}=\frac{f \lambda}{v \lambda^{\prime}}=\frac{\lambda}{\lambda^{\prime}}\) (∵frequency f = constant)]

(9) Which mirror has large vision area?
Answer:
Convex mirror

(10) Absolute refractive index of carbon disulphide is 1.63. Then, explain this statement in context to speed of light.
Answer:
From n = \(\frac { c }{ v }\), v = \(\frac { c }{ n }\) = \(\frac{3 \times 10^8}{1.68}\)
= 1.84 x 10⁸ m s^-1
Hence speed of light in carbon disulphide would be 1.84 x 10⁸ m s^-1.

(11) Between which two points on the principal axis of a concave mirror should an object be placed so that the magnification produced by the mirror is m = – 3?
Answer:
Here, m = – 3. So, the image would be real, inverted and magnified. So, the object should be placed between F and C.

(12) What is mirror?
Answer:
Mirror means smooth and polished surface which reflects (approximate 95 %) most of the light incident on it.

(13) Which lens is used as a magnifying lens?
Answer:
Convex lens

Question 2.
Fill in the blanks :

1. The centre of a spherical mirror is called the ………………… and the centre of a lens is called the …………………
2. The power of a convex lens having focal length 10 cm is ………………… D.
3. When a ray of light travels from water to air, its speed …………………
4. Focal length of a convex lens is …………………
5. ………………… image can be obtained on a screen.
6. Magnification produced by a spherical mirror is positive. So, the type of image is …………………
7. Magnification produced by a lens is negative. So, the type of image is ………………… and …………………
8. A ray of light is incident at an angle of 30° to the surface of a plane mirror. Then the angle between the incident ray and reflected ray must be …………………
9. If a virtual, erect and magnified image of i an object is formed by a convex lens; the object must be at a distance …………………
10. In case of a mirror or lens, the ratio of the image height to object height is called …………………
11. A ………………… lens is thicker at its edges than at its centre.
12. ENT specialist doctor uses a ………………… of small aperture.
13. A ………………… is used as a magnifying glass.
14. If the power of the correcting lens used in spectacles is – 0.4 D, then the type of lens must be …………………
15. ………………… is the instrument used to measure the power of a lens.

Answer:

1. Pole, Optical centre
2. 10
3. increases
4. positive
5. Real
6. erect and virtual
7. inverted and real
8. 120°
9. less than the focal length of the lens
10. magnification
11. concave
12. concave mirror
13. convex lens
14. concave
15. Dioptre meter

Question 3.
State whether the following statements are true or false:

1. A virtual image can be obtained on a screen.
2. In case of irregular reflection, the beam of reflected light remains parallel in a specific direction.
3. Laws of reflection are applicable to any type of mirror.
4. Laws of reflection are not applicable to a rough and irregular surface.
5. The diameter of the reflecting surface of a spherical mirror is known as the aperture of the mirror.
6. At least, four rays are required to obtain the position of an image formed by a mirror.
7. If an object is kept at infinite distance from concave mirror, the image is formed at the centre of curvature of the mirror.
8. Air is optically denser than water.
9. When a ray of light enters from an optically denser medium to an optically rarer medium, its speed decreases.
10. The central part of a thin lens made of glass works as a rectangular glass slab.
11. According to the New Cartesian sign convention for every mirror and lens, Object distance is taken as negative.
12. For a mirror, m = \(\frac { v }{ u }\) and for a lens m = – \(\frac { v }{ u }\)
13. A lens has power P = 5 D. So, its focal length is f = 0.2 cm.

Answer:

1. False
2. False
3. True
4. False
5. True
6. False
7. False
8. False
9. False
10. True
11. True
12. False
13. False

Question 4.
Match the following :
(1)

Column ‘A’	Column ‘B’
1. Concave mirror	p. Negative power
2. Convex lens	q. Always m = positive
3. Convex mirror	r. Always m = negative
4. Concave lens	s. Used for shaving
	t. Simple microscope

Answer:
(1 – s), (2 – t), (3 – q), (4 – p).

(2)

Column ‘A’	Column ‘B’
1. Concave mirror	p. It is used by doctors
2. Convex lens	q. It is used as a rear-view mirror in vehicles
3. Convex mirror	r. Power is negative
4. Concave lens	s. It is used in a compound microscope

Answer:
(1 – p), (2 – s), (3 – q), (4 – r).

Question 5.
Choose the correct option from those given below each question:
1. What is the wavelength range of visible light?
A. 4 x 10^-7 m to 8 x 10^-7 m
B. 4 x 10^-9 m to 8 x 10^-9 m
C. 4 x 10^-5 m to 8 x 10^-5 m
D. 4 x 10^-6 m to 8 x 10^-6 m
Answer:
A. 4 x 10^-7 m to 8 x 10^-7 m

2. What is the relation between the radius of curvature (K) and the focal length (f) of a spherical mirror?
A. R = f/ 2
B. R = f
C. R = 2f
D. R = 3f
Answer:
C. R = 2f

3. Through which of the following points, will a ray passing through the centre of curvature (C) and reflected by a concave mirror pass?
A. Focus
B. Centre of curvature
C. Pole
D. All
Answer:
B. Centre of curvature

4. Where, in front of a concave mirror, should an object be placed to get its virtual and erect image?
A. At principal focus F
B. At centre of curvature C
C. Between the principal focus and the pole
D. Beyond the centre of curvature
Answer:
C. Between the principal focus and the pole

5. The magnification produced by a plane mirror is always
A. more than 1
B. 1
C. less than 1
D. zero
Answer:
B. 1
Hint: m = \(\frac { h’ }{ h }\) = 1 (∵ Here, h’ = h)
or
m = – \(\frac { v }{ u }\)
Here, | v | = | u | and u is negative
while v is positive.
∴ We get, m = 1

6. The focal length of a plane mirror is ……………
A. zero
B. infinity
C. equal to half the object distance f
D. equal to the object distance
Answer:
B. infinity
Hint: In case of plane mirror | v | = | u |, also according to the New Cartesian sign convention, v is positive and u is negative.
Now, from the mirror formula \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), for a plane mirror, as |u| = [u| and u is negative while v is positive, we get \(\frac { 1 }{ f }\) = 0 ∴ f = ∞ (infinity)

7. The distance between the object at 2 m from a plane mirror and its image is ………………..
A. 4 m
B. 1 m
C. 2 m
D. 3 m
Answer:
A. 4 m
Hint: The distance between the object and its image = | u | + | v |
= 2 + 2 (∵ Here, | u | = | v |)
= 4 m

8. Where should an object be placed to obtain its image as real, inverted and of the same height as the object by a convex lens?
A. At the principal focus
B. Between the principal focus and centre of curvature
C. At the centre of curvature
D. Between the optical centre and principal focus
Answer:
C. At the centre of curvature

9. Which of the following materials has maximum optical density?
A. Glass
B. Water
C. Carbon disulphide
D. Diamond
Answer:
D. Diamond
Hint: From the given substances, diamond has highest refractive index.

Question 10.
The absolute refractive index of any medium is …………………
A. 1
B. > 1
C. < 1
D. zero
Answer:
B. > 1

11. Which of the lenses with focal length 10 cm, 20cm, 25cm and 50cm has maximum power?
A. 50 cm
B. 25 cm
C. 20 cm
D. 10 cm
Answer:
D. 10 cm

Hint: From power P = \(\frac { 1 }{ f }\), the lens with less focal length f has more power P. So, the lens with focal length 10 cm has maximum power.

12. What is the focal length of a convex lens having power + 5.0 D?
A. – 10 cm
B. – 20 cm
C. + 10 cm
D. + 20 cm
Answer:
D. + 20 cm
Hint : P = \(\frac { 1 }{ f }\)
∴ f = \(\frac { 1 }{ p }\) = \(\frac { 1 }{ 5 }\) = 0.2 m = + 20 cm

13. If the absolute refractive indices of water, benzene and sapphire are 1.33, 1.50 and 1.77 respectively, then which medium has maximum relative refractive index?
A. Sapphire relative to water
B. Sapphire relative to benzene
C. Benzene relative to water
D. Water relative to benzene
Answer:
A. Sapphire relative to water
Hint:
\(n_{21}=\frac{n_2}{n_1}\)
∴ n₂₁ is maximum, when n₂ is maximum and n₁ is minimum.
Hence, it is obvious that in the present case, medium 1 must be water and medium 2 must be sapphire.
∴ Refractive index of sapphire with respect to water would be maximum.

14. Which type of image is formed by a plane mirror?
A. Real and inverted
B. Real and erect
C. Virtual and erect
D. Virtual and inverted
Answer:
C. Virtual and erect

15. If the absolute refractive indices of water and glass are \(\frac { 4 }{ 3 }\) and \(\frac { 3 }{ 2 }\) respectively, then what is the ratio of the speed of light in water to that in glass?
A. 2
B. \(\frac { 8 }{ 9 }\)
C. \(\frac { 9 }{ 8 }\)
D. \(\frac { 1 }{ 2 }\)
Answer:
C. \(\frac { 9 }{ 8 }\)
Hint: Taking water as medium 1 and glass as medium 2, speed of light in water would be v₁ and speed of light in glass would be v₂.
Here, n₁ = \(\frac { 4 }{ 3 }\) and n₂ = \(\frac { 3 }{ 2 }\)
Now, n₂₁ = \(\frac{n_2}{n_1}=\frac{v_1}{v_2}\)
∴ \(\frac{3 / 2}{4 / 3}=\frac{v_1}{v_2}\)
∴ \(\frac{v_1}{v_2}=\frac{9}{8}\)

16. The absolute refractive indices of water, glass and diamond are 1.33, 1.50 and 2.42 respectively. Which medium has maximum optical density ?
A. Water
B. Glass
C. Diamond
D. Data not enough to draw any conclusion
Answer:
C. Diamond
Hint: The greater the absolute refractive index, the greater is the optical density.

17. Which of the following always form virtual image ?
A. Concave mirror and convex lens
B. Convex mirror and concave lens
C. Convex mirror and convex lens
D. Concave mirror and concave lens
Answer:
B. Convex mirror and concave lens

18. What is the angle of refraction for a light ray incident normal to the surface of a medium?
A. 90°
B. 60°
C. 30°
D. 0°
Answer:
D. 0°
Hint: According to general form of Snell’s law n₁ sin i = n₂ sin r. As i = 0°, then r = 0° because value of n₁ and n₂ are never zero.

19. No matter how far you stand from a mirror, your image appears always erect. The mirror is likely to be ………..
A. plane only.
B. concave only.
C. convex only.
D. either plane or convex.
Answer:
D. either plane or convex.
Hint: In case of both plane and convex mirror for every position of an object its image would be always erect.

20. A virtual image of an object is formed by a mirror, then …………….
A. it would be a convex mirror.
B. it would be a concave mirror.
C. it would be a plane mirror.
D. it would be any one of the mirrors given above.
Answer:
D. it would be any one of the mirrors given above.

21. An object is kept at the centre of curvature of a concave mirror. The distance between the image of the object and pole of the mirror ……………..
A. would be equal to f.
B. would be less than 2f but more than f.
C. would be equal to 2f.
D. would be more than 2f.
Answer:
C. would be equal to 2f.

22. A ray of light travelling in air is obliquely incident on undisturbed horizontal surface of pond water. Inside the water, it…
A. will travel without deviation.
B. will bend away from the normal drawn at the point of incidence.
C. will bend towards the normal drawn at the point of incidence.
D. will turn back following the original path.
Answer:
C. will bend towards the normal drawn at the point of incidence.

23. As shown in the following figure, a light ; ray travels from medium A to medium B. Then refractive index of medium B relative? to medium A would be ……………

A. \(\frac{\sqrt{3}}{\sqrt{2}}\)
B. \(\frac{\sqrt{2}}{\sqrt{3}}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\sqrt{2}\)
Answer:
A. \(\frac{\sqrt{3}}{\sqrt{2}}\)
Hint: From Snell’s law, refractive index of medium B relative to medium A.
n_BA = \(\frac{\sin 60^{\circ}}{\sin 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\)

24. As shown in the following figure, a light ray travels from medium A to medium B. Then refractive index of medium A relative to medium B would be

Answer:
B. < 1

25. Magnification produced by a rear-view mirror fitted in vehicles ………….
A. is always less than 1.
B. is always equal to 1.
C. is always more than 1.
D. can be more than or less than 1 depending upon the position of the object in front of the mirror.
Answer:
B. is always equal to 1.
Hint: Convex mirror is used as rear-view mirror in vehicles. Image formed by it is always virtual, erect and diminished in size compared to the

26. The absolute refractive indices of three substances A, B and C are 1.33, 1.65 and 1.46 respectively. In which substance is the speed of light maximum?
A. A
B. B
C. C
D. Would be the same in A, B and C
Answer:
A. A
Hint: n_m = \(\frac { c }{ v }\)

27. An object placed in front of a convex mirror is moved from far-off distance towards the pole of the mirror. Then size of the image …
A. increases.
B. decreases.
C. remains constant.
D. first increases and then decreases.
Answer:
A. increases.

28. An object placed in front of a convex mirror is moved from far-off distance towards the pole of the mirror. Then the image ……………..
A. moves towards the pole.
B. moves away from the pole.
C. first moves away from the pole and then towards the pole.
D. first moves towards the pole and then away from the pole.
Answer:
A. moves towards the pole.

29. What is the centre of a spherical mirror called?
A. Principal focus
B. Centre of curvature
C. Principal axis
D. Pole
Answer:
D. Pole

30. What is the diameter of the reflecting circular surface of a spherical mirror called?
A. Aperture
B. Radius of curvature
C. Focal length
D. Principal axis
Answer:
A. Aperture

31. Where should the object be placed to obtain its real image of the same size as the object using a concave mirror?
A. At infinite distance from the mirror
B. At the centre of curvature of the mirror
C. At the principal focus of the mirror
D. Between the principal focus and the centre of curvature of the mirror
Answer:
B. At the centre of curvature of the mirror

32. Where should the object be placed to obtain a real and diminished image using a concave mirror?
A. Between the principal focus and the pole of the mirror
B. At the centre of curvature of the mirror
C. Between the principal focus and the centre of curvature of the mirror
D. Beyond the centre of curvature of the mirror
Answer:
D. Beyond the centre of curvature of the mirror

33. Where should the object be placed to obtain a real and magnified image using a concave mirror?
A. Between the pole and the principal focus of the mirror
B. At the principal focus of the mirror
C. Between the principal focus and the centre of curvature of the mirror
D. At the centre of curvature of the mirror
Answer:
C. Between the principal focus and the centre of curvature of the mirror

34. Which type of image cannot be obtained by a concave mirror?
A. Real and enlarged
B. Real and diminished
C. Virtual and enlarged
D. Virtual and diminished
Answer:
D. Virtual and diminished

35. Which type of image is formed by a convex mirror?
A. Real, inverted and diminished
B. Virtual, erect and diminished
C. Real, inverted and enlarged
D. Virtual, erect and enlarged
Answer:
B. Virtual, erect and diminished

36. What happens to an obliquely incident ray travelling from glass to air?
A. It bends towards the normal.
B. It bends away from the normal.
C. It does not undergo refraction.
D. It turns back following the original path.
Answer:
B. It bends away from the normal.

37. A point object emits rays in all directions. Consider one ray that the object emits which is parallel to the principal axis of a concave mirror as shown in the following figure.

Select the point in the figure given here, through which the reflected ray passes.
A. C
B. F
C. A
D. P
Answer:
B. F

38. Which type of image of an object cannot be obtained by a convex lens?
A. Real and diminished
B. Real and enlarged
C. Virtual and diminished
D. Virtual and enlarged
Answer:
C. Virtual and diminished

39. From which of the following letters, we cannot obtain its laterally inverted image?
A. N
B. O
C. P
D. Q
Answer:
B. O

40. From which of the following letters, can we obtain its laterally inverted image?
A. W
B. X
C. Y
D. Z
Answer:
D. Z

41. A boy is running with speed u towards a plane mirror, then at what speed, his image will be running towards him?
A. 2v
B. > v
C. v
D. < v
Answer:
A. 2v

42. If a spherical mirror is immersed in water, then its focal length…
A. changes.
B. does not change.
C. may or may not change
D. becomes zero.
Answer:
B. does not change.

Question 6.
Answer the following questions in very short as directed (Miscellaneous) :
(1) Write Snell’s law in mathematical form.
Answer:
If i is the angle of incidence and r is the angle of refraction, then
\(\frac { sin i }{ sin r }\) = constant
This constant is called the refractive index of the second medium with respect to the first medium.

(2) There are five different transparent media, which are in physical contact with one another. Then, write the formula for the refractive index of the fifth medium with respect to the first medium.
Answer:
n₅₁ = \(\frac{n_5}{n_1}\)

(3) If the radius of curvature of a spherical mirror is 40 cm, what is its focal length in metre?
Answer:
f = \(\frac{R}{2}=\frac{40 \mathrm{~cm}}{2}\) = 20 cm = 0.2 m

(4) A concave mirror produces a real image of an object, placed at 10 cm from the mirror. If the image is enlarged by a factor of four relative to object, find the focal length of the mirror.
Answer:
Here, u = – 10 cm
As image formed is real and four times larger than object m = – 4
Now, m = – \(\frac { v }{ u }\)
∴ – 4 = – \(\frac { v }{ (-10) }\) ∴ v = – 40 cm
Using the mirror formula
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
We get
\(\frac{1}{f}=\frac{1}{(-10)}+\frac{1}{(-40)}\)
∴ f = \(\frac{-40 \times-10}{-40-10}=\frac{400}{-50}\)
∴ f = – 8 cm

(5) The magnification of an image formed by a convex lens is – 1. Find the object distance in terms of its focal length.
Answer:
Here, m = – 1 and u = – u
So, from m = \(\frac { v }{ u }\)
– 1 = \(\frac { v }{ (-u) }\)
∴ v = + u
Now, \(\frac{1}{f}=\frac{-1}{u}+\frac{1}{v}\)
∴ \(\frac{1}{f}=\frac{-1}{(-u)}+\frac{1}{(+u)}\)
∴ \(\frac{1}{f}=\frac{2}{u}\)
∴ u = 2 f

(6) An object is kept at 40 cm from a concave lens of focal length 60 cm. Find the image distance.
Answer:
Here, u = – 40 cm; f = – 60 cm
Using the lens formula,
\(\frac{1}{f}=\frac{-1}{u}+\frac{1}{v}\)
We get
∴ \(\frac{1}{-60}=\frac{-1}{-40}+\frac{1}{v}\)
∴ \(\frac{1}{v}=\frac{-1}{40}-\frac{1}{60}\)
∴ \(\frac{1}{v}=-\left(\frac{1}{40}+\frac{1}{60}\right)\)
∴ v = \(-\left(\frac{60 \times 40}{40+60}\right)=-\frac{2400}{100}\) = – 24 cm

(7) What is diffused reflection ?
Answer:
When a parallel beam of light falls on a rough surface, the light is reflected in different directions. This type of reflection is known as diffused reflection.

(8) The magnifying power of convex lens is ……………….
A. > 1
B. < 1
C. = 1
D. cannot be determined
Answer:
A, B and C

(9) A pencil, when dipped obliquely in water in a transparent glass tumbler, appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use some other liquid like kerosene or turpentine? Support your answer with reason.
Answer:
Yes.
The angle of refraction depends on the refractive index of the medium. So, bending will vary in different liquids. It will be less for a liquid with low refractive index and more for a liquid with high refractive index.

The refractive index of kerosene as well as that of turpentine oil is more than that of water. Hence, the bending will be more for both the oils than in case of water.

(10) Draw a ray diagram showing the path of a ray of light when it enters at oblique incidence from air into water.
Answer:

(11) Draw a ray diagram showing the path of a ray of light when it enters at oblique incidence from water into air.
Answer:

(12) A student traces the path of a ray of light passing through a rectangular glass slab and marks the angle of incidence i, angle of refraction r and angle of emergence e, as shown in the figure :

Which angle is correctly marked?
Answer:
r.

(13) What is the angle of reflection, when a ray of light falls normally on the surface of a plane mirror?
Answer:
r = 0°

(14) A ray of light strikes a plane mirror at an angle of 50° to the mirror surface. What will be the angle of reflection?
Answer:
40°

(15) Name the spherical mirror which has virtual principal focus.
Answer:
Convex mirror

(16) What is the angle of incidence for a ray of light passing through the centre of curvature of a concave mirror?
Answer:
0°

(17) What sign [+ve (plus) or -ve (minus)] has been given to the following on the basis of the New Cartesian sign convention?
(a) Height of a real image and
(b) Height of a virtual image?
Answer:
(a) -ve (minus)
(b) +ve (plus)

(18) What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is (a) + 3 (b) – 2 ?
Answer:
(a) Virtual
(b) Erect

(19) An object is placed at a very long distance in front of a convex mirror of radius of curvature 30 cm. State the position of its image.
Answer:
At the principlal focus; 15 cm behind the convex mirror.

(20) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to the glass is 4/6. Do you agree with this statement?
Answer:
Yes

(21) Name the following:
A point inside a lens through which the light passes undeviated.
Answer:
Optical centre.

