Class 10

Jharkhand Board JAC Class 10 Science Solutions Chapter 16 Management of Natural Resources Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 16 Management of Natural Resources

Jharkhand Board Class 10 Science Management of Natural Resources Textbook Questions and Answers

Question 1.
What changes would you suggest in your home in order to be environment-friendly?
Answer:
Certain changes can be applied in our daily routine at home to be environment- friendly.

• Check the wastage of water. Not letting the water run while brushing, soaping or washing.
• Turning off lights and fans when not in use.
• Avoid wastage of water, food and energy.
• Reusing the materials whenever possible.
• Use solar water heater and cookers.
• Reduce the garbage.
• Make small rainwater harvesting system if possible.

Question 2.
Can you suggest some changes in your s school which would make it environment-friendly?
Answer:
A school can become environment-friendly by following ways:

• Growing plants and trees all around the play ground.
• Making a rainwater harvesting system.
• Arranging solar cell pannel if possible.
• Making compost of biomass waste collected e.g., food waste, fallen leaves, etc.

Question 3.
We saw in this chapter that there are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management of forest produce? Why do you think so?
Answer:
Out of the four main stakeholders, the local people living near the forest areas should given the authority to decide the management of forest produce.

Because these people know the traditional methods to use the natural resources in sustainable manner. These local people have been using the forest and wildlife resources since the ancient times without causing any damage to environment.

Question 4.
How can you as an individual contribute or make a difference to the management of (a) Forests and wildlife, (b) Water resources and (c) Coal and petroleum?
Answer:
Contribution of an individual to the management of:
(a) Forests and wildlife : Do not waste paper, use less paper, recycle the waste paper, minimise the wooden furniture, etc. help into save trees. Any animal products like fur, skin, tusk, horn, etc. should not be used by killing them.

(b) Water resources : Stop the wastage of water in daily routine. Instead of shower, use buckets to take bath, to wash the car. Close the taps properly. Do not let the water run while brushing, soaping or washing.

(c) Coal and petroleum : To walk or to use a bicycle over a short distance, use of public transport switching of unnecessary electrical appliances.

Question 5.
What can you as an individual do to reduce your consumption of the various natural resources?
Answer:
At an individual level, I will do following activities that can help to reduce the consumption of the various natural resources :

• Switch off electric appliances when not in use.
• Turn off tap when not in use.
• Minimum use of auto-vehicles.
• No wastage of food.
• No wastage of paper.
• Say no to plastic.

Question 6.
List five things you have done over the last one week to –
(a) conserve our natural resources.
(b) increase the pressure on our natural resources.
Answer:
(a) To conserve our natural resources:

• Turned off tap while brushing, soaping, etc.
• Walking to the nearby places.
• Switched off lights when not in use.
• Reusing envelopes by turning them.
• Using cracked crockery for growing the plant.

(b) To increase the pressure on our natural resources :

• Frequent use of plastic bags and throwing these anywhere.
• Often wastage of food
• Buying leather belt, purse and shoes.
• Leaving the light / lamp on even when not needed.
• Tearing the pages of notebook.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your lifestyle in a move towards a sustainable use of our resources?
Answer:
The Ganga Action Plan was launched in 1985 to improve the quality of water in Ganga and remove the pollution caused by disease causing microorganisms. Faecal coliform bacteria were found in Ganga water indicating contamination.

Jharkhand Board Class 10 Science Management of Natural Resources InText Questions and Answers

Question 1.
What changes can you make in your habits to become more environment-friendly?
Answer:
We can make following changes in our habits to become more environment-friendly:

• Use of paper bags, jute bags instead of plastic bags.
• Walk or cycle to cover short distance.
• Do not throw garbage anywhere.
• Use of renewable resources and biodegradable substances.
• Switching off electrical appliances when not required.
• Do not pollute and waste water.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
Answer:
The advantages of exploiting resources with short-term aims are as follows :

• We will able to meet current basic human needs.
• It will be beneficial for the present generation.
• There will be rapid industrial growth, agricultural growth and hence, economic development.

Question 3.
How would these advantages differ from the advantages of using a long-term perspective in managing our resources?
Answer:
Using a long-term perspective is to reap the profit in a sustainable manner so that the natural resources will last for generations to come and will not merely be exploited to hit short-term gains. All the sections of society should be made aware about conservation and protection of the environment.

Question 4.
Why do you think that there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
There should be equitable distribution s of resources so that both rich as well as poor, will get benefitted. Powerful and rich people take advantage of their influence and get more benefit as compared to weak and poor people. Money and power are important factors which work against the proper distribution of resources.

Question 5.
Why should we conserve forests and wildlife?
Answer:
We should conserve forests because they,

• Provide raw materials for various industries.
• Provide fruits, vegetables, fodder, grass, etc.
• Provide medicines, herbs, gum, resin, catechu, etc.
• Provide habitat to animals.
• Prevent soil erosion and flood.
• Provide sources for economic and social growth.
• Play an important role in maintaining CO₂ – O₂ balance in the atmosphere. Regulate earth’s average temperature.

Importance of wildlife conservation :

• To maintain forest ecosystem and ecological balance in nature.
• It helps into maintain forests by facilitating growth of plants in different places by dispersing seeds.
• Flow of energy in trophic levels maintained and biodiversity increases.

Question 6.
Suggest some approaches towards conservation of forests.
Answer:
Approaches towards conservation of forests are :

• Indiscriminate felling of trees for the purpose of timber must be reduced.
• The forest ecosystem must be protected from fuel starved villages, fodder-starved cattles and commercial exploitation.
• Replantation of trees and also plantation of indiginous species to develop forests in all available land.
• Participation of local people and villagers must be taken in conservation of forests.
• Scientific research, monitoring and spreading awareness about conservation of forests through education.

Question 7.
Find out about the traditional systems of water harvesting / management in different regions of India.
Answer:

Region	Traditional water harvesting system
Rajasthan	Khadins, tanks and nadis
Maharashtra	Bandharas and tals
Madhya Pradesh and Uttar Pradesh	Bundhis
Bihar	Ahars and pynes
Himachal Pradesh	Kulhs
Kandi belt of Jammu	Ponds
Tamil Nadu	Eris (tanks)
Karnataka	Kattas
Kerala	Surangams

Question 8.
Compare the traditional water harvesting system with the probable systems in hilly mountainous area or plains or plateau regions.
Answer:
Traditional water harvesting system in hilly and mountainous area is different from plains and from plateau region.

Example: In hilly areas like Himachal Pradesh there is a local system of irrigation called Kulhs. The water flowing in the streams is diverted into man-made channels which takes this water to numerous villages down the hillside.

Whereas water in plains is collected in check dams or tanks, tals or bundhis.

Question 9.
Find out the source of water in your region / locality. Is water from this source available to all people living in that area?
Answer:
The source of water in our region is municipal corporation supply of water and from ground water source. There is scarcity of water during summer and most people have to wait in long queues at the nearest municipal water tap to collect water for their daily consumption.

Activity 16.1 [T. B. Pg. 266]

To find out the international norms to regulate the emission of carbon dioxide.

• Carbon dioxide (CO₂) is the primary greenhouse gas emitted through human activity.
• The international norms to regulate the emission CO₂ are based on Kyoto Protocol.
• This protocol was negotiated in December, 1997 at the city of Kyoto, Japan and came into force on February 16th, 2005.
• As of December 2006, a total of 169 countries have signed the agreement.
• Under this protocol, industrialised countries must reduce their collective emissions of CO₂ and other greenhouse gases to an average of 5 % against 1990 levels.

Questions:

Question 1.
Why is it necessary to regulate the emission of carbon dioxide?
Answer:
Carbon dioxide is a major greenhouse gas that causes global warming. Which results into climate change. So, it is necessary to regulate the emission of carbon dioxide.

Question 2.
Which is the simplest way to regulate the CO₂ level in environment?
Answer:
Plantation of trees and conservation of green cover on the earth is the simplest way to regulate the CO₂ level in environment.

Question 3.
What is global warming?
Answer:
Global warming is the increase in average temperature of the earth’s environment due to increase in amount of greenhouse gases.

Activity 16.2 [T. B. Pg. 267]

To find out the names of organisations involved in spreading awareness about the conservation of natural resources.

NGO (Non Government Organisations) such us SEACOLOGY and Mera Desh Foundation spread awareness about the environment and promote activities that lead to conservation of natural resources.

Some other organisations that work for the conservation of the environment.

• CSE – The Centre for Science and Environment
• TER – The Energy and Resources Institute
• WTI – Wildlife Trust of India.

Questions:

Question 1.
Which organisation(s) is/are active to spread awareness about conservation of our environment and natural resources in your village / town / city?
Answer:

• Natural resource management
• Core environmental NGOs
• Centre for Environment and Social Concerns.

Question 2.
How you can contribute towards the conservation of our environment and natural resources.
Answer:
We can contribute towards the conservation of our environment and natural resources by following:

• Limit the use personal vehicle, instead use public transport.
• Minimise the wooden furniture at our home.
• Do not waste paper and other natural resources.
• Do not pollute water reservoirs.
• Do not use insecticides / pesticides.
• Do not cut trees, do not clear forests.

Activity 16.3 [T. B. Pg. 268]

To check the pH of tap water and compare it with pH of the water in the local water body.

Apparatus: Test tubes or beaker

Materials : Universal pH indicator / litmus paper

Procedure:

• Take water sample from tap of your home in a test tube.
• Add few drops of the universal indicator into the test tube containing tap water.
• Observe the water sample in the test tube for colour change.
• Test pH using a litmus paper, in the water sample placed in a beaker.
• Dip the litmus paper into the water in the local water body.
• Observe colour changes of the litmus paper.

Observation:

• Test tube containing tap water, turns yellow colour with addition of pH indicator.
• Litmus paper does not show colour change with tap water.
• Water sample collected from pond shows red colour with pH indicator.
• Litmus paper turns red as it is dipped in water of local water body.

Conclusion:
The uncontaminated tap water is neutral while water of local water body is acidic.

Questions:

Question 1.
What is the cause for acidic water in water body?
Answer:
Pollution is the main cause for acidic water in water body.

Question 2.
Can aquatic life survive in acidic water of water body?
Answer:
No, aquatic life cannot survive in acidic water.

Question 3.
Can you say whether the water is polluted or not on the basis of your observations?
Answer:
Yes, tap water is clean and non-polluted. Water of water body is polluted.

Activity 16.4 [T. B. Pg. 269]

To visit a town or village after a few years of absence, note down the major developmental changes there.

Major developmental changes in a village are

• New roads have been built
• New houses have been constructed.
• Factory and new market have been setup.

Questions :

Question 1.
Make a list of the materials for making roads and buildings with their probable sources.
Answer:

Materials	Probable sources
Granite	Rocks
Cement	Factories. Raw materials obtained from nature.
Bricks	Soil
Steel	Mined from the soil
Wood	Forest

Question 2.
What are the ways in which the materials used in construction can be reduced?
Answer:

• Roads can be built with cement or with polymer plastic mix instead of coaltar.
• By using cement and iron to make beams and by using aluminium or fibreglass to make windows and doors, we can reduce the usage of wood.

Activity 16.5 [T. B. Pg. 270]

To prepare a report on traditional practices for conservation of nature in day-to-day life.

Traditional practices for conservation of nature :

• Tulsi, Calotropis, Asopalav, Ficus (Pipal), Khejdo, Banyan trees and various other trees are planted, which are considered sacred and worshipped by the people in . majority of Indian villages and towns.
• Several birds, cow and even snakes have been considered sacred. Therefore, they are protected.
• The concept of cultural landscape such as sacred forests, sacred corridors and a variety of ethno-forestry practices are the means for protection of nature and natural resource. They are observed by people.
• Several rivers have been considered sacred.

Activity 16.6 [T. B. Pg. 271]

To know about our dependency on forest resources.

Questions :

Question 1.
Make a list of forest products that we commonly use.
Answer:
Wood, bamboo, fuel, timber, fruits, herbs, paper, medicine and many other products that we commonly use come from forests.

Question 2.
What do you think a person living near a forest would use?
Answer:
A person living near a forest might use wood, fodder, dry leaves, bamboo, fruits, vegetables, herbs, etc.

Question 3.
What do you think a person living in a forest would use?
Answer:
A person living in a forest is almost entirely dependent on the forest for all basic needs, i.e., fire wood, small timber, fodder and grass, bamboo, nuts and medicines, food, clothing, shelter, fishing, hunting, etc.

Question 4.
How these needs depend upon the person’s area in which they live?
Answer:
People who live in a village or town are less dependent on forest than those who live in or near the forest.

Activity 16.7 [T. B. Pg. 272]

To name any two products of forest used in an industry and find out a sustainable alternative for such products.

• Tendu leaves are used in Bidi industry. Essential oils are used in the manufacture of soaps and cosmetic industry.
• Over exploitation of forests is not sustainable in the long run. We need to control our consumption. Plantation of trees is necessary.

Activity 16.8 [T. B. Pg. 275]

To debate following topics which are concerned with the damage caused to forests:

• Building rest houses for tourists in national parks.
• Grazing domestic animals in national parks.
• Tourists throwing plastic bottles / covers and other litter in national parks.

Conclusion:
(a) The rest houses made in national park for tourists become hub of activities, which damage ecology of forests. Tourists see forest as travel destination. They cause pollution and disturb natural environment of forests.

(b) Over grazing of domestic animals leaves little or no grass for the herbivores animals that live in forests. Thus their number slowly decreases, in turn disturbs the upper trophic levels of the food chains in the forests.

(c) Tourists throwing plastic wastes in national parks cause pollution. Plastic is non-biodegradable wastes.

Decomposers cannot decompose it and such dumped waste piles up. Biotic components of national parks are adversely affected.

Activity 16.9 [T. B. Pg. 275]

In Maharashtra, the village suffering from chronic water shortage surround a water theme park to get water for their needs.

Debate whether they get optimum use of available water or they should search for alternative rain water harvesting.

Conclusion:

• No, the construction of a water theme park in the region of water scarce area in wrong decision.
• Water is a basic need for all living organisms. If people are suffering due to lack of drinking water and water for other uses too, how will they amuse themselves with a water park.
• This is unequal distribution of water resources. People should oppose this and the authorities should first provide adequate supply of water to people. People should also search for alternative rain water harvesting.

Activity 16.10 [T. B. Pg. 275]

Study the pattern of rainfall in various parts of India from atlas.

Do by yourself.

Conclusion :
Rainfall more than 200 cm in North-East part of India and the regions of water scarcity where rainfall is less than 25 cm are part of Rajasthan and Gujarat.

Activity 16.11 [T. B. Pg. 279]

To find ways to reduce our consumption of coal and petroleum.

Some ways to reduce our consumption of coal and petroleum are as follows :

• Take a public transport instead of using your personal vehicle.
• Use CFL instead of bulbs.
• Install solar water heater.
• Walk or cycle to nearby places instead of using motorised vehicles.
• Switch off vehicles at red lights.
• Use pressure or solar cookers to cook food, keep proper air pressure in tires, etc.
• Wearing an extra sweater instead of using heater or sigri in cold days.
• Taking the stairs instead of using the lift up to 3-4 floors.
• Using efficient engines that ensures complete combustion of fuels.

Activity 16.12 [T. B. Pg. 279]

To find out the norms for emission from vehicles.

• Euro norms refer to the permissible emission level from petrol and diesel vehicles, which have been implemented in Europe.
• These norms require manufacturers to reduce pollution emission levels in an efficient manner by making technical changes in the vehicles they manufacture.
• Under the Euro norms, emissions of carbon monoxide (CO), burnt hydrocarbons (HCs), oxides of nitrogen (NOx) and suspended particulate matter (SPM) are regulated.
• The European Union has been upgrading emission norms as well as the fuel quality standards in stages :
  Norms Implemented
  Euro I → 1992 – 93
  Euro II → 1996 – 97
  Euro III → 2000
  Euro IV → 2005
• The enforcement of these norms help in reducing air pollution due to vehicular s emissions.

Jharkhand Board JAC Class 10 Science Solutions Chapter 9 Heredity and Evolution Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 9 Heredity and Evolution

Jharkhand Board Class 10 Science Heredity and Evolution Textbook Questions and Answers

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as …………..
A. TTWW
B. TTww
C. TtWW
D. TtWw
Answer:
A. TTWW

Question 2.
An example of homologous organs is …………..
A. our arm and a dog’s foreleg.
B. our teeth and an elephant’s tusks.
C. potato and runners of grass.
D. all of the above
Answer:
D. all of the above

Question 3.
In evolutionary terms, we have more in common with …………..
A. a Chinese school-boy.
B. a chimpanzee.
C. a spider.
D. a bacterium.
Answer:
A. a Chinese school-boy.

Question 4.
A study found that children with light coloured eyes are likely to have parents with light coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
No, we cannot say anything about whether the light eye colour trait is dominant or recessive on the basis of given information because information regarding cross between two traits, i.e., light colour with black eye colour is essential to determine it. In general population light coloured eyes are in much less proportion as compared to dark eyes. This indicates that it may be recessive trait.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:
One can work out the evolutionary relationships of the species by identifying hierarchies of characteristics between them.
Similarities among organisms will allow us to group them together. If the characteristics between two species are more common, then they are more closely related and they had a recent common ancestor.

Thus, build up small group of species with recent common ancestors, then super group with more distant common ancestors and so on. By this one can interlink the study of evolution and classification.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:
Analogous organs : Organs with different structure and components but have common function.
Examples: The wing of a bat and the wing of a bird.

Homologous organs : Organs with similar basic structure but modified to perform different functions.
Examples : Fore limbs of frog, lizard, bird and human.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:

So, black coat is dominant trait in dog. [If in F₁ generation all progeny are with white coat then white coat can be considered as dominat trait.]

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Fossils are the remains or impressions of the dead animals and plants that lived in the past. Fossils are the direct evidence of evolution. If the fossils are found closer to the surface then they are more recent than the fossils found in deeper layers. One can find evolutionary relationships between ancient and present organisms by study of fossils. We also come to know the time period of evolutionary process by studying fossils.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
Stanley L. Miller and Harold C. Urey:
In 1953, they assembled an experiment in which primitive atmosphere was simulated. It had compounds like ammonia, methane, hydrogen and water vapour but no oxygen. This was maintained at a temperature just below 100°C and electric sparks were passed through the mixture of gases to simulate lightning. At the end of a week, 15% of the carbon had been converted to simple compounds of carbon including amino acids which are monomers of protein molecules.

Question 10.
Explain how sexual reproduction gives rise to more visible variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
(1) Sexual reproduction gives rise to more visible variations than asexual reproduction because in sexual reproduction, each new generation is the combination of the DNA copies from two pre-existing individuals.