(22) A 2 cm high object is placed at a distance of 2f from a convex lens. What is the height of the image formed?
Answer:
m = \(\frac { v }{ u }\) = \(\frac { h’ }{ h }\)
∴ h’ = h(\(\frac { v }{ u }\))
Here, u = – 2f and also v = – 2f
∴ The height of the image formed,
h’ = 2 cm\(\left(\frac{-2 f}{-2 f}\right)\) (∵ h = 2 cm)
= 2 cm

(23) Where should an object be placed in order to use a convex lens as a magnifying glass?
Answer:
At a distance less than the focal length of the convex lens.

(24) Draw the given diagram in your answer book and complete the path of the ray of light on refraction by the lens.

Answer:

(25) Copy and complete the diagram given below to show what happens to the ray of light when it passes through the concave lens :

Answer:

(26) A concave lens produces an image at 20 cm from the lens of an object placed at 30 cm from the lens. Find the focal length of the lens.
Answer:
Using the lens formula,

(27) Which causes more bending (or more refraction) of light rays passing through it: a convex lens of long focal length or a convex lens of short focal length?
Answer:
A convex lens of short focal length

(28) Which has more power : a thick convex lens or a thin convex lens made of the same material (glass)? Give reason for your choice.
Answer:
Thick convex lens because it has shorter focal length and hence more power.

(29) For a converging lens, the power is positive while for a diverging lens, the power is negative. Agree / Disagree?
Answer:
Agree

(30) The optician’s prescription for a pair of spectacles is :
Right eye : – 3.50 D, Left eye : – 4.00 D
(a) Are these lenses thinner at the middle or at the edges?
(b) Which is the weaker eye?
Answer:
(a) Thinner at the middle
(b) Left eye

(31) A person got his eyes tested by an optician. The prescription for the spectadle lenses to be made reads :
Right eye : + 2.00 D, Left eye : + 2.50 D
(a) State whether these lenses are thicker in the middle or at the edges.
(b) State whether these spectacle lenses will converge light rays or diverge light rays.
Answer:
(a) Thicker in the middle
(b) Converge light rays

(32) Match the following columns correctly:

Column I	Column II
1. Converging mirror	p. It is used as rear view mirror in vehicles
2. Diverging mirror	q. It is used in the design of solar furnaces
	r. It is used in a dressing-table

Answer:
(1 – q), (2 – p).

(33) Match the following columns correctly:

Column I	Column II
1. Converging lens	p. It Is a concave lens
2. Diverging lens	q. It acts as a magnifier
	r. It is used as a reflector

Answer:
(1 – q), (2 – p).

(34) State any two factors on which the lateral displacement of an emergent ray from a rectangular glass slab depend.
Answer:

1. Angle of incidence
2. Thickness of the slab

(35) What is the minimum distance between an object and its real image in case of a concave mirror?
Answer:
0 (zero). [When the object is at centre of curvature C of the mirror.]

(36) When a light ray passes obliquely through the atmosphere in the upward direction, how does its path generally change?
Answer:
Light will bend away from its normal direction.

(37) Explain briefly, why a ray of light passing through the centre of curvature of a concave mirror, gets reflected along the same path.
Answer:
The ray of light passing through the centre of curvature of a concave mirror is incident on the mirror along its normal. So, i = r = 0. Therefore, the ray retraces its path.

(38) A ray of light is incident on a concave mirror as shown in the adjoining figure. Redraw the diagram and complete the path of the ray after reflection from the mirror.

Answer:

(39) A ray of light is incident on a convex mirror as shown in the adjoining figure. Redraw the diagram and complete the path of the ray after reflection from the mirror.

Answer:

(40) What is the unit of refractive index?
Answer:
Refractive index as no unit as it is the ratio of two similar quantities.

(41) State the conditions for no bending of light travelling from one medium to another.
Answer:

1. Light incident normally.
2. Equal refractive index of the two different media.

(42) Why is the refractive index of atmosphere different at different altitudes?
Answer:
The refractive index of atmosphere is different at different altitudes because the air density changes with altitude.

(43) For the same angle of incidence in media P Q and R, the angles of refraction are 45°, 35° and 30° respectively. In which medium is the speed of light minimum? Give reason for your answer.
Answer:
In medium R, the speed of light is minimum.
In the usual notation, as n = \(\frac { c }{ v }\) = \(\frac { sin i }{ sin r }\).
For medium R, r is minimum and hence the speed v is minimum.

(44) Observe the following incomplete ray diagram, where the image A’B’ is formed after refraction from a convex lens :

On the basis of the above information, fill in the blanks :
(1) The position of object AB (not shown in the figure) must be
(2) The size of the object must be than the size of the image.
Answer:
(1) beyond 2F₁
(2) greater.

(45) Raju focussed the image of a candle flame on a white screen using a convex lens. He noted the position of the candle as 26.0 cm, the position of the convex lens as 50.0 cm and the position of the screen as 74.0 cm. What is the focal length of the convex lens?
Answer:
Here, u = the distance of the candle from the convex lens = – (50.0 – 26.0) = – 24.0 cm
v = the distance of the screen from the convex lens = (74.0 – 50.0) = 24.0 cm
From the lens formula,
∴ \(\frac{1}{f}=\frac{-1}{u}+\frac{1}{v}\)
∴ \(\frac{1}{f}=\frac{-1}{(-24)}+\frac{1}{24}=\frac{2}{24}\)
∴ f = 12 cm

(46) Study the given ray diagram. Name the device Y.

Answer:
Device Y is a concave mirror.

(47) Which property of a concave mirror is utilized for using it as a shaving mirror ?
Answer:
When an object is held between the pole and the principal focus of a concave mirror, a virtual, erect and magnified image of the object is formed. This property of a concave mirror is utilized for using it as a shaving mirror.

(48) How do you draw a normal to a spherical mirror at a particular point on the mirror?
Answer:
Join that point on the mirror to the centre of curvature of the mirror and extend the line if required.

(49) The magnification produced by a concave mirror is -1. What is the position of the object?
Answer:
The object must be at the centre of curvature of the concave mirror.
(The image formed is real, inverted and of the same size as the object. That is why magnification = -1).

(50) The magnification produced a spherical mirror is ±2. What kind of mirror can it be?
Answer:
The mirror must be a concave mirror. (Only then magnification can be positive or negative).

(51) Bharat claims to have obtained an image twice the size of the object with a concave lens. Is he correct? Give reason for your answer.
Answer:
Bharat’s claim is not correct.
This is because the image formed by a concave lens is always smaller in size than the object.

(52) Out of alcohol with n= 1.36 and carbon disulphide with n = 1.63 which is denser optically?
Answer:
Carbon disulphide with greater refractive index is optically denser than alcohol.

(53) What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is +3?
Answer:
The image must be virtual, erect, enlarged (three times in size relative to the object) and formed behind the mirror.

(54) With respect to air, the refractive index of ice is 1.31 and that of rock salt is 1.54. Calculate the refractive index of rock salt with respect to ice.
Answer:
n_ia = 1.31 and n_ra = 1.54, n_ri = ?
Now, n_ri = \(\frac{n_{\mathrm{ra}}}{n_{\mathrm{ia}}}=\frac{1.54}{1.31}\) = 1.17

(55) What is the minimum number of rays required for locating the image formed by a concave mirror?
Answer:
At least two rays are required for locating the image formed by a concave mirror.

(56) By changing the object distance the focal length of a given spherical mirror cannot be changed but the linear magnification produced by the mirror can be changed. Agree / Disagree?
Answer:
Agree

(57) Write the correct sequence of angle of incidence, angle of emergence, angle of refraction and lateral displacement shown in the following diagram by numbers 1, 2, 3 and 4.

Answer:
2, 1, 4, 3.

(58) What is the relation between n1 and n2 in the following three ray diagrams (figures)?

Answer:
(a) n₁ < n₂
(b) n₁ = n₂
(c) n₁ > n₂

Important note: In case of a concave lens the corresponding relation between n₁ and n₂ along with, the ray diagrams (figures) is shown below:

(59) Which factors determine the focal length of a lens?
Answer:
The focal length of a lens depends on:

• the radii of curvature of the surfaces of the lens.
• the nature of the material of the lens.
• the nature of the medium in which the lens is placed.

[Note : It also depends upon the frequency of the incident light.]

Value Based Questions With Answers

Question 1.
Two fast friends named Anjalee and Shweta spend most of the time with each other. One S day, Anjalee observed that Shweta has pain in gums while eating her school-lunch. Anjalee’s S father is a dentist. So she advised Shweta to come with her to her father’s clinic.

Anjalee’s father examined Shweta’s mouth and teeth with a mirror and light and advised her not to eat too many chocolates and drink soft drinks, Then she started taking care of her mouth as she washed her mouth properly every time after having food and also started taking calcium-rich diet.

1. Which type of mirror is used by the dentist?
2. Name the phenomenon of light by which the doctor is able to examine Shweta.
3. What values are shown by Anjalee?

Answer:

1. Concave mirror
2. Reflection of light
3. Friendship
4. Concern for other
5. Helping nature

Question 2.
In a small town-fair, Amey took his friend and showed him a mirror in which his image showed the upper half of the body very fat and lower half of the body very thin. Amey’s friend got upset but Amey explained him by showing his similar image in the mirror.

1. Name the two mirrors used in this fair shop.
2. Name the mirror in which the size of image is small.
3. What values are displayed by Amey?

Answer:

1. Concave and convex mirror
2. Convex mirror
3. Compassion and empathy.

Question 3.
Prashant was going to his office in his car. While driving his car, Prashant saw a man behind him on a motorcycle through his rear-view mirror. A woman was also sitting behind the man on the motercycle. Through his rear-view mirror, Prashant noticed that the saree of the woman was almost touching the spokes of the motorcycle wheel. He signalled the motorcyclist to stop and alerted the woman. She tied her saree properly and thanked Prashant for the s alert.
(1) What type of mirror is used as a rear-view mirror?
(2) State two characteristics of the image formed by such a mirror.
(3) What values are displayed by Prashant in this incident?
Answer:
(1) Convex mirror

(2)

• Virtual and erect
• diminished (smaller than the object)

(3) The values displayed by Prashant in this incident are …

• Vigilant (because he kept careful watch for a likely mishap around him)
• Concern about others (here, he was concerned about the safety of the woman)
• Responsible citizen (because, he stopped the motorcycle and alerted the woman about an impending danger to her life).

Question 4.
On the way from Ahmedabad to Vadodara there were four friends in a car. Sanjay was driving the car. He saw from his side mirror that the car which was behind their car had met with an accident.

He suddenly applied the brakes, even after his friends asked him to leave the situation as it is. But still Sanjay did not agree and got down from the car and persuaded his friends to help the injured. All of them took the injured person to the nearest hospital. After taking first-aid from the hospital, the victim thanked them for saving his life.
(1) Name the type of mirror from which Sanjay saw the accident.
(2) Why is this type of mirror used as a side mirror in vehicles?
(3) What can you learn from Sanjay’s character?
Answer:
(1) A convex mirror

(2) Convex mirrors have a wider field of view, as they are curved outwards. Therefore, a convex mirror enables the driver to view much larger area.

(3) Sanjay is of helping nature.
We learn to help needy people.

Question 5.
At one of the blind ends of a road in a colony, lot of accidents used to take place. One day, Shalini moved into a house near ; the blind end. On a certain day, she saw an accident taking place at the blind end.

She, at once rushed to the spot and helped the victims with the first-aid. She then telephoned for an ambulance. Later on she met the president of the welfare committee of the colony and requested him to install huge convex mirrors at both the ends of the blind end. After the installation of the mirrors the accident rate suddenly dropped.

1. Why did Shalini advise the installation of huge convex mirrors at both ends of the blind end?
2. What according to you are the values displayed by Shalini?
3. What according to you are the values displayed by president of the welfare committee of the colony?

Answer:

1. A convex mirror has a wide range of view. As a result, it is capable of showing the traffic on the other side or the blind end. Since, she was aware of this fact, Shalini advised for their installation.
2. The values displayed by Shalini are concern for others and right use of knowledge.
3. The values displayed by the president of the welfare committee of the colony are understanding nature, sense of duty and concern for the safety of others.

Practical Skill Based Questions With Answers

Question 1.

(1) In the figure, u = ……………. and f = ……………..
(2) For the image of the object AB, v = ……………. and h’ = …………….
(3) Find magnification m.
Answer:
(1) – 30 cm, +15 cm
because from the given figure,
u = 2f₁ = – 2 x 15 = – 30 cm and
f = + 15 cm as the given lens is convex.

Question 2.

(1) In the figure, f = …………….. and u = ………………
(2) State the type of image.
(3) Will the value (modulus) of magnification be 1 or less than 1 or more than 1?
Answer:
(1) – 12 cm, – 24 cm
From the given figure, f = – 12cm.
(∵ the given lens is concave) and u = – 2f₁ = – 2 x 12 = – 24 cm
(∵ (Real) object distance is always negative.)

(2) Type of image : virtual, erect and diminished

(3) Less than 1

Question 3.

As shown in the figure, three different reflected rays are obtained from three different mirrors. Carefully study the given diagrams and answer the following questions:

1. Which one of the given mirrors is convex?
2. Which of the mirrors will (i) reflect light as a parallel beam of light (ii) converge the parallel beam when a parallel beam of light is incident on it?
3. Which one of the given mirrors will always produce an image equal in size to that of the object?
4. Which mirror will form an image smaller than object size?
5. Which mirror will form an image smaller or bigger than the object depending on the object distance?
6. Which mirror is used (i) in a dressing table, (ii) as a rear-view mirror in vehicles (iii) by a dentist?

Answer:

1. Mirror B
2. Mirror A will reflect light as a parallel beam of light and mirror C will converge a parallel beam of light.
3. Mirror A
4. Mirror B
5. Mirror C
6. Mirror A is used in a dressing table, mirror B is used as a rear-view mirror in vehicles and mirror C is used by a dentist.

Question 4.
For a spherical mirror, obtain the equation for magnification m in terms of (1) f and u only (2) f and v only.
Answer:
(1) Mirror formula : = \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
Multiply both the sides by – u
∴ \(\frac{-u}{f}=\frac{-u}{v}+\frac{-u}{u}\)
∴ \(\frac{-u}{f}=\frac{-u}{v}\) – 1
∴ \(\frac{-u}{v}=1-\frac{u}{f}\)
∴ \(\frac{-u}{v}=\frac{f-u}{f}\)
∴ \(\frac{-v}{u}=\frac{f}{f-u}\)
∴ m = \(\frac{f}{f-u}\) (∵ m = – \(\frac{v}{u}\))

(2) Mirror formula : \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Multiply both the sides by – v
∴ \(\frac{-v}{f}=\frac{-v}{v}+\frac{-v}{u}\)
∴ \(\frac{-v}{f}=-1-\frac{v}{u}\)
∴ \(\frac{-v}{u}=1-\frac{v}{f}\)
∴ \(\frac{-v}{u}=\frac{f-v}{f}\)
∴ \(\frac{-v}{u}=\frac{f-v}{f}\)
∴ m = \(\frac{f-v}{f}\) (∵ m = – \(\frac{v}{u}\))

Question 5.
For a lens, obtain the equation for magnification m in terms of (1) f and u only (2) f and v only.
Answer:
(1) Lens formula : \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Multiply by u on both the sides,
∴ \(\frac{u}{f}=\frac{u}{v}-\frac{-u}{u}\)
∴ \(\frac{u}{f}=\frac{u}{f}\) + 1
∴ \(\frac{u}{v}=\frac{u+f}{f}\)
∴ \(\frac{v}{v}=\frac{f}{f+u}\)
∴ m = \(\frac{f}{f+u}\) (∵ m = \(\frac{v}{u}\))

(2) Lens formula : \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Multiply by v on both the sides,
∴ \(\frac{v}{f}=\frac{v}{v}-\frac{v}{u}\)
∴ \(\frac{v}{f}= 1-\frac{v}{u}\)
∴ \(\frac{v}{u}=1-\frac{v}{f}\)
∴ \(\frac{v}{u}=\frac{f-v}{f}\)
∴ m = \(\frac{f-v}{f}\) (∵ m = \(\frac{v}{u}\))

Question 6.
In case of a convex mirror, if an object is kept at a distance equal to its focal length, then the height of the image is half the height of the object. Prove this result from the formula for magnification only.
Answer:
Here u = – f
Now, m = \(\frac { f }{ f-u }\)
∴ m = \(\frac { f }{ f-(-f)}\)
m = \(\frac { f }{ 2f }\)
= \(\frac { 1 }{ 2 }\)
Now, m = \(\frac { h’ }{ h }\)
∴ \(\frac { h’ }{ h }\) = \(\frac { 1 }{ 2 }\)
∴ h’ = \(\frac { h }{ 2 }\)

Question 7.
The focal length of a convex lens is +10 cm and the focal length of a concave lens is -10 cm. If two objects of equal height are kept on the principal focus of both the lenses, then find the position of the image formed in each case.
Answer:
In case of a convex lens, f = + 10 cm,
u = – 10 cm
Now, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
∴ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)
\(\frac{1}{+10}+\frac{1}{-10}\)
= 0
∴ v = ∞
Hence, the image formed by the convex lens would be at infinite distance from the lens.
In case of concave lens, f = – 10cm, u = – 10 cm
Now, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
∴ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)
= \(\frac{1}{-10}+\frac{1}{-10}\)
= \(\frac{(-10)+(-10)}{-10 \times-10}\)
= \(\frac { -20 }{ 100 }\)
= – \(\frac { 1 }{ 5 }\)
Hence, the image formed by the concave lens would be exactly between principal focus F and optical centre O of the concave lens on the same side as that of the object.

Question 8.
If p, q and r denote the object distance, image-distance and the radius of curvature of a spherical mirror, then prove, r = \(\frac{2 p q}{p+q}\).
Answer:

Question 9.
A real, inverted image of the size of the object is to be formed by holding the object at 1 metre from a convex lens. What should be the focal length of the lens? Draw the ray diagram.
Answer:
When the size of the image = the size of the object, the distance of the object from a convex lens, u = 2f where f is focal length of a convex lens.
Here, u = 2f = 1 metre (given).
∴ f = \(\frac { 1 }{ 2 }\) m = 0.5 m = 50 cm

The course of rays in this case is as shown in the figure above.

Question 10.
AB and CD, two spherical mirrors, form parts of a hollow spherical ball with its centre at O, as shown in the diagram. If arc AB = \(\frac { 1 }{ 2 }\) arc CD,

(1) What is the ratio of their focal lengths?
(2) State which of the two mirrors will always form a virtual image of an object placed in front of it and why?
Answer:
(1) f = \(\frac { R }{ 2 }\) The radius of curvature is the same in two cases. Hence, numerically, the focal length |f| will be the same. But the focal length of a convex mirror (AB) is positive while that of a concave mirror (CD) is negative. Hence the ratio of the focal lengths will be – 1.
[Note: Ratio of numerical value of the focal lengths will be 1.]

(2) Mirror AB will always form a virtual image as mirror AB is a diverging / convex mirror.

Question 11.
What is lateral displacement? Show that a ray incident obliquely on a rectangular glass slab, emerges parallel to the incident ray.
Answer:

1. Place a glass slab on a plane sheet of paper fixed on a drawing board.
2. Draw its boundary.
3. Fix two pins A and B on one side of the slab.
4. Look through the opposite side of the slab and fix two pins C and D in such a way that all the four pins
5. A, B, C and D appear to be in a straight line.
6. CD is the emergent ray. Let it subtend an angle e with the normal.
7. In the first refraction from medium a (air) to medium b (glass), i is the angle of incidence and r is the angle of refraction.
   

Refraction through a rectangular glass slab of thickness: A lateral displacement In the second refraction at E, the angle of incidence in medium b is r and the angle of refraction in medium a is e.
n_ab = \(\frac{\sin r}{\sin e}\) … (2)
From (1) and (2), we have
n_ba x n_ab = \(\frac{\sin i}{\sin r} \times \frac{\sin r}{\sin e}\)
Butn_ba x n_ab = 1, according to the principle of reversibility.
Hence, 1 = \(\frac{\sin i}{\sin e}\)
∴ sin i = sin e
∴ i = e
Thus, the angle of emergence is equal to the angle of incidence. Therefore, incident ray produced (or extended) i.e., AOd is parallel to emergent ray CD.

Thus, the emergent ray is parallel to the incident ray, but displaced sideways as shown in the figure. This sideways displacement FE is known as lateral displacement.

Question 12.
Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case.

1. An object is placed between the device and its principal focus. The image formed is enlarged and behind the device.
2. The object is placed between the principal focus and device. The image formed is enlarged and on the same side as that of the object.
3. The object is placed at finite distance from the device (i.e., between infinity and device). The image formed is diminished and lies between the principal focus and optical centre and on the same side as that of the object.
4. The object is placed at finite distance from the device (i.e., between infinity and device). The image formed is diminished and lies between the pole and principal focus as well as behind it.

Answer:

1. A concave mirror.
2. A convex lens.
3. A concave lens.
4. A convex mirror.

Question 13.
Sudha finds that a sharp image of the s windowpane of her science laboratory is formed at 15 cm from the lens. She now tries to focus the building visible to her outside the? window instead of the windowpane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of s the building? What is the approximate focal length of this lens?
Answer:
To obtain a sharp image of the building instead of the windowpane, Sudha will have to move the screen slightly towards the lens. The approximate focal length of this lens is 15 cm.

Question 14.
Under what condition in an arrangement of two plane mirrors, will an incident ray and the reflected ray be always parallel to each other? (whatever may be the angle of incidence) Show the same with a diagram.
Answer:
When two plane mirrors are placed at 90° to each other, then the incident and reflected rays will always be parallel to each other, whatever may be the angle of incidence. The course of rays is shown in Fig. for i = 45°.

[Note: Students can repeat this activity for different angles of incidence to verify the answer given above.]

Question 15.
Study the diagram given below carefully and answer the following questions :

1. Due to which phenomenon does the path of a ray of light get deviated?
2. Name the device X.
3. What is point A called ?
4. Which quantity is the inverse of the distance between device X and point A?
5. State the SI unit of the physical quantity mentioned as the answer to question (4).

Answer:

1. Refraction
2. Convex lens
3. Principal Focus
4. Power of a lens
5. D (dioptre)

Memory Map:

Students must go through these JAC Class 10 Science Notes Chapter 10 Light Reflection and Refraction to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 10 Light Reflection and Refraction

→ Light: Light is an electromagnetic radiation, which produces the sensation of sight in our eyes.

• Light is a form of energy that travels in the form of waves.
• These waves do not require a material medium for their propagation. They are non-mechanical waves.
• These waves travel at a speed of 3 x 10⁸ m s^-1 in vacuum.
• The wavelength of visible light ranges from 4 x 10^-7m to 8 x 10^-7m.

→ Image: When a number of rays starting from a point on an object, after reflection or refraction, meet at another point or appear to meet (i.e., give illusion of meeting at another point), then the second point is called the image of the first point. A group of such rays of light is called a beam of light.

Ray of light: A straight-line path joining one point to another in the direction of propagation of light is known as a ray of light.

• The image formed is called a real image, if the rays actually meet after reflection or refraction. It can be obtained on a screen.
• If the rays do not meet actually at a point, but appear to meet, after reflection or refraction, then the image is called a virtual image. It cannot be obtained on a screen.
• When a parallel beam of light is incident on a shining plane or smooth surface, the beam remains parallel after reflection in a specific direction. Such a reflection is called regular reflection.
• Example: the reflection of light by a plane mirror.
• In case of irregular reflection, the reflected beam of light does not remain parallel in specific direction but spreads over a wide region.
• Example : reflection from a book, a table, a chair, etc.

→ Laws of reflection:

• The angle of incidence is equal to the angle of reflection, i.e., i = r.
• The incident ray, the normal to the mirror at a point of incidence and the reflected ray, all lie in the same plane.

→ Image formation by a plane mirror:

• The image formed by a plane mirror is always virtual and erect.
• The image formed is as far behind the mirror, as the object is in front of it.
• The size of the image is equal to that of the object.
• The image formed is laterally inverted, i.e., the left side of the object seems to be the right side of the image and vice versa.

(The phenomenon, by which the left side of the object becomes the right side of the image and right side of the object becomes the left side of the image, is called lateral inversion.)

→ Spherical mirrors: A mirror whose reflecting surface is a part of a hollow sphere is called a spherical mirror. The reflecting surface of a spherical mirror is curved inwards or outwards.
The curved mirrors are of two types:

•  Concave mirror and
•  Convex mirror.

→ Concave mirror: A spherical mirror with reflecting surface curved inwards is called a concave mirror. Concave mirror forms either a real or virtual image of the object depending on the object distance.

→ Convex mirror: A spherical mirror with reflecting surface curved outwards is called a convex mirror. Convex mirror forms a virtual image of the object for all object distances.

→ Terminology used in respect of spherical mirrors:
(1) Pole (P): The centre of the reflecting surface of a sperical miror is a point called the pole (P) of the mirror.

(2) Centre of curvature (C) : The centre of curvature of a spherical mirror is the centre of the hollow sphere, of which the reflecting surface of the spherical mirror forms a part.

(3) Radius of curvature (R): The radius of curvature of a spherical mirror is the radius of the hollow sphere, of which the reflecting surface of the spherical mirror forms a part.

(4) Aperture : The diameter of the edge of the reflecting surface of a spherical mirror is called the aperture of the mirror.

(5) Principal axis : The imaginary straight¬line passing through the pole (P) and the centre of curvature (C) of a spherical mirror is called the principal axis of the mirror.

(6) Principal focus (F): The point on the principal axis of a concave mirror, at which rays of light incident on the mirror in the direction parallel to the principal axis (actually) meet / intersect after reflection from the mirror is called the principal focus (F) of the concave mirror. When rays parallel to the principal axis are incident on a convex mirror, the reflected rays appear to come from a point on the principal axis. This point is called the principal focus of the convex mirror.

(7) Focal length (f): The distance between the pole (P) and the principal focus (F) of a spherical mirror is called the focal length (f).

→ Selection of rays for locating the image formed by a spherical mirror:

• A ray parallel to the principal axis, after reflection, will pass through the principal focus (F) in case of a concave mirror or appear to diverge from the principal focus (F) in case of a convex mirror.
• A ray passing through the principal focus (F) of a concave mirror or a ray which is directed towards the principal focus of a convex mirror, after reflection, will emerge parallel to the principal axis.
• A ray passing through the centre of curvature of a concave mirror or directed towards the centre of curvature of a convex mirror, after reflection, is reflected back along the same path.
• A ray incident obliquely to the principal axis, towards a point P (pole of the mirror), on the concave mirror or a convex mirror is reflected obliquely, following the laws of reflection.

Out of these four types of rays, only two types of rays are necessary to locate the position of an image of a point object.

→ Image formation by a Concave mirror:

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

→ Image formation by a Convex mirror:

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F, behind the mirror	Highly diminished, point-sized	Virtual and erect
Between infinity and the pole P of the mirror	Between P and F, behind the mirror	Diminished	Virtual and erect

→ New Cartesian sign convention for reflection by spherical mirrors:

• The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.
• All distances parallel to the principal axis are measured from the pole of the mirror.
• All the distances measured to the right of the origin (along + X-axis) are taken as positive while those measured to the left of the origin (along-X-axis) are taken as negative.
• The distances measured perpendicular to and above the principal axis of the mirror (along + Y-axis), are taken as positive.
• The distances measured perpendicular to and below the principal axis (along -Y-axis), are taken as negative.

→ Magnification prpduced by a spherical mirror: The ratio of the height of the image to the height of the object is called the magnification (m). It is the relative extent to which the image of the object is magnified with respect to the size of the object.

→ Refraction: When a ray of light travels obliquely from one transparent medium to another, its speed changes. Therefore, at the boundary separating the two media, there occurs a change in its direction of propagation. This phenomenon is called refraction of light.

→ Laws of refraction:
(1) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

(2) The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is known as Snell’s law of refraction. (This is true for angle 0° < i° < 90°)
If i is the angle of incidence and r is the angle of refraction,
\(\frac { sin i }{ sin r }\) = constant = n₂₁
where, n₂₁ is known as the refractive index of medium 2 with respect to medium 1.
The above equation is the mathematical representation of the Snell’s law.

→ Absolute Refractive Index: The ratio of the speed of light in vacuum (c) to the speed of light in medium (v), is called the absolute
refractive index of the medium.
n_m = \(\frac { c }{ v }\)
Absolute refractive index of any medium is always more than 1.

→ Relative Refractive Index: The ratio of the speed of light v1 in medium 1, to the speed of light v₁ in medium 2, is called the relative refractive index of medium 2 with respect to medium 1 and is represented by the symbol n₂₁. v₁ and v₂ correspond to the same frequency of light.
n₂₁ = \(\frac{v_1}{v_2}\)
Also, n₂₁ = \(\frac{n_2}{n_1}\) where, n₂ = \(\frac{c}{v_2}\) and n₁ = \(\frac{c}{v_1}\)
Here, v₁ is the speed of light in medium 1, v₂ is the speed of light in medium 2, n₁ is the absolute refractive index of medium 1 and n₂ is the absolute refractive index of medium 2, for the given frequency (given colour) of light.

→ Lateral shift: When a ray of light is refracted at two parallel refracting surfaces, the ray is shifted sideward. This sideward displacement of a ray of light is called lateral shift.

The amount of lateral shift depends upon the perpendicular distance between two parallel refracting surfaces as well as upon the angle of incidence and the refractive index of the second medium with respect to the first medium.

→ Terminology used in respect of lens:
(1) Centre of curvature (C): The centre of a transparent (glass) sphere, of which the curved surface of a lens forms a part is called the centre of curvature C of the respective spherical surface.
Lens has two centres of curvatures C₁ and C₂.

(2) Principal axis : The imaginary straight-line passing through the two centres of curvature C₁ and C₂ of a lens, is called the principal axis of the lens.

(3) Radius of curvature (R): The radius of a transparent (glass) sphere of which the curved surface of a lens forms a part is called the radius of curvature R of the respective spherical surface of the lens. Lens has two radii of curvature R₁ and R₂.

(4) Optical centre (O): The central point of a lens on the principal axis of the lens is called an optical centre O of the lens.

(5) Principal focus (F): When the rays parallel to the principal axis of a convex lens are refracted through the lens, they converge at a point on the principal axis. This point is called the principal focus F of the convex lens.

When the rays parallel to the principal axis of a concave lens are refracted through the lens, they appear to diverge from a point on the principal axis. This point is called the principal focus of the concave lens. A lens has two principal foci F₁ and F₂ on either side of the lens.

(6) Focal length (f): The distance of the principal focus from the optical centre of a lens is called the focal length f of the lens.

(7) Aperture : The effective diameter of the circular outline of a spherical lens is called the aperture of the lens.

→ Selection of rays for image formation by lens:

• A ray of light from the object, parallel to the principal axis, after refraction through a convex lens, passes through the principal focus on the other side of the lens.
• A ray of light from the object, parallel to the principal axis, after refraction through a concave lens, appears to diverge from the principal focus located on the same side of the lens.
• A ray of light passing through the principal focus, after refraction through a convex lens, will emerge parallel to the principal axis. A ray of light directed towards the principal focus on the other side of a concave lens, after refraction, will emerge parallel to the principal axis.
• A ray of light passing through the optical centre of a lens will emerge without any deviation. (This is true for a convex lens as well as a concave lens.)

From the above three types of rays any two types of rays will locate the position of the image.

→ Image formation by a Convex Lens:

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F2	Highly diminished, point-sized	Real and inverted
Beyond 2F1	Between F2 and 2F2	Diminished	Real and inverted
At 2F1	At 2F2	Same size	Real and inverted
Between F1 and 2F1	Beyond 2F2	Enlarged	Real and inverted
At focus F1	At infinity	Infinitely large or Highly enlarged	Real and inverted
Between focus F1 and optical centre O	On the same side of the lens as the object	Enlarged	Virtual and erect

→ Image formation by a Concave Lens:

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F1	Highly diminished, point-sized	Virtual and erect
Between infinity and optical centre O of the lens	Between focus F1 and optical centre O	Diminished	Virtual and erect

→ Sign convention for spherical lens : It is similar to that followed for spherical mirrors, but the optical centre of a lens is chosen as the origin of the co-ordinate system.

→ Magnification produced by a lens: The magnification (m) produced by a lens is defined as the ratio of the height of the image to the height of the object.

→ Power of a lens : The reciprocal of the focal length of a lens is called the power of the lens (P).
P = \(\frac { 1 }{ f }\)
SI unit: The dioptre (D)
1 D = 1 m^-1
The power of a convex lens is positive and that of a concave lens is negative.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
1. 2x³ + x² – 5x + 2; \(\frac{1}{2}\), 1, -2
2. x³ – 4x² + 5x – 2; 2, 1, 1
Solution:
1. Let p(x) = 2x³ + x² – 5x + 2

Hence, \(\frac{1}{2}\) is a zero of
p(x) = 2x³ + x² – 5x + 2.
Again,
p(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
Hence, 1 is a zero of p(x) = 2x³ + x² – 5x + 2.
Again,
p(-2) = 2(-2)² + (-2)² – 5(-2)+2
= -16 + 4 + 10 + 2 = 0
Hence, -2 is a zero of
p(x) = 2x³ + x² – 5x + 2.
Now, for p(x) = 2x³ + x² – 5x + 2;
a = 2, b = 1, c = -5 and d = 2.
The zeroes of p (x) are α = \(\frac{1}{2}\), β = 1 and γ = -2.
Now, α + β + γ = \(\frac{1}{2}\) + 1 + (-2)
= \(-\frac{1}{2}=\frac{-(1)}{2}=\frac{-b}{a}\)
αβ + βγ + γα = (\(\frac{1}{2}\)) (1) + (1) (-2) + (-2) (\(\frac{1}{2}\))
= \(\frac{1}{2}\) – 2 – 1 = \(\frac{-5}{2}=\frac{c}{a}\) and
αβγ = (\(\frac{1}{2}\))(1)(-2) = -1 = \(\frac{-(2)}{(2)}=\frac{-d}{a}\)

2. Let p (x) = x³ – 4x² + 5x – 2.
Then, p (2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
Hence, 2 is a zero of p (x) = x³ – 4x² + 5x – 2.
Again,
p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
Hence, 1 is a repeated zero of p(x) = x³ – 4x² + 5x – 2.
Now, for p(x) = x³ – 4x² + 5x – 2, a = 1, b = -4, c = 5 and d = -2.
The zeroes of p (x) are α = 2, β = 1 and γ = 1.
Now,
α + β + γ = 2 + 1 + 1 = 4 = \(\frac{-(-4)}{1}=\frac{-b}{a}\)
αβ + βγ + γα = (2) (1) + (1) (1) + (1) (2)
= 2 + 1 + 2 = 5 = \(\frac{5}{1}=\frac{c}{a}\) and
αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}=\frac{-d}{a}\)

Question 2.
Find a cubic polynomial with the sum of its zeroes, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, 7, 14 respectively.
Solution:
Let p (x) = ax³ + bx² + cx + d, a ≠ 0, be the required cubic polynomial and its zeroes be α, β and γ
Then, as given in the data,
α + β + γ = 2 ∴ \(\frac{-b}{a}\) = 2
αβ + βγ + γα = -7 ∴ \(\frac{c}{a}\) = -7
αβγ = \(\frac{-d}{a}\) = -14
So, if a = 1, then b = -2, c = -7 and d = 14.
Hence, the required polynomial is p(x) = x³ – 2x² – 7x + 14.

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Solution:
For the given polynomial x³ – 3x² + x + 1.
A = 1, B = -3, C = 1 and D = 1.
Its zeroes are given to be a – b, a and a + b.
Now, sum of zeroes = (a – b) + a + (a + b) = 3a
From the polynomial,
Sum of zeroes = \(\frac{-B}{A}=\frac{-(-3)}{1}=3\)
Hence, 3a = 3 ∴ a = 1
Product of zeroes = (a – b) × a × (a + b)
= a(a² – b²)
From the polynomial.
Product of zeroes = \(\frac{-\mathrm{D}}{\mathrm{A}}=\frac{-1}{1}=-1\)
Hence, a (a² – b²) = -1
1(1² – b²) = -1 ∴ 1 – b² = -1
∴ 1 + 1 = b² ∴ b² = 2
∴ b = ±\(\sqrt{2}\)
Thus, a = 1 and b = ±\(\sqrt{2}\).

Question 4.
It two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.
Solution:
Let p (x) = x⁴ – 6x³ – 26x² + 138x – 35
2 + \(\sqrt{3}\) and 2 – \(\sqrt{3}\) are zeroes of p (x).
∴ (x – 2 – \(\sqrt{3}\))(x – 2 + \(\sqrt{3}\)) = (x – 2)² – (\(\sqrt{3}\))²
= x² – 4x + 4 – 3
= x² – 4x + 1
is a factor of p(x).

Then, Quotient = x² – 2x – 35
= x² – 7x + 5x – 35
= x(x – 7) + 5(x – 7)
=(x – 7)(x + 5)
x – 7 = 0 gives x = 7 and x + 5 = 0 gives x = -5.
Hence, the other zeroes of the given polynomial are 7 and -5.

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:

But, the remainder is given to be x + a.
Hence, comparing the coefficients of x and the constant term, we get
2k – 9 = 1 and k² – 8k + 10 = a
Now, 2k – 9 = 1
∴ 2k = 10
∴ k = 5
and a = k² – 8k + 10
∴ a = (5)² – 8(5) + 10
∴ a = 25 – 40 + 10
∴ a = -5
Thus, k = 5 and a = -5.

Jharkhand Board JAC Class 10 Science Important Questions Chapter 14 Sources of Energy Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 14 Sources of Energy

Additional Questions and Answers

Question 1.
Distinguish between :
(1) Solar Cooker and Solar Cell Panel
Answer:

Solar Cooker	Solar Cell Panel
1. It converts solar energy into heat.	1. It converts solar energy into electricity.
2. It is mainly designed for cooking purpose, i.e., for domestic use	2. It is designed for delivering enough electricity for commercial use.
3. It consists of mirror, glass sheet, black wooden box.	3. It consists of large number of solar cells and silicon is used in solar cells.
4. It is useful for cooking only at certain times of the day.	4. The limitation of using solar energy is overcome by using it.

(2) Fossil fuels and Bio fuels
Answer:

Fossil Fuels	Bio Fuels
1. It is a conventional source of energy mainly used in industries and vehicles.	1. It is a conventional source of energy mainly for domestic use.
2. Coal, petroleum and natural gas are fossil fuels.	2. Wood, cow-dung cakes, biogas, etc. are bio fuels.
3. The fossil fuels are non-renewable and exhaustible source of energy.	3. Bio fuels are renewable source of energy in the form of biomass.
4. Fossil fuels are the major fuels used all over the world.	4. Bio fuels are mainly used in rural area.

Question 2.
Give scientific reasons for the following statements:
(1) From environmental point of view, we should reduce the use of fossil fuels.
Answer:
The fossil fuels are non-renewable source of energy and in limited reserves.

• The growing demand for energy at global level are largely meet by fossil fuels.
• Excessive burning of fossil fuels – coal and petroleum increases the level of air pollution. example:
• Carbon dioxide, sulphur dioxide and oxides of nitrogen are released in excess in atmosphere causing such air pollution.
• Carbon dioxide is a main greenhouse gas that causes global warming.
• Sulphur dioxide and oxides of nitrogen lead to acid rain which affects water and soil resources.

So, to reduce the adverse effects on environment, we should reduce the use of fossil fuels.

(2) Though hydropower is renewable source of energy, it creates certain problems.
Answer:
Hydropower is renewable source of energy because water gets refilled in reservoir during raining.

• High-rise dams are constructed to generate hydroelectricity.
• Such dams are constructed in hilly areas causing environmental changes.
• Large eco-systems are submerged under water and destroyed.
• The submerged vegetation decaying anaerobically releases large amount of greenhouse gas, i.e., methane.
• Large areas of agricultural land is also submerged. Many natives are thus displaced.
• Rural-tribal human races may lose their habitat.

So, Hydropower plant construction creates certain problems.

(3) The use of dung as raw material for biogas production is profitable than using it only as fertiliser or fuel.
Answer:
Animal dung is used as raw material c for production of biogas. It provides a safe and i efficient method of waste disposal.

• Besides it, by supplying biogas as a fuel gas, cooking, lighting, etc. we can save natural gas.
• The residual slurry can be used as excellent manure that is rich in nitrogen and phosphorous.

This natural fertiliser maintains soil fertility and agriculture practice becomes more economical.

(4) Inspite of the high cost, solar cells are widely used.
Answer:
A special grade silicon is used for making solar cells and silver is used for interconnection of cells in panel.

• This makes the solar cells expensive.
• But it converts solar energy into electricity and about 0.7 W of electricity is produced when exposed to the Sunlight.
• Solar cells and panels are widely used in artificial satellite, space probes like Mars orbiters, wireless transmission systems, radio or TV relay stations, traffic signals, calculators and in many toys.
• At domestic level, maximum use of solar energy can be achieved by using solar roof top thus saving electricity.
• So, solar cells are widely used.

(5) Generation of nuclear energy is hazardous.
Answer:
Nuclear fission and nuclear fusion are two processes used to generate nuclear energy.

• Radioactive heavy atoms such as uranium, plutonium or thorium are used in a nuclear fission process whereas hydrogen isotopes are used in a nuclear fusion.
• There is a risk of accidental leakage of nuclear radiation.
• There is a possibility of high risk of environmental contamination i.e., radioactive pollution.
• Radioactive waste is hazardous to human health and environment. Radiations which are emitted from the radioactive waste can cause mutation (Change in DNA) in living organisms.
• It creates various disorders and cancer in human beings.

So, generation of nuclear energy is hazardous.

Question 3.
Carefully observe the given diagram / chart and answer the questions based on it.
(1) Observe the following pie-chart showing the major sources of energy.

Questions:
(1) Identify ‘a’ and state where is it used as a fuel to generate electric energy through conversion of heat energy.
Answer:
Coal. It is used as a fuel in thermal power plant.

(2) Label ‘b’ and give the name of processes to generate it.
Answer:
Nuclear energy. The processes to generate nuclear energy are: nuclear fission and nuclear fusion.

(3) Label ‘c’ and state the name of any two place where such power plants are established.
Answer:
Hydropower. Hydropower plants are established at Tehri Dam on the Ganga river and Sardar Sarovar Project on the Narmada river.