(2) Even during gamete formation new combination of genes occurs during meiosis. From more visible variations during sexual reproduction, certain favourable variation can be selected by natural selection.

Whereas in asexual reproduction generally offsprings are exact copies of their single parent.

Question 11.
How is equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Male and female parents produce male gametes (Sperms) and egg cells (Ovum) respectively through a process of meiosis. It reduces the number of chromosomes and amount of DNA to half in gametes.

Such germ cells from two individuals, i. e., male and female parents are fused to form zygote. This ensures the equal genetic contribution of male and female parents in the progeny that develop from zygote.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, because of useful inherited / genetic variations of an individual organism can adapt and survive. Moreover they leave behind more offsprings and populations with such genetic variations will survive.

Jharkhand Board Class 10 Science Heredity and Evolution InText Questions and Answers

Question 1.
If a trait A exists in 10 % of a population of an asexually reproducing species and a trait B exists in 60 % of the same population, which trait is likely to have arisen earlier?
Answer:
A trait B which exist in 60 % of population of an asexually reproducing species, must have arisen earlier than trait A.

Because in asexually reproducing species there are chances of appearance of very few new traits due to small inaccurancies during DNA copying. So, populations having trait A have arisen later.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
The creation of variations in a species is either due to inaccuracies in DNA copying or during sexual reproduction.

• Depending on the nature of variations, different individuals have different kinds of advantages.
• The individuals with useful variations can adapt to the prevailing environment and show better survival.
• The individuals with useful variations then increase in numbers.

Question 3.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel’s conducted, cross fertilisation between pure tall plant (TT) and pure short plant (tt) which resulted in all tall (Tt) plant in F₁ generation.

This shows that single copy of T is enough to make the plant tall. It shows that one trait which is expressed in the presence of its contrasting form. This is dominant trait and the other remains unexpressed in the presence of its contrasting form is recessive trait.

Question 4.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel performed dihybrid experiment on pea plants. A tall plant with round seeds was crossed with a short plant with wrinkled seeds. Fx progeny plants were all tall with round seeds. F₁ progeny are used to generate F₂ progeny by self-pollination. Along with parental combinations, F₂ progeny showed new combinations too. Some of them were tall with wrinkled seeds while some others were short with round seeds.

It means factors (genes) controlling for seed shape and height of plant recombine to form new combinations in F₂ offsprings. Thus, tall/short trait and the round seed / wrinkled seeds trait are inherited independently.

Question 5.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits blood group A or O is dominant? Why or why not?
Answer:
No, the given information is not enough to tell us whether the trait of blood group A or blood group O is dominant.

• Blood group trait is controlled by genes and inherited from parents.
• Daughter has blood group O and two copies of genes as it inherited one each from the father and the mother.

For Information:
Trait blood group A is dominant and blood group O is recessive. This fact derived from the study of genotype.
Gene I^A for blood group trait A
Gene i recessive for blood group trait O.

Question 6.
How is the sex of the child determined in human beings?
Answer:
The sex of the child will be determined in human beings by the sex chromosome inherited from father. Father has XY sex chromosomes in each of his cell. Two types of gametes (sperms) are produced based on chromosomes, 50 % sperms are with X chromosome and 50 % sperms are with Y chromosome.

When sperm bearing X chromosome fertilise the egg, girl is born and when sperm bearing Y chromosome fertilise the egg, son is born.

Question 7.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
A particular trait may increase in a population by following ways:

• Natural selection – directing evolution with a survival advantage.
• Genetic drift – provides diversity without any adaptations.

Question 8.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Traits acquired during the life-time of an individual may not be inherited because they may be changes in non-reproductive tissues which cannot be passed on to the DNA of the germ cells and hence cannot be passed on its progeny.

Question 9.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
The small numbers of surviving tigers is a cause of worry from the point of view of genetics because if they become extinct then the genes of this species will be lost forever. There will be no chance of getting this species back again to life in future.

Question 10.
What factors could lead to the rise of a new species?
Answer:
Factors that can lead to the rise of a new species are –

• Gene flow
• Genetic drift
• Natural selection and
• Reproductive isolation.

Question 11.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?
Answer:
No, geographical isolation will not be a major factor in the speciation of a self-pollinating plant species because single parent is involved in it. There is no gene flow between two geographically isolated population.

Question 12.
Will geographical isolation be a major factor in the speciation of an organism that reproduce asexually? Why or why not?
Answer:
No, geographical isolation will not be a major factor to the speciation that reproduces asexually because single parent is involved in asexual reproduction due to which variations are not formed.

Question 13.
Give an example of characteristics S being used to determine how close two species? are in evolutionary terms.
Answer:
Homologous organs are the characteristics < being used to determine two species are close in evolutionary terms.

Example: The basic structure of the limbs of mammals, birds, reptiles and amphibians is similar though it has been modified to perform different functions.

Question 14.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
No. they are not considered homologous organs because the wing of a butterfly and the wing of a bat are similar in function but the designs of two wings, their structure and components are very different. So, they can be considered as analogous organs and not homologous.

Question 15.
What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are the remains or impressions of the dead animals and plants that lived in the past.

Fossils are the direct evidence of evolution. If the fossils are found closer to the surface then they are more recent than the fossils found in deeper layers. One can find evolutionary relationships between ancient and present organisms by study of fossils.

We also come to know the time period of evolutionary process by studying fossils.

Question 16.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
All human beings look so different from each other in size, colour and looks due to environmental factors, new combination of genes during reproduction.

All they belong to Homo sapiens and have descended from a common ancestor in Africa. They have capacity of interbreeding which is an important criteria to categorize them as one species.

Question 17.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a better body design? Why or why not?
Answer:
In evolutionary terms, we can say that chimpanzees have a better developed body design among remaining three because chimpanzees are highly complex as compared to remaining three.

Activity 9.1 [T. B. Pg. 143]

To suggest rule for inheritance of earlobe type.

Procedure / Method:

• The lowest part of the ear pinna called the earlobe, is closely attached to the side of the head, i.e., attached or not, i.e., free.
• Observe the earlobes of all the students In – the class. Prepare a list of students. Enter < the data about the earlobes whether they are free or attached.
• Find out about the earlobes of the parents of each student in the class.
• Correlate earlobe type of each student with that of their parents.
• In the column write F for free earlobe and A for attached earlobe.
  

No.	Details	Free Earlobe	Attached Earlobe
1.	Name of student….
	Mother
	Father
2.	Name of student….
	Mother
	Father

Answer the following questions on the basis of the collected data:

Questions :

Question 1.
Which expression of earlobe is observed more in number in your class?
Answer:
Expression of free earlobe is observed more in number in our class.

Question 2.
Are the types of earlobes hereditary?
Answer:
Yes, the type of earlobe is hereditary.

Question 3.
From your collected data state which expression is dominant and which one is recessive for earlobe.
Answer:
Free earlobe is dominant and attached earlobe is recessive expression.

Question 4.
Determine the percentage of free earlobe and attached earlobe in the students of your classroom.
Answer:
There are 60 students in the classroom. Out of which 51 students have free earlobe and 9 students have attached earlobe.
Percentage of free earlobe = \(\frac { 51 }{ 60 }\) x 100 = 85 %
Percentage of attached earlobe = \(\frac { 9 }{ 60 }\) x 100
= 15 %

Activity 9.2 [T. B. Pg. 144]

What experiment would we do to confirm that F₂ generation did infact have a 1:2:1 ratio of TT, Tt and tt trait combination?
We do Mendel’s monohybrid cross-experiment:

Jharkhand Board JAC Class 10 Science Solutions Chapter 15 Our Environment Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 15 Our Environment

Jharkhand Board Class 10 Science Our Environment Textbook Questions and Answers

Question 1.
Which of the following groups contain only biodegradable items?
A. Grass, flowers and leather
B. Grass, wood and plastic
C. Fruit peels, cake and lime juice
D. Cake, wood and grass
Answer:
Grass, flowers and leather; Fruit peels, cake and lime juice; Cake, wood and grass.

Question 2.
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:
Grass, goat and human

Question 3.
Which of the following are environment-friendly practices?
A. Carrying cloth-bags to put purchases in while shopping
B. Switching off unnecessary lights and fans
C. Walking to school instead of getting your mother to drop you on her scooter.
D. All of the above
Answer:
All of the above

Question 4.
What will happen if we kill all the organisms in one trophic level?
Answer:
If we kill all the organisms in one trophic level then organisms of next trophic level will not get food (chemical energy) and the entire food chain gets disturbed. All the organisms which are dependent on these are died.

On the other hand, the organisms at the lower trophic level will increase in abundance. Due to this, ecosystem will be in imbalance.

Question 5.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:
The impact of removing all the organisms in a trophic level will be different for different trophic levels. Removal of producers will affect all the organisms of successive trophic levels. It will be a threat the survival. The removal of organisms at higher trophic level will lead to increase in organisms of lower trophic level. Removal of organisms of any trophic level will cause the damage to the ecosystem.

Question 6.
What is biological magnification? Will the levels of this magnification be different at different level of the ecosystem?
Answer:
Successively increasing concentration of some substance (e.g., pesticides) at various trophic levels of a food chain of organisms is known as biological magnification.

The level of biological magnification will be different at different trophic levels of the ecosystem. Maximum concentration will be at the third and fourth trophic level and concentration of chemical will be less at lower trophic levels.

Question 7.
What are the problems caused by the non-biodegradable wastes that we generate?
Answer:
The problems caused by the generated non-biodegradable wastes are as follows :

• It causes biological magnification.
• They keep on accumulating in nature causing pollution.
• They prevent growth of vegetation when dumped underground.
• They may be inert and simply persist in the environment for a long time and may harm various members of the ecosystem.
• There is imbalance of the food chains causing problems in ecosystem.

Question 8.
If all the wastes we generate is biodegradable, will this have no impact on the environment?
Answer:
If all the waste we generate is biodegradable and if properly allowed it to decompose, then it will have no impact on the environment. They should be properly managed.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Answer:
Ozone layer absorbs ultraviolet radiation of the sun which is very harmful to living organisms.

Damage to the ozone layer is a cause for concern because depletion of ozone layer allows harmful ultraviolet radiation to reach to the surface of earth, which may lead to skin cancer, cataract, etc.

To reduce the damage to the ozone layer, use of chlorofluorocarbons has been minimised. In 1987, the UNEP-United Nations Environment Programme has passed an agreement to freeze CFC production at 1986 levels. This will protect the ozone layer and subsequent effects of radiation.

Jharkhand Board Class 10 Science Our Environment InText Questions and Answers

Question 1.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:
Successive levels of nourishment in the food chain are known as trophic levels.

• It shows transfer of energy in an ecosystem.
• Food chain is a sequential list of prey-predator
  

Question 2.
What is the role of decomposers in the ecosystem?
Answer:
Decomposers feed on the excretory substances as well as dead bodies of plants and animals.
Bacteria and fungi are decomposers.

• They breakdown the complex organic substances into simple inorganic substances.
• Such simple inorganic substances are used up by the plants again.
• So, they play an important role in cyclic pathway of the elements.

Question 3.
Why are some substances bio¬degradable and some non-biodegradable?
Answer:
Some substances such as paper, vegetables, peels, etc. are acted upon by decomposers and get converted into simple form are called biodegradable. Biodegradable substances are natural substances.

Some synthetic substances such as plastic, polythene, etc. cannot be degraded by microbial activity are called non-biodegradable.

Question 4.
Give any two ways in which biodegradable substances would affect the environment.
Answer:

• Biodegradable substances get degraded by microbial activity releasing simple component back to nature. These components help to sustain the life of other organisms.
• Diming the biodegradation certain gases may be released in atmosphere causing pollution.

Question 5.
Give any two ways in which non-biodegradable substances would affect the environment.
Answer:

• Non-biodegradable substances such as pesticides cause soil and water polluton. It may cause biological magnification.
• Non-biodegradable substances block the functions of an ecosystem, i.e., transfer of energy and elements may stop.

Question 6.
What is ozone and how does it affect any ecosystem?
Answer:
Ozone is a molecule formed by three atoms of oxygen in the presence of UV (Ultraviolet) rays. Ozone performs an essential function at the higher levels of the atmosphere. However, at ground level it is a deadly poison. Ozone absorbs shorter wavelength UV rays from the sun. Thus it protects the living system on earth.

Question 7.
How can you help in reducing the problem of waste disposal? Give any two methods.
Answer:
We can help in reducing the problem of waste disposal by following methods :

• Biodegradable domestic wastes such as left-over food, fruit and vegetable peels, dry leaves and other wastes of gardens, etc. can be hurried in a pit. They are converted into compost and used as manure.
• Waste materials such as tin, cans, paper, glass, metallic articles are recycled. Through the process of recycling such materials are reused to form new products.

Activity 15.1 [T. B. Pg. 256-257]

To make an aquarium.

Materials:
Large jar of glass, water, pebbles, fish food, aerator (oxygen pump), small fishes, aquatic plant

Procedure:

• Take a large jar of glass. Keep some pebbles in it.
• Fill the jar with water.
• Add few aquatic plants such as algae in the jar.
• Add few small fishes in it.
• Arrange oxygen pump in such a way that we can provide oxygen in the jar.
• Provide fish food regularly which is available in the market.
• Add few small aquatic animals other than fishes in the jar.

Questions :

Question 1.
How can an aquarium become self-sustaining system by adding a few aquatic plants and animals?
Answer:
Aquatic plants performs photosynthesis. They are producers on which animals depend for their nutrition. Plants release oxygen during photosynthesis. Oxygen is utilised by animals in respiration and release carbon dioxide in the water. Carbon dioxide is available to plants for photosynthesis. Thus, aquarium becomes self-sustaining system.

Question 2.
Can we leave the aquarium as such after it is set up?
Answer:
No, we cannot leave the aquarium as it is because metabolic wastes are released in water make it polluted. Therefore, finally change of water is needed.

Question 3.
Why does aquarium have to be cleaned once in a while?
Answer:
Metabolic wastes produced by aquatic organisms make the water polluted. So, it should be cleaned once in a while.

Question 4.
Do we need to clean lakes or ponds in the same manner? Why or why not?
Answer:
Yes, because sometimes excretory wastes accelerate the growth of algae. The lake or pond gets covered with algal growth. Some toxic substances are released and dissolved 02 in water body will be depleted. This will lead to death of all aquatic life and such ecosystem may destructed.

Activity 15.2 [T. B. Pg. 257]

To find more about aquarium.
Material:
An aquarium

Questions:

Question 1.
While creating an aquarium did you take care not to put an aquatic animal which would eat others? What would have happened otherwise?
Answer:
Yes, while creating an aquarium care was taken predator aquatic animals were not used. Otherwise such predators can feed on other organisms and destroy. All small aquatic animals will be consumed by carnivores which later would all die.

Question 2.
Write the aquatic organisms in order of who eats whom and form a chain of at least three steps.
Answer:

Question 3.
Would you consider any one group of organisms to be of primary importance? Why or why not?
Answer:
Yes, plants (producers) should be given primary importance because they form first trophic level and all the consumers directly or indirectly depend on plants for their food (energy) requirement.

Question 4.
Explain the components of an ecosystem.
OR
Explain the groups of organisms based on their role in an ecosystem.
Answer:
Each ecosystem consists of two main components:
(1) Abiotic components : All the non-living constituents of an ecosystem are included in the abiotic components.

Abiotic components are the physical factors like temperature, rainfall, wind, soil, light, minerals, etc.

(2) Biotic components : All living organisms of an ecosystem are included in the biotic components.
Organisms can be grouped as producers, consumers and decomposers according to their food habit. Their mode of sustenance forms the trophic relationship in the environment.

(i) Producers : Those organisms which can make organic compounds like sugar and starch from inorganic substances using solar energy in presence of chlorophyll are called producers.
Example: Certain bacteria, various kinds of algae and all green plants.

(ii) Consumers : Those organisms which consume the food produced either directly from
producers or indirectly by feeding on other consumers are called consumers.

Example: Non-chlorophyllous and heterotrophic organisms.
Consumers can be divided into four categories.

(iii) Decomposers : The microorganisms which breakdown the complex organic substances into simple inorganic substances are called decomposers.
Example: Certain bacteria and fungi, breakdown the dead remains and waste products of organisms.

Question 5.
Write a short note on : Consumers
Answer:
Consumers : Those organisms which consume the food produced either directly from producers or indirectly by feeding on other consumers are called consumers.

Example: Non-chlorophyllous and heterotrophic organisms.
Consumers can be divided into four categories

Activity 15.3 [T. B. Pg. 260]

Newspaper reports about pesticides level in ready-made food items are often seen these days and some states have banned these products.

Questions :

Question 1.
What would be the source of pesticides in ready-made food items?
Answer:
Pesticides are non-biodegradable substances. They are excessively used for controlling the pest in crop field. They enter in food items through food chain.

Question 2.
Could pesticides get into our bodies from food products other than ready-made food?
Answer:
Yes, pesticides can get into our bodies from the food grains, vegetables, fruits, milk, etc. that contain residues of pesticides.

Question 3.
What methods could be applied to reduce our intake of pesticides?
Answer:

1. Pesticides in the crop fields should be s judiciously used.
2. Monitoring of pesticides level in agricultural products should be tested at regular interval.

Question 4.
Why some states have banned on some ready-made food products?
Answer:
Some states have banned on some ready-made food products because the level of pesticides is high in it, thus causing hazard to our health.

Activity 15.4 [T. B. Pg. 261]

To find harmful chemicals for ozone layer.
Materials:
Library, internet, newspaper reports

Questions:

Question 1.
Which chemicals are responsible for the depletion of the ozone layer?
Answer:
Ozone depleting substances such as chlorofluorocarbons (CFCs), hydrofluoro-carbons (HFCs) and oxides of nitrogen are resonsible for depletion of the ozone layer.

Question 2.
Find out if the regulations to control the emission of ozone depleting chemicals have succeeded in reducing the damage to the ozone layer. Has the size of the hole in the ozone layer changed in recent years?
Answer:
In 1987, the United Nations Environment Programme (UNEP), succeeded in forging an agreement to freeze CFC production at half of the 1986 levels.

Yes, by reducing the continuous use of ozone depleting chemicals, the size of the hole in the ozone layer has reduced in recent years.

Activity 15.5 [T. B. Pg. 261]

Collect waste materials from your home. (kitchen wastes, waste paper, torn clothes and its pieces, empty cartons, milk packets, empty bottles, its lid, used tea leaves, empty medicine bottles/strips/bubble packs, broken footwear, etc.)

• Bury these materials in a pit near your home.
• Keep these moist by spraying water on it.
• Fill the pit with moist clay and cover the waste.
• Dig the pit and observe at 15-day intervals.

Questions:

Question 1.
What are the materials that remain unchanged over long periods of time?
Answer:
The materials that remain unchanged over long periods of time are empty medicine bottles, bubble packs, milk packets, broken plastic footwear.