(4) Refer to the pie-chart above and arrange the sources of energy in an ascending order.
Answer:
Nuclear < hydro < petroleum and natural gas < coal

Questions:

(1) Label a, b and c in given diagram.
Answer:
a – Digester, b – Slurry, c – Gas outlet

(2) How is biogas obtained from slurry?
Answer:
Slurry consists of bio-wastes / biomass and biogas can be obtained through decomposition by anaerobic microorganisms.

(3) Why is biogas considered as an excellent fuel?
Answer:
Biogas is considered as an excellent fuel because it burns without smoke, leaves no ash residue and has high heating capacity.

(4) Which are the gases present in biogas?
OR
What are components of biogas?
Answer:
Biogas mainly consists gases like methane, carbon dioxide, hydrogen and hydrogen sulphide.

(5) Which are abundant nutrients in the slurry left behind in a biogas plant?
Answer:
Nitrogen and phosphorous are abundant nutrients in the slurry left behind in a biogas plant.

Questions :
(1) Which metal is used in the panel for the interconnection of cells?
Answer:
Silver

(2) Which energy conversion occurs by using solar cells?
Answer:
Solar energy is converted into electricity.

(3) Write any two uses of solar cells for space science.
Answer:
Solar cells are used as a main energy source in an artificial satellites and space probes like Mars orbiters.

(4) If 12 solar cells are interconnected in the panel. How many watt of electricity can be produced from it when exposed to Sun?
Answer:
About 8.4 W

(5) Why is silicon used in a solar cells?
Answer:
Silicon is a natural semiconductor, easily obtained and easily modified to convert solar energy into electricity.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) Name the form of energy obtained from sea.
Answer:
The forms of energy obtained from sea are tidal energy, wave energy and ocean thermal energy.

(2) Which fuel is considered as a cleaner source for vehicles?
Answer:
CNG (Compressed Natural Gas)

(3) State the name of fuels that we are using.
Answer:
We are using wood, coal, petrol, diesel, CNG, kerosene, LPG, etc. as fuels.

(4) What is a source of energy?
Answer:
A source of energy is one which is capable of providing useful energy in sufficient amount.

(5) Which energy conversion occurs in a hydropower plant?
Answer:
In a hydropower plant, potential energy of stored water is converted into electrical energy.

(6) Which is a main constituent of biogas? What is its content?
Answer:
Methane gas is a main constituent of biogas and it is about 75 %.

(7) What are the conventional sources of energy?
Answer:
Fossil fuels, i.e., coal and petroleum are the conventional source of energy, which are used on a large scale.

(8) Name two elements that are used in manufacturing solar cell panels.
Answer:
Silicon, Silver

(9) Name the devices to harness solar energy.
Answer:
Solar cell, solar cooker, solar water heater.

(10) Where are the fossil fuels used directly?
Answer:
In gas stoves, vehicles and thermal power plants, fossil fuels are used directly.

(11) Why is charcoal used as a fuel?
Answer:
Charcoal is used as a fuel because it burns without flames is comparatively smokeless s and has a higher heat generation efficiency.

(12) Why is the term thermal power plant used?
Answer:
The term thermal power plant is used because in such power plants fuel is burnt to produce heat energy which is converted into electrical energy.

(13) State the name of any two greenhouse gases.
Answer:
CO₂ and CH₄

(14) Which fuels leave residue like ash on burning?
Answer:
Wood, charcoal and coal leave residue like ash on burning.

(15) How anaerobic microorganisms help in formation of fuel?
Answer:
Anaerobic microorganisms (Methanogenic bacteria) produce biogas by decomposition of complex compounds that present in cow-dung slurry.

(16) State two advantages of wind energy.
Answer:
Advantages of wind energy :

• It is an environment-friendly energy
• It is efficient source of renewable energy.

(17) What is the age of the sun? What will be its expected lifespan?
Answer:
The sun is 5 billion years old and another 5 billion years will its expected lifespan.

(18) Why is our demand for energy increasing day by day?
Answer:
Our demand for energy is increasing day by day because –

• We use machines to do more and more of our tasks, which needs more energy.
• Industrialisation improves our living standards. The energy requirement therefore constantly rises.

(19) Where is solar cell panels are mounted for domestic use? What is its advantages?
Answer:
The solar cell panels are mounted on specially designed inclined rooftops. More solar rays are incident over it, producing more electric
energy.

(20) Name the two oxides that causes acid rain.
Answer:
Oxides of nitrogen and sulphur cause acid rain.

(21) What is the minimum speed of wind required by a windmill to maintain the necessary speed of turbine in electric generator?
Answer:
More than 15 km/h

(22) Which are the renewable sources of energy?
Answer:
Hydropower, biomass, solar energy, energy from the sea, wind energy, etc. are renewable sources of energy.

Question 2.
Define: OR Explain the terms :
(1) Nuclear fission
Answer:
The process of splitting a heavy nucleus into smaller nuclei along with the release of large amount of energy is called nuclear fission.

(2) Nuclear fusion
Answer:
The process of joining lighter nuclei to make a heavier nucleus along with release of tremendous amount of energy is called nuclear fusion.

(3) Solar constant
Answer:
The solar energy reaching unit area on outer surface of the earth’s atmosphere which is exposed perpendicularly to the rays of sun at the average distance between the sun and earth is called solar constant.

(4) Geothermal energy
Answer:
The heat energy obtained in the form of steam from hot spots region is called geothermal energy.

(5) Biogas
Answer:
A fuel gas that is produced anaerobically from biomass is called biogas.

(6) Conventional sources of energy
Answer:
Those sources of energy which meet our major energy requirement are known as conventional sources of energy.

(7) Non-conventional sources at energy
Answer:
Those sources of energy which meet only a limited energy requirement are known as non-conventional sources of energy.

(8) Wind energy farm
Answer:
A number of windmills erected over a large area for the commercial use of wind energy is known as wind energy farm.

(9) Fossil fuels
Answer:
The fuels such as coal, petroleum and natural gas, that were formed millions of years ago from dead and fossiliged animals or plants in the ground are called fossil fuels.

Question 3.
Fill in the blanks:

1. Biogas obtained from decomposition of biomass by the activity of ……………… microorganisms.
2. Petroleum products are ……………… fuel.
3. Hydropower plants convert the ……………… of falling water into electricity.
4. When wood is burnt in a limited supply of oxygen, ……………… is left behind as a residue.
5. Manure obtained from biogas plant is rich in ……………… and ………………
6. ……………… plant is an efficient method of bio waste disposal.
7. ……………… is the best process to capture solar energy and convert it into biomass.
8. Use of ……………… mirror would be best for in solar cooker.
9. ……………… is abundant in nature but availability of its special grade is limited.
10. The hydrogen bomb is based on ……………… reaction.
11. India is ranked ……………… in harnessing wind-energy for the production of electricity.
12. The total ……………… of the universe remains constant because it can neither be created nor destroyed.

Answer:

1. anaerobic
2. fossil
3. potential energy
4. charcoal
5. nitrogen, phosphorous
6. Biogas
7. Photosynthesis
8. concave
9. Silicon
10. thermonuclear fusion
11. fifth
12. energy

Question 4.
State whether the following sentences are true or false:

1. The air pollution is caused by burning of fossil fuels.
2. The potential energy of flowing water gets, transformed into kinetic energy by collecting the water in dam.
3. In solar cookers, a plain mirror is used to converge the sunlight.
4. Fossil fuels are the major fuels used for generating electricity.
5. Submerged vegetation rots under anaerobic condition producing large amounts of methane.
6. Biogas is derived totally from animal biomass.
7. The level of water in the sea rises and falls due to the gravitational pull of the moon on the spinning earth.
8. Nuclear energy is the best alternative of fossil fuel because of its environment friendly nature.
9. The nuclear bomb is embedded in a deuterium and lithium containing substance.
10. The assembly of the solar cell may be pollution free but the actual operation of it may cause some environmental harm.
11. The energy sources that can be regenerated are called renewable energy sources.
12. Nuclear fusion reactions are the source of energy in the sun and other stars.

Answer:

1. True
2. False
3. False
4. True
5. True
6. False
7. True
8. False
9. True
10. False
11. True
12. True

Question 5.
Graph / diagram based question :
Which of the following graph is correct for Burning of fossil fuel → level of air pollution?

Answer:

Question 6.
Match the following:
(1)

Column I	Column II
1. Silicon	a. nuclear reactor
2. Hot-springs	b. leaves residue like ash
3. Fission of uranium atom	c. used in solar cells
4. Charcoal	d. geothermal energy

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

(2)

Column I	Column II
1. Fossil fuel	a. Plutonium, thorium
2. Biomass	b. coal, petroleum
3. Nuclear fuel	c. CNG
4. Clean fuel	d. wood, cow-dung cake

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

(3)

Column I	Column II
1. Geothermal energy plant	a. Traffic signals
2. Solar cell panels	b. New Zealand
3. Wind energy farm	c. Tehri dam
4. Hydropower plant	d. Kanyakumari

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

Question 7.
Select the correct alternative from those given below each questions:
1. Which of the following is a main component of biogas?
A. Methane
B. Hydrogen
C. Hydrogen sulphide
D. Oxygen
Answer:
A. Methane

2. Which fuel is used in thermal power plant?
A. Biomass
B. Fossil fuel
C. Wood
D. Charcoal
Answer:
B. Fossil fuel

3. Which of the following is the adverse effect caused by burning of fossil fuels?
A. Acid rain
B. Greenhouse effect
C. Both A and B
D. Agricultural land submerged
Answer:
C. Both A and B

4. Find greenhouse gases.
A. Hydrogen and hydrogen sulphide
B. Sulphure dioxide and nitrogen
C. Nitrogen and hydrogen
D. Methane and carbon dioxide
Answer:
D. Methane and carbon dioxide

5. Which of the following fuel does not leave ash-like residue while burning?
A. Wood
B. Charcoal
C. Biogas
D. Coal
Answer:
C. Biogas

6. Which one of the following is a non-renewable source of energy?
A. Bio fuel
B. Fossil fuel
C. Wind power
D. Tidal energy
Answer:
B. Fossil fuel

7. Which of the following exploits energy due to temperature difference at surface water and water at depth?
A. Tidal energy
B. Wave energy
C. Ocean thermal energy
D. All of the given
Answer:
C. Ocean thermal energy

8. Statement A: We need to look for more and more source of energy.
Reason R: There are only limited reserves of fossil fuels.
Which option is correct for statement A and reason R?
A. Both A and R correct, R is explanation of A.
B. Both A and R correct, R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
A. Both A and R correct, R is explanation of A.

9. Charcoal is considered as a better fuel compared to wood because …
A. it burns without flames.
B. it is comparatively smokeless.
C. it has a higher heat generation efficiency.
D. all of the given.
Answer:
D. all of the given.

10. From which of following, electricity can be generated without the use of turbine?
A. Solar energy
B. Tidal energy
C. Geothermal energy
D. Wave energy
Answer:
A. Solar energy

11. Which is the fundamental reaction in nuclear weapon for destructive purposes?
A. Fusion chain
B. Fission chain
C. Radiation chain
D. Thermal chain
Answer:
B. Fission chain

12. By which of the following greenhouse effect can be achieved in a solar cooker?
A. Black surface
B. Concave mirror
C. Convex mirror
D. Covered glass plate
Answer:
D. Covered glass plate

13. Which one of following is used as a main source of energy in artificial satellite?
A. Fossil fuel
B. Uranium
C. Solar cells
D. Gravitation pull
Answer:
C. Solar cells

14. Statement X: Ocean thermal energy conversion plant can operate if the temperature difference is 20°C or more between the water at surface and water at depth. Statement Y: Wind energy farm can generate electricity only if the wind speed should be higher than 15 km/h.
Which is correct option for statement X and Y?
A. X is correct, Y is incorrect.
B. Both X and Y are correct.
C. X is incorrect, Y is correct.
D. Both X and Y are incorrect.
Answer:
B. Both X and Y are correct.

15. Which pf the following is the ultimate source of energy on the earth?
A. Solar
B. Wind
C. Ocean
D. Fossil fuel
Answer:
A. Solar

16. Which of the following is an environmental friendly source of energy?
A. Wind energy
B. Fossil fuel
C. Nuclear energy
D. Solar energy
Answer:
A. Wind energy, Solar energy

17. Statement A: Many of the sources ultimately derive energy from the sun.
Reason R: Nuclear fusion reactions are the source of energy in the sun.
Which option is correct for statement A and reason R?
A. Both A and R correct, R is explanation of A.
B. Both A and R correct, R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
B. Both A and R correct, R is not explanation of A.

Question 8.
Answer as directed (Miscellaneous):
(1) Write full form of CNG, OTEC; MeV
Answer:
CNG – Compressed Natural Gas
OTEC-Ocean Thermal Energy Conversion
MeV – Mega-electron Volts

(2) State the units of Solar constant, Electricity, Energy.
Answer:
Solar constant – kW / m²
Electricity – MW
Energy – eV or Joules

(3) Find mismatched pair :
I. Fossil fuel – Air pollution
II. Wind energy farm – Large area of land
III. Biogas – Aerobic microorganisms
IV. Geothermal energy – Hot springs
Answer:
III. Biogas – Aerobic microorganisms

(4) Find correct sequence for following events :
I. Winds to blow
II. Kinetic energy of huge waves near the sea-shore
III. Solar radiation
IV. Electricity generated
V. Waves generated
Answer:
III → IV → II → IV

(5) Find mismatched pair :
I. Bio fuel-wood, cow-dung cake
II. Solar cell panel – Silicon, copper
III. Windmill – Large fan, turbine of generator
IV. Nuclear energy – Fission and Fusion reaction
Answer:
II. Solar cell panel – Silicon, copper

(6) Sequentially arrange the events that occur in biogas plant.
p. activity of anaerobic microbes
q. The slurry left behind is used as manure
r. cow-dung and water is mixed and this slurry is fed into the digester
s. Methane gas is generated
t. Breakdown of complex compounds of the cow-dung slurry.
Answer:
r → p → t → s → q

Value Based Questions With Answers

Question 1.
Your father decided to fit CNG kit in his petrol car. Your mother is a working woman. Your father has managed their working time in such a way that he took your mother with him.

Your school is just 3 km away from your home. Your father also insists you to use bicycle instead of fuel using vehicle for going to school.
Questions:

1. Why CNG is prefered as a fuel?
2. Why we need to conserve fossil fuels?
3. What value is reflected in the above family?
4. State any two disadvantages of petrol as a fuel.

Answer:

1. CNG is a cleaner fuel than petrol and it is economical. It is less polluting.
2. The fossil fuels are non-renewable and exhaustible sources of energy. So we need to conserve them.
3. This family shows the values of responsibility to environment and they contribute their efforts to save fuels.
4. Disadvantages of petrol as a fuel: (i) It is expensive, (ii) It causes air pollution.

Question 2.
A group of students visited to a village. They noticed that village people uses dung cakes, agricultural wastes, etc. as a fuel. Students found smoke and ash particles in the environment. They requested the sarpanch to install a biogas plant and insist the people to collect the waste to produce biogas.

Questions:

1. What is the use of slurry left behind?
2. State the two uses of biogas.
3. What are the advantages of biogas plant?

Answer:

1. Slurry is used as nitrogen and phosphorous rich manure.
2. Two uses of biogas : Cooking, lighting
3. Biogas plant is a safe and efficient method for waste-disposal besides supplying energy and manure.

Question 3.
In high rised residential flats, maintainance expense per month per flat is high. Some members suggested to install solar panel and solar water heater. They explained that this would provide some electricity to light the building and may reduce electric bill amount.
Questions:

1. Which energy conversion is achieved by installation of solar panel and solar water heater?
2. What is the disadvantage of appliances which run on solar energy?
3. How much electric power is generated by a solar cell?

Answer:

1. Solar energy is converted to electricity by solar panel and in solar water heater, solar energy is converted to heat.
2. Solar appliances do not work during cloudy and rainy days.
3. 0.7 W

Practical Skill Based Questions With Answers

Question 1.
Two cars are parked In open parking area. One car is of black colour with black film on window glass. Other car is of white colour and without black film on window glass.
You just open the door and sit for 1-2 minute in each car.
Questions:

1. Which car is more hot inside?
2. Do both the cars have warm internal environment? Yes or no, why?
3. Why is black car more hotter than a white s car?
4. What are your suggestions to keep top floor cool in summer?

Answer:

1. Black car
2. Due to absorbing heat the cars are heated, making the inside very warm.
3. A black surface absorbs more heat as compared to a white surface.
4. Suggestions to keep top floor cool in summer:

• • • Paint white colour on the outer I side wall.
    • Mount china mosaic on the s terrace.
    • Prepare terrace garden.

Question 2.

A picture to demonstrate windmill and its function.
Questions :

1. Which energy conversion is shown in the picture?
2. What function does windmill show in the picture?
3. Which is the structure similar to windmill?

Answer:

1. Kinetic energy into mechanical energy.
2. To lift water from a well.
3. The structure of windmill is simillar to electric fan.

Memory Map:

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.4

Question 1.
Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution :
ΔABC ~ ΔDEF

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Solution :

In trapezium ABCD, AB || CD and diagonals AC and BD intersect at O.
Then, in ΔAOB and ΔCOD.
∠OAB = ∠OCD (Alternate angles)
∠OBA = ∠ODC (Alternate angles)
∠AOB = ∠COD (Vertically opposite angles)
∴ By AAA criterion, ΔAOB ~ ΔCOD.

Question 3.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar (ABC) / ar (DBC) = \(\frac{AO}{DO}\)
Solution :

Draw AM ⊥ BC and DN ⊥ BC.
Then, ar (ABC) = \(\frac{1}{2}\) × BC × AM and
ar (DBC) = \(\frac{1}{2}\) × BC × DN

In ΔAMO and ΔDNO
∠AMO = ∠DNO (Right angles)
∠AOM = ∠DON (Vertically opposite angles)
∴ By AA criterion, ΔAMO ~ ΔDNO.
∴ \(\frac{AM}{DN}=\frac{AO}{DO}\) ……………….(2)
From (1) and (2), ar (ABC) / ar (DBC) = \(\frac{AO}{DO}\)

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution :
Given: ΔABC ~ ΔPQR and ar (ABC) = ar (PQR)
To prove : ΔABC ≅ ΔPQR

Proof: ΔABC ~ ΔPQR

∴ AB = PQ, BC = QR and CA = RP
∴ By SSS criterion for congruence of triangles, ΔABC = ΔPOR.

Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.
Solution :

In ΔABC, D, E and F are the mid-points of sides AB, BC and CA respectively.
Then, EF || AB and DE || AC
∴ EF || AD and DE || AF
∴ ADEF is a parallelogram.
∴ ∠A = ΔDEF
(Opposite angles of parallelogram)
Similarly, we can prove that ∠B = ∠EFD and ∠C = ∠EDF
Now, in ΔABC and ΔEFD,
∠A = ∠E, ∠B = ∠F and ∠C = ∠D
∴ By AAA criterion, ΔABC ~ ΔEFD.
∴ ar (DEF) / ar(ABC) = (\(\frac{EF}{AB}\))² ………………(1)
In ΔABC, E and F are the mid-points of BC and CA respectively.
∴ EF = \(\frac{1}{2}\)AB ………………(2)
(Alternately: ADEF is a parallelogram.
∴ EF = AD = \(\frac{1}{2}\)AB)
From (1) and (2),

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution :
Given: ΔABC ~ ΔPQR, AD and PM are medians of triangles ABC and PQR respectively.

Proof :
In ΔABC, AD is a median ∴ BD = \(\frac{1}{2}\)BC
In ΔPQR, PM is a median ∴ QM = \(\frac{1}{2}\)QR.
ΔABC ~ ΔPQR
∴ ∠B = ∠Q and \(\frac{AB}{PQ}=\frac{BC}{QR}\)
∴ ∠ABD = ∠PQM and \(\frac{AB}{PQ}=\frac{BD}{QM}\)
So, by SAS criterion ΔABD ~ ΔPQM.
∴ \(\frac{AB}{PQ}=\frac{AD}{PM}\) ……………(1)
Now, ΔABC ~ ΔPQR
∴ ar (ABC) / ar (PQR) = (\(\frac{AB}{PQ}\))²
∴ ar (ABC) / ar (PQR) = (\(\frac{AD}{PM}\))² [By (1)]

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilateral triangle described on one of its diagonals.
Solution :

ABCD is a square. PAB, is an equilateral triangle described on side AB and QAC is an equilateral triangle described on diagonal AC.
In ΔABC, ∠B = 90° and AB = BC (Properties of a square)
Then, AC² = AB² + BC² (Pythagoras theorem)
∴ AC² = AB² + AB²
∴ AC² = 2AB²
∴ \(\frac{\mathrm{AB}^2}{\mathrm{AC}^2}\) = \(\frac{1}{2}\)
∴ (\(\frac{18}{36}\))
ΔPAB is an equilateral triangle.
∴ ∠P = ∠A = ∠B = ∠60°
ΔQAC is an equilateral triangle.
∴ ∠Q = ∠A = ∠C = 60°
Thus, in ΔPAB and ΔQAC,
∠P = ∠Q and ∠A = ∠A and ∠B = ∠C
∴ By AAA criterion, ΔPAB ~ ΔQAC.
∴ ar (PAB) / ar(QAC) = (\(\frac{AB}{AC}\))²
∴ ar (PAB) / ar(QAC) = \(\frac{1}{2}\) [By (1)]
∴ ar (PAB) = \(\frac{1}{2}\)ar (QAC)
Thus, the area of an equilateral triangle described on a side of a square is half the area of an equilateral triangle described on one of its diagonals.

Tick the correct answer and justify:

Question 8.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution :
The correct answer is (C) 4 : 1.
ΔABC and ΔBDE are equilateral triangles.
Hence, any of their correspondences is a similarity.

Question 9.
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution :
The correct answer is (D) 16 : 81.
By theorem 6.6.
Ratio of areas of two similar triangles
= (Ratio of their corresponding sides)²
= (4 : 9)²
= (\(\frac{4}{9}\))²
= 16 : 81

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the age of Ani be x years and the age of Biju be y years.
Then, from the given,
x – y = 3 ………….(1)
or y – x = 3 ………(2)
Dharam is twice as old as Ani.
∴ Age of Dharam = 2x years
Biju is twice as old as his sister Cathy. Hence, the age of Cathy is half the age of Biju.
∴ Age of Cathy = \(\frac{y}{2}\) years
Naturally, Dharam is older than Cathy by 30 years.
2x – \(\frac{y}{2}\) = 30
∴ 4x – y = 60 ………..(3)
1. We solve equation (1) and equation (3).
Subtracting equation (1) from equation (3),
we get
(4x – y) – (x – y) = 60 – 3
∴ 3x = 57
∴ x = 19
Substituting x = 19 in equation (1), we get
19 – y = 3
∴ 19 – 3 = y
∴ y = 16
Hence, the age of Ani is 19 years and the age of Biju is 16 years.

2. We now solve equation (2) and equation (3). Adding equations (2) and (3), we get
(y – x) + (4x – y) = 3 + 60
∴ 3x = 63
∴ x = 21
Substituting x = 21 in equation (2), we get
∴ y – 21 = 3
∴ y = 24
Hence, the age of Ani is 21 years and the age of Biju is 24 years.
Thus, the ages of Ani and Biju are 19 years and 16 years respectively or 21 years and 24 years respectively.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? (From the Bijaganita of Bhaskara II) [Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
Solution:
Let the amount with the first person (say A) be ₹ x and the amount with the second person (say B) be ₹ y.
If B gives ₹ 100 to A. then A will have ₹(x + 100) and B will have ₹(y – 100).
Then, from the first condition, we get
x + 100 = 2(y – 100)
∴ x + 100 = 2y – 200
∴ x – 2y = -300 ………..(1)
If A gives ₹ 10 to B, then A will have ₹(x – 10) and B will have ₹(y + 10).
Then, from the second condition, we get
y + 10 = 6(x – 10)
∴ y + 10 = 6x – 60
∴ 10 + 60 = 6x – y
∴ 6x – y = 70 ………..(2)
Multiplying equation (2) by 2, we get
12x – 2y = 140 ………..(3)
Subtracting equation (1) from equation (3),
we get
(12x – 2y) – (x – 2y) = 140 – (-300)
∴ 11x = 440
∴ x = 40
Substituting x = 40 in equation (1), we get
40 – 2y = -300
∴ 40 + 300 = 2y
∴ 2y = 340
∴ y = 170
Thus, the first person has got ₹ 40 and the second person has got ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have. been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the regular uniform speed of the train be x km/h and regular time taken by the train be y hours. Then, the total distance of the journey = speed × time = xy km.
Now, according to the first information, the new speed = (x + 10) km/h and new time = (y – 2) hours. Then, speed × time = distance gives
(x + 10) (y – 2) = xy
∴ xy – 2x + 10y – 20 = xy
∴ -2x + 10y = 20 …………..(1)
Again, according to the second condition. the new speed = (x – 10) km/h and new time = (y + 3) hours.
Then, (x – 10) (y + 3) = xy
∴ xy + 3x – 10y – 30 = xy
∴ 3x – 10y = 30 …………..(2)
Adding equations (1) and (2), we get
(-2x + 10y) + (3x – 10y) = 20 + 30
∴ x = 50
Substituting x = 50 in equation (1), we get
-2(50) + 10y = 20
∴ -100 + 10y = 20
∴ 10y = 120
∴ y = 12
Now, the total distance covered by the train = xy = 50 × 12 = 600 km.
Thus, the distance covered by the train is 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of students in each row be x and the number of rows be y. Then total number of students = xy.
From the first condition, number of students in each row = x + 3 and the number of rows = y – 1.
∴ (x + 3)(y – 1) = xy
∴ xy – x + 3y – 3 = xy
∴ -x + 3y = 3 …………..(1)
From the second condition, number of students in each row = x – 3 and the number of rows = y + 2.
∴ (x – 3) (y + 2) = xy
∴ xy + 2x – 3y – 6 = xy
∴ 2x – 3y = 6 ……….. (2)
Adding equations (1) and (2), we get
(-x + 3y) + (2x – 3y) = 3 + 6
∴ x = 9
Substituting x = 9 in equation (1), we get
-9 + 3y = 3
∴ 3y = 12
∴ y = 4
Now, total number of students = xy = 9 × 4 = 36. Thus, the number of students in the class is 36.

Question 5.
In a ΔABC, ∠C = 3∠B = 2 (A + B). Find the three angles.
Solution:
In ΔABC, we have
∠A + ∠B + ∠C = 180°
∴ ∠A + ∠B + 3∠B = 180° (∵ ∠C = 3∠B)
∴ ZA + 4∠B = 180° …………..(1)
Again, ∠A + ∠B + ∠C = 180°
∠A + ∠B + 2(∠A + ∠B) = 180° (∵ ∠C = 2 (∠A + ∠B))
∴ 3(∠A + ∠B) = 180°
∴ ∠A + ∠B = 60° ……….(2)
Subtracting equation (2) from equation (1),
we get
(∠A + 4∠B) – (∠A + ∠B) = 180° – 60°
∴ 3∠B = 120°
∴ ∠B = 40°
Substituting ∠B = 40° in equation (2), we get
∠A + 40° = 60°
∴ ∠A = 20°
Substituting ∠B = 40° in ∠C = 3∠B, we get
∠C = 3(40°)
∴ ∠C = 120°
Thus, in ΔABC, ∠A = 20°; ∠B = 40°; ∠C = 120°
Note: Replacing ∠A + ∠B = \(\frac{\angle \mathrm{C}}{2}\) in ∠A + ∠B + ∠C = 180°, we get a simple equation in one variable is \(\frac{3}{2}\)∠C = 180°.
After solving it for ∠C, ∠B and ∠A can be found easily.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
5x – y = 5 gives y = 5x – 5

x	1	2
y	0	5

3x – y = 3 gives y = 3x – 3

x	1	2
y	0	3

Now, we draw the graphs of both the equations.

From the graph, we observe that the co-ordinates of the vertices of the triangle formed by the graphs of 5x – y = 5 and 3x – y = 3 and the y-axis are (1, 0), (0, -3) and (0, -5).

Question 7.
Solve the following pair of linear equations:
1. px + qy = p – q
qx – py = p + q
2. ax + by = c
bx + ay = 1 + c
3. \(\frac{x}{a}-\frac{y}{b}=0\)
ax + by = a² + b²
4. (a – b)x + (a + b) y = a² – 2ab – b²
(a + b) (x + y) = a² + b²
5. 152x – 378y = -74
-378x + 152y = -604
Solution:
1. px + qy = p – q ……….(1)
qx – py = p + q ……….(2)
Multiplying equation (1) by p and equation (2) by q, we get
p²x + pqy = p² -pq ……….(3)
q²x – pqy = pq + q² ……….(4)
Adding equations (3) and (4), we get
(p²x + pqy) + (q²x – pqy) = (p² – pq) + (pq + q²)
∴ x(p² + q²) = p² + q²
∴ x = 1
Substituting x = 1 in equation (1), we get
p(1) + qy = p – q
∴ qy = -q
∴ y = -1
Thus, the solution of the given pair of linear equations is x = 1, y = -1.

2. ax + by = c ……….(1)
bx + ay = 1 + c ……….(2)
Multiplying equation (1) by a and equation (2) by b, we get
a²x + aby = ac ……….(3)
b²x + aby = b + bc ……….(4)
Subtracting equation (4) from equation (3), we get
(a²x + aby) – (b²x + aby) = ac – (b + bc)
x(a² – b²) = ac – b – bc
x = \(\frac{c(a-b)-b}{a^2-b^2}\)
Substituting x = \(\frac{c(a-b)-b}{a^2-b^2}\) in equation (1), we get

Thus, the solution of the given pair of linear equations is
x = \(\frac{c(a-b)-b}{a^2-b^2}\), y = \(\frac{c(a-b)+a}{a^2-b^2}\)

3. \(\frac{x}{a}-\frac{y}{b}=0\) ……….(1)
ax + by = a² + b² ……….(2)
From equation (1), we get
\(\frac{x}{a}=\frac{y}{b}\)
y = \(\frac{b}{a}\)x
Substituting y = \(\frac{b}{a}\)x in equation (2), we get

Substituting x = a in y = \(\frac{b}{a}\)x, we get
y = \(\frac{b}{a}\)(a)
∴ y = b
Thus, the solution of the given pair of linear equations is x = a, y = b.

4. (a – b)x + (a + b)y = a² – 2ab – b² ……….(1)
(a + b) (x + y) = a² + b²
∴ (a + b)x + (a + b)y = a² + b² ……….(2)
Subtracting equation (1) from equation (2), we get
[(a + b)x + (a + b)y] – [(a – b)x + (a + b)y] = (a² + b²) – (a² – 2ab – b²)
∴ x(a + b – a + b) = a² + b² – a² + 2ab + b²
∴ x (2b) = 2ab + 2b²
∴ x = \(\frac{2 b(a+b)}{2 b}\)
∴ x = a + b
Substituting x = a + b in equation (1), we get
(a – b)(a + b) + (a + b)y = a² – 2ab – b²
∴ a² – b² + (a + b) y = a² – 2ab – b²
∴ (a + b)y = -2ab
∴ y = \(-\frac{2 a b}{a+b}\)
Thus, the solution of the given pair of linear equations is x = a + b, y = \(-\frac{2 a b}{a+b}\)

5. 152x – 378y = -74 ……….(1)
-378x + 152y = -604 ……….(2)
Adding equations (1) and (2), we get
-226x – 226y = -678
∴ x + y = 3 (Dividing by -226) ……….(3)
Subtracting equation (2) from equation (1),
we get
(152x – 378y) – (-378x + 152y) = (-74) – (-604)
∴ 530x – 530y = 530
x – y = 1 (Dividing by 530) ……….(4)
Adding equations (3) and (4), we get
2x = 4
∴ x = 2
Substituting x = 2 in equation (3), we get 2 + y = 3
∴ y = 1
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

Question 8.
ABCD is a cyclic quadrilateral (See the given figure). Find the angles of the cyclic quadrilateral.

Solution:
ABCD is a cyclic quadrilateral.
∠A + ∠C = 180° and ∠B + ∠D = 180°
∠A + ∠C = 180° gives 4y + 20° – 4x = 180°
∴ 4y – 4x = 160
∴ y – x = 40° (Dividing by 4) …………..(1)
∠B + ∠D = 180° gives 3y – 5° – 7x + 5° = 180°
∴ 3y – 7x = 180° …………..(2)
From equation (1), we get y = x + 40°
Substituting y = x + 40° in equation (2), we get
3(x + 40°) – 7x = 180°
∴ 3x + 120° – 7x = 180°
∴ -4x = 60°
∴ x = -15°
Substituting x = -15° in equation (1), we get
y – (-15) = 40°
∴ y + 15° = 40°
∴ y = 25°
Now, ∠A = 4y + 20° = 4(25°) + 20° = 120°,
∠B = 3y – 5° = 3 (25°) – 5° = 70°,
∠C = -4x = -4(-15) = 60° and
∠D = -7x + 5° = -7(-15°) + 5° = 110°
Thus, in the given cyclic quadrilateral ABCD.
∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°.

Jharkhand Board JAC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

Jharkhand Board Class 10 Science Light Reflection and Refraction Textbook Questions and Answers

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay
[Hint: The material of a lens must be transparent but clay is opaque.]

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length of the lens
(c) At infinity
(d) Between the optical centre of the lens and its principal focus
Answer:
(b) At twice the focal length of the lens

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of – 15 cm. The mirror and the lens are likely to be …
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer:
(a) both concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be _______.
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Answer:
(d) either plane or convex.
[Hint: Both give erect images. The size of image in plane mirror is same as the object but it is diminished in convex mirror.]

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer:
(c) A convex lens of focal length 5 cm [Hint: Since a convex lens gives a magnified image of the object and the smaller the focal length, the more the magnifying power.]

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Solution:
When the object is placed between the pole and the principal focus of a concave mirror, an erect, virtual and enlarged image is formed. This image can be viewed in the mirror itself, not on the screen. In short, image is formed behind the mirror as shown in the figure 10.57.

Therefore, the range of distance of the object from the mirror must be greater than zero and less than 15 cm (focal length of the concave mirror).

Question 8.
Name the type of mirror used in the following situations:
(a) Headlights of a car
(b) Side / rear-view mirror of a vehicle
(c) Solar furnace
Support your answer with reason.
Answer:
(a) For headlights of a car, a concave mirror is used.
The light source is kept at the focus s of the mirror. On reflection, a strong? parallel beam of light emerges out.

(b) For side / rear-view mirror of a vehicle, s convex mirror is used.
This is because its field of view is very large and it forms a virtual, erect and diminished image of the object behind the vehicle, which enables a driver to see most of the traffic behind him/her.

(c) For a solar furnace, a concave mirror is used.
Light from the Sun, on reflection from t the mirror, is concentrated at the focus of s the mirror, producing heat and temperature increases sharply up to 180°C-200°C

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes.
Even when one-half of a convex lens is covered with a black paper, the lens produces a complete (or full) image of the object.

Here, the intensity (brightness) of image will become one-forth as that with complete lens exposed, because less number of light rays can be passed / refracted through the lens.

The nature, size and location of the image will be the same since light from all parts of the object reach the exposed part of the lens.

It can be understood by the following two cases (Experimental verification) :
(1) When the upper half of the lens is covered :
In this case, the ray of light coming from the object will be refracted by the lower half of the lens.

These rays meet at the other side of the lens to form the image of the given object.

(2) When the lower half of the lens is covered:
In this case, the ray of light coming from the object will be refracted by the upper half of the lens.

These rays meet at the other side of the lens to form the image of the given object.

[Note: Covering one-half of the lens reduces the brightness of the image.]

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Solution:
Here, Object size h = + 5 cm
Object distance u = – 25 cm
Focal length of lens f = + 10 cm
(∵ Converging, i.e., Convex lens)
Image distance v = ?
Image height h’ = ?
As, \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) + \(\frac { 1 }{ u }\)
= \(\frac{1}{(+10)}+\frac{1}{(-25)}=\frac{5-2}{50}=\frac{3}{50}\)
= 16.67 cm
Image distance v is positive. This shows that the image formed is real and on the other side of the lens, at 16.67 cm from the lens as shown in the figure 10.60:

Now, Magnification m = \(\frac { h’ }{ h }\) = \(\frac { v }{ u }\)
∴ h’ = h\(\frac { v }{ u }\)
= \(\frac { 1 }{ f }\)
= – \(\frac { 10 }{ 3 }\) cm
= – 3.3 cm
The negative (minus) sign shows that the image is inverted, real, diminished (3.3 cm).
Figure 10.60 show the position, size and nature of the image formed.

[Note: The ray diagram can be drawn without calculating v and h’. Once, f, u and h are fixed, v and h’ have definite values.

Question 11.
A concave lens of focal length 15 cm forms an image at 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Solution:
Here, Focal length of the lens f = – 15cm (∵ Concave lens)
Image distance v = – 10 cm
(∵ In case of a concave lens the image is formed on the same side of the object)
Object distance u = ?

The negative (minus) sign shows that the object is placed on the left side of the lens.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Solution:
Here, Object distance u = – 10 cm
Here, Object distance u = – 10 cm
Focal length of the mirror f = + 15 cm (∵ Convex mirror)
Image distance v = ?
Using the mirror formula \(\frac { 1 }{ f }\) = \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\)
We have
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) – \(\frac { 1 }{ u }\)
= \(\frac{1}{(+15)}-\frac{1}{(-10)}\)
= \(\frac{1}{15}+\frac{1}{10}\)
= \(\frac { 2+3 }{ 3 }\)
= \(\frac { 5 }{ 30 }\)
= \(\frac { 1 }{ 6 }\)
∴ v = 6 cm
The positive (plus) sign of v indicates that the image is formed behind the mirror.
Now, magnification m = – \(\frac { v }{ u }\) = – \(\frac { 6 }{ – 10 }\) = + 0.6
Positive value of magnification indicates that the image is virtual and erect.
The magnitude of magnification is 0.6, which is less than 1. This shows that the S image is diminished.

Question 13.
The magnification produced by a plane mirror is + 1. What does this mean?
Solution:
Here, m = + 1
As, m = \(\frac { h’ }{ h }\)
\(\frac { h’ }{ h }\) = + 1
∴ h’ = h
So, the size of the image is equal to the size of object.
Further, the positive (plus) sign of m indicates that the image is erect and hence virtual.
Again, m = – \(\frac { v }{ u }\) and m = + 1
∴ – \(\frac { v }{ u }\) = 1
∴ v = – u
This shows that the image is formed behind the mirror and the distance of the image from the mirror equals that of the object from the mirror.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Solution:
Here, Size of object h = + 5.0 cm
Object distance u = – 20 cm
Radius of curvature R = + 30 cm (∵ Convex mirror)
Focal length f = \(\frac { + 30 cm }{ 2 }\) = + 15 cm
Image distance v = ?
Image size h’ = ?
Mirror formula: \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)

The positive (plus) sign of u indicates that the image is behind the mirror or on the right-hand side of the mirror.

Positive value of image height indicated that the image is virtual and erect.
Moreover, h’ < h
So, the image is smaller in size than the object.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Solution:
Here, Object size h = 7.0 cm
Object distance u = – 27 cm
Focal length f = – 18 cm
(∵ Concave mirror)
Image distance u = ?
Image size h’ = ?
From mirror formula :
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)

The negative (minus) sign of v indicates that ’ the image is formed on the same side of the object. So, screen should be held in front of the mirror at a distance of 54 cm from the mirror. The image can be obtained on the screen and hence, it is real.
Now, magnification m = \(\frac { h’ }{ h }\) = – \(\frac { v }{ u }\)
∴ h’ = – h(\(\frac { v }{ u }\))
= – 7.0 (\(\frac { -54 }{ -27 }\))
= – 14 cm
The negative (minus) sign of h’ shows that the image is inverted and hence would be real.

Question 16.
Find the focal length of a lens of power -2.0D. What type of lens is this?
Solution :
Here, Focal length of the lens f = ?
Power P = – 2.0 D = – 2.0 m^-1
Now,
As, P = \(\frac { 1 }{ f }\)
f = \(\frac { 1 }{ p }\)
= \(\frac { 1 }{ -2D }\)
= – 0.5 cm
As the power of the lens is negative (given), the lens must be concave.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Solution:
Here, Power P = + 1.5 D = + 1.5 m^-1
Focal length f = ?
From, f = \(\frac { 1 }{ p }\)
f = \(\frac { 1 }{ 1.5 D }\)
= 0.67 m
= 67 cm
As the power of the lens is positive (given), it is a converging lens, i.e., a convex lens.

Jharkhand Board Class 10 Science Light Reflection and Refraction InText Questions and Answers

Question 1.
Define the principal focus of a concave mirror.
Answer:
The point on the principal axis of a concave mirror, at which rays of light incident on the mirror in a direction parallel to the principal axis (actually) meet / intersect after reflection from the mirror is called the principal focus (F) of the concave mirror.

[Note : In general, the point on the principal axis where rays of light incident parallel to the principal axis converge to or appear to diverge from after reflection is called the principal focus of the spherical mirror.]

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Solution:
Here, R = 20 cm; f =?
As f = \(\frac { R }{ 2 }\), f = \(\frac { 20 }{ 2 }\)
= 10 cm
[Note: According to the New Cartesian sign convention, if the given spherical mirror is a convex mirror, then f = + 10cm and if the given spherical mirror is a concave mirror, then f = – 10 cm.]

Question 3.
Name a mirror that can give (form) an erect and enlarged image of an object.
Answer:
Concave mirror:
A concave mirror produces an erect and enlarged image of an object, when an object is placed between the pole and principal focus of the concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
This is because a convex mirror (always) forms an erect, virtual and diminished image of an object, wherever the object may be located.

Also, a convex mirror has a wider field of view, relative to a plane mirror, as it is curved outwards. Thus, a convex mirror fitted on the side of a vehicle enables the driver to view much larger area than that would be possible with a plane mirror, enabling the driver to see traffic behind him/her to facilitate safe driving.

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Solution:
Here, the radius of curvature R = 32 cm
We know that, the focal length f = \(\frac { R }{ 2 }\)
∴ f = \(\frac { 32 }{ 2 }\) = 16 cm

Question 6.
A cancave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where 5 is the image located?
Solution:
Here, the linear magnification m = – 3
(negative sign for a real image, which is inverted)
The object distance u = – 10 cm
(object distance (real) is always negative)
The image distance v =?
As m = – \(\frac { v }{ u }\)
v = – mu
∴ v = – (-3) (- 10)
= – 30 cm
Thus, the real image is located at 30 cm in front of the mirror (on the same side as the object).