Question 2.
What are the materials which change their form and structure over time?
Answer:
The materials that change their form and structure are food, vegetable peels, used tea leaves, empty cartons, waste paper, torn clothes, broken leather footwear.

Question 3.
What are the materials that change the faster?
Answer:
The materials that are changed faster are vegetable peels, used tea leaves, spoilt food, etc.

Activity 15.6 [T. B. Pg. 262]

To find more about biodegradable and non-biodegradabie substances.

Questions:

Question 1.
How long are various non-biodegradable substances expected last in our environment?
Answer:
Non-biodegradable substances such as plastic wastes can be acted upon by physical factors such as heat and pressure. But under the ambient conditions found in our environment, non-biodegradable wastes persist for a long time.

Question 2.
Find out whether biodegradable plastic do or do not harm the environment.
Answer:
Polymer fabrics and dental implants are examples of biodegradable plastics. Biodegradable plastics do not cause any harm to the environment.

Activity 15.7 [T. B. Pg. 263]

To find about the disposal of waste generated at home,

Questions

Question 1.
Find out what happens to the waste generated at home? Is there a system in place to collect this waste?
Answer:
The waste generated at home is collected in dustbins daily. The municipal corporation of our city has set up a system to remove such garbage on daily basis. Large collection containers are set up at specific sites to collect the household waste.

In some areas, municipal corporation has set up a system to collect household waste from door to door.

Question 2.
Find out how the local body (panchayat, municipal corporation, resident welfare association) deals with the waste.
Are there mechanisms in place to treat the biodegradable and non-biodegradable wastes separately?
Answer:
Biodegradable and non-biodegradable wastes are separated at the source. In some villages, biogas plants have been established to use biodegradable waste to generate biogas and manure.

In cities, municipal corporation collects waste and finally send to specific vacant site on the extreme edge of the city and dump it there.

Biodegradable wastes should not be burnt as burning causes air pollution. They are converted into manure by composting. Non-biodegradable wastes such as plastics, glass, metals are collected separately and send to respective recycling units.

Question 3.
Calculate how much waste is generated at home in a day. How much of this waste is biodegradable?
Answer:
Large amount of waste is generated at home in a day. Most of it is biodegradable. Some of it is non-biodegradable.

Question 4.
Calculate how much waste is generated in a classroom in a day. How much of this waste is biodegradable?
Answer:
In a classroom, large amount of waste which is mostly biodegradable waste generated.

Question 5.
Suggest the ways of dealing with the waste.
Answer:
Biodegradable wastes should be burried in a pit. After few days they change into manure / compost due to action of decomposers. It can be used in garden.

Activity 15.8 [T. B. Pg. 263]

To find about how the sewage is treated.

Questions:

Question 1.
Find out how the sewage in your locality is treated. Are there mechanisms in place to ensure that local water bodies are not polluted by untreated sewage?
Answer:
We have underground drainage system in our city. The interconnected drainage lines take away sewage to distantly place where sewage treatment plant is located. Here it is treated.

Sewage is not allowed to pollute the local water bodies. It is first treated in sewage treatment plant, during this the treated water is disinfected through chlorination, water is then used for irrigation.

Question 2.
Find out how the local industries in your locality treat their wastes.
Are there mechanisms in place to ensure that the soil and water are not polluted by this water?
Answer:
Local industries are legaly bound to treat their industrial waste before releasing/discarding them in local water bodies.

But, these industries do not follow the rules and the norms set for them. Due to it, the industrial wastes that come out from industries have still pollutants and pollute our water bodies.

Activity 15.9 [T. B. Pg. 264]

To find out hazardous materials in disposed electronic items.
Materials:
Internet, books from library.

Questions:

Question 1.
What hazardous materials have to be dealt with while disposing the electronic items? How would these materials affect the environment?
Answer:
When we dispose off electronic wastes, the hazardous materials in them include plastics and electronic chips made of silicon. Such materials are non-biodegradable and will remain unchanged for a long time in the environment.

Question 2.
How are plastics recycled? Does the recycling process have any impact on the environment?
Answer:
Plastic wastes are isolated from garbage and sent to recycling unit. Plastics are melted in the unit and then remoulded to form various plastic items for reuse.

The recycling process reduces plastic wastes from the environment. Recently, plastic wastes are being used to construct polymer plastic road.

Jharkhand Board JAC Class 10 Science Solutions Chapter 3 Metals and Non-metals Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 3 Metals and Non-metals

Jharkhand Board Class 10 Science Metals and Non-metals Textbook Questions and Answers

Question 1.
Which of the following pairs will give displacement reactions?
(a) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal
Answer:
AgNO₃ solution and copper metal

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) All of the above
Answer:
Applying a coating of zinc

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be…
(a) calcium
(b) carbon
(c) silicon
(d) iron
Answer:
calcium

Question 4.
Food cans are coated with tin and not with zinc because …
(a) zinc is costlier than tin.
(b) zinc has a higher melting point than tin.
(c) zinc is more reactive than tin.
(d) zinc is less reactive than tin.
Answer:
zinc is more reactive than tin.

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch.
(a) How could you use them to distinguish between samples of metals and non-metals?
(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.
Answer:
(a) Metals can be hammered Into thin sheets hence, metal possesses property of malleability while non-metals cannot be beaten into thin sheets.
Arrange the battery, bulb, wires and switch in a proper circuit and by passing the electric current, if the bulb glows, then it must be a metal, because metal is a good conductor of electricity. But if the bulb does not glow, then it Is a sample of non-metal, because non-metal is a non-conductor of electricity.

(b) First experiment shows that metal possesses property of malleability and ductility, while second experiment justifies that metals are good conductors of electricity while non-metals are non-conductor of electricity.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
Metal oxides which react with both acids and bases to forms salt and water are called amphoteric oxides.
Examples:

• Aluminium oxide (Al₂O₃)
• Zinc oxide (ZnO)

Aluminium oxide reacts with an acid and a base in the following manner :
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Zinc oxide reacts with an acid and a base in the following manner :
ZnO + 2HCl → ZnCl₂ + H₂O

Question 7.
Name two metals which will displace hydrogen from dilute acids and two metals which will not.
Answer:
Two metals which displace hydrogen from dilute acids are :

• zinc (Zn) and
• aluminium (Al).

Two metals which cannot displace hydrogen from dilute acids are :

• copper (Cu) and
• mercury (Hg).

Question 8.
In the electrolytic refining of a metal M, What would you take as an anode, the cathode and the electrolyte?
Answer:
In the electrolytic refining, Impure metal (M) is taken as an anode and thin strip of pure metal (M) Is taken as a cathode and solution of soluble salt of metal (M) Is taken as an electrolyte.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it as shown in figure below:
(a) What will be the action of gas on
(i) dry litmus paper?
(ii) moist litmus paper?
(b) Write a balanced chemical equation for the reaction taking place.

When sulphur powder is heated in air, it forms sulphur dioxide, which is acidic in nature and its aqueous solution is called sulphurous acid (H₂SO₃).
(a) Action of gas :

• There will be no effect of gas on dry litmus paper.
• Moist blue litmus paper changes its colour to red due to H⁺ ions present in the aqueous solution of H₂SO₃ formed from SO₂ obtained after burning sulphur.

(b) Balanced chemical equation for the above activity is as given below:
S(s) + O₂(g) → SO₂(g)

Question 10.
State two ways to prevent the rusting of iron.
Answer:
The rusting of iron can be prevented by painting, oiling, greasing, chrome plating, anodising or by making alloys.

Question 11.
What type of oxides are formed when non¬metals combine with oxygen?
Answer:
Non-metals combines with oxygen to form acidic oxides. For example, SO₂, SO₃, CO₂, Cl₂O₇, etc.

Question 12.
Give reasons :
(a) Platinum, gold and silver are used to make jewellery.
Answer:
Platinum, gold and silver are used to make jewellery, because these metals possesses lustre. They are malleable and ductile, as a result, jewellery of different shapes can be made. Moreover, these metals do not chemically react with water or air. Due to these properties platinum, gold and silver are used to make jewellery.

(b) Sodium, potassium and lithium are stored under oil.
Answer:
Sodium, potassium and lithium are highly reactive metals. They release hydrogen gas, when they react with air or moisture in air. Hydrogen gas is highly combustible and it catches fire. To prevent accidental fire, lithium, sodium and potassium are stored under oil.

(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
Answer:
Aluminium is highly reactive metal. It reacts with oxygen of air to form aluminium oxide, which gets deposited as thin layer on the surface of aluminium. This layer acts as protective layer and prevents further reaction of aluminium with oxygen. Moreover it is light in weight and a good conductor of heat. Also, it’s manufacturing cost is very low in comparison to other metals. Therefore, most of the utensils for cooking are made from aluminium.

(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.
Answer:
It is easier to obtain a metal from its oxide. Hence, it is essential to convert carbonate and sulphide ores into oxide ores during the extraction of metal. Extraction of metal by reduction of oxide ores is easier than extraction from its carbonate or sulphide ores.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
Copper vessels are tarnished and corroded due to formation of copper oxide layer on its surface. The green coating on copper vessels is due to formation of basic copper carbonate. On cleaning with lemon or tamarind juice, the citric acid present in them neutralises the basic copper carbonate and dissolves the layer, there by helps to regain the lustre of copper utensils.

Question 14.
Distinguish between metal and non-metal on the basis of their chemical properties.
Answer:

Metals	Non-metals
1. They are electro-positive elements.	1. They are electro-negative elements.
2. Aqueous solutions of metal oxides are basic.	2. Aqueous solutions of non-metal oxides are acidic.
3. Hydrogen gas is evolved when metals reacts with dilute acid.	3. When non-metal react with dilute acid hydrogen gas is not evolved.
4. Oxides of metals are basic in nature. For example, Na2O	4. Oxides of non-metals are acidic in nature. For example, SO2, CO
5. Atoms of metal possess one, two or three electrons in their outermost shell.	5. Atoms of non-metals possesses more than three electrons in their outermost shell.

Question 15.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
Answer:
That man was using a solution of aqua regia; which is a mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3 : 1 by volume. Gold dissolves in aqua regia.

Question 16.
Give reason : Why copper is used to make hot water tanks and not steel (an alloy of iron.)?
Answer:
Copper does not react with cold and hot water nor it reacts with steam of water. Hence copper can be used to make hot water tanks.

But, steel is an alloy of an iron and it reacts with steam of water, hence, iron in steel is slowly corroded.
Therefore, copper is used for making hot water tanks and not steel.

Jharkhand Board Class 10 Science Metals and Non-metals InText Questions and Answers

Question 1.
Give an example of a metal which …
(i) is a liquid at room temperature.
(ii) can be easily cut with a knife.
(iii) is the best conductor of heat.
(iv) is a poor conductor of heat.
Answer:
(i) Mercury is a liquid at room temperature.
(ii) Sodium, potassium can be easily cut with a knife.
(iii) Silver and copper are good conductor of heat.
(iv) Lead is a poor conductor of heat.

Question 2.
Explain the meaning of malleable and ductile.
Answer:

1. Malleable : Malleable indicates a tendency of a metal to be brought in sheet form by hammering.
2. Ductile: Ductile indicates a tendency of a s metal to be brought in thin wire form.

Question 3.
Why is sodium kept immersed in kerosene oil?
Answer:
Sodium is highly reactive metal. It reacts with oxygen of air at room temperature. This reaction is highly exothermic. Thus, to prevent the reaction of sodium with oxygen, it is kept immersed in kerosene oil.

Question 4.
Write equations for the reactions of (i) Iron with steam.
(ii) Calcium and potassium with water.
Answer:
(i) Iron with steam:
4H₂O(g) + 3Fe(s) → Fe₃O₄(s) + 4H₂(g)

(ii) Calcium and potassium with water :

Question 5.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows:

Metal	Iron (II) sulphate	Copper (II) sulphate	Zinc sulphate	Silver nitrate
A	No reaction	Displacement
B	Displacement		No reaction
C	No reaction	No reaction	No reaction	Displacement
D	No reaction	No reaction	No reaction	No reaction

Use the table above to answer the following questions about metals A, B, C and D:
(1) Which is the most reactive metal?
(2) What would you observe if B is added to a solution of copper (II) sulphate?
(3) Arrange the metals A, B, C and D in the order of decreasing reactivity.
Answer:
(1) Metal B is the most reactive metal.

(2) Blue colour of copper (II) sulphate solution disappears and reddish brown copper metal is < deposited on the metal B.

(3) Decreasing order of reactivity is s B > A > C > D.

Question 6.
Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H₂SO₄.
Answer:
When a reactive metal reacts with dilute hydrochloric acid, it forms hydrogen gas. The reactive metal displaces the hydrogen from acid releasing hydrogen gas.

Question 7.
What would you observe when zinc is added to a solution of iron (II) sulphate?
Write the chemical reaction that takes place.
Answer:
Zinc (Zn) is more reactive than iron (Fe). Hence, when it is added to iron (II) sulphate, it displaces the iron metal. As a result, the colour of the solution fades from green to colourless due to formation of zinc sulphate, and the greyish black coloured iron metal gets displaced.
Zn(s) + FeSO₄(aq) → ZnSO₄(aq) + Fe(s)

Question 8.
Name two metals which displace hydrogen from dilute acid and name two metals which cannot displace hydrogen from dilute acids.
Answer:
Two metals which displace hydrogen from dilute acids are:

• zinc (Zn) and
• aluminium (Al).

Two metals which cannot displace hydrogen from dilute acids are:

• copper (Cu)
• mercury (Hg).

Question 9.
Explain the electronic configuration of noble gases (He, Ne, Ar); metals (Na, Mg, Al, K, Ca) and non-metals (N, O, F, E S, Cl).
Answer:
Electronic Configurations of Some Elements

Question 10.
What are ionic compounds or electrovalent compounds? Explain with example.
OR
Explain the formation of sodium chloride (NaCl).
Answer:
The compounds formed by the transfer of electrons from a metal to a non-metal are known as Ionic compounds or electrovalent compounds. Atomic number of sodium is 11. Sodium atom has one electron in its outermost shell (M-shell). Sodium atom loses the electron from its M-shell and forms sodium cation (Na⁺) and acquires stable complete octet structure of noble gas neon (Ne).

Similarly, atomic number of chlorine is 17. Chlorine atom has seven electrons in its outermost shell (M-shell). Chlorine atom gains one e^– which is lost by sodium atom and forms chloride anion (Cl^–) and acquires stable complete octet structure of noble gas argon (Ar).

Sodium cation (Na⁺) and chloride anion (Cl^–) being oppositely charged attract each other and are held by strong electrostatic forces of attraction and exists as sodium chloride (NaCl).

Sodium chloride exists as a group of oppositely charged ions.

Question 11.
State the formation of magnesium chloride by the transfer of electrons.
Answer:
Atomic number of magnesium is 12. It loses its two electrons from outermost shell and acquires complete octet stable structure.

Similarly, atomic number of chlorine is 17. It gains one electron and acquires complete octet stable structure.

Thus, two electrons lost by magnesium atom are gained by two chlorine atoms (each one gets one electron) and forms magnesium chloride.

Question 12.
Explain the electronic configuration of noble gases (He, Ne, Ar); metals (Na, Mg, Al, K, Ca) and non-metals (N, O, F, P, S, Cl).
Answer:
Electronic Configurations of Some Elements

Question 13.
What are ionic compounds or electrovalent compounds? Explain with example.
OR
Explain the formation of sodium chloride (NaCl).
Answer:
The compounds formed by the transfer of electrons from a metal to a non-metal are known as Ionic compounds or electrovalent compounds. Atomic number of sodium is 11. Sodium atom has one electron in its outermost shell (M-shell). Sodium atom loses the electron from its M-shell and forms sodium cation (Na⁺) and acquires stable complete octet structure of noble gas neon (Ne).

Similarly, atomic number of chlorine is 17. Chlorine atom has seven electrons in its outermost shell (M-shell). Chlorine atom gains one e^– which is lost by sodium atom and forms chloride anion (Cl^–) and acquires stable complete octet structure of noble gas argon (Ar).

Sodium cation (Na⁺) and chloride anion (Cl^–) being oppositely charged attract each other and are held by strong electrostatic forces of attraction and exists as sodium chloride (NaCl).

Sodium chloride exists as a group of oppositely charged ions.

Question 14.
State the formation of magnesium chloride by the transfer of electrons.
Answer:
Atomic number of magnesium is 12. It loses its two electrons from outermost shell and acquires complete octet stable structure.

Similarly, atomic number of chlorine is 17. It gains one electron and acquires complete octet stable structure.

Thus, two electrons lost by magnesium atom are gained by two chlorine atoms (each one gets one electron) and forms magnesium chloride.

Activity 3.1 [T. B. Pg. 37]

Aim : To study the lustrous properties of metals.

Activity:

• Take samples of iron, copper, aluminium and magnesium. Note the appearance of surface of each sample.
• Clean the surface of each sample by rubbing them with sand paper and note their appearance again.

Questions :

Question 1.
How does the surface of samples of metal appear?
Answer:
The surface of samples of metal appear dull.

Question 2.
What happens when the surface of sample of metal is rubbed with sand paper?
Answer:
When the surface of sample of metal is rubbed with sand paper, it appears shiny.

Question 3.
How does the surface of metal appear in their pure form?
Answer:
Metals have shining surface in their pure form.

Question 4.
Name the shining property of metal.
Answer:
The property of shining surface of metal is known as metallic lustre.

Activity 3.2 [T. B. Pg. 37]

Aim : To study the hardness of metals.

Caution : Always handle sodium metal with care.

Activity:

• Take small pieces of iron, copper, aluminium and magnesium.
• Try to cut these metals with a sharp knife and note your observations.
• Hold a piece of sodium metal with a pair of tongs, put it on a watch-glass and try to cut it with a knife.
  What do you observe?

Questions:

Question 1.
Can we cut the metals such as an iron, copper, aluminium and magnesium with a knife?
Answer:
No, we can’t cut.

Question 2.
Can sodium metal be cut with a knife?
Answer:
Yes, we can cut sodium metal into pieces.

Question 3.
Which property of metal is seen in the above activity?
Answer:
Metals are hard and their hardness varies from metal to metal.

Activity 3.3 [T. B. Pg. 38]

Aim : To study that metals are malleable.

Activity:

• Take pieces of iron, zinc, lead and copper?
• Place any one metal on a block of iron and strike it four to five times with a hammer.
• What do you observe?
• Repeat the same with other metals.
• Record the change in the shape of these < metals.

Questions:

Question 1.
Which property of metal is observed in this activity?
Answer:
Metals can be hammered into thin sheets.

Question 2.
What is malleability of metals?
Answer:
The property of metals to be converted into thin sheets when hammered is known as ‘Malleability of metals’.

Question 3.
Name the metals which are most malleable in nature.
Answer:
Gold and silver.

Activity 3.4 [T. B. Pg. 38]

Aim: To study that metals are ductile.

Activity:
Take samples of metals such as iron, copper, aluminium and lead.

Questions:

Question 1.
What is meant by ductility?
Answer:
The ability of metals to be drawn into? thin wires (filaments) is called ductility.

Question 2.
Which metal is most ductile?
Answer:
Gold is most ductile metal.

Question 3.
What is the maximum length of wire which (can be drawn from one gram of gold?
Answer:
2 km length

Question 4.
Why the metals can be given different shapes?
Answer:
Metals can be given different shapes due to their properties like malleability and ductility.

Activity 3.5 [T. B. Pg. 38]

Aim : To show that metals are good conductors of heat and possesses high melting point.

Activity:

• Take an aluminium or copper wire. Clamp this wire on a stand as shown in the figure 3.1.
• Fix a pin to the free end of the wire using wax.
• Heat the wire with a spirit lamp, candle or a burner near the place where it is clamped.
• What do you observe after some time?
• Note your observations.
• Does the metal wire melt?
  

Questions:

Question 1.
What happens to a pin attached to the wire?
Answer:
Due to heating, metal wire expands and the pin is displaced forward. This indicate that heat flows through the wire and melts the wax but wire does not melt.

Question 2.
Which property of metal is observed in the above activity?
Answer:
The above activity shows that metals are good conductors of heat and have high melting point.

Question 3.
Which metal is used as the best conductor of heat?
Answer:
Silver and copper are best conductors of heat.

Question 4.
Which metals are poor conductors of heat?
Answer:
Lead and mercury are poor conductors of heat.

Activity 3.6 [T. B. Pg. 39]

Aim: To study the property of electrical conductivity of metals.

Activity:

• Set up an electric circuit as shown in the figure 3.2.
• Place the metal to be tested in the circuit between terminals A and B as shown.
  

Questions :

Question 1.
Which property of metal is observed in the above activity?
Answer:
It is observed that metals are good conductors of electricity.

Question 2.
Which coating is used on electric wire?
Answer:
The coating of polyvinyl chloride is formed on electric wire.

Question 3.
Does the bulb glow? What does it indicate?
Answer:
The bulb glows. It indicate that electric current flows through the metal.

Activity 3.7 [T. B. Pg. 39]

Aim : To study the properties of non-metals.

Activity:

• Collect the samples of carbon (coal or graphite), sulphur and iodine.
• Carry out the activities 3.1 to 3.4 and 3.6 with these non-metals and record your observations.

Questions:

Question 1.
State the physical states of non-metals.
Answer:
Non-metals exist in solid, liquid and gaseous states at room temperature.

Question 2.
State the common properties of non-metals based on the above activities.
Answer:

• Non-metals are non-conductors of heat and electricity.
• Boiling points of non-metals are comparatively lower.
• Non-metals do not possess property of malleability and ductility.

Activity 3.8 IT. B. Pg. 40]

Aim: To test the acidity and basicity of oxides.

Activity:

• Take a magnesium ribbon and some sulphur powder.
• Burn the magnesium ribbon. Collect the ashes formed and dissolve it in water.
• Test the resultant solution with red and blue litmus paper.
• Is the product formed on burning magnesium ribbon acidic or basic?
• Now, burn sulphur powder. Place a test tube over the burning sulphur to collect the fumes produced.
• Add some water to the above test tube and shake.
• Test this solution with blue and red litmus paper.
• Is the product formed on burning sulphur acidic or basic?
• Can you write equations for these reactions?

Questions :

Question 1.
What is formed on burning magnesium?
Answer:
Magnesium oxide (MgO) is formed on burning magnesium.

Question 2.
Is the magnesium oxide acidic or basic?
Answer:
Magnesium oxide is basic in nature.

Question 3.
Which product is obtained on burning sulphur?
Answer:
Sulphur dioxide (SO₂) is obtained.

Question 4.
Is the product formed on burning sulphur acidic or basic?
Answer:
It is acidic.

Question 5.
Write the equations of chemical reactions which occur in the activity 3.8.
Answer:

• 2Mg(s) + O₂(g) → 2MgO(s)
• MgO(s) + H₂O(l) → Mg(OH)₂(aq)
• S(s) + O₂(g) → SO₂(g)
• SO₂(g) + H₂O(l) → H₂SO₃(aq)

Question 6.
Is the sulphur dioxide acidic or basic?
Answer:
It is an acidic.

Activity 3.9 [T. B. Pg. 41]

Aim: To study the burning of metals in air.

Caution:

• This activity needs the teacher’s assistance.
• It would be better, if students wear goggles for eye protection.

Activity:

• Take samples of metals such as aluminium, copper, iron, lead, magnesium, zinc and? sodium.
• Hold any sample of metal taken above with a pair of tongs and try burning over a flame.
• Repeat the same with the other metal samples.
• Collect the product, if formed.
• Let the products and the metal surface cool down.
• Which metals burn easily?
• What flame colour did you observe when the metal burnt?
• How does the metal surface appear after burning?
• Arrange the metals in the decreasing order of their reactivity towards oxygen.
• Are the products soluble in water?

Questions:

Question 1.
Which metal burns easily in air?
Answer:
Magnesium burns easily in air.

Question 2.
What colour does Na, Mg, Cu and Al impart to the oxidising flame?
Answer:

• Na → Yellow flame
• Mg → Dazzling white flame
• Cu → Greenish blue flame
• Al → White flame

Question 3.
How does the metal surface appear after burning?
Answer:
Silvery white.

Question 4.
What is the solubility of product in water obtained by the reaction of Cu, Fe, Zn, Al with oxygen?
Answer:
The products obtained by the reaction of Cu, Fe, Zn and Al with oxygen are insoluble in water.

Question 5.
Arrange the metals such as Al, Cu, Fe, Pb, Mg, Zn and Na in the decreasing order of their reactivity towards oxygen.
Answer:
Na > Mg > Al > Zn > Fe > Pb > Cu.

Question 6.
Which amongst the given metals forms water soluble product after heating it?
Answer:
Amongst the given metals, only sodium metal on heating forms oxide which is soluble in water.

Activity 3.10 [T. B. Pg. 42]

Aim : To study the reaction of water with metals.

Caution :
This activity requires the teacher’s help.

Activity:

• Take samples of metals such as aluminium, copper, iron, lead, magnesium, zinc, calcium, gold, silver, sodium and potassium.
• Put small pieces of the samples separately in beakers half filled with cold water.
• Which metals reacted with cold water? Arrange them in the increasing order of their reactivity with cold water.
• Did any metal produce fire on water?
• Does any metal start floating after some time?
• Put the metals that did not react with cold water in beakers half filled with hot water.
• For the metals that did not react with hot water, arrange the apparatus as shown in the figure 3.3 and observe their reaction with steam.
• Which metals did not react even with steam?
• Arrange the metals in the decreasing order of reactivity with water.
  

Questions:

Question 1.
Which metals react with cold water?
Answer:
Sodium, potassium and calcium react with cold water.

Question 2.
Which metals produce fire on water?
Answer:
Sodium and potassium produce fire on water.

Question 3.
Which metal floats on water?
Answer:
Calcium and magnesium float on water.

Question 4.
Which metal does not react with cold water, but reacts with hot water?
Answer:
Magnesium does not react with cold water but it reacts with hot water.

Question 5.
Which metal does not react with either cold or hot water but reacts with steam?
Answer:
Aluminium, iron and zinc do not react with cold and hot water, but they react with steam.

Question 6.
Arrange the given metals (in activity) in the decreasing order of their reactivity with water.
Answer:
Decreasing order of the reactivity of metals with water is given as follows:
K > Na > Ca > Mg > A1 > Zn > Fe
Pb, Cu, Ag and Au does not react with water.

Question 7.
Arrange sodium, potassium and calcium metals in the increasing order of their reactivity with water.
Answer:
Ca < Na < K Question 8. Which metals do not react with cold water and steam? Answer: Pb, Cu, Ag and Au do not react with cold water and steam.

Activity 3.11 [T. B. Pg. 44]

Aim : To study the reaction of metal with an acid.

Activity: Collect the samples of pieces of metals such as magnesium, aluminium, zinc, iron and copper. Put the samples separately in test tubes containing dilute hydrochloric acid. Suspend thermometers in the test tubes. Which metals reacted vigorously with dilute hydrochloric acid? With which metal did you record the highest temperahire? Arrange the metals in the decreasing order of their reactivity with dilute acids.

Questions:

Question 1.
Which metal reacts vigorously with dilute hydrochloric acid?
Answer:
Magnesium metal reacts vigorously with dilute hydrochloric acid.

Question 2.
Which metal does not react with hydrochloric acid?
Answer:
Copper (Cu) metal do not react with hydrochloric acid.

Question 3.
With which metal the rise of temperature is maximum during the reaction?
Answer:
The rise of temperature is maximum in case of magnesium during the reaction.

Question 4.
Arrange the metals Mg, Al, Zn and Fe in the decreasing order of their reactivity towards dilute hydrochloric acid.
Answer:
Mg > Al > Zn > Fe

Activity 3.12 [T. B. Pg. 44 – 45]

Aim : To study the reaction of metal with solutions of other metal salts.

Activity:

• Take a clean wire of copper and an iron nail.
• Put the copper wire in a solution of iron sulphate and the iron nail in a solution of copper sulphate taken separately in test tubes (figure 3.4).
• Record your observations after 20 minutes.
• In which test tube did you find that a reaction has occurred?
• On what basis can you say that a reaction has actually taken place?
• Write a balanced chemical equation for the reaction that has taken place.
• Name the type of reaction.
  

Questions:

Question 1.
In which test tube, reaction has occurred? Write the balanced chemical equation for the reaction.
Answer:
The following reaction occurred in a test tube containing iron nail dipped in a copper sulphate solution :
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
In this reaction, more reactive Fe displaces less reactive Cu.

Question 2.
On what basis can you say that a reaction has actually taken place?
Answer:
Blue colour of copper sulphate solution turns green during reaction. This change indicates that reaction has occurred.

Question 3.
Name the type of reaction occurring in the above activity.
Answer:
Displacement reaction occurs in the above acitivity.

Question 4.
Why does the reaction not occurred in the test tube containing copper wire dipped in a iron sulphate solution?
Answer:
Because, iron (Fe) is more reactive than copper (Cu); the reaction does not take place.

Activity 3.13 [T. B. Pg. 48]

Aim : To study the properties of ionic compounds, s

Activity:

• Take samples of sodium chloride, potassium s iodide and barium chloride.
• What is the physical state of these salts?
• Take a small amount of a sample on a metal ; spatula and heat it directly on the flame (figure 3.5). Repeat with other samples.
• What did you observe? Did the samples s impart any colour to the flame?
• Try to dissolve the samples in water, petrol and kerosene. Are they soluble?
• Make a circuit as shown in figure 3.6 and insert the electrodes into a solution of one salt, s
• What did you observe? Test the other samples too in this manner.
• What is your inference about the nature of these compounds?
  

Answers of the questions asked in the above activity are given in the following table :

Jharkhand Board JAC Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World Important Questions and Answers.

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

Additional Questions and Answers

Question 1.
Give four points of difference between the following terms / quantities :
(1) Near-sightedness and Far-sightedness
Answer:

Near-sightedness	Far- sightedness
1. The eye lens does not become thin as required but remains thick.	1. The eye lens does not become thick as required but remains thin.
2. The light rays from objects at far distances are focused short of the retina. As a result, the distant objects cannot be seen clearly.	2. The light rays from objects nearby the eyes are focused behind the retina. As a result the nearby objects cannot be seen clearly.
3. The light rays from objects nearby eyes are focused on the retina. As a result, the nearby objects are seen clearly.	3. The light rays from distant objects are focused on the retina. As a result the distant objects are seen clearly.
4. This defect can be corrected by using concave lens of appropriate focal length.	4. This defect can be corrected by using convex lens of appropriate focal length.

(2) Near point and Far point
Answer:

Near point	Far point
1. The minimum distance at which the object can be seen clearly without contraction of eye lens is called the near point of an eye.	1. The farthest distance upto which the eye can see objects clearly is called far point of an eye.
2. For young adult with normal vision, this value is 25 cm.	2. For young adult with normal vision, this value is infinite.
3. For a person having defect of far-sightedness value of the near point is at a distance more than 25 cm.	3. For a person having defect of near-sightedness value of the far point is at a distance less than infinite distance.
4. For a person having defect of far-sightedness value of the near point is obtained at a distance of 25 cm using convex lens of appropriate focal length.	4. For a person having defect of near-sightedness value of the far point is obtained at distance of infinite distance using concave lens of appropriate focal length.

Question 2.
Give scientific reasons for the following statements:
(1) To rectify the defect of near-sightedness or myopia, concave lens of suitable focal length is used as corrective lens.
Answer:
The eye lens of a person having defect of near-sightedness or myopia, does not become thin as per the requirement and so the rays after being refracted by the eye lens get focussed at a position short of retina. So distant objects cannot be seen clearly.

If a person having defect of near-sightedness, uses concave lens of suitable focal length, then light rays become slightly divergent. Hence, the image can be formed on retina.

Thus, if the image is formed on retina, then the distant object may be seen clearly.

(2) To rectify the defect of far-sightedness or hypermetropia, convex lens of suitable focal length is used as corrective lens.
Answer:
The eye lens of a person having defect of far-sightedness as hypermetropia, does not become thick as per the requirement and so the rays after being refracted by the eye lens get focussed behind retina. So nearby objects cannot be seen clearly.

If a person having defect of far-sightedness, uses convex lens of suitable focal length, then light rays become slightly convergent. Hence, the image can be formed on retina.

Thus, if the image is formed on retina, then the nearby object may be seen clearly.

(3) A rainbow is visible in the sky only after rain shower.
Answer:
In rainy season there are many clouds in the sky having tiny water droplets. When the sunlight is incident on tiny water droplets, due to these tiny water droplets, refraction, dispersion, internal reflection and at the end again refraction of sunlight take place.

Due to which a band is created which contains seven colours in the sky which is known as rainbow.

In other seasons there are no clouds in the sky. Therefore there are no tiny water droplets s in the sky. Hence, no rainbow is formed in the sky in other seasons and so it is not visible in? the sky.

(4) The sunrise is experienced two minutes early and the sunset is experienced two minutes delayed.
Answer:
As the altitude (height) from the earth’s surface gradually increases, the earth’s atmosphere becomes rarer. Hence, the refractive index of the air decreases continuously.

So a ray of light coming from the Sun towards an observer continuously passes from an optically rarer to optically denser medium and bends towards the normal. Thus, its direction of propagation changes continuously.

The actual sunrise or sunset begins when the Sun reaches the horizon.
1. In figure 11.12, S₁ is the actual position of the Sun a little below the horizon.

2. In this case, the light rays coming from S₁, continuously get refracted in the earth’s atmosphere (atmospheric refraction) and reach the observer as shown in the figure.

3. The tangent drawn to the curved path of the ray at point P passes through S₂, above the horizon.

4. S₂ is the apparent position of the Sun.

5. Thus, during the sunrise, the Sun is seen even though it is little below the horizon. Similarly, during the sunset it is seen for sometime even though it is little below the horizon.

6. Taking the refractive index of air as 1.00029 the apparent angular shift in the position of the Sun is found to be approximately (\(\frac { 1 }{ 2 }\))°.
Now, for angular displacement of 180° of the Sun, time required is 12 hours. Hence for angular displacement of (\(\frac { 1 }{ 2 }\))° of the Sun, time required is
= \(\frac{\left(\frac{1}{2}\right)^{\circ} \times 12 \text { hours }}{180^{\circ}}\)
= 0.03333 hour
= 1.9998 minutes
≈ 2 minutes
Due to advance sunrise and delayed sunset the duration of a day increases by about 4 minutes.

(5) The clear sky appears blue in colour.
Answer:

The molecules of air and other fine particles in the atmosphere are smaller in size than the wavelength of light in the visible region.

• The wavelength of red light is about 1.8 times that of blue light.
• When the sunlight passes through the atmosphere, the fine particles in the air scatter blue light more strongly than red light.
• At this time if the observer looks upwards, he finds sky blue.
• If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark.

Important Note:
The sky appears dark to passengers flying at very high altitudes, as scattering of light is not prominent at such heights.

(6) The danger signal lights are red in colours.
Answer:
The red light gets scattered least by fog or smoke because of its longer wavelength relative to light of any other colour. So, it can be seen even from a long distance. Therefore it is used in signals showing danger.

(7) The sun appears reddish at sunrise and sunset.
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.]

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence :

Question 1.
What is dispersion of white light?
Answer:
The phenomenon of a splitting of white light into its constituent colours is called the dispersion of white light.

Question 2.
What happens to the image-distance in the normal eye, when we increase the distance of an object from the eye?
Answer:
The image distance always remains constant.

Question 3.
What can be said about the focal length of the eye lens if its curvature increases?
Answer:
decreases

Question 4.
What can be said about the curvature of the eye lens if it becomes thin?
Answer:
decreases

Question 5.
For normal eye vision what is the object-distance and image-distance when the object is placed at a near point? (Take the distance between the eye lens and the retina as 2.3cm.)
Answer:
u = – 25 cm, v = + 2.3 cm

Question 6.
For normal eye vision what is the object-distance and image-distance when the object? is placed at a far point? (Take the distance between the eye lens and the retina as 2.3 cm.)
Answer:
u = – 00, a = + 2.3 cm

Question 7.
State the type of image of an object formed s on the retina.
Answer:
real, inverted and diminished

Question 8.
Write the name of the most front part of human eye.
Answer:
cornea

Question 9.
State the function of the iris.
Answer:
Iris can control the amount of light entering into the eye as well as size of pupil can be controlled by it.

Question 10.
State the function of light sensitive cells present in retina.
Answer:
Light rays falling on the retina are converted into the electrical signals.

Question 11.
Write the function of optic nerves.
Answer:
The work of optic nerves is to sent electrical signals to the brain.

Question 12.
Write use of bifocal lens.
Answer:
To rectify/remove the eye-defect known as presbyopia.