Question 7.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
For oblique incidence, the light ray bends towards the normal, on entering water. It happens because water is optically denser than air.

The speed of light is higher in an optically rarer medium than in an optically denser medium. So, a ray of light travelling from air to water slows down and bends towards the normal.
OR

Thus, the angle of refraction is less than the angle of incidence. It implies that the light ray bends towards the normal.

Question 8.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 x 10⁸ m s^-1.
Solution:
Speed of light in vaccum c = 3 x 10⁸ m s^-1.
Refractive index of glass n_g = 1.50
Speed of light in glass v = ?
Now, absolute refractive index of glass,
n_g = \(\frac { c }{ v }\)
∴ The speed of light in the glass,
v = \(\frac{c}{n_{\mathrm{g}}}=\frac{3 \times 10^8}{1.50}\)
= 2 x 10⁸ m s^-1.

Question 9.
Find out from Table 2 (Given with Q. 42), the medium having highest optical density. Also, find the medium with lowest optical density.
Answer:
The higher the refractive index, the higher is the optical density. Diamond has the highest optical density as it has the highest refractive index, 2.42 and air has the lowest optical density as it has the lowest refractive index, 1.0003.

Question 10.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 2.
Answer:
From Table 2,
refractive index of kerosene = 1.44
refractive index of turpentine = 1.47 and
refractive index of water = 1.33
Here, from given liquids, turpentine has the highest refractive index and water has the lowest refractive index.
Now, the lower the refractive index of a medium, the higher is the speed of light in the medium.
Hence, out of the given liquids the light travels fastest in water.

Question 11.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
Refractive index of diamond
= \(\frac{\text { Speed of light in vacuum / air }}{\text { Speed of light in diamond }}\)
∴ Speed of light in diamond
= \(\frac{\text { Speed of light in vacuum / air }}{\text { Refractive index of diamond }}\)
This expression states that the speed of light in diamond is \(\frac { 1 }{ 2.42 }\) times the speed of light in vacuum / air.
Speed of light in diamond = \(\frac{3 \times 10^8}{2.42}\)
= 1.24 x 10⁸ms^-1
In other words, it can also be said that, the ratio of the speed of light in vacuum/air to the speed of light in diamond is equal to 2.42.

Question 12.
Define 1 dioptre of power of a lens.
OR
Define the dioptre.
Answer:
1 dioptre is the power of a lens of focal length 1 metre.

Question 13.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Solution:
A convex lens forms a real, inverted image of the same size as that of the object, if the object is placed at 2F₁.
In this case v = + 50 cm and m = – 1

Thus, the needle is placed at 50 cm from the convex lens of power 4 D.

Question 14.
Find the power of a concave lens of focal length 2 m.
Solution :
Here, focal length f = – 2 m (∵ Concave lens)
∴ Power P = \(\frac { 1 }{ f(m) }\)
= \(\frac { 1 }{ -2m }\)
= – 0.5 m^-1
= – 0.5 dioptre
= – 0.5 D

Activity 10.1 [T. B. Pg. 161]

To determine the nature of the inner and outer curved surface of a spoon.

Procedure :
1. Take a large shining spoon. Try to view your face in its inner curved surface.

• Do you get the image?
• Is it smaller or larger?

2. Move the spoon slowly away from your face. Observe the image.

• How does it change?

3. Reverse the spoon and repeat the activity.

• How does the image look like now?

4. Compare the characteristics of the image on the two surfaces.

Observation :

• Yes.
• The image formed is enlarged and erect when the spoon is closer to the face.
• As the spoon is moved slowly away from the face, the image becomes large and inverted. When the face is very far off, highly diminished and inverted image of our face is seen.
• When the spoon is reversed, the image formed by the outer surface of the spoon is virtual and diminished in size.
• Now, if we move the spoon away from our face, the image moves away, however the image continues to be virtual and diminished in size.
• Finally, at last the image is diminished to almost point-size.

Conclusion:
The inner surface of the spoon acts as a concave mirror, whereas the outer surface acts as a convex mirror.

Activity 10.2 [T. B. Pg. 162]

To show the converging nature of a concave mirror and find its focal length.

Procedure:

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Hold the mirror and the paper in the same position for a few minutes.
• What do you observe? Why?
  

Observation:
The paper initially turns blackish, then burns producing smoke. Eventually it catches fire. The light from the Sun is converged at a point, as a sharp bright spot by the mirror. In fact, this spot of light is the image of the Sun on the sheet of paper.

This point is the principal focus of the concave mirror. The heat produced due to concentration of sunlight ignites the paper.

Conclusion:

• The distance of this image of the Sun from the pole of the mirror gives the approximate value of the focal length of the concave mirror.
• It is also concluded that when a parallel beam of light from a far off object falls on a concave mirror, a real, inverted and point-ized image is formed at the focus of the concave mirror.

Activity 10.3 [T. B. Pg. 163]

To locate the image formed by a concave mirror for different positions of the object.

Procedure :
1. Take a concave mirror.

2. Find out its approximate focal length in the way described in Activity 10.2.

3. Note down the value of the focal length.

4. Mark a straight line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that the pole of the mirror lies over the line.

5. Draw with a chalk two more straight lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror.

These lines will now correspond to the positions of the points P F and C respectively.
(Remember : For a spherical mirror of small aperture, the principal focus F lies mid-way between the pole P and centre of curvature C.)

6. Keep a bright object, say a burning candle, at a position far beyond C.
Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.

7. Observe the image carefully. Note down its nature, position and relative size with respect to the object size.

8. Repeat the activity by placing the candle –

• just beyond C
• at C
• between F and C
• at F
• between P and F

In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Where would you look to observe the image of that object?

9. Note down and tabulate your observations.

Observation:
1. Image formation by a concave mirror for different positions of the object.

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Just beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

2. When an object (the burning candle in the present case) is placed between P and F, its image cannot be obtained on the screen.

3. Hence, to observe its image, in this case one has to look in the mirror itself because the image is virtual.

Conclusion:
The nature, position and size of the image formed by a concave mirror depends on the position of the object in relation to the points E F and C.

Activity 10.4 [T. B. Pg. 166]

To locate the image formed by a concave mirror for different positions of an object using ray diagrams.

Procedure:

• Draw a neat ray diagram for each position of the object as discussed in the Activity 10.3. You may take any two of the rays (mentioned in the previous point 10.2.2), for locating the image.
• Observe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observation:
The following figures illustrate the ray diagrams for the formation of image by a concave mirror for various positions of the object:

Conclusion:
Image formation by the concave mirror for different positions of an object using ray diagrams is tabulated as follows :

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Just beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

Activity 10.5 [T. B. Pg. 167]

To locate the image formed by a convex mirror for different positions of an object.

Procedure:

• Take a convex mirror. Hold it in one hand.
• Hold a pencil in the upright position in the other hand.
• Observe the image of the pencil in the mirror.
• Is the image erect or inverted? Is it diminished or enlarged?
• Move the pencil away from the mirror slowly.
• Does the image become smaller or larger?
• Repeat this activity carefully.
• State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror.

Observation:
1. When we hold a pencil in the upright position in front of a convex mirror, the image of the pencil is observed on the back side of the mirror, i.e., in the mirror itself.
The image is erect and virtual.
The image is diminished in size relative to the object.

2. When the pencil is moved away from the mirror slowly, the image becomes smaller and smaller, moving away from the mirror.

3. On repeating this activity, we find that as the object is moved away from the mirror, the image moves closer to the focus of the mirror.

Conclusion:
This activity confirms all the characteristics of the images formed by a convex mirror for different positions of the object and it is tabulated as follows :

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F, behind the mirror	Highly diminished, point-sized	Virtual and erect
Between infinity and the pole P of the mirror	Between P and F, behind the mirror	Diminished	Virtual and erect

Activity 10.6 [T. B. Pg. 167]

To demonstrate the wide field of view nature of a convex mirror.

Procedure:

• Observe the image of a distant object, say a distant tree in a plane mirror.
• Could you see a full-length image?
• Try with plane mirrors of different sizes.
• Did you see the entire object in the image?
• Repeat this activity with a concave mirror.
• Did the concave mirror show full-length image of the object?
• Now, try using a convex mirror.
• Did you succeed?
• Explain your observation with reason.

Observation :
No, we can not see the full-length image of an object in plane mirror.

When we try with plane mirrors of different sizes, we find that the entire object in the image is seen when the size of the plane mirror is at least half of the size of the object.

No, concave mirror did not show full-length S image of the object (here a distant tree).

When we use a small convex mirror, we succeed in seeing the full-length image of an object, wherever the object may be located.

The reasons for these observations are as follows :
(1) In a plane mirror, size of image is S always equal to the size of the object.

(2) In case of a concave mirror, when object is between P and F, then only its enlarged, virtual and erect image is seen behind the concave mirror.
(i.e., full-length image of an object can be seen).
But here our object is far away from the concave mirror, e.g., distant tree; so its full- length image can not be seen in the concave mirror itself (i.e., behind the concave mirror).

(3) While, in a convex mirror, the image is always virtual, erect and smaller than the object, wherever the object may be located.

Conclusion:
Out of plane mirror, concave mirror and convex mirror, only a convex mirror can show a full-length image of a tall object in all its positions.
OR
A convex mirror has a wide field of view and the image formed by it is always virtual, erect and shorter than the object, wherever the object may be located.

Activity 10.7 [T. B. Pg. 172]

To demonstrate refraction of light.
OR
To show that the apparent depth of the bucket filled with water is less than the real depth of the bucket.

Procedure:

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go.
• Did you succeed in picking up the coin?
• Repeat the Activity.
• Why did you not succeed in doing it in one go?
• Ask your friends to do this.
• Compare your experience with theirs.

Observation:
No.
When you try picking up the coin with your eye to a side above water, you do not succeed in one go.

The reason for this can be explained as follows :
As shown in the figure 10.31, a coin is at point ‘O’ at the bottom of a bucket filled with water.

When we view this coin with our eye to a side above water, we observe the image I of the coin, which is above ‘O’. When we try to pick up this coin in one go, we do not succeed in picking up the coin. This is because we move our hand up to I, where the coin is being observed, but actually the coin lies below at O, the bottom of the bucket.

When our friends try the same way, they also fail to pick up the coin.

Conclusion:
In going from water to air, rays of light bend away from the normal and the bottom of the bucket appears to be raised, i.e., the apparent depth of the bucket is less than the real depth of the bucket.

Activity 10.8 [T. B. Pg. 172]

To show that the object in water appears to be raised due to refraction.
OR
Apparent depth of an object in the large shallow bowl filled with water increases as its real depth increases.

Procedure:

• Place a large shallow bowl filled with water on a table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Keep looking for the coin from your position.
• Does the coin becomes visible again from your position?
• How could this happen?

Observation:

O = Real position of the coin
I = Initial apparent position of the coin
I’ = New apparent position of the coin

As shown in the figure 10.32, in a large shallow bowl filled with water, the real position of the coin is ‘O’.
On account of refraction of light the coin appears raised from O to I. Image I of the coin appears to be raised which is original apparent position of a coin.

When we move our eye slowly away from the bowl as shown by an arrow in the figure, the coin disappears from our sight. This happens because rays starting from the coin O fail to enter our eye, after refraction at water surface AB.

When a friend pours water gently into the bowl without disturbing the coin, the coin becomes visible again as it appears raised further to position I’.

This happens because on adding water, the real- depth of the coin increases. As the apparent depth of a coin is equal to the real depth divided by the refractive index (n) of water, the apparent depth of the coin also increases. Hence, the coin appears raised from I to I’. Therefore, the coin becomes visible from the displaced position (new position) of our eye.

Conclusion:
The coin becomes visible again and slightly raised above its actual position on pouring water into the bowl. This is because of refraction of light.

In other words, apparent depth of a coin in the large shallow bowl filled with water increases as its real depth increases.

Activity 10.9 [T. B. Pg. 172]

To show that refraction does not occur for normal incidence.

Procedure:

1. Draw a thick straight line in ink, over a sheet of white paper placed on a table.

2. Place a glass slab over the line in such a way that one of its edges makes an angle with the line.

3. Look at the portion of the line under the slab from the sides.

• What do you observe?
• Does the line under the glass slab appear to be bent at the edges?

4. Next, place the glass slab such that it is normal to the line.

• What do you observe now?
• Does the part of the line under the glass slab appear bent?

5. Look at the line from the top of the glass slab.

• Does the part of the line, beneath the slab, appear to be raised?
• Why does this happen?

Observation :

• When we look at the portion of the line under the slab from sides, careful observation shows that line AB is bent at edge B and line BC at edge C. [figure 10.33 (a)]
• Yes. The line under the glass slab appears bent at edges B and C.
• On placing the glass slab normal to the line, we observe that the part of the line under the glass slab does not appear bent, [figure 10.33 (c)]
• No. The line under the glass slab does not appear bent at edges B and C.
• Yes. The part of the line, beneath the slab, appears raised while looking at the line from the top of the glass slab.
• This is due to refraction of light.

Conclusion:
The object appears raised due to refraction of light and for normal incidence, refraction of light does not occur.

Activity 10.10 [T. B. Pg. 173]

To show the refraction of light and lateral displacement through a rectangular glass slab.

Procedure:

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil and name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge.
• Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight-line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly,
• join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
• Join O and O’. Also, produce EF up to P as shown by a dotted line in the following figure.
• At O, draw NN’ perpendicular to AB. At O’, draw MM’ perpendicular to CD. Also, draw O’L perpendicular to OP.
  

[Note : (1) In figure 10.34, face AB is air-glass interface and face CD is glass-air interface. ( 2 ) Reflection of light is not shown in the figure.]

Observation :

• At O, light ray along EF enters from air into glass. It bends towards the normal NN’. This is the first refraction.
• At O’, the light ray enters from glass into air. It bends away from the normal MM’ and travels 5 along GH. This is second refraction.
• Here, angle of emergence r₂ is equal to angle of incidence i₁, i.e., the emergent ray is parallel to the original direction of the incident ray. This is because there is identical medium, air in this case, adjacent to both edges AB and CD.
• However, the light ray is shifted sideways slightly. This is lateral displacement and is represented by O’L.

Conclusion:

• The ray of light travelling from a rarer medium (air in this case) to a denser medium (glass in this case) bends towards the normal.
• The ray of light travelling from a denser medium (glass in this case) to a rarer medium (air in this case) bends away from the normal.
• The emergent ray is parallel to the incident ray but is slightly displaced sideways.

Activity 10.11 [T. B. Pg. 177]

To show the converging nature of a convex? lens and find its focal length.

Procedure :
1. Hold a convex lens in your hand. Direct it towards the Sun.

2. Focus the light from the Sun (through lens) on a sheet of paper. Obtain a sharp bright image? of the Sun.

3. Hold the paper and the lens in the same position for a while.
Keep observing the paper.

• What happened?
• Why?
  

Observation :

• When the paper and the lens were held in the same position for sometime, the paper began to burn producing smoke. After sometime, it caught fire.
• Because, the light from the Sun constitutes parallel rays of light. These rays were converged by the lens at the sharp bright spot formed on the paper and so heat is produced by these Sun rays as they concentrate on the spot. This caused the paper to burn.

Conclusion:

• When a parallel beam of light from a far-off object like the Sun falls on a convex lens, a real, inverted, point size image of the Sun is formed at the focus of the lens.
• The distance between the position of the lens and the position of the image of the Sun gives the approximate focal length of the lens.

[Note : The distance between the lens and the image of the Sun on the paper is the approximate focal length of the lens.]

Activity 10.12 [T. B. Pg. 178]

To locate the position and examine the nature of the image formed by a convex lens for different positions of the object.

Procedure:

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel lines, using chalk, on a long table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂ and 2F₂ respectively.
• Place a burnig candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing the object just beyond 2F₁, at 2F₁, between F₁ and 2F₁, at F₁, between F₁ and O.
• Note down and tabulate your observations.

Observation:
Position, relative size and the nature of the image formed by a convex lens for various positions of the object.

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F2	Highly diminished, point-sized	Real and inverted
Beyond 2F1	Between F2and 2 F2	Diminished	Real and inverted
At F1	At 2F2	Same size	Real and inverted
Between F1and 2F1	Beyond 2 F2	Enlarged	Real and inverted
At focus F1	At infinity	Infinitely large or Highly enlarged	Real and inverted
Between focus F1 and optical centre O*	On the same side of the lens as the object	Enlarged	Virtual and erect

Conclusion:
The nature, position and relative size of the image formed by a convex lens for various positions of the object depends on the position of the object in front of the lens.

Activity 10.13 [T. B. Pg. 179]

To locate the position and examine the nature of the image formed by a concave lens for different positions of the object.

Procedure:

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible.
• If not, observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the the image.
• Move the candle away from the lens. Note the change in the size of the image.
• What happens to the size of the image when the candle is placed too far away from the lens?
• What conclusion can you draw from this Activity?

Observation:
The image formed by a concave lens, when a burning candle is kept too far away from the lens is highly diminished, i.e., point-sized.
Position, relative size and the nature of the image formed by a concave lens for various position of an object

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F1	Highly diminished, point-sized	Virtual and erect
Between infinity and optical centre O of the lens	Between focus F1 and optical centre O	Diminished	Virtual and erect

Conclusion:
The image formed by a concave lens is always virtual, erect and diminished irrespective of the position of the object.

Jharkhand Board JAC Class 10 Science Solutions Chapter 11 Human Eye and Colourful Worlds Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

Jharkhand Board Class 10 Science Human Eye and Colourful World Textbook Questions and Answers

Question 1.
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to ……………
(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.
Answer:
(b) accommodation.
[Hint : Accommodation is the ability of the eye lens to focus both nearby and distant objects by adjusting its focal length.]

Question 2.
The human eye forms the image of an object at its ……………..
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Answer:
(d) retina.
The retina is the light sensitive surface of the eye on which the image is formed.

Question 3.
The least distance of distinct vision for a young adult with normal vision is about ……………..
(a) 25 m
(b) 2.5 m
(c) 25 m
(d) 2.5 m
Answer:
(c) 25 cm.
The minimum distance at which an object can be seen most distinctly without any strain (i.e., comfortably) is 25 cm.

Question 4.
The change in focal length of an eye lens is caused by the action of the…
(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.
Answer:
(c) ciliary muscles.
The ciliary muscles contract and expand, in order to change the curvature of the eye lens for focussing the image of an object at varying distances at the retina.

Question 5.
A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptres. What is the focal length of the lens required for correcting (1) distant vision and (2) near vision?
Solution :
(1) For distant vision, f = ?
P = – 5.5, D = – 5.5 m^-1
Now, f = \(\frac { 1 }{ p }\)
∴ f = \(\frac{1}{-5.5 m^{-1}}\)
= – 0.182m
= – 18.2 cm (concave lens)

(2) For near vision, f = ?
P = + 1.5, D = + 1.5 m^-1
Now, f = \(\frac { 1 }{ p }\)
∴ f = \(\frac{1}{+1.5 m^{-1}}\)
= 0.667 m
= 66.7 cm (Convex lens)

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Solution:
1. The defect of an eye called myopia (short-sightedness or near-sightedness) is corrected by using spectacles containing concave lens of appropriate focal length.

2. Here the far point of myopic person is 80 cm. (while far point of normal person is infinity oo)

3. This means that this person can see the distant object (kept at infinity) clearly if the image of this distant object is formed at his own far point (which is 80 cm here).
So, in this case:
Object distance u = – ∞ (Normal far point)
Image distance v = – 80 cm (Far point of this defective eye in front of lens)
Focal length f = ?
Now,
By the lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
∴ \(\frac{1}{f}=\frac{1}{-80}-\frac{1}{-\infty}\)
∴ \(\frac{1}{f}=-\frac{1}{80}\)
∴ f = – 80 cm
= – 0.8 m (Concave lens)
Now,
Power of the lens,
P = \(\frac { 1 }{ f }\)
∴ P = \(\frac { 1 }{ -0.8m }\)
= – \(\frac { 10 }{ 8 }\)m^-1
= – 1.25 D
A concave lens of power -1.25D is required to correct the problem.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Solution:
1. The defect of an eye called hypermetropia (long-sightedness or far-sightedness) is corrected by using spectacles containing convex lenses of appropriate focal length.

[N = Near point of a hypermetropic eye and
N’ = Near point of a normal eye]
[11.16 : (a) Hypermetropic eye (b) corrected eye]

2. Here the near point of the hypermetropic eye is 1 m = 100 cm
(while near point of normal eye is 25 cm)

3. This means that this person can see the nearby object (kept at 25 cm) clearly if the image of this nearby object is formed at his own near point (which is lm = 100 cm here).
So, in this case :
Object distance u = – 25 cm (Normal near point)
Image distance u = – 1 m = – 100 cm
(Near point of this defective eye in front of lens)
Focal length f = ?
Now,

Now,
Power of the lens,
P = \(\frac { 1 }{ f }\)
= \(\frac{1}{\left(\frac{1}{3} \mathrm{~m}\right)}\)
= 3D

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
For seeing nearby objects, the ciliary muscles contract to make the eye lens thicker near the middle, so as to reduce the focal length of the eye lens.