Question 13.
How much duration in second increases per day due to early sunrise and delayed sunset?
Answer:
240s

Question 14.
Due to which effect does the smoke emitted by the combustion of the engine oil in a motorcycle sometimes appears blue in colour?
Answer:
The Tyndall effect

Question 15.
Which effect is developed commercially to determine the size and density of aerosol and other colloidal particles?
Answer:
The Tyndall effect

Question 16.
Wavelength of red colour is approximately how many times the wavelength of violet colour?
Answer:
1.8

Question 2.
Fill in the blanks :

1. The type of image formed by the eye lens is ………………. and ……………….
2. A triangular glass prism has ………………. triangular bases and ………………. rectangular surfaces.
3. Light enters our eye through the ……………….
4. A person suffering from far-sightedness or hypermetropia cannot see clearly ………………. objects.
5. A ………………. corrective lens is used to rectify near-sightedness.
6. An old person suffering from near-sightedness and a far-sightedness uses ………………. to rectify his vision.
7. While passing through the prism the light ray travelling from air to glass bends towards the ……………….
8. In a glass prism ………………. light propagates with maximum speed.
9. At night stars are seen slightly at a higher position than their actual position because of the ……………….
10. For a light ray passing through the prism, the angle between the incident ray and the emergent ray is known as the ……………….
11. At the time of sunrise the sun appears ………………. in colour.
12. The fine particles in air scatter ………………. light more strongly.
13. ………………. light, while passing through a prism, does not disperse.
14. Stars behave like ………………. sources of light and planets behave like ………………. sources of light.
15. Danger signals are red in colour because red light is ……………….
16. In the spectrum of white light, ………………. and ………………. colours seen at the two ends.
17. The diameter of the human eyeball is approximately ………………. cm.
18. The distance between the eye lens and the retina is known as the ……………….
19. In the normal situation in the relaxed position of the ciliary muscles, the eye lens is ……………….
20. At night as we move up in the atmosphere of the earth, the refractive index ………………. continuously.

Answer:

1. real, inverted
2. two, three
3. cornea
4. nearby
5. concave / diverging
6. spectacles with bi-focal lenses
7. normal
8. red
9. atmospheric refraction
10. angle of deviation
11. reddish
12. blue
13. Monochromatic (It means light of single frequency, i.e., of single colour.
14. point, extended
15. scattered the least
16. violet, red
17. 2.3
18. size of the eyeball
19. thin
20. decreases

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

1. The near point of every person is 25 cm.
2. The splitting of white light into its constituent colours is called the scattering of light.
3. Far-sightedness can be rectified by using a concave lens of suitable power.
4. In the eye of myopic person, the image of a distant object is formed behind the retina.
5. Near-sightedness arises due to more curvature of the cornea or due to the eye lens remaining thick permanently.
6. The speed of light decreases as it passes from an optically denser medium to an optically rarer medium.
7. A myopic person has the far point nearer than infinity.
8. A hypermetropic person has near point farther away from the normal near point (25 cm).
9. The construction of the human eye can be compared with that of a camera.
10. A rainbow is formed due to refraction taking place twice, one internal reflection and dispersion of sunlight by water droplets in the sky.
11. Planets twinkle.
12. When the sunlight passes through a canopy of dense forest, tiny water droplets in the mist, scatter the light. This effect is known as the Tyndall effect.

Answer:

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

Question 4.
Match the following:
(1)

Column I	Column II	Column III
1. Myopia	p. The focal length of the eye lens increase	a. Bifocal lens
2. Hypermetropia	q. The focal length of the eye lens decrease	b. Concave lens
3. Presbyopia	r. The power of accommodation of the eye decrease with ageing	c. Convex lens

Answer:
(1 – q – b), (2 – p – c), (3 – r – a).

(2)

Column I	Column II
1. TWinkling of stars	p. Tiny water droplets present (or suspended) in the atmosphere
2. Blue coloured sky	q. Band of colours
3. Rainbow	r. Scattering of light
4. Spectrum	s. Uneven atmosphere

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

(3)

Column I	Column II
1. Human eye or eyeball	a. It controls and regulates the amount of light entering the eye.
2. Self operated accommodation power of eye	b. Delicate membrane having large number of light sensitive cells.
3. Retina	c. Works as a photographic camera.
4. Ciliary muscles	d. Not able to see the nearby objects.
5. Myopia	e. Electrical signals related to image are sent to the brain.
6. Cataract	f. A circular muscular diaphragm which can control the size of pupil.
7. Presbyopia	g. Milky and cloudy layer is formed on the eye lens.
8. Iris	h. The capacity of eye to see objects clearly between 25 cm and infinite distance.
9. Pupil	i. Increases or decreases curvature of eye lens.
10. Optic nerves	j. Image of object at infinite distance is formed in front of retina.
11. Hypermetropia	k. Accommodation power of eye decreases with ageing.

Answer:
(1 – c), (2 – h), (3 – b), (4 – i), (5 – j), (6 – g), (7 – k), (8 – f), (9 – a), (10 – e), (11 – d).

Question 5.
Choose the correct option from those given below each question:
1. Splitting of white light into its seven constituent colours is called ……………
A. refraction
B. reflection
C. dispersion
D. interference
Answer:
C. dispersion

2. Which colour of light deviates maximum in the dispersion of white light by a prism?
A. Violet
B. Blue
C. Green
D. Red
Answer:
A. Violet

3. In the human eye, the image of an object is formed at the ……………….
A. iris
B. pupil
C. retina
D. cornea
Answer:
C. retina

4. The focal length of the eye lens is changed due to the action of the ……………….
A. pupil
B. retina
C. ciliary muscles
D. cornea
Answer:
C. ciliary muscles

5. A ………………. lens is used to correct presbyopia.
A. convex
B. concave
C. bi-focal
D. contact
Answer:
C. bi-focal

6. Out of the following, which phenomenon does not play a role in the formation of a rainbow?
A. Reflection
B. Refraction
C. Dispersion
D. Absorption
Answer:
D. Absorption

7. Where is the image formed in the eye of a person suffering from near-sightedness?
A. On the retina
B. Behind the retina
C. In front of the retina
D. On the pupil
Answer:
C. In front of the retina

8. Which phenomenon is responsible for the twinkling of stars?
A. Atmospheric reflection
B. Atmospheric refraction
C. Reflection
D. Total internal reflection
Answer:
B. Atmospheric refraction

9. The phenomenon of ………………. of light by the colloidal particles gives rise to the Tyndall effect.
A. reflection
B. refraction
C. scattering
D. dispersion
Answer:
C. scattering

10. What is the time difference between actual sunset and apparent sunset?
A. 2 seconds
B. 20 seconds
C. 2 minutes
D. 20 minutes
Answer:
C. 2 minutes

11. Which light gets scattered maximum due to atmosphere?
A. Blue
B. Yellow
C. Green
D. Red
Answer:
A. Blue

12. Which light has minimum speed in glass (prism)?
A. Red
B. Green
C. Blue
D. Violet
Answer:
D. Violet

13. When an eye is focussed on a distant object, the focal length of the eye lens is ……………….
A. maximum
B. minimum
C. half of its minimum
D. half of its maximum
Answer:
A. maximum

14. How many surfaces does a triangular prism have?
A. 3
B. 4
C. 5
D. 6
Answer:
C. 5

15. The wavelengths of violet, yellow and red light are λ_v, λ_y and λ_r respectively, then ……………………..
A. λ_v > λ_y > λ_r
B. λ_v < λ_y < λ_r
C. λ_y < λ_v < λ_r
D. λ_y < λ_r < λ_v
Answer:
B. λ_v < λ_y < λ_r

16. For normal vision, the far point is at ……………… distance.
A. 25 cm
B. 1 cm
C. 1 m
D. infinite
Answer:
D. infinite

17. For normal vision, the near point is ………………
A. 25 cm
B. 25 m
C. zero
D. infinite
Answer:

18. Which phenomenon can explain the advance sunrise and the delayed sunset?
A. Dispersion of light
B. Scattering of light
C. Tyndall effect
D. Atmospheric refraction
Answer:
D. Atmospheric refraction

19. Which of the following phenomena cannot be explained by scattering of light?
A. The red light used for signal lights for danger.
B. Blue colour of clear sky
C. White colour of clouds
D. Early sunrise
Answer:
D. Early sunrise

20. The base of an equilateral triangle ABC is BC. When it is arranged in four different situations and white light is incident on it, then in which of the following arrangements of the prism, the third colour from the top is the colour of clear sky in dispersion of light is produced?

A. (i)
B. (ii)
C. (iii)
D. (iv)
Answer:
B. (ii)
Hint: The dispersion of white light is shown in arrangement (ii) of prism

The third colour of light from the top is blue which is the colour of the clear sky.

21. The Sun appears white in afternoon. The reason is …
A. less scattering of light.
B. more scattering of all the colours of white light.
C. more scattering of blue colour.
D. more scattering of red colour.
Answer:
A. less scattering of light.
Hint: White light coming from the Sun has to travel less distance in the atmosphere before reaching the observer. So less scattering of light takes place and as a result the Sun appears white.

22. Sea water at more depth appears blue. The reason is …
A. presence of some plants in sea water.
B. the image of the sky appears in water.
C. scattering of light.
D. light is absorbed by sea water.
Answer:
C. scattering of light.

23. When the ciliary muscles are relaxed, the eye lens becomes …………….. and its focal length ……………… This enables us to see distant objects clearly.
A. thin, increases
B. thin, decreases
C. thick, increases
D. thick, decreases
Answer:
A. thin, increases

24. When the ciliary muscles contract, the eye lens becomes …………….. and its focal length ……………..
This enables us to see nearby objects clearly.
A. thick, decreases
B. thick, increases
C. thin, increases
D. thin, decreases
Answer:
A. thick, decreases

25. The rainbow on the moon ……………..
A. is not possible.
B. is rare.
C. is observed with the reverse order of colours.
D. is of two types.
Answer:
A. is not possible.

26. In dispersion of white light due to a triangular glass prism, the deviation of red colour is less compared to violet colour. The reason ………………
A. is n_v > n_r.
B. is n_r > n_v.
C. is n_v = n_r.
D. does not depend on n.
Answer:
A. is n_v > n_r.
Hint: In glass medium, speed of violet colour is less as compared to speed of red colour. So, n_v > n_r from n = \(\frac { c }{ v }\)

27. Which lens from the following, should a person suffering from near-sightedness use?
A. A convex lens
B. A concave lens
C. A cylindrical lens
D. A bi-focal lens
Answer:
B. A concave lens

28. Which lens is used by a person suffering from far-sightedness?
A. A convex lens
B. A concave lens
C. A cylindrical lens
D. A bi-focal lens
Answer:
A. A convex lens

29. Which of the following is true for near-sightedness?
A. Nearby objects cannot be seen clearly.
B. Distant objects cannot be seen clearly.
C. The eye lens cannot become thick as required.
D. This defect can be rectified using spectacles of convex lenses.
Answer:
B. Distant objects cannot be seen clearly.

30. Which of the following is true for far-sightedness?
A. Nearby objects cannot be seen clearly.
B. Distant objects cannot be seen clearly.
C. The eye lens cannot become thin as required.
D. This defect can be rectified using spectacles of concave lenses.
Answer:
A. Nearby objects cannot be seen clearly.

31. Where is the image formed in the eye of a person suffering from far-sightedness?
A. On the retina
B. Behind the retina
C. On the pupil
D. In front of the retina
Answer:
B. Behind the retina

32. A person has a defect of eye vision. His near point is 40 cm. It means …
A. he cannot clearly see objects at a distance more than 40 cm from the eye.
B. he can clearly see objects at a distance of 40 cm only.
C. he can clearly see objects at a distance equal to 40 cm or more from the eye.
D. he can clearly see objects at a distance less than 40 cm e.g., 25 cm from the eye.
Answer:
C. he can clearly see objects at a distance equal to 40 cm or more from the eye.
Hint: Here the person suffers from far-sightedness.

33. A person has a defect of vision. His far point is 1.5m. It means …
A. he cannot clearly see objects at a distance more than 1.5 m from the eye.
B. he can clearly see objects at a distance more than 1.5 m from the eye.
C. he cannot clearly see objects at a distance less than 1.5 m from the eye.
D. he suffers from far-sightedness.
Answer:
C. he cannot clearly see objects at a distance less than 1.5 m from the eye.
Hint: Here the person suffers from near-sightedness.

34. Out of the following, which light is deviated minimum in the dispersion of white light through a glass prism?
A. Green
B. Violet
C. Yellow
D. Dispersion of the given three colours is the same.
Answer:
C. Yellow

35. Which light has maximum speed in glass?
A. Violet
B. Blue
C. Green
D. Red
Answer:
D. Red

36. Which ray of light is present exactly at the middle of the spectrum obtained from white light?
A. Green
B. Yellow
C. Red
D. Violet
Answer:
A. Green

37. The lens in human eye is a …………….
A. convex mirror
B. convex lens
C. concave mirror
D. concave lens
Answer:
B. convex lens

38. For persons suffering from near-sightedness the power of the lens used in spectacles is …………….
A. positive
B. zero
C. negative
D. infinite
Answer:
C. negative

39. For a person suffering from ……………. the power of the lens used in spectacles is positive.
A. far-sightedness
B. near-sightedness
C. presbyopia
D. cataract
Answer:
A. far-sightedness

40. Which phenomenon/phenomena of light is/ are involved in the formation of a rainbow?
A. Refraction
B. Dispersion
C. Internal reflection
D. All of the given above
Answer:
D. All of the given above

41. ……………. light from the following is least scattered by fog, dust and smoke.
A. Violet
B. Blue
C. Red
D. Yellow
Answer:
C. Red

42. Which of the following controls the amount of light entering into the human eye?
A. Ciliary muscles
B. Pupil
C. Cornea
D. Iris
Answer:
D. Iris

43. The refractive index of glass is maximum for ……………. light.
A. violet
B. green
C. blue
D. red
Answer:
A. violet

Question 6.
Answer the following questions in very short as directed (Miscellaneous) :
(1) What is the power of a lens that can be used to correct the eye defect of a person who cannot see the objects distinctly kept beyond 2m?
Answer:
0.5 D
(∵ For a myopic eye the focal length of the corrective lens is equal to the far point of the myopic person, i.e., f = – 2m).

(2) Why does the Sun appear white at noon?
Answer:
The Sun appears white at noon because white light (sunlight) is least scattered by the atmosphere.

(3) Why is the eye lens not perfectly solid?
Answer:
If it becomes solid, its focal length would be fixed. Then, we would not be able to focus the objects lying at different distances on the retina. In short, accommodation of the eye would be reduced to zero.

(4) What is the focal length of plain goggles?
Answer:
Infinity

(5) What happens when elasticity of the crystalline lens is reduced to zero?
Answer:
Power of accommodation would be almost zero.

(6) Which defect of the eye occurs due to distortion of cornea?
Answer:
Astigmatism occurs due to the distortion of cornea.

(7) How the defect of astigmatism can be corrected?
Answer:
Astigmatism is corrected by the use of a cylindrical lens.

(8) What is colour blindness? How can it be cured?
Answer:
It is a defect of the eye in which a person is unable to distinguish between certain colours due to insufficient or no cone shaped cells on retina. It cannot be cured.

(9) In hypermetropia, how does the size of eyeball change?
Answer:
In hypermetropia, the eyeball becomes too small (flat).

(10) What change occurs in the focal length, when our eye lens becomes thick?
Answer:
The focal length will decrease when the eye lens becomes thick.

(11) What are rods and cones?
Answer:
Sensitive portion of retina has large number of cells, rod shaped and cone shaped cells. Rod shaped cells are sensitive to the intensity or brightness of the light whereas cone shaped cells are sensitive to colours.

(12) What is cataract?
Answer:
Sometimes, the eye lens of a person becomes hazy or even opaque resulting in reduced or total loss of vision. This is called cataract.

(13) What would have been the colour of the sky, had there been no atmosphere?
Answer:
Black

(14) Due to which phenomenon is the colour of water in deep sea blue?
Answer:
Due to scattering of light.

(15) What is the cause for presbyopia?
Answer:
The eye lens becomes less elastic or non-elastic after 40 years as the, accommodation power of eye becomes less.

(16) Give the relationship between wavelength of light and its angle of deviation, when it is passed through a prism.
Answer:
Wavelength λ = \(\frac{1}{\text { Angles of deviation } \delta}\)

(17) For which colour has the glass larger refractive index – violet or green?
Answer:
Violet

(18) Which part of human eye is also known as ‘white of the eye’?
Answer:
Sclera

(19) Why is blind spot so called?
Answer:
Blind spot is a point at which the optic nerve leaves the eye. It contains no rods or cones. So an image formed at this point, is not sent to the brain.

(20) Which liquid is filled in the space between the eye lens and the retina?
Answer:
Vitreous humour (It is a transparent jelly.)

(21) What happens to the pupil of the eye when the light is (a) very bright and (b) very dim?
Answer:
(a) In very bright light, the size of the pupil becomes very small, so that less amount of light enters into the eye.
(b) In very dim light, the size of the pupil increases, so that more amount of light enters into the eye.

(22) A man is wearing spectacles of focal length +1 m. What can be the defect in the eye?
Answer:
As the focal length of spectacles is positive, the man is using convex lenses. So, he is suffering from hypermetropia.

(23) Which portion of a bi-focal lens is
(a) a concave lens
(b) a convex lens?
Answer:
(a) Upper portion
(b) Lower portion.

(24) When sunlight enters into a room filled with dark smoke, its path becomes visible. Name the phenomenon responsible for this.
Answer:
The Tyndall effect.

(25) What is the function of the iris?
Answer:
The iris controls the size of the pupil.

(26) What are light sensitive cells?
Answer:
Rods and cones

(27) What type of signals are generated and sent to the brain by light sensitive cells in the retina?
Answer:
Electrical signals.

(28) What holds the crystalline lens in the human eye?
Answer:
Ciliary muscles.

(29) Which part of the human eye helps in changing the thickness of the eye lens?
Answer:
Ciliary muscles.

(30) What is dispersion of white light?
Answer:
The splitting of white light into its various components (i.e., 7 colours) is called dispersion of white light.

(31) Give the main difference between the lens of the human eye and the lens of a camera.
Answer:
The lens of the human eye has flexible aperture, so that its focal length can be changed, while in a camera the focal length of the lens is fixed.

(32) The image formed on the retina is inverted, but we see the object erect. Why?
Answer:
The inverted image formed on the light sensitive cells (called rods and cones of retina), generates electrical signals. These signals reach the brain via the optic nerve. It is the brain? that interprets this image and while processing the image it helps in perceiving the objects as they are.

(33) The absolute refractive index of a medium is 2.0. The speed of light in vacuum/air is 3 x 10⁸ms^-1. Find the speed of light in the medium.
Answer:
Absolute refractive index of medium is,
n = \(\frac { c }{ v }\)
∴ v = \(\frac { c }{ n }\)
= \(\frac{3 \times 10^8}{2}\)
= 1.5 x 10⁸ms^-1

(34) Match the column properly :

Column I (eye defect)	Column II (correcting lens)
1. Myopia	p. Bi-focal lens
2. Hypermetropia	q. Concave lens
	r. Convex lens

Answer:
(1 – q), (2 – r).

(35) Match the following column :

Column I (eye defect)	Column II (correcting lens)
1. Astigmatism	p. Convex lens
2. Presbyopia	q. Cylindrical lens
	r. Concave lens

Answer:
(1 – q), (2 – p).

(36) The far point of a myopic eye is 100 cm. What is the focal length of the lens required to see very distant (normal far point) objects clearly?
Answer:
As the focal length of concave lens used for correcting the myopic eye is equal to distance of far point of the myopic eye, the focal length of correcting lens required is f = – x = – 100 cm = – 1 m.

(37) The near point of a hypermetropic eye is 75 cm. What is the focal length of the lens required to see clearly an object placed at 25 cm from the eye (normal near point)?
Answer:
The focal length of convex lens used for correcting the hypermetropic eye is given by
(where x’ = defective near point = 75 cm and d = normal near point = 25 cm)
f = \(\frac{x^{\prime} d}{x^{\prime}-d}\)
∴ f = \(\frac{75 \times 25}{75-25}\)
= 37.5 cm
= 0.375 m

(38) The eye lens of human eye is a double convex lens. Agree or Disagree?
Answer:
Agree.

(39) Cone-shaped retinal cells respond to the brightness or intensity of light. Agree or Disagree?
Answer:
Disagree.

(40) Which property of vision is used in cinematography?
Answer:
Property of persistence of vision is used in cinematography.

(41) What is aqueous humour?
Answer:
Aqueous humour is a transparent viscous liquid in the space between the cornea and the eye lens.

(42) What is the maximum power of accommodation of a normal eye?
Answer:
The maximum power of accommodation of a normal eye
= \(\frac{1}{\text { near point of the normal eye (in metre) }}\)
= \(\frac { 1 }{ 0.25 m }\)
= \(\frac { 100 }{ 25 m}\)
= 4 m^-1
= 4 D

(43) What is meant by scattering of light?
Answer:
It is the phenomenon of change in the direction of light on striking a scatterer.

(44) What is the basic cause of atmospheric refraction?
Answer:
The basic cause of atmospheric refraction is variation in the refractive index (i.e., optical density) of different layers of the earth’s atmosphere with altitude.

(45)

In the above figure a narrow beam of white light is shown to pass through a triangular glass prism. After passing through the prism, it produces a spectrum XY on a screen. State the colour seen at (i) X and (ii) Y.
Answer:
The colour seen at X is violet and that seen at Y is red.

(46)

In the above figure, which angles are correctly marked?
Answer:
∠A and ∠e are correctly marked.

(47)

In the above figure (ray diagram), state the angle of incidence and the angle of deviation.
Answer:
∠PQN = i = Angle of incidence
∠P’OR = D = Angle of deviation

(48)

In the above figure (ray diagram), state angle of incidence, angle of emergence and angle of deviation.
Answer:
∠p = Angle of incidence
∠y = Angle of emergence
∠z = Angle of deviation

(49) What is the principle of the working of s the human eye?
Answer:
It is like a camera having a lens system forming an inverted, real image on the light |> sensitive screen (retina) inside the eye.

(50) On which factor does the colour of the scattered white light depend?
Answer:
Size of the particles of the medium : through which it is passing.

(51) Give the scientific names of the following parts of the eye :
(a) Carrying signals from an eye to the brain.
(b) A small opening (hole) in the middle of the iris.
Answer:
(a) Optic nerve
(b) Pupil

(52) A near-sighted person has a near point 25 cm and a far point 50 cm. Can he see clearly an object at a distance of: (i) 5 cm, (ii) 25 cm, (iii) 60 cm. Write ‘Yes’ or ‘No’ only.
Answer:

1. No
2. Yes
3. No

(53) The near point of a far-sighted person is 50 cm.
Can the person see clearly an object at a distance of:
(i) 20 cm
(ii) ∞ (infinity)
Answer:
(i) No
(ii) Yes

(54) How much is our horizontal field of view (a) with one eye open
(b) with both eyes open?
Answer:
(a) about 150°
(b) about 180°.

(55) Which of the following have a wider s field of view?
(a) Animals having two eyes on the opposite 5 sides of their head or
(b) Animals having two eyes at the front of their head.
Answer:
(a) Animals having two eyes on the opposite 5 sides of their head or

Value Based Questions With Answers

Question 1.
Maull and Vlshva are best friends and they study in 4th grade. Recently, Mauli has been facing difficulty in reading the blackboard text from the last desk/bench. Vishva wonders why Mauli avoids sitting on the last desk/bench. On observation, Vishva found that Mauli often carries junk food in her lunch. Vishva has started sharing her lunch full of green vegetables and fruits with her. Mauli is now better and has also started taking a “balanced diet”.

1. Name the eye defect Mauli is suffering from.
2. What are two possible deformities related s to her eye defect?
3. What values do you learn from Vishva and Mauli?

Answer:

1. Myopia (near-sightedness)
2. Lens defect (increased thinness i.e., excessive curvature of the eye lens) and Eyeball defect (elongation of the eyeball).
3. Friendship, concern for each other, s importance of a balanced diet.

Question 2.
An eye camp was organised by the doctors in a village. They found that the eyes of imaged people in the village have the near point receded and the far point also gets reduced. Often aged people suffer from both myopia s and hypermetropia. Doctors (opthalmologists) provide these people spectacles of bi-focal lenses to correct the defects. The people were happy and grateful to the doctors.
(1) Name the eye defect from which the people were suffering.
(2) Give any two causes of this defect.
(3) What were benefits to organise such camps > in rural areas? Give two suggestions.
Answer:
(1) Presbyopia

(2) (a) Weakening of ciliary muscles
(b) Reducing ability of the lens to change the curvature.

(3) (a) To make people aware of eye diseases
(b) To tell people to take proper and balanced diet.

Question 3.
Four friends went to a picnic. The weather was pleasant. They played various games and then had snacks. Suddenly, Raju, one of them, observed seven colours in the sky. He said to others, “wow what a rainbow” !

Ram, one of them, asked him “What is a rainbow?” Raju then explained to all about its formation. After that everyone in the group thanked Raju for the knowledge, he had given to them.
(1) When Raju was facing the rainbow, where was the Sun?
(2) Which device can be used to obtain such a phenomenon?
(3) What moral value do you learn from Raju?
Answer:
(1) The Sun was behind Raju.
(2) A small prism. (Water droplets present in the atmosphere act like small prisms.)
(3) Knowledge increases by sharing, friendship, love and affection with nature.

Question 4.
In a beautiful valley, there was a village. When trains passed from the village, the whistle and the sound of train, mixed with the sound of waterfall, seemed to be very pleasant to everyone. Hence, children of that village used to play near the railway track. Once on a very light foggy day, a group of children found that a fish plate was missing from the track. As such, all the villagers were worried.

Prashant, one of the children, suddenly put his ear to the line and tried to know whether a train is coming or not. He knew a train is coming. He asked his friends to inform the railway cabin crew and he himself put off his red shirt and started running s towards the train, waving his red shirt. The driver and cabin man got the alert signal in time and thus a major accident was averted.?
(1) Name the two physical phenomena of science used by Prashant.
(2) Why did Prashant use his red shirt instead of any other coloured shirts?
(3 ) What moral values do you learn from Prashant?
Answer:
(1) (a) Sound travels through a medium.
(b) Scattering of light

(2) The red light is least scattered by fog or smoke so it can be seen from a large distance.

(3) (a) Proper knowledge and its application.
(b) Concern for others.

Question 5.
Millions of people of the developing countries of the world are suffering from corneal blindness. They can be cured by replacing the defective cornea with the cornea of a donated eye. A charitable society of your city has organised a campaign in your neighbourhood in order to create awareness about this fact.
(1) State the objective of organising such campaigns.
(2) Write one argument which you would give s to motivate the people to donate their eyes after death.
(3) List two values which could be developed in the persons who actively participate and < contribute in such programme.
Answer:
(1) The objective of organising such campaign is to help those people who are suffering from corneal blindness and they can be cured by replacing their defective cornea with the cornea of a donated eye.

(2) Come forward to participate in this compaign because, if someone gets his vision through your eyes, it is an incredible help as the eye is one of the most valuable sense organs through which an individual can achieve so many things in his / her life.

(3) The persons who actively participate and contribute in such programme are

• strong hearted
• very much helpful for the people living in such situations.

Question 6.
Mr Bharat’s 65 year old mother was complaining about blurred vision in both the eyes due to which she could not see things clearly. Mr Bharat took his mother to an eye hospital. The doctor examined her eyes carefully and concluded that she has a medical condition which could not be corrected by using any type of spectacle lenses and it required surgery. Her eyes were operated upon and she could then see once again properly.
(1) What could be the defect in the eyes of Mr Bharat’s mother?
(2) What happens to the eye lens during this defect? What is done during surgical operation of the eyes to restore the correct vision?
(3) What values do you learn from Mr Bharat in this episode?
Answer:
(1) The defect in the eyes of Mr Bharat’s mother is known as cataract.

(2) During the development of cataract, a membrane is gradually formed over both the eye lenses making the eye lenses cloudy. This makes the vision blurred. During surgical operation, the cloudy eye lenses are removed from the eyes and suitable artificial lenses are inserted in their place.

(3)

• Awareness and knowledge about the eye defects that can be cured by eye specialist doctors.
• Desire to mitigate the suffering of others (here mother).
• Sense of responsibility.

Question 7.
Amit is a domestic help (or maid) working at ; Mr Dave’s house. One day Amit complained to Mr Dave that he had difficulty in reading the letter which he had received from his s parents. Mr Dave, realising that Amit had an eye defect, took him to an eye specialist doctor.

The doctor tested his eyes carefully and told Amit to wear spectacles containing certain type of lenses having specified power, Mr Dave bought the required spectacles for Amit. By wearing this spectacle, Amit could read and write easily. He was very happy and thanked Mr Dave.
(1) What could be the eye defect Amit was suffering from?
(2) What could be the two possible reasons responsible for his eye defect? What type of lenses do you think the doctor recommended for Amit’s spectacles?
(3) What values are displayed by Mr Dave in this episode??
Answer:
(1) Amit was suffering from an eye defect called hypermetropia (far-sightedness) in ; which a person cannot see the nearby objects clearly though he can see the distant objects clearly.

(2) (a) Low converging power of the eye lens (because of the eye lens being less convex or less thick)
(b) Eyeball being too small (flat) (because of which the distance of the retina from the eye lens is less than normal).

The doctor recommended convex lenses for the spectacles of Amit.

(3) Mr Dave displayed the values of –

• awareness, which means having s knowledge of a situation or facts.
• concern for others (to mitigate their suffering).
• kindness and generosity.

Question 8.
Rohit is a car driver working for Mr Joshi. One day Rohit complained that he had difficulty in driving the car because he could not see the distant traffic clearly though he could see the nearby objects clearly. Mr Joshi took Rohit to an eye hospital. The eye specialist doctor checked and tested his eyes with various machines and gave him the name and power of the lenses to be worn as spectacles.

Mr Joshi paid for the required spectacles for the driver. By wearing these spectacles, the driver could now see even the distant vehicles and people on the road clearly. He thanked Mr Joshi for this.
(1) Name the eye defect Rohit is suffering from.

(2) What could be the two possible reasons for his eye defect? What type of lenses do you think the doctor recommended for Rohit’s spectacles?

(3) What values (or qualities) do you learn from Mr Joshi in this episode?
Answer:
(1) Myopia (near-sightedness) in which a person cannot see distant objects clearly though he can see nearby objects clearly.

(2) (a) High converging power of the eye lens (because of the eye lens being too convex or too thick).
(b) Eyeball may be too long (elongated) (because of which the distance of the retina from the eye lens is more than normal).
The doctor recommended concave lenses for the spectacles of Rohit.

(3) (a) General awareness (that an eye defect can usually be corrected by wearing spectacles containing suitable lenses).
(b) Concern for others (because Mr Joshi wanted to mitigate or remove the suffering of driver).
(c) Kindness and generosity.

Practical Skill Based Questions With Answers

Question 1.
Dispersion is caused by refraction and not by reflection. Why?
Answer:
White light is composed of seven colours having different wavelengths. And speed of different colours is the same in vacuum / air while in different media it is different.

Now, for a given angle of incidence, the angle of reflection is the same for all the wavelengths of white light, while the angle of refraction is different for different wavelengths.

Question 2.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 as shown in the diagram.

(1) The colours at positions marked 3 and 5 are similar to the colour of the sky and the colour of gold (metal), respectively. Is the above statement made by a student correct or incorrect. Justify.
(2) Which of the positions shown above correspond approximately to the colour of (a) a brinjal, (b) danger signal, (c) neel (applied to clothes), (d) orange.
Answer:
(1) No, because 3 refers to yellow and 5 to blue colour of the spectrum.

(2) (a) 7 (b) 1 (c) 6 (d) 2.

Question 3.
When a beam of white light is passed through a triangular glass prism, it gets dispersed into its component colours. Why do we get these colours? In the given figure, the colours X and Y represent the extreme components of the spectrum. Identify X and Y.

Answer:
(1) Different colours of light while passing through a prism bend through different angles with respect to the incident ray, as they travel with different speeds, this leads to dispersion of light.

(2) X – violet, Y – red

Question 4.
A narrow beam PQ of white light passes through a glass prism ABC as shown in the diagram.

Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.
(1) Write the name and cause of the phenomenon observed.
(2) Where else in nature is this phenomenon observed?
(3) Based on this observation, state the conclusion which can be drawn about the constituents of white light.
Answer:

(1) The phenomenon of splitting of white light into its constituent colours is called dispersion of light. It occurs because different constituent colours of light travel with different speeds in a material medium (other than air /vacuum) and hence bend through different angles.

(2) In nature, this phenomenon is observed in formation of a rainbow.

(3) Based on the phenomenon of dispersion, we can conclude that –

• White light consists of seven colours.
• Violet light suffers maximum deviation and red light suffers minimum deviation.

Question 5.
(a) A narrow beam of white light is incident on three glass objects as shown below. Comment on the nature of the behaviour of the emergent beam in all three cases.

(b) There is a similarity between two of the emergent beams. Identify the two.
Answer:
(a) (1) The incident beam of light after refraction through the glass slab emerges parallel to the incident beam but is laterally shifted. No dispersion takes place in this case.

(2) The incident beam of light after refraction through the prism splits into a band of seven colours which are violet, indigo, blue, green, yellow, orange and red. These coloured rays emerge out of the prism along different directions and become distinct. Thus, in this case, dispersion of white light takes place.

(3) When the incident beam passes through the first prism, the prism splits it into a band of seven colours. These coloured rays are then incident on an identical inverted prism.

Then the recombination of the coloured rays takes place. The emergent beam of light is parallel to the incident beam but slightly shifted outward.

(b) The emergent beam in the cases (1) and (3) are similar. In both the cases, the emergent beam is parallel to the incident beam and is laterally shifted.

Memory Map:

Students should go through these JAC Class 10 Maths Notes Chapter 2 बहुपद will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 10 Maths Notes Chapter 2 बहुपद

भूमिका :
पिछली कक्षाओं में हमने एक व्यंजक बहुपद तथा उनकी घातों, गुणनखंड तथा गुणक के बारे में पढ़ा है। स्मरण रहे कि P(x), x चर में एक बहुपद है P(x) में x की उच्चतम घात बहुपद की घात कहलाती है। उदाहरण के लिए, 4y² – 5y + 9, y चर में एक बहुपद है, जिसकी घात 2 है।
इस अध्याय में हम रैखिक तथा द्विघातीय बहुपदों के ज्यामितीय निरूपण (Geometrical Representation) और उनके शून्यकों के ज्यामितीय अर्थ के साथ-साथ बहुपद के गुणांकों और शून्यांकों के बीच सम्बन्धों का अध्ययन करेंगे।

बहुपद के प्रकार :
→ एकपदी (Monomial): ऐसे बहुपद को जिसमें केवल एक पद हो, एकपदी (monomial) कहते हैं:
जैसे : x², ax, 3x², a²x³, \(\frac{1}{2}\)x⁴ इत्यादि।

→ द्विपदी (Binomial): ऐसे बहुपद को जिसमें केवल दो पद हों, द्विपदी (binomial) कहते हैं; जैसे: ax + b, 5x² + 3x, a²xⁿ + b इत्यादि ।

→ त्रिपदी (Trinomial) ऐसे बहुपद को जिसमें केवल तीन पद हों, त्रिपदी (trinomial) कहते हैं: जैसे 3x² + 5x – 7, ax² + bx + c, ….. इत्यादि।

→ शून्य बहुपद (Zero polynomial): यदि किसी बहुपद में सभी पदों के गुणांक शून्य हों तो वह शून्य बहुपद कहलाता है। जैसे : p(x) = 0.
एक घात वाले बहुपद को रैखिक बहुपद, दो घात वाले बहुपद को द्विघात बहुपद और तीन घातों वाले बहुपद को त्रिघात बहुपद कहते हैं।

→ बहुपद की घात (Power of polynomial) चर x के बहुपद p(x) में x की उच्चतम घात को बहुपद की घात कहते हैं।
उदाहरण के लिए: (i) बहुपद p(x) = x⁵ + 4x³ – 3x² + 2x – 3 में चर राशि x की उच्चतम घात का पद है। जिसका घातांक 5 है। अतः बहुपद की घात 5 होगी।
(ii) बहुपद p(y) = 3y³ + 4y² – y + 8 में चर राशि की उच्चतम घात का पद 3y3 है, जिसका घातांक 3 है। अतः बहुपद P(y) की घात 3 होगी।

→ अचर बहुपद (Costant Polynomial): शून्य घात वाला बहुपद नियतांक या अचर बहुपद कहलाता है।
जैसे: p(x) = 7, g(x) = –\(\frac{3}{2}\), h(y) = 2, p(t) = 1 इत्यादि में किसी भी चर की घात शून्य होगी। अत: इस प्रकार के बहुपद अचर बहुपद कहलाते हैं।

→ रैखिक बहुपद (Linear Polynomial): घात 1 वाला बहुपद रैखिक बहुपद कहलाता है।
जैसे: 2x + 3, \(\sqrt{3}\)x + 5, y + \(\sqrt{2}\), x – \(\frac{2}{11}\), 3t + 4 इत्यादि सभी रैखिक बहुपद हैं।