But ciliary muscles cannot be contracted beyond a certain limit and hence we cannot see clearly the objects closer than 25 cm from the eye.

In other words, a normal eye is not able to see clearly the objects placed closer than 25 cm because all its power of accommodation has already been exhausted.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
Answer:
For a normal eye, the image distance (u) in the eye is fixed = distance of the retina from the eye lens ≈ 2.3 cm
When we increase the distance of the object (u) from the eye, the focal length of eye lens is changed on account of the accommodating power of the eye so as to keep the image distance (u) constant according to the relation \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\).

Question 10.
Why do stars twinkle?
Answer:
Twinkling of stars is due to atmospheric refraction of starlight.

• Starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth.
• As the optical density of air increases towards surface of the earth, light from the star travels from rarer to denser layers, bending every time towards the normal.
• On producing the final refracted ray backwards as shown in the following figure, we find that the apparent position (B) of a star is higher than the actual position (A) of the star as shown in figure 11.11.
  
• The star appears somewhat higher (above) than its actual position when viewed near the horizon.
  Further, this apparent position of the star is not stationary, but keeps on changing slightly, since physical conditions of the earth’s atmosphere are not stationary.
• Since the stars are very distant, they appear as point-sized sources of light.
• Due to a continuous change in the direction of propagation of light, the apparent position of the star fluctuates all the time and the amount of starlight entering the eye flickers, i.e., the brightness of the star changes continuously (the star sometimes appears brighter and at some other time fainter). This is called the twinkling s of a star.

Question 11.
Explain why the planets do not twinkle.
Answer:
Planets are much closer to the earth relative to stars. Therefore, planets appear bigger > than stars. Stars are far away from the earth. Therefore they appear smaller.

• So, stars can be considered as point-sized sources and planets can be considered as a ? collection of a large number of point-sized sources of light, i.e., extended sources of light.
• As planet is a collection of a large number of point-sized sources of light, the total variation in amount of light entering our eye from all the individual point-sized sources averages out to zero, thereby nullifying the twinkling effect. That is why planets do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:

• In figure 11.14, the situation at the sunrise is shown.
• Here, the white light coming from the Sun near the horizon, passes through thick layers of air and covers larger distance in the earth’s atmosphere before reaching the observer. During this, more scattering of blue light and shorter wavelengths take place. Hence, the reddish light reaches the observer and the Sun appears reddish.
• The same thing occurs at the sunset.

[Note : Rising and setting of the full moon from the horizon appears reddish due to this reason.]

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears dark to the astronaut because in outer space, there is no atmosphere to scatter the sunlight. Since there is no scattering of the blue component of the white sunlight which can reach the eye of an astronaut in outer space, the sky appears dark to the astrounaut, instead of blue.

Jharkhand Board Class 10 Science Human Eye and Colourful World InText Questions and Answers

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length, so that nearby as well as distant objects can be focussed on the retina and hence seen comfortably and distinctly is called the power of accommodation of the eye.

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
A person with a myopic eye should use a concave lens of suitable focal length or power to restore proper vision.
[Here, the person with defect of myopia has the far point nearer than infinity at a distance of 1.2 m from the eye.
So, v = – 1.2m; u = – ∞; f = ?
So, from the lens formula
\(\frac{1}{f}=\frac{1}{-u}-\frac{1}{v}\)
∴ \(\frac{1}{f}=\frac{1}{-(-\infty)}+\frac{1}{-1.2}\)
∴ f = – 1.2 m
∴ P = \(\frac { 1 }{ – 1.2 }\) = – 0.83 D
∴ A concave lens of focal length 1.2 m should be used to restore proper vision.]

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
For the human eye with normal vision, f the far point is at infinity and the near point is at 25 cm from the eye.

Question 4.
A student has difficulty in reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
As the child cannot see distant objects 5 clearly it means that he is suffering from myopia or near-sightedness. In this case the child can S see nearby objects clearly but cannot see far off objects distinctiy as their images are formed s before the retina.

To correct this defect, the child has to use spectacles with concave lens of suitable focal length.

Activity 11.1 [T. B. Pg. 192]

To study the refraction of light through a s triangular glass prism.

Procedure:
1. Fix a sheet of white paper on a drawing board ; using drawing pins.

2. Place a glass prism on it in such a way that ; it rests on its triangular base. Trace the outline of the prism using a pencil.

3. Fix two pins, say at points P and Q, on the line PE as shown in the following figure :

Refraction of light through a triangular glass prism
PE – Incident ray
EE – Refracted ray
FS – Emergent ray
∠A – Angle of the prism
∠i – Angle of incidence
∠r – Angle of refraction
∠e – Angle of emergence
∠D – Angle of deviation

4. Look for the images of the pins, fixed at P and Q, through the other face AC.

5. Fix two more pins, at points R and S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.

6. Remove the pins and the glass prism.

7. Join the points P and Q and extend the line till it meets the nearest boundary of the prism. Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F respectively. Join E and E

8. Extend PE and RS such that they meet at point G. Mark the angle of deviation (∠D) as shown in figure.

9. Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F respectively.

10. Mark the angle of incidence (∠i), the angle of refraction (∠r) and the angle of emergence (∠e) as shown in the figure.

Compare the angle of incidence and the angle of refraction at each refracting surface of the prism.
Bending of rays PE, EF and FS are similar to the kind of bending that occurs in a glass slab.

Observation:
1. A ray of light suffers two refractions while passing through a prism.

2. The first refraction occurs at E on the surface AB. The incident ray PE, enters from air into glass E. It is refracted along EF bending towards normal NN’ on face AB at E.

3. The second refraction occurs at F on the surface AC. The initial refracted ray EF travelling in glass emerges in air at F It emerges along FS, bending away from normal MM’ on face AC at F.

4. At the first refracting surface AB, angle of refraction (r) is smaller than the angle of incidence (i).

But at the second refracting surface AC, angle of refraction (e) is larger than angle of incidence (∠EFM’).

5. Bending of rays PE, EF and FS is of the same kind of bending that occurs in a glass slab. The net deviation in a rectangular glass slab is zero and there is lateral shift. However, due to the peculiar shape of the prism, net diviation in passing through a prism is not zero and the prism makes the emergent ray bend at an angle to the direction of the incident ray. This angle is called the angle of deviation, i.e., D = ∠HGS, It is the angle through which the incident ray is deviated as it passes through the prism.

6. The angle of deviation depends on the angle of incidence, the angle of the prism and the nature of the material of the prism as it depends on the refractive index of the material.

Conclusion :
While passing through a prism, a ray of light undergoes two refractions as it does in case of rectangular glass slab.

However due to peculiar shape of the prism, net deviation through prism is never zero, though the net deviation is zero in case of rectangular glass slab.

In short, deviation suffered by a ray of light on passing through a prism is non-zero.

Activity 11.2 [T. B. Pg. 193]

To show that white light is made up of seven colours and different colours suffer different deviations on passing through a prism.

Procedure:
1. Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.

2. Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light.

3. Now, take a glass prism and allow the light from the slit to fall on one of its faces as shown in the following figure:

4. Turn the prism slowly until the light that comes out of it appears on a nearby screen.

• What do you observe?
• Why does this happen? OR How could white light of the Sun give us various colours of the rainbow?
• What is the sequence of colours that you observe on the screen?

Observations:

• We observe beautiful band of seven colours (VIBGYOR) on the screen. The deviation suffered by the violet light is maximum and that by the red light is minimum. Hence, the violet colour is at the lower end and the red colour is at the upper end of the screen.
• This happens because prism itself splits the incident white light into a band of colours.
• The sequence of colours seen from the lower end of the screen is violet (V), indigo (I), blue (B), green (G), yellow (Y), orange (O) and red (R).

Conclusion :
White light is made up of seven colours and different colours suffer different deviations.

Activity 11.3 [T. B. Pg. 196]

To observe scattering of light by colloidal particles.

Procedure:
1. Place a strong source (S) of white light at the focus of a convex (converging) lens (L₁) as shown in the figure 11.15 to produce a parallel beam of light.

2. Allow the light beam to pass through a transparent glass tank vessel (T) containing clean water.

3. Now, allow the beam of light to pass through the circular hole (C) made in a cardboard and obtain a sharp image of the circular hole on screen (MN) using a second convex (converging) lens (L₂) as shown in figure.

4. Dissolve 200g of sodium thiosulphate (hypo) – (Na₂S₂O₃) in about 2 L of clean water taken in the tank. Add 1 to 2 mL of concentrated sulphuric acid (H₂SO₄) in the water.
What do you observe?

Observation:
In the arrangement shown in the figure, we find that fine microscopic sulphur particles precipitate in water (from solidum thiosulphate) in a couple of minutes.

Blue colour is seen from three sides of the tank due to scattering of short wavelengths by minute sulphur particles. We see transmitted light from the fourth side of the tank facing the hole (C) of the cardboard.

We first observe orange-red colour and then bright crimson red colour on the screen (MN) because the transmitted light contains mainly longer wavelengths only.

Conclusion:
Very fine particles of liquid, mainly scatter blue light of smaller wavelength and the red light containing longer wavelength passes straight (without scattering) from the vessel.

Jharkhand Board JAC Class 10 Science Solutions Chapter 12 Electricity Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 12 Electricity

Jharkhand Board Class 10 Science Electricity Textbook Questions and Answers

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \(\frac { R }{ R }\) is
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25
Hint Here, the resistance of a piece of wire is R. This piece of wire is cut into five equal parts, so the resistance of each part would be \(\frac { R }{ 5 }\).
Now, these five parts each having resistance of \(\frac { R }{ 5 }\) are connected in parallel.
The equivalent resistance R’ of this combination is given by
\(\frac{1}{R^{\prime}}=\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}\)
∴ \(\frac{1}{R^{\prime}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}=\frac{25}{R}\)
∴ \(\frac { R }{ R’ }\) = 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) \(\frac{V^2}{R}\)
Answer:
(b) IR²
Hint: Electric power P = VI = (IR) I = I²R
Thus, the three expressions given in (a), (c) and (d) represent power while expression given in (b) does not.

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be ……………..
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W
Hint: Here, the power rating of the electric bulb is P = 100 W and the voltage rating is V = 220 V
So, the resistance of the filament of the bulb is
R = \(\frac{V^2}{P}\)
= \(\frac{(220)^2}{100}\)
= 484 Ω

Now, the power consumed by the bulb when it is operated at 110 V is given by
P = \(\frac{V^{\prime 2}}{R}\)
= \(\frac{(110)^2}{484}=\frac{110 \times 110}{484}\)
= 25 W

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4
Hint: As both the wires are made of the same material and are of equal lengths and equal diameters, they have the same resistance R.
1. When they are connected in series their equivalent resistance Rs would be –
R_s = R + R = 2R
and when they are connected in parallel their equivalent resistance R_p is given by
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\)
∴ R_p = \(\frac { R }{ 2 }\)

2. Now, heat produced in time t is H = \(\frac{V^2 t}{R}\). As here source voltage V is same,
When wires are connected in series,
H_s = \(\frac{V^2 t}{R_{\mathrm{s}}}=\frac{V^2 t}{2 R}\)
When wires are connected in parallel
∴ H_p = \(\frac{V^2 t}{R_{\mathrm{p}}}=\frac{V^2 t}{\left(\frac{R}{2}\right)}=\frac{2 V^2 t}{R}\)
The ratio of heat produced,
\(\frac{H_{\mathrm{s}}}{H_{\mathrm{p}}}=\frac{V^2 t}{2 R} \times \frac{R}{2 V^2 t}\) = \(\frac { 1 }{ 4 }\)
∴ H_s : H_p = 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is (always) connected in parallel across the points in the circuit between which the potential difference is to be measured.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Solution:
We are given,
the diameter of the wire d = 0.5 mm = 0.5 x 10^-3 m = 5 x 10^-4 m
Resistivity of copper ρ = 1.6 x 10^-8 Ω m
Required resistance R= 10 Ω
Length l = ?

R ∝ \(\frac { 1 }{ d² }\) (If there is no change in p and L)
So, when diameter d is doubled, then resistance R becomes one-fourth of its original value.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

I (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
The graph between V and I plotted by using the given data is shown below:

Calculation of the resistance of given resistor :

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution:
Here, V = 12 V; I = 2.5 mA = 2.5 m A = 2.5 x 10^-3 A; R = ?
The resistance of the resistor R = \(\frac { V }{ I }\)
= \(\frac { 1 }{ 2 }\)
= 4800 Ω
= 4.8 k Ω

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω respectively. How much current would flow through the 12 Ω resistor?
Solution:
Since all the given resistors are connected in series, their equivalent resistance R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω The current through the circuit,
I = \(\frac{V}{R_{\mathrm{s}}}=\frac{9}{13.4}\) = 0.67 A
In a series combination, the same current I flows through all the resistors, so the current flowing through 12 Ω resistor = 0.67 A.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5A on a 220V line?
Solution :
Here, I = 5 A; V = 220 V
So, the total resistance of the given circuit is
R_total = \(\frac { V }{ I }\) = \(\frac { 220 }{ 5 }\) = 44 Ω
i.e., when 44 Ω resistance is connected with 220 V line, 5 A current would flow through the given circuit.
Now, suppose ‘n’ resistors, each of resistance R, are required to be connected in parallel, so that the total resistance R_total becomes 44 Ω.
Hence, \(\frac{1}{R_{\text {total }}}=\frac{1}{R}+\frac{1}{R}\) + … n times
= \(\frac{1+1+\ldots n \text { times }}{R}=\frac{n}{R}\)
∴ R_total = \(\frac { R }{ n }\)
Now, R_total = 44 Ω and R = 176 Ω
So, 44 = \(\frac { 176 }{ 44 }\)
∴ n = \(\frac { 176 }{ 44 }\) = 4
Thus, 4 resistors each of 176 Ω connected in parallel will result in total resistance of 44 Ω causing a current of 5 A to flow when connected to 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Solution:
(i) In order to get a resistance of 9 Ω from three resistors, each of resistance 6 Ω, we connect two 6 Ω resistors in parallel and this parallel combination is connected in series with the third 6 Ω resistor as shown in the following figure:

(ii) In order to get a resistance of 4 Ω from three resistors, each of resistance 6 Ω, we connect two 6 Ω resistors in series and this series combination is connected in parallel to the third 6 Ω resistor as shown in the following figure:

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line are rated 10W. How many bulbs can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Solution:
Here, the voltage rating of each bulb is 220 V and the power rating of each bulb is 10W.
So, the resistance of each bulb
R = \(\frac{V^2}{\rho}=\frac{220^2}{10}=\frac{220 \times 220}{10}\) = 4840 Ω
Now, here, V = 220 V and I = 5 A given.
So, the total resistance of the given circuit is
R_total = \(\frac { V }{ I }\) = \(\frac { 220 }{ 5 }\) = 44 Ω
i.e., When 44 Ω resistance is connected with 220 V line, 5 A current would flow through the given circuit.
Now, when ‘n’ bulbs each of resistance R, are connected in parallel, their equivalent
resistance P_total = \(\frac { R }{ n }\)
Hence, 44 = \(\frac { 4840 }{ 44 }\)
∴ n = \(\frac { 4840 }{ 44 }\) = 110
Thus, 110 bulbs each of resistance 4840 Ω connected in parallel will result in total resistance of 44 Ω causing a current of 5 A to flow, when connected to the 220 V line.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Solution:
Here, the potential difference V = 220 V
The resistance of each coil R_A = R_s = 24 Ω
(1) When each of the coils A or B is connected separately, the current through each coil is
I = \(\frac{V}{R_{\mathrm{A}}}\) or \(\frac{V}{R_{\mathrm{B}}}\)
= \(\frac { 220 }{ 24 }\)
= 9.166 A

(2) When the coils A and B are connected in series, the equivalent resistance of the circuit R_s = R_A + R_B = 24 + 24 = 48 Ω
So, the current through the series combination
I_s = \(\frac{V}{R_{\mathrm{s}}}\)
= \(\frac { 220 }{ 48 }\)
= 4.58 A ≈ 4.6 A

(3) When the coils A and B are connected in parallel, the equivalent resistance R_p of the circuit is given by
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_{\mathrm{A}}}+\frac{1}{R_{\mathrm{B}}}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}\)
∴ R_p = 12 Ω
So, the current through the parallel combination
I_p = \(\frac{V}{R_{\mathrm{p}}}=\frac{220}{12}\) = 18.33 A

Question 14.
Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Solution:
(i) As 1 Ω resistor and 2 Ω resistor are connected in series, the equivalent resistance R_s = 1 + 2 = 3 Ω
Now, the voltage of the battery V = 6 V
So, the current flowing through the circuit,
I_s = \(\frac{V}{R_{\mathrm{s}}}=\frac{6}{3}\) = 2 A
In a series combination the same current 2 A flows through each resistor. Hence the current flowing through 2 Ω resistor is also 2 A.
∴ Power used in 2Ω resistor,
P₁ = I²_sR
= (2)² x 2 = 8 W

(ii) As 12 Ω resistor and 2 Ω resistor are connected in parallel and 4 V battery is connected in parallel with this parallel combination of resistors,
The p.d. across 2 Ω resistor will also be 4 V.
∴ Power used in 2 Ω resistor,
P₂ = \(\frac{V^2}{R}=\frac{4^2}{2}=\frac{16}{2}\) = 8 W
In order to compare the power used in 2 Ω resistor in two different circuits, find the ratio of P₁ and P₂.
So, \(\frac{P_1}{P_2}=\frac{8}{8}\) = 1
∴ P₁ = P₂
Hence, 2 Ω resistor uses equal power, i.e., 8 W in both the circuits.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Solution :
The resistance of 100 W lamp,
R₁ = \(\frac{V^2}{P_1}=\frac{220^2}{100}=\frac{220 \times 220}{100}=\frac{4840}{10}\)Ω = 484 Ω
The resistance of 60 W lamp,
R₂ = \(\frac{V^2}{P_2}=\frac{220^2}{60}=\frac{220 \times 220}{60}=\frac{4840}{6}\)Ω = 806.7 Ω
When, these lamps are connected in parallel, their equivalent resistance R_p is given by
\(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_2+R_1}{R_1 R_2}\)
∴ R_p = \(\frac{R_1 R_2}{R_1+R_2}\)
= \(\frac{484 \times 806.7}{484+806.7}=\frac{390442.8}{1290.7}\) = 302. 6 Ω
The current drawn from the line,
I = \(\frac{V}{R_{\mathrm{p}}}=\frac{220}{302.6}\) = 0.7270 ≈ 0.73 A

Question 16.
Which uses more energy, a 250 W TV set in 1 h, or a 1200 W toaster in 10 minutes?
Solution:
For TV set:
Power P = 250 W = 250 \(\frac { J }{ s }\)
Time t = 1 h = 3600 s
Electric energy (used by TV set)
= P x t
= 250 \(\frac { J }{ 2 }\) x 3600 s
= 900000
= 900 kJ
For Toaster:
Power P = 1200 W = 1200\(\frac { J }{ s }\)
Time t = 10 minute = 10 x 60 = 600 s
Electric energy (used by toaster)
= P x t
= 1200 \(\frac { J }{ s }\) x 600 s = 720000 = 720 kJ s
From above calculations, it is clear that a 250 W TV set in 1 h consumes more electric energy than a 1200 W toaster in 10 minute.

Question 17.
An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution :
Here, I = 15 A; R = 8 Ω; t = 2 h
The rate at which heat is developed in the
heater means its electric power,
P = I² R
= (15)² x 8 = 225 x 8 = 1800 W = 1800 \(\frac { J }{ s }\)

Question 18.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(a) Because tungsten has a high melting point (3380 °C), It does not melt at high temperature. It retains as much of heat generated, so that it becomes very hot and emits light without melting away.

Moreover, tungsten has high flexibility and low rate of evaporation at high temperature. That is the reason why tungsten is used as filament of electric lamps.

(b) The coils of electrical heating devices such as electric toasters and electric irons are made of an alloy, e.g., Nichrome rather than a pure metal because :

• the resistivity of an alloy e.g., Nichrome is much higher than that of its constituent metals, c
• alloys do not oxidise (i.e., burn) readily at high temperature (i.e., when it is red hot at 800 °C)
  alloy has a high melting point.

(c) (1) In a series circuit the current is constant throughout the electric circuit. So it s is obviously impracticable to connect an electric
bulb and an electric heater in series because they need currents of widely different values to operate properly.

(2) In a series circuit when one component (or electrical appliance) fails due to some defect, the circuit is broken and none of the components (or electrical appliances) works.

(3) In a series circuit all the electrical appliances have only one switch due to which they cannot be turned ON or OFF separately.

(4) In a series circuit electrical appliances of different power ratings do not get the same voltage (220V) as that of the power supply line because the voltage is shared by all the appliances. The appliances get less voltage and hence do not work properly.

(d) The resistance of a wire is inversly proportional to its area of cross-section
i.e., R ∝ \(\frac { 1 }{ A }\)
Thus, if the wire is thick (large area of cross-section), then its resistance is less. If the wire is thin (less area of cross-section), then its resistance is large.

(e) Because:

• Copper and aluminium have low electric resistivity, so they conduct electric current without heavy heat losses (due to which they are known as very good conductors of electricity).
• Copper and aluminium are much more easily and cheaply available as compared to metal like silver.
  They can be easily made into wires due to high malleability.