→ द्विघात बहुपद (Quadratic Polynomial): घात 2 वाला बहुपद द्विघात बहुपद कहलाता है। द्विघात (Quadratic) शब्द क्वाड्रेट (quadrate) शब्द से बना है जिसका अर्थ वर्ग अर्थात् घात 2 है।
जैसे : f(x) = 2x² + 3x – \(\frac{4}{5}\), g(y) = 2y² – 3
h(u) = 2 – u² + \(\sqrt{3}\)u, p(v) = \(\sqrt{3}\)v² – \(\frac{4}{3}\)v + \(\frac{1}{2}\), इत्यादि द्विघात बहुपद हैं जिनके गुणांक वास्तविक संख्याएँ हैं।
व्यापक रूप से चर x में कोई द्विघात बहुपद f(x) = ax² + bx + c के रूप में होता है, जहाँ a, b, c वास्तविक संख्याएँ हैं, जहाँ a ≠ 0 है।

→ त्रिघात बहुपद (Cubic Polynomial) : घात 3 का बहुपद त्रिघात बहुपद कहलाता है।
जैसे : f(x)= \(\frac{9}{5}\)x³ – 2x² + \(\frac{7}{3}\)x – \(\frac{1}{5}\), g(x) = 2 – x³ तथा h(y) = 3y³ – 2y² + y + 1
चर x में एक त्रिघात बहुपद का व्यापक रूप निम्न है :
f(x) = ax³ + bx² + cx + d, जहाँ a, b, c, d वास्तविक संख्याएँ हैं जहाँ a ≠ 0 है।

बीजीय व्यंजक (Algebraic Expression) : कुछ निश्चित चर तथा अचर राशियों के योग, अन्तर, गुणन, भाग इत्यादि के संयोग से बने पद को बीजीय व्यंजक कहते हैं।
उदाहरणार्थ : f(x) = x³ – 6x² + x + 9, f(x) = -3x² + 2x – 1 तथा f(x) = 4x + 3 इत्यादि
इस प्रकार के बीजीय व्यंजकों को बहुपद (Polynomial) कहते हैं।
बहुपद : बीजीय व्यंजक के बहुपद होने के लिए निम्नलिखित शर्तें पूर्ण होनी चाहिए :
1. चर राशि का घातांक एक धनात्मक पूर्णांक हो।
2. पदों की संख्या निश्चित (सीमित) हो।
3. प्रत्येक पद में चर का गुणांक एक वास्तविक संख्या हो।
यदि x एक चर, प्राकृत संख्या और a₀, a₁, a₂, a₃, ….. a_n वास्तविक सख्याएँ हैं तो
p(x) = a₀ + a₁x + a₂x² + …… +a_nxⁿ
p(x) को चर x में एक बहुपद कहते हैं।
जहाँ a₀, a₁x, a₂x², a₃x³, …… इसके पद (Term) कहलाते हैं और a₀, a₁, a₂, a₃….. उनके गुणांक (Coefficient) कहलाते हैं।
उदाहरण के लिए:
(i) p(x) = 3x – 2 [चर x में एक बहुपद है।]
(ii) q(y) = 3y² – 2y + 4 [चर y मेँ एक बहुपद है।]
(iii) f(u) = \(\frac{1}{2}\)u³ – 3u² + 2u – 4 [चर u में एक बहुपद है।]
क्योंकि इन सभी (i), (ii) एवं (iii) की घात धनात्मक पूर्णांक है तथा प्रत्येक पद में चर राशि का गुणांक एक वास्तविक संख्या है।

निम्न व्यंजकों पर ध्यान दीजिए:
(i) p(x) = 2x² – 3\(\sqrt{x}\) यह बहुपद नहीं है क्योंकि \(\sqrt{x}\) या x^1/2 में x की घात \(\sqrt{2}\) है जो कि पूर्णांक नहीं है।
(ii) f(x) = \(\frac{1}{2 x^2-2 x+5}\) में भी x के घात धन पूर्णांक नहीं हैं अतः यह बहुपद नहीं है।
(iii) q(u) = x³ – \(\frac{1}{x^2}\) + 3 में \(\frac{1}{x^2}\) या x^-2 में x की घात -2 है, जो कि धन पूर्णाक नहीं है। अतः बहुपद नहीं है।

बहुपद का मान (Value of Poynomial): यदि f(x) चर x में एक बहुपद है और कोई वास्तविक संख्या है तो f(x) में x के स्थान पर α का मान रखने से प्राप्त वास्तविक संख्या बहुपद f(x) का x = α पर मान होगा और इसे f(α) द्वारा व्यक्त करते हैं।
जैसे: (i) f(x) = 2x² – 3x – 2 के x = 1 और x = -2 पर मान ज्ञात कीजिए ।
x = 1 पर, f(1) = 2(1)² – 3(1) – 2 = 2 × 1 – 3 × 1 – 2 = 2 – 3 – 2 = -3
x = -2 पर, f(-2) = 2(-2)² – 3(-2) – 2 = 2 × (+4) + 6 – 2 = 8 + 6 – 2 = 12
(ii) यदि बहुपद f(x) = x³ – 6x² + 11x – 6 है, x = -1 पर बहुपद का मान ज्ञात कीजिए।

बहुपद के शून्यक (Zeroes of Polynomial) : यदि किसी बहुपद p(x) में चर x के स्थान पर a (एक वास्तविक संख्या) प्रतिस्थापित करने पर p(a) = 0 हो, तो को बहुपद p(x) का शून्यक (Zero) कहते हैं। अर्थात् किसी भी बहुपद के शून्यकं ज्ञात करने का अर्थ होता है समीकरण p(x) = 0 को हल करना।
जैसे : (i) p(x) = 2x³ + 3x² + 3x + 2 का शून्यक ज्ञात करना है।
x = 0 पर, p(0) = 2(0)³ + 3(0)² + 3(0) +2
P(0) = 2 अर्थात p(0) ≠ 0
अत: शून्य, बहुपद p(x) का शून्यक नहीं है।
x = -1 पर, p(-1) = 2(-1)³ + 3(-1)² + 3(-1) + 2 = – 2 + 3 – 3 + 2 = 0
P(-1) = 0
अत: -1 बहुपद p(x) का एक शून्यक है।

विशेष :
(i) प्रत्येक रैखिक बहुपद का एक और केवल एक ही शून्यक होता है।
(ii) द्विघात बहुपद के दो शून्यक होते हैं।
(iii) त्रिघात बहुपद के तीन शून्यक होते हैं।
(iv) प्रत्येक बहुपद के वास्तविक शून्यक नहीं होते हैं।
जैसे : x² + a, x² + 2 तथा y² + y + 1 का कोई भी वास्तविक शून्यक नहीं है।
p(x) = x² + 6x + 15 का कोई शून्यक नहीं होता।
हल: माना कि p(x) = x² + 6x + 15
∴ p(x) = {x² + 2. (3). x + 9} + 6 = (x + 3)² + 6
यहाँ हम देखते हैं कि x के प्रत्येक वास्तविक मान के लिए (x + 3)² कभी भी ऋणात्मक मान ग्रहण नहीं कर सकता। अत: (x + 3)² का मान सदैव शून्य से बड़ा ही होगा। परिणामस्वरूप f(x) का मान भी 6 या उससे अधिक होगा।
इसलिए p(x) का कोई शून्यक विद्यमान नहीं है।

बहुपद के आलेख एवं शून्यकों का ज्यामितीय अर्थ :
बहुपद p(x) के ज्यामितीय आलेख को x अक्ष जिन बिन्दुओं पर प्रतिच्छेद करता है, उन बिन्दुओं के भुज या x-निर्देशांक बहुपद p(x) के शून्यक (zeroes) के रूप में जाने जाते हैं।
रैखिक बहुपद : व्यापक रूप में एक रैखिक बहुपद f(x) = ax + b, a ≠ 0 के लिए ग्राफ एक सरल रेखा प्राप्त होती है, जो x अक्ष को ठीक एक बिन्दु \(\left(-\frac{b}{a}, 0\right)\) पर काटती है।
अत: रैखिक बहुपद p(x) = ax + b, a ≠ 0 का केवल एक शून्यक होगा क्योंकि बहुपद का आलेख x-अक्ष पर केवल एक बिन्दु पर काटता है।

(ii) द्विघात बहुपद : किसी द्विघात समीकरण ax² + bx + c, a ≠ 0 के लिए ग्राफ दो में से किसी एक आकार का हो सकता है या तो ऊपर की ओर खुला (∪ की तरह) अथवा नीचे की ओर खुला (∩ की तरह) का होता है जो कि a > 0 या a < 0 पर निर्भर करता है।
इन वनों को परवलय (Parabola) कहते हैं।
बहुपद ax² + bx + c, जहाँ a ≠ 0 के शून्यक ठीक-ठीक उन बिन्दुओं के x-निर्देशांक होते हैं जहाँ बहुपद y = ax² + bx + c को निरूपित करने वाला ग्राफ (परवलय) x-अक्ष को प्रतिच्छेद करता है।
y = ax² + bx + c, के ग्राफ के आकार (परवलय) की तीन स्थितियाँ सम्भव हो सकती हैं।

स्थिति (I) : जब बहुपद ax² + bx + c के दो अलग-अलग गुणनखण्ड हैं:
इस स्थिति में ax² + bx + c का ग्राफ x-अक्ष को दो भिन्न-भिन्न बिन्दुओं A और A’ पर प्रतिच्छेद करता है तो इन बिन्दुओं के x निर्देशांक बहुपद ax² + bx + c के दो शून्यक होते हैं।
परवलय y = ax² + bx + c के शीर्ष के निर्देशांक \(\left(\frac{-b}{2 a}, \frac{-D}{4 a}\right)\) हैं, जहाँ D = b² – 4ac है।

किसी बहुपद के शून्यकों और गुणांकों में सम्बन्ध :
व्यापक रूप में, यदि द्विघात बहुपद p(x) = ax² + bx + c, जहाँ a ≠ 0 के शून्यक α और β हों, तो हम जानते हैं कि (x – α) और (x – β), p(x) के गुणनखण्ड होते हैं।
अतः ax² + bx + c = k(x – α) (x – β),
जहाँ k एक अचर है
= k[x² – (α + β)x + αβ]
ax² + bx + c = kx² – k(α + β)x + kαβ
दोनों पक्षों से x², x के गुणांकों तथा अचर पदों की तुलना करने पर,
a = k, b = – k (a + B) और c = kαβ
इससे प्राप्त होता है, α + β = \(\frac{-b}{a}\), αβ = \(\frac{c}{a}\)

व्यापक रूप में, यदि त्रिघात बहुपद ax³ + bx² + cx + d के शून्यक α, β और γ हों, तो हम जानते हैं कि (x – α), (x – β) और (x – γ) बहुपद के गुणनखण्ड होते हैं। अतः
ax³ + bx² + cx + d = k(x – α) (x – β) (x – γ), जहाँ k एक अचर है।
= k[{x² – (α + β)x + αβ} (x – γ)]
= k[x³ – (α + β + γ)x² + (αβ + βγ + γα)x – αβγ]
= kx³ – k(α + β + γ)x² + k (αβ + βγ + γλ) x – kλβγ
दोनों पक्षों से x³, x², x के गुणांकों और अचर पदों की तुलना करने पर,
a = k, b = -k(α + β + γ)
तथा c = k(αβ + βγ + γλ), d = -kαβγ
इससे प्राप्त होता है :

बहुपदों के लिए यूक्लिड की विभाजन एल्गोरिथ्म :
माना कि p(x) और g(x) कोई दो बहुपद हैं, जहाँ g(x) ≠ 0 हो तो बहुपद q(x) और r(x) ऐसे प्राप्त किए जा सकते हैं कि
p(x) = g(x) × q(x) + r(x)
जहाँ r(x) = 0 है अथवा r(x) की घात < g(x) की घात है।
उक्त परिणाम बहुपदों के लिए विभाजन एल्गोरिथ्म कहलाता है।

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 1 वास्तविक संख्याएँ Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 1 वास्तविक संख्याएँ

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
सिद्ध कीजिए कि प्रत्येक तीन क्रमागत धनात्मक पूर्णांकों में से एक 3 से विभाज्य है।
हल :
माना कि n, n + 1, n + 2 तीन क्रमागत धनात्मक पूर्णांक हैं।
∵ हम जानते हैं कि n, 3q या 3q + 1 या 3q + 2 के रूप का होता है।
∴ निम्नलिखित तीन स्थितियाँ सम्भव हैं।
स्थिति 1. जब n = 3q, जो कि 3 से विभाज्य है।
n + 1 = 3q + 1, 3 से विभाज्य नहीं है।
n + 2 = 3q + 2, 3 से विभाज्य नहीं है।
इस स्थिति में n, 3 से विभाज्य है परन्तु n + 1 और n + 2, 3 से विभाज्य नहीं हैं।
स्थिति 2. जब n = 3q + 1
इस स्थिति में n + 2 = 3q + 1 + 2 = 3(q + 1) जो कि 3 से विभाज्य है परन्तु तथा n + 1, 3 से विभाज्य नहीं है।
स्थिति 3. जब n = 3q + 2
इस स्थिति में n + 1 = 3q + 1 + 2 = 3(q + 1), 3 से
विभाज्य है परन्तु n या n + 2, 3 से विभाज्य नहीं है।
अत: n, n + 1 तथा n + 2 में से एक 3 से विभाज्य है।

प्रश्न 2.
सिद्ध कीजिए कि \(\sqrt{3}\) एक अपरिमेय संख्या
हल :
माना कि \(\sqrt{3}\) एक परिमेय संख्या है।
हम ऐसी सह अभाज्य (Co-prime) संख्याएँ a और b ज्ञात करते हैं कि
\(\sqrt{3}\) = \(\frac{a}{b}\) [जहाँ b ≠ 0]
दोनों पक्षों का वर्ग करने पर
3 = \(\frac{a^2}{b^2}\)
⇒ a² = 3b²
अतः 3, a² को विभाजित करता हैं
⇒ 3, a को विभाजित करेगा।
माना कि a = 3c (जहाँ c कोई पूर्णांक है)
⇒ a² = 9c²
⇒ 3b² = 9c² [∵ a² = 3b²]
⇒ b² = 3c²
अत: 3, b² को विभाजित करता है।
⇒ 3, b को विभाजित करेगा।
अतः a और b में कम से कम एक उभयनिष्ठ गुणनखण्ड 3 है।
परन्तु यह इस तथ्य का विरोध करता है कि a और b में 1 के अतिरिक्त कोई उभयनिष्ठ गुणनखण्ड नहीं है। अतः हमारी परिकल्पना गलत है।
अतः \(\sqrt{3}\) एक अपरिमेय संख्या है। इति सिद्धम् ।

प्रश्न 3.
संख्याओं 180, 72 व 252 का HCF और LCM ज्ञात कीजिए।
हल :
संख्याओं के गुणनखण्ड करने पर

प्रश्न 4.
अभाज्य गुणनखण्ड विधि द्वारा पूर्णांक 375 और 675 का HCF ज्ञात कीजिए।
हल :
अभाज्य गुणनखण्ड विधि द्वारा

375 = 3 × 5 × 5 × 5 = 3 × 5³
675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5²
अंतः म. स. 3 × 5² = 3 × 25 = 75

प्रश्न 5.
सिद्ध कीजिए \(\sqrt{6}\) एक अपरिमेय संख्या है।
हल :
माना \(\sqrt{6}\) एक परिमेय संख्या है।
∴ हम दो पूर्णांक r और s ऐसे ज्ञात कर सकते हैं कि हो तथा s ≠ 0 हो। यहाँ और सह अभाज्य हैं, अर्थात् इनमें 1 के अतिरिक्त कोई उभयनिष्ठ गुणनखण्ड नहीं है।
अब \(\sqrt{6}\) = \(\frac{r}{s}\)
दोनों पक्षों का वर्ग करने पर
6 = \(\frac{r^2}{s^2}\)
⇒ 6s² = r²
⇒ s² = \(\frac{r^2}{6}\)
⇒ 6, r² को विभाजित करता है।
⇒ इसलिए 6, r को भी विभाजित करता है।
⇒ 6, r का एक गुणनखण्ड है। ………..(A)
माना r = 6m
⇒ 6s² = (6m)²
⇒ 6s² = 36m²
⇒ s² = 6m²
⇒ m² = \(\frac{s^2}{6}\)
⇒ 6, s² को विभाजित करता है।
⇒ इसलिए 6, s को भी विभाजित करता है।
⇒ 6, s का एक गुणनखण्ड है। …(B)
समीकरण (A) और (B) अवलोकन करने पर हम पाते हैं कि r और s में 1 के अतिरिक्त एक अन्य उभयनिष्ठ गुणनखण्ड 6 है। परन्तु इससे इस तथ्य का विरोधाभास प्राप्त होता है कि r और s में 1 के अतिरिक्त कोई उभयनिष्ठ गुणनखण्ड नहीं है। इसका तात्पर्य है कि हमने एक त्रुटिपूर्ण कल्पना की थी।
अत: \(\sqrt{6}\) एक अपरिमेय संख्या है।

प्रश्न 6.
दो पूर्णांक संख्याओं का HCF व LCM क्रमशः 12 और 36 हैं, यदि एक पूर्णांक 48 है, तो दूसरा पूर्णांक ज्ञात कीजिए। (मा. शि. बो. राज. 2017 )
हल :
दिया है, HCF = 12, LCM = 336, एक पूर्णांक संख्या = 48
∴ HCF × LCM = पहली संख्या × दूसरी संख्या
⇒ 12 × 36 = 48 × दूसरी संख्या
अतः दूसरी संख्या = \(\frac{12 \times 36}{48}\) = 9

प्रश्न 7.
किसी परेड़ में 612 सदस्यों वाली एक सेना की टुकड़ी को 48 सदस्यों वाले एक बैंड के पीछे मार्च करना है। दोनों समूहों को समान संख्या वाले स्तम्भों में मार्च करना है। उन स्तम्भों की अधिकतम संख्या, जिसमें वे मार्च कर सकते हैं, क्या है?
हल :
स्तम्भों में अधिकतम संख्या ज्ञात करने के लिए में 612 और 48 का म.स. ज्ञात करना है।

∴ म. स. (612, 48) = 12
अतः अधिकतम स्तम्भों की संख्या 12 है।

प्रश्न 8.
वह सबसे छोटी संख्या लिखिए जो 306 तथा 657 से पूर्णतया विभाजित हो ।
हल :
सबसे छोटी संख्या जो 306 और 657 दोनों से पूर्णतया विभाजित होगी वह है :
ल. स. (657, 306)
657 = 3 × 3 × 73
306 = 3 × 3 × 2 × 17
अत: ल. स. (657, 306) = 3 × 3 × 73 × 2 × 17
= 22338

प्रश्न 9.
\(\sqrt{2}\) तथा \(\sqrt{3}\) के बीच में स्थित एक परिमेय संख्या ज्ञात कीजिए।
हल:
\(\sqrt{2}\) = 1.414
\(\sqrt{3}\) = 1.732
\(\sqrt{2}\) तथा \(\sqrt{3}\) के बीच परिमेय संख्या है,
= 1.5 (∵ 1.414 < 1.5 < 1.732)