Therefore, copper and aluminium wires are usually employed for electricity transmission.

Jharkhand Board Class 10 Science Electricity InText Questions and Answers

Question 1.
What does an electric circuit mean?
Answer:
An electric circuit is a continuous and closed path of an electric current.
OR
A continuous and closed path consisting of conducting wires and other electrical components along which an electric current flows is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
The SI unit of current is called an ampere (A).
If 1 coulomb charge flows through any cross-section of a conductor in 1 second, then the electric current flowing through the conductor is
said to be 1 ampere.
i.e., 1 A = \(\frac { 1C }{ 1s }\) = 1C s^-1

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Solution :
We know that one electron possesses a negative charge of 1.6 x 10^-19 C.
i.e., charge on 1 electron = – 1.6 x 10^-19 C

∴ No. of electron constituting – 1 C of charge
= \(\frac{-1 C \times 1}{-1.6 \times 10^{-19} \mathrm{C}}\)
= \(\frac { 10 }{ 1.6 }\) x 10¹⁸ = 6.25 x 10¹⁸
Thus, 6.25 x 10¹⁸ electrons taken together constitute 1 coulomb of charge.
In other words, the SI unit of electric charge coulomb (C) is equivalent to the charge contained in 6.25 x 10¹⁸ electrons.

Question 4.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
An electric cell or battery is a device that helps to maintain a potential difference across a conductor.

Question 5.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
The potential difference between two points (in a electric field) is said to be 1 volt if 1 J of work is done to move a charge of 1 C from one point to another point.
1 V = \(\frac { 1 J}{ 1 C }\)

Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Solution:
Here the term, ‘each coulomb’ means ‘every 1 coulomb’. So Q = 1 coulomb, potential difference V = 6 volt, Energy = work done, W = ?
Now, W = VQ
= 6V x 1C
= 6 J
Since the work done on each coulomb of charge is 6 J, 6 J energy is given to each coulomb of charge passing through a 6V battery.

Question 7.
On what factors does the resistance of a conductor depend?
Answer:
The resistance R of a conductor depends on :

• its length l as R ∝ l
• its area of cross-section (i.e., thickness of the conductor) as R ∝ \(\frac { 1 }{ A }\)
• the nature of the material of the conductor (i.e., resistivity of material of the conductor)
• its temperature

Question 8.
Will current can flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
The current will flow more easily through a thick wire than through a thin wire of the s same material and having the same length when connected to the same source.

• Because, the resistance of a wire is inversly proportional to its area of cross-section, A thick wire has more area of cross-section and hence less resistance compared to a thin wire provided the two wires have the same length.
• Hence, a current can flow more easily through a thick wire compared to a thin wire when connected to the same source.

Question 9.
Let the resistance of an electrical component remain constant while the protential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
According to Ohm’s law, I = \(\frac { V }{ R }\)
Here, as R = constant, I ∝ V.
So, when the potential difference across the two ends of the electrical component decreases to half of its former value, the current through it will also decreases to half of its former value.

Question 10.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The coils of electrical heating devices such as electric toasters and electric irons are made of an alloy, e.g., Nichrome rather than a pure metal because :

• the resistivity of an alloy e.g., Nichrome is much higher than that of its constituent metals.
• alloys do not oxidise (i.e., burn) readily at high temperature (i.e., when it is red hot at 800 °C)
• alloy has a high melting point.

Question 11.
Use the data in Table 2 to answer the following:
(a) Which is a better conductor, iron or mercury?
(b) Which material is the best conductor?
Answer:
(a) The electrical resistivity of iron is 10.0 x 10^-8 Ω m whereas that of mercury is 94.0 x 10″8fim. As the resistivity of iron is less than that of mercury, iron (Fe) is a better conductor than mercury (Hg).

(b) Silver metal has the lowest electrical resistivity of 1.60 x 10^-8 Ω m, therefore silver metal is the best conductor of electricity.

Question 12.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
A schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series is shown in figure 12.12

Question 13.
Redraw the circuit of putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would he the readings in the ammeter and the voltmeter?
Answer:
The circuit diagram of above with an ammeter A and a voltmeter V across 12 Ω resistor is shown in figure 12.13.

Calculation of current flowing in the circuit:
Equivalent resistance of the circuit,
R_s = 5 Ω + 8 Ω + 12 Ω = 25 Ω
Potential difference V = 6 volt
In a series combination, the current flowing through each resistor is the same and equal to the total current flowing through the circuit.
∴ Current through the resistors,
I = \(\frac { V }{ R }\)
= \(\frac { 6 }{ 25 }\)
= 0.24 A
Since the current in the circuit and the current through each resistor is the same, the ammeter will show a reading of 0.24 A.
Calculation of potential difference across 12 Ω resistor:
As total current 0.24 A flows in the circuit, the same current 0.24 A would also flow through the 12 Ω resistor which is connected in series.
∴ The potential difference across the 12 Ω resistor,
V = IR
= 0.24 x 12 = 2.88 V
Thus, the p.d. across 12 Ω resistor is 2.88 volt.
So, the voltmeter will show a reading of 2.88 V.

Question 14.
Judge the eΩuivalent resistance when the following are connected in parallel:
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω, 10³ Ω and 10⁶Ω
Answer:
(a) Less than 1 Ω (but approximately 1 Ω)
(b) less than 1 Ω (but approximately 1 Ω)
Explanation: When resistors are connected in parallel, the equivalent resistance is less than the least resistance (here in both the cases it is 1 Ω) connected in the combination.
or
(a) Here, R₁ = 1 Ω and R₂ = 10⁶ Ω
The equivalent resistance R_p of the parallel combination is given by,
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_1}+\frac{1}{R_2}\)
∴ R_p = \(\frac{R_1 \times R_2}{R_1+R_2}\)
= \(\frac{1 \times 10^6}{1+10^6}\)
= \(\frac { 1000000 }{ 1000001 }\) ≈ 1 Ω

(b) Here, R₁ = 1 Ω, R₂ = 10³ Ω and R₃ = 10⁶
The equivalent resistance Rp of the parallel combination is given by,

Question 15.
An electric lamp of 100 Ω, a toaster of resistance 50 Cl, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current through it?
Solution :
Here, Resistance of an electric lamp R₁ = 100 Ω
Resistance of a toaster R₂ = 50 Ω
Resistance of a water filter R₃ = 500 Ω.
Equivalent resistance R_p of three resistors R₁, R₂ and R₃ connected in parallel is given by,

Total current through the circuit (i.e., through all the three appliances)
I = \(\frac{V}{R_p}\)
= \(\frac { 220 }{ 31.25 }\)
= 7.04 A
Since the electric iron connected to the same source (i.e., 220 V) takes as much current as taken by all the three appliances (i.e., f = 7.04A), its resistance must be equal to R_p.
So, the resistance of the electric iron = 31.25 Ω
and the current through the electric iron = 7.04 A

Question 16.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
The advantages of connecting electrical devices in parallel with the battery instead of connecting them in series are as follows :
(1) A parallel combination / circuit is helpful when each device has different resistance and requires different current for its operation, as in this case the total resistance of the circuit is reduced due to which the current from the battery is high and hence each electrical device can draw the required amount of current.

This is not so in a series combination / circuit because total resistance increases too much in series circuit due to which the current from the battery is low and the same low current flows through all the devices, irrespective of their resistances and hence devices cannot work properly.

(2) In a parallel combination / circuit each electrical device gets the same p.d. (potential difference) across it as that of the battery due to which all the devices work properly.

While in case of a series combination / circuit the devices do not get the same p.d. as that of the battery because the p.d. is shared by all the devices connected in series.

(3) In a parallel combination / circuit if one electrical device stops working due to some defect, then other devices are not affected. They continue to work without any problem.

On the other hand, in a series combination/ circuit if one electrical device stops working due to some defect, then all other devices also stop working as the whole circuit is broken.

(4) In a parallel combination / circuit each electrical device has its own switch due to which it can be turned ‘ON’ or turned ‘OFF’ independently without affecting other devices.

But in a series combination / circuit all the electrical devices have only one switch due to which they cannot be turned ‘OFF’ or turned ‘ON’ independently.

Question 17.
How can three resistors of resistances 2 Ω, 3 Ω and G Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer:
(a) In order to obtain a total resistance of 4 Ω from three resistors of 2 Ω, 3Ω and 6Ω…
1. First connect the two resistors of 3 Ω and 6 Ω in parallel to get a total resistance of 2 Ω. This is because in parallel combination,

2. Now, above parallel combination of 3 Ω and 6 Ω resistors is connected in series with the remaining 2 O resistor to get total resistance of 4 Cl. This is because in series combination.
R_s = R_p + R₃
= 2 + 2
= 4 Ω
Hence, the arrangement of three resistors 2 Ω, 3 Ω and 6 Ω which gives total resistance 4 Ω can be represented as follows :

(b) In order to obtain a total resistance of 1 Ω from three resistors of 2 Ω, 3 Ω and 6 Ω, all the three resistors should be connected in parallel. This is because in a parallel combination,

Hence, the arrangement of three resistors 2 Ω, 3 Ω and 6 Ω which gives total resistance 1 Ω can be represented as follows :

Question 18.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω,
8 Ω, 12 Ω, 24 Ω?
Solution:
(a) The highest resistance can be secured (or obtained) by connecting all the four coils in series. In this case,
R_s = R₁+R₂ + R₃ + R₄
= 4 + 8 + 12 + 24 = 48 Ω
Thus, the highest resistance which can be secured is 48 Ω.

(b) The lowest resistance can be secured by connecting all the four coils in parallel. In this cae,

Thus, the lowest resistance which can be secured is 2 Ω.

Question 19.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The heating element of an electric heater is made of an alloy (such as Nichrome) which has high resistance (partly due to high resistivity) whereas the cord is made of copper metal which has very low resistance (partly due to low resistivity).

• Now, the heating element of an electric heater made of Nichrome glows because it becomes red-hot due to the large amount of heat (according to H = I²Rt) is produced on passing current.
• On the other hand, the connecting cord of the electric heater made of copper does not glow because relatively very less (negligible) heat (according to H = I²Rt) is produced in it by passing the same current.

Question 20.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Solution:
Here, charge, Q = 96000 C,
time t = 1 h = 60 x 60 = 3600 s,
potential difference V = 50 V
Now,
Heat generated H = VIt
= V(\(\frac { Q }{ t }\))t (∵ I = \(\frac { Q }{ t }\))
= 50 x 96000
= 4800000 J
= 4.8 x 10⁶ J
or
= 4.8 x 10³ x 10³ J
= 4.8 x 10³ kJ
= 4800 kJ

Question 21.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat devloped in 30 s.
Solution :
Here, current I = 5 A,
resistance R = 20 Ω, time t = 30 s
Now,
Heat produced H = I²Rt
= (5)² x 20 x 30
= 25 x 20 x 30
= 15000 J
= 15 kJ

Question 22.
What determines the rate at which energy is delivered by a current?
Answer:
The electric power of the source determines the rate at which energy is delivered by the current to the load / appliance.
[Whereas the electric power of an appliance determines the rate at which energy delivered by a current Is consumed by appliance.]

Question 23.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2h.
Solution:
Here, I = 5A, V = 220V,
t = 2h = (2 x 60 x 60)s = 7200s
Power P = VI = 220 x 5 = 1100W = 1100J/s
Now,
Energy consumed W = Pt = 1100J/s x 7200s
= 7920000
= 7.92 x 106J
OR
Energy consumed W = Pt = 1100 w x 2 h
= 2200 Wh
= 2.2 x 10³ Wh
= 2.2 k Wh

Activity 12.1 [T. B. Pg. 203]

To verify Ohm’s law.
OR
To find the relationship between potential difference (V) and current (I).

Procedure:
1. Set up a circuit as shown in figure 12.3, consisting of a +Nichrome wire XY of length say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each.

2. First use only one cell as the source in the circuit.
Note the reading of ammeter I, for the current and reading of the voltmeter V for the potential difference across the Nichrome wire XY in the circuit.
Tabulate them in the Table given.

3. Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the Nichrome wire and potential difference across the Nichrome wire.

4. Repeat the above steps using three cells and then four cells in the circuit separately.
Calculate the ratio V to I for each pair of potential difference V and current I.

Observation table :

Sr. No.	Number of cells used in the circuit	Potential difference across the Nichrome wire V (volt)	Current through the Nichrome wire I (ampere)	\(\frac { V }{ I }\) (volt / ampere)
1.	1	1.5	0.1	15
2.	2	3	0.2	15
3.	3	4.5	0.3	15
4.	4	6	0.4	15

Plot a graph between V and I and observe the nature of the graph.
(Take the voltage (V) on X-axis and the current (J) on Y-axis. Draw the graph of I-V taking proper scale.)

Observation:
When V increases I also increases linearly, i.e., I ∝ V. The ratio V/I is found to be (approximately) the same, i.e., 15V/A.
The graph between V and 1 is a straight line passing through the origin O.

Conclusion:
The electric current flowing through a metallic wire is directly proportional to the potential difference across its ends (J ∝ V) and V/I is a constant ratio in particular case.

Activity 12.2 [T. B. Pg. 205]

To show that the strength of an electric current in a circuit depends on resistance used in it.

Procedure:
1. Take a Nichrome wire, a torch bulb, a low bulb (10W) and an ammeter (0-5A range), a plug key and some connecting wires.

2. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in figure 12.5.

3. Complete the circuit by connecting the Nichrome wire in the gap XY. Plug the key.
Note down the ammeter reading. Take out the key from the plug after measuring the current through the circuit.
[Note: Always take out the key from the plug after measuring the current through the circuit.]

4. Replace the Nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.

5. Now repeat the above step with the 10 W bulb in the gap XY.

• Are the ammeter readings different for different s components connected in the gap XY?
• What do the above observations indicate?

6. You may repeat this Activity by keeping any material component in the gap.
Observe the ammeter readings in each case. Analyse the observations.

Observation :

• Yes, the ammeter readings are different for different components (Nichrome wire, a torch bulb and a low bulb) when connected in the gap XY.
• If we take any other material component in the gap XY, the reading of the ammeter may be different.
• This is because certain components offer an easy path for the flow of electric current while the others resist the flow.
• i.e., the current through an electric component depends on its resistance.

Conclusion :
The strength of an electric current in a circuit depends on the resistance used in it.

• A component of a given size that offers low resistance is called a good conductor.
• A component of the same size which offers appreciable resistance is called a resistor.
• A component of the same size that offers a higher resistance is called a poor conductor or a bad conductor.
• A component of the same size that offers very high resistance is called an insulator.

Activity 12.3 [T. B. Pg. 206]

To study the factors on which resistance of conducting wire depends.

Procedure:

• Complete an electric circuit consisting of a cell, an ammeter, a Nichrome wire of length l [say, marked (1)] and a plug key, as shown in figure 12.6.
  
• Now, plug the key. Note the current in the ammeter.
• Replace the Nichrome wire by another Nichrome wire of same thickness but twice the length, that is 21 [marked (2) in the figure],
• Note the ammeter reading.
• Now replace the wire by a thicker Nichrome wire, of the same length l [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
• Instead of taking a Nichrome wire, connect a copper wire [marked (4) in the figure] in the circuit.
• Let the wire be of the same length and same area of cross-section as that of the first Nichrome wire [marked (1)]. Note the value of the current.
• Notice the difference in the current in all cases,
• Does the current depend on the length of the conductor ?
• Does the current depend on the area of s cross-section of the wire used ?

Observations :

• Ammeter shows the current I flows in the Nichrome wire marked (1).
• When the length of Nichrome wire marked (2) is doubled keeping its same thickness (i.e., when the Nichrome wire marked (1) is replaced by the Nichrome wire marked (2) the ammeter reading decreases to half, i.e. current becomes 1/2.
• When the thickness of Nichrome wire marked (3) is increased keeping its length same i.e., when the Nichrome wire marked (3) is S connected in the circuit), the ammeter reading increases, i.e., current I increases.
• When Nichrome wire is replaced by a copper s wire marked (4) of same length and same cross-sectional area as marked (1), the ammeter reading increases, i.e., current I increases.

Conclusion :

• Resistance of the conducter depends on its s length. Here resistance of wire R is directly proportional to length l, i.e., R ∝ l.
• Resistance of the conductor depends on its area of cross-section. Here resistance of wire R is inversely proportional to area of cross-section A, i.e., R ∝ \(\frac { 1 }{ 2 }\)
• Copper is a better conductor than Nichrome. Resistance of wire depends on the nature of S its material.

Activity 12.4 [T. B. Pg. 210]

To study the series combination of resistors.

Procedure :

• Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in figure 12.9.
• You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω, etc. and a battery of 6 V for performing this Activity.
• Plug the key. Note the ammeter reading. [part (1)]
• Change the position of the ammeter to anywhere inbetween the resistors. Note the ammeter reading each time, [part (2)]
• Do you find any change in the value of current through the ammeter?
  

Observation :

• The ammeter reading is nearly 1 A.
• When the position of the ammeter is changed to anywhere inbetween the resistors, its reading remains the same, i.e., the current flowing through it and through each part of the circuit, i.e., through each resistor is the same.
• No, current flowing through the ammeter does not change.

Conclusion :

• The value of the current in the ammeter is the same, independent of its position in the electric circuit.
• In a series combination of resistors the current is the same in every part of the circuit, i.e., the same current flows through each resistor. This is due to the reason that, the current has only one path to flow.

Important note:
(1) In discussion of Activity 12.4, ammeter and voltmeter are considered as ideal meters. (2) Ideally, an ammeter should have zero resistance and a voltmeter should have infinite resistance. But in practice, these conditions cannot be realized and hence practically the voltmeter draws some current from the main branch and so the ammeter reading in part (2) is slightly less than that in part (1).

Activity 12.5 [T. B. Pg. 211]

To show that in a series combination of resistors, the total potential difference across the combination divides itself across the individual resistors.

Procedure:

• In Activity 12.4, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in figure 12.7.
• Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V.
• Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown
  
• Plug the key and measure the potential difference across the first resistor. Let it be V₁
• Similarly, measure the potential difference across the other two resistors, separately. Let these values be V₂ and V₃, respectively.
• Deduce a relationship between V, V₁, V₂ and V₃.

Observation:

• The p.d. across the series combination is equal to the p.d. across the two terminals of the battery.
• p.d. across the first resistor R₁ is found out to be V₁ using voltmeter.
• Similarly with the help of voltmeter, p.d. across resistors R₂ and R₃ separately are found out to be V₂ and V₃ respectively.
• It is found out that V = V₁ + V₂ + V₃.

Conclusion:

• In a series combination of different resistors, the potential difference across different resistors is different, i.e., If R₁ ≠ R₂ ≠ R₃, V₁ ≠ V₂ ≠ V₃.
• The total p.d. across the combination is equal to the sum of p.d. across each of the resistors, i.e., total p.d. across the combination divides itself across the individual resistors, i.e., V is equal to V₁ + V₂ + V₃.

Important note:
In Activity 12.5 it is assumed that, R (ammeter) = zero and R (voltmeter) = ∞, so V₁ + V₂ + V₃, In practice, these conditions cannot be realized, hence practically V is (very nearly) equal to V₁ + V₂ + V₃.

Activity 12.6 [T. B. Pg. 213]

To study the parallel combination of resistors.
OR
To study the relationship between potential difference and respective resistance in parallel circuit as well as the relationship between s current and respective resistance in parallel circuit.

Procedure:
1. Make a parallel combination XY of three resistors having resistances R₁, R₂ and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in figure 12.14. Also connect a voltmeter in parallel with the combination of resistors.

2. Plug the key and note the ammeter reading. (Let the current be I).

3. Also take the voltmeter reading. (It gives the potential difference V, across the combination).

4. Connect a voltmeter across the resistor with resistance R₁ and note the voltmeter reading V₁ (see figure 12.14).

5. Similarly connect a voltmeter across the resistors with resistances R₂ and R₃ separately and note s the corresponding voltmeter readings V₂ and V₃.

6. Take out the plug from the key. Remove the ammeter and voltmeter from the circuit.

7. Insert the ammeter in series with the resistor s with resistance R₁, as shown in figure 12.15. Note the ammeter reading, I₁.

8. Similarly, measure the current through the resistors with resistances R₂ and R₃. Let these be I₂ and I₃, respectively.
What is the relationship between I, I₁, I₂ and I₃?

Observations :

• Ammeter showed the reading I when connected as shown in figure 12.14.
• Voltmeter showed reading V when connected as shown in figure 12.14.
• It is found that the p.d. across each resistor is the same (i.e., V), as across the combination of resistors, i.e., V₁ = V₂ = V₃ = V
• Ammeter showed reading I₁, when connected with the resistor R₁ as shown in figure 12.15.
• Similarly ammeter showed reading I₂ and I₃ measured separately through R₂ and R₃ respectively.
• The relationship between I, I₁, I₂ and I₃ is:
  I = I₁ + I₂ + I₃
  Where, I = Total current through the combination of resistors in parallel
  I₁, I₂, I₃ = Currents through R₁, R₂ and R₃ respectively.

Conclusion:

• In parallel combination of resistors, the potential difference across different resistances will be same.
• Total current I in the given parallel circuit is divided amongst different resistors.