प्रश्न 10.
दर्शाइए कि प्रत्येक विषम धनपूर्णांक (4q + 1) अथवा (4q + 3) के रूप का होता है, वहाँ q कोई पूर्णांक है।
हल :
मान लीजिए कि ‘a’ एक विषम धन पूर्णांक है तथा b = 4 है।
a तथा b में विभाजन एल्गोरिथ्म का प्रयोग करने पर,
a = bq + r, r < b
∴ a = 4q + r
∵ 0 ≤ r ≤ 4 इसलिए संभावित शेषफल 0, 1, 2, 3 हैं।
∵ a = 4q, 4q + 1, 4q + 2, 4q + 3; जहाँ q भागफल है।
∵ a एक विषम धन पूर्णांक 4q + 1, 4q + 3 के प्रकार का होगा।

प्रश्न 11.
दो धनपूर्णांक a तथा b जहाँ a = x³y² तथा b = xy³ के रूप में लिखे जा सकते हैं, जहाँ x, y अभाज्य संख्याएँ हैं, तो ल.स. (LCM) (a, b) का मान ज्ञात कीजिए ।
हल :
दिया है, a = x²y², b = xy³, और
म. स. (HCF) = xy²
∵ म. स. × ल. स. = a × b
xy² × ल. स. = x³y² × xy³

प्रश्न 12.
छोटी से छोटी अभाज्य संख्या तथा छोटी-से-छोटी भाज्य संख्या का म.स. (HCF) क्या है?
हल :
छोटी से छोटी अभाज्य संख्या = 2 तथा छोटी से
भाज्य संख्या = 4

अत: म. स. (HCF) = 2

प्रश्न 13.
404 तथा 96 का म. स. (HCF) तथा ल. स. (LCM) ज्ञात कीजिए तथा निम्न का सत्यापन कीजिए :
HCF × LCM = दोनों दी गई संख्याओं का गुणनफल
इल :
404 व 96 के अभाज्य गुणनखण्ड :

404 = 2 × 2 × 101 × 1
96 = 2 × 2 × 2 × 2 × 3 × 1
म. स. = 2 × 2 × 1 = 4
ल. स. = 2 × 2 × 101 × 2 × 2 × 2 × 3
= 9696
सत्यापन:
∵ HCF × LCM – दोनों दी गई संख्याओं का गुणनफल
⇒ 4 × 9696 = 404 × 96
⇒ 38784 = 38784
अतः दिया कथन सत्यापित हुआ। इति सिद्धम्

रिक्त स्थानों की पूर्ति कीजिए-

प्रश्न (क)

1. यूक्लिड विभाजन प्रमेयिका पूर्णांकों की ……………. के संबंध में है।
2. वह सबसे बड़ा धनात्मक पूर्णांक जो दिए हुए दो धनात्मक पूर्णांकों को पूर्णत: विभाजित कर देता है, संख्याओं का ………………… समापवर्तक कहलाता है।
3. सबसे छोटी भाज्य सम संख्या ………………… तथा सबसे छोटी भाग्य विषम संख्या …………………. है।
4. जिन संख्याओं का दशमलव प्रसार विभाजन प्रक्रिया के परिमित चरणों के बाद समाप्त हो जाता है ……………… दशमलव कहलाती है।
5. ऐसा दशमलव प्रसार जिसका अंत नहीं होता तथा भागफल में अंकों का एक पुनरावृत्ति खंड प्राप्त होता, …………… प्रसार कहलाता है।

हल :

1. विभाज्यता,
2. महत्तम,
3. 4, 9,
4. सांत,
5. असांत आवर्ती

निम्न में सत्य / असत्य कथन बताइए :

प्रश्न (ख)

1. प्रमेयिका एक सत्य कथन है, जिसका प्रयोग अन्य कथनों को सिद्ध करने के लिए किया जाता है।
2. संख्या 2 सबसे छोटी अभाज्य संख्या है।
3. दो या दो से अधिक संख्याओं का ल.स. वह छोटी से ‘छोटी संख्या होती है, जो दी गई संख्याओं से पूर्णतया विभाजित होती है।
4. असांत अनावर्ती दशमलव प्रसार वाली संख्याओं को परिमेय संख्या कहते हैं।
5. यदि किसी परिमेय संख्या को किसी अन्य शून्येत्तर परिमेय संख्या से भाग दिया जाए, तो प्राप्त संख्या हमेशा परिमेय होगी।

हल :

1. सत्य
2. सत्य
3. सत्य
4. असत्य
5. सत्य

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
दिया है, HCF (156, 78) = 78, तो LCM (156, 78) का मान है :
(A) 156
(B) 78
(C) 156 × 78
(D) 156 × 2
हल :
∵ HCF × LCM = संख्याओं का गुणनफल
⇒ 78 × LCM = 156 × 78
⇒ LCM = \(\frac{156 \times 78}{78}\) = 156
अत: सही विकल्प (A) है।

प्रश्न 2.
\(\frac{1095}{1168}\) का सरलतम रूप है :
(A) \(\frac{17}{26}\)
(B) \(\frac{25}{26}\)
(C) \(\frac{13}{26}\)
(D) \(\frac{15}{16}\)
हल :
1095 और 1168 का म. स. = 73

∴ सही विकल्प (D) है।

प्रश्न 3.
निम्न में से कौन-सी परिमेय संख्या को सांत दशमलव के रूप में व्यक्त किया जा सकता है?
(A) \(\frac{124}{165}\)
(B) \(\frac{131}{30}\)
(C) \(\frac{2027}{625}\)
(D) \(\frac{1625}{462}\)
हल :
\(\frac{124}{165}\) = \(\frac{124}{3^1 \times 5^1 \times 11^1}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में नहीं है।
\(\frac{131}{30}\) = \(\frac{131}{2^1 \times 3^1 \times 5^1}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में है।
\(\frac{2027}{625}\) = \(\frac{2027}{2^0 \times 5^4}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में नहीं है।
\(\frac{1625}{462}\) = \(\frac{1625}{2^1 \times 3^1 \times 7^1 \times 11^1}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में नहीं है।
अत: सही विकल्प (C) है।

प्रश्न 4.
वह बड़ी से बड़ी संख्या, जिससे 245 तथा 1029 को भाग देने पर क्रमशः 5 एवं शेष बचे है:
(A) 15
(B) 60
(C) 9
(D) 5
हल :
∵ प्रत्येक स्थिती में 5 शेष बचता है।
∴ 245 – 5 = 240 तथा 1029 – 9 = 1020
अभीष्ट संख्या 240 और 1020 का म. स. है ।

∴ म.स. = 5
अत: सही विकल्प (D) है।

प्रश्न 5.
225 को निम्न रूप में व्यक्त किया जा सकता है:
(A) 5 × 3².
(B) 5² × 3
(C) 5² × 3²
(D) 5 × 3
हल :
अभाज्य गुणनखण्ड विधि से

∴ 225 = 3 × 3 × 5 × 5 = 3² × 5²
अत: सही विकल्प (c) है।

प्रश्न 6.
\(2 . \overline{35}\) है एक :
(A) पूर्णांक
(B) परिमेय संख्या
(C) अपरिमेय संख्या
(D) प्राकृत संख्या
हल :
\(2 . \overline{35}\) का प्रसार असांत आवर्ती है।
∴ यह एक परिमेय संख्या है।
अतः विकल्प (B) सही है।

प्रश्न 7.
144 तथा 198 का महत्तम समापवर्तक है:
(A) 9
(B) 18
(C) 6
(D) 12
हल :

∴ म. स. = 18
अत: विकल्प (B) सही है।

प्रश्न 8.
दो संख्याओं का म. स. 27 है तथा उनका ल.स. 162 है। यदि एक संख्या 54 है, तो दूसरी संख्या है :
(A) 36
(B) 35
(C) 9
(D) 81
हल :
∵ म. स. × ल. स. = पहली संख्या × दूसरी संख्या
⇒ 27 × 162 = 54 × दूसरी संख्या
⇒ दूसरी संख्या = \(\frac{27 \times 162}{54}\) = 81
अत: सही विकल्प (D) है।

प्रश्न 9.
2\(\sqrt{3}\) एक :
(A) पूर्णांक है
(B) परिमेय संख्या है
(C) अपरिमेय संख्या है
(D) एक पूर्ण संख्या है.
हल :
एक परिमेय संख्या (2) और अपरिमेय संख्या (\(\sqrt{3}\)) का गुणनफल एक अपरिमेय संख्या होती है। अत: सही विकल्प (c) है।

प्रश्न 10.
संख्या \(\frac{3}{2^5 \times 5^2}\) का दशमलव प्रसार दशमलव के कितने स्थानों के बाद सांत होगा ?
(A) 2
(B) 4
(C) 5
(D) 1
हल :
यहाँ हर = 2⁵ × 5²
∴ दशमलव प्रसार 5 दशमलव स्थानों के बाद सांत होगा।
अतः सही विकल्प (c) है।

प्रश्न 11.
एक अभाज्य संख्या के कुल गुणनखण्डों की संख्या है :
(A) 1
(B) 0
(C) 2
(D) 3
हल :
दो 1 और स्वयं वही संख्या।
अत: सही विकल्प (c) है।

प्रश्न 12.
12, 21, 15 का म. स. और ल. स. है:
(A) 3, 140
(B) 12,420
(C) 3, 420
(D) 420, 3
हल :
यहाँ,
12 = 2 × 2 × 3
21 = 3 × 7
15 = 3 × 5
∴ म. स. = 3
ल. स. = 22 × 3 × 5 × 7
= 420
अत: सही विकल्प (C) है।

प्रश्न 13.
संख्या 196 के अभाज्य गुणनखण्ड में अभाज्य गुणनखण्डों की घातों का योग है :
(A) 3
(B) 4
(C) 5
(D) 2
हल :

196 = 2 × 2 × 7 × 7= 2² × 7²
घातों का योग = 2 + 2 = 4
अत: सही विकल्प (B) है।

प्रश्न 14.
यूक्लिड विभाजन प्रमेविका के अनुसार दो धनात्मक पूर्णांकों a और b के लिए ऐसी अद्वितीय पूर्ण संख्याएँ q और r विद्यमान है कि a = bq + r है तथा
(A) 0 < r < b
(B) 0 < r < b
(C) 0 ≤ r < b
(D) 0 ≤ r ≤ b
हल :
सही विकल्प (C) है।

प्रश्न 15.
संख्या n² – 1, 8 से विभाज्य होती है, यदि n है एक:
(A) पूर्णांक
(B) प्राकृत संख्या
(C) विषम संख्या
(D) सम संख्या
हल :
सही विकल्प (C) है।

प्रश्न 16.
यदि 65 और 117 के H.C.F. को 65m – 117 रूप में व्यक्त किया जा सके, तो m का मान है :
(A) 4
(B) 2
(C) 1
(D) 3

हल :
65 = 5 × 13
117 = 3 × 3 × 13
H.C.F. = 13
⇒ 65m – 117 = 13
[∵ H.C.F. = 65m – 117]
⇒ 65m = 13 + 117
⇒ 65m = 130
⇒ m = \(\frac{130}{65}\) = 2
अत: सही विकल्प (B) है।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 रचनाएँ Ex 11.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 11 रचनाएँ Exercise 11.2

निम्नलिखित में से प्रत्येक के लिए रचना का औचित्य भी दीजिए:

प्रश्न 1.
6 सेमी त्रिज्या का एक वृत्त खींचिए । केन्द्र से 10 सेमी दूर एक बिन्दु से वृत्त पर स्पर्श रेखा -युग्म की रचना कीजिए और उनकी लम्बाइयाँ मापकर लिखिए।
हल :
दिया है : 6 सेमी त्रिज्या का एक वृत्त और उसके केन्द्र बिन्दु O से 10 सेमी की दूरी पर एक बिन्दु P है।

रचना के चरण :

1. सर्वप्रथम बिन्दु O को केन्द्र मानकर 6 सेमी त्रिज्या का एक वृत्त खींचा।
2. वृत्त के केन्द्र O से 10 सेमी की दूरी पर एक बिन्दु P लिया।
3. OP को मिलाया और OP की समद्विभाजक रेखा खींची जो OP को M बिन्दु पर प्रतिच्छेद करती है।
4. बिन्दु M को केन्द्र मानकर PM त्रिज्या का एक वृत्त खींचा जो O केन्द्र वाले वृत्त को T₁ तथा T₂ बिन्दुओं पर काटता है।
5. PT₁ और PT₂ को मिलाया, जो वृत्त की अभीष्ट स्पर्श रेखाएँ हैं।

औचित्य (उपपत्ति) : हम जानते हैं कि किसी बिन्दु पर स्पर्श रेखा, उस बिन्दु पर त्रिज्या पर लम्ब होती है।
∴ ∠OT₁P = ∠OT₂P = 90°
अब OT₁ और OT₂ को मिलाया, वृत्त OT₁PT₂ में OP व्यास है।
∴ ∠OT₁P अर्द्धवृत्त में बना कोण है।
∴ ∠OT₁P = 90°
इसी प्रकार ∠OT₂P = 90°
अत: PT₁ तथा PT₂ वृत्त की स्पर्श रेखाएँ हैं।
स्पर्श रेखा की लम्बाई नापने पर,
PT₁ = PT₂ = 8.0 सेमी

प्रश्न 2.
4 सेमी त्रिज्या के एक वृत्त पर 6 सेमी त्रिज्या के एक संकेन्द्रीय वृत्त के किसी बिन्दु से एक स्पर्श रेखा की रचना कीजिए और उसकी लम्बाई मापिए परिकलन से इस माप की जाँच भी कीजिए।
हल :

दिया है: 4 सेमी त्रिज्या का एक वृत्त और 6 सेमी त्रिज्या का एक संकेन्द्रीय वृत्त जिस पर एक बिन्दु P दिया है।
रचना के चरण :

1. 4 सेमी त्रिज्या लेकर केन्द्र O वाला एक वृत्त खींचा।
2. केन्द्र O से 6 सेमी त्रिज्या पर एक संकेन्द्रीय वृत्त खींचा और इस पर एक बिन्दु P लिया।
3.  रेखाखण्ड OP खींचा और इसका लम्ब समद्विभाजक खींचा जो OP को बिन्दु M पर काटता है।
4. बिन्दु M को केन्द्र मानकर MP त्रिज्या का एक वृत्त खींचा जो केन्द्र O के 4 सेमी त्रिज्या वाले वृत्त को T₁ और T₂ बिन्दुओं पर काटता है।
5. PT₁ और PT₂ को मिलाया जो वृत्त की अभीष्ट स्पर्श रेखाएँ है।

औचित्य (उपपत्ति) : ∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∴ ∠OT₁P = ∠OT₂P = 90°
OT₁ तथा OT₂ को मिलाया, OP वृत्त का व्यास है ।
∠OT₁P तथा ∠OT₂P अर्द्धवृत्त के कोण हैं।
∴ ∠OT₁P = 90° तथा ∠OT₂P = 90°
∴ OT₁ ⊥ PT₁ तथा OT₂ ⊥ PT₂
अत: PT₁ तथा PT₂ अभीष्ट स्पर्श रेखाएँ हैं।
स्पर्श रेखा की लम्बाई मापने पर,
PT₁ = PT₂ = 4.7 लगभग ।

प्रश्न 3.
3 सेमी त्रिज्या का एक वृत्त खींचिए। इसके किसी बढ़ाए गए व्यास पर केन्द्र से 7 सेमी की दूरी पर स्थित दो बिन्दु और 2 लीजिए। इन दोनों बिन्दुओं से वृत्त पर स्पर्श रेखाएं खींचिए ।
हल :
दिया है : 3 सेमी त्रिज्या का एक वृत्त है जिसका केन्द्र O है। AOB वृत का व्यास है जिसको इस प्रकार बिन्दुओं P व Q तक बढ़ाया गया है कि वृत्त के केन्द्र O से प्रत्येक बिन्दु P व Q की दूरियाँ OP व OQ, 7 सेमी हैं।
रचना के चरण :

1. O केन्द्र वाला 3 सेमी त्रिज्या का एक वृत्त खींचा।
2. वृत्त का व्यास AB खींचकर इसे दोनों ओर क्रमशः P व Q तक इस प्रकार बढ़ाया कि OP = OQ = 7 सेमी ।
3. OP और OQ के लम्ब समद्विभाजिक खींचे जो OP को M₁ तथा OQ को M₂ पर काटते हैं।
4. बिन्दु M₁ को केन्द्र मानकर M₁O त्रिज्या का वृत्त खींचा जो O केन्द्र वाले वृत्त को T₁ व T₂ पर काटता है।
5. PT₁ और PT₂ को मिलाया ।
6. बिन्दु M₂ को केन्द्र मानकर M₂O त्रिज्या का वृत्त को खींचा जो O केन्द्र वाले वृत्त को S₁ व S₂ पर काटता है।
7. QS₁ और QS₂ को मिलाया।

अत: PT₁, PT₂, QS₁ और QS₂ अभीष्ट स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) : केन्द्र O वाले वृत्त की त्रिज्याएँ OT₁, OT₂, OS₁ व OS₂ खींचीं।
∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
अतः ∠OT₁P – ∠OT₂P = 90° तथा
∠OS₁Q = ∠OS₂Q = 90°
∵ केन्द्र M₁ वाले वृत्त में ∠OT₁P व ∠OT₂P अर्द्धवृत्तों में स्थित कोण है।
∴ ∠OT₁P व ∠OT₂P समकोण हैं जो क्रमश: त्रिज्याओं OT₁ OT₂ के सिरों T₁ व T₂ पर स्थित हैं।
∴ PT₁ व PT₂, केन्द्र O वाले वृत्त की स्पर्श रेखाएँ हैं। इसी प्रकार, QS₁ व QS₂ भी केन्द्र O वाले वृत्त की स्पर्श रेखाएँ हैं।

प्रश्न 4.
5 सेमी त्रिज्या के एक वृत्त पर ऐसी दो स्पर्श रेखाएँ खींचिए, जो परस्पर 60° के कोण पर झुकी हों।
हल :
दिया है एक वृत्त का केन्द्र O है तथा त्रिज्या 5 सेमी है।
रचना के चरण :
1. O को केन्द्र मानकर 5 सैमी त्रिज्या का एक वृत खींचा।
2. वृत्त का व्यास AB खींचा।
3. बिन्दु O पर OB त्रिज्या से 60° का कोण बनाती हुई एक OP रेखा खींची जो वृत्त को P बिन्दु पर काटती है।
4. बिन्दु A पर OA से 90° कोण बनाती हुई AX स्पर्श रेखा खींची।

5. इसी प्रकार बिन्दु P से स्पर्श रेखा PY खींची।
AX तथा PY एक दूसरे को T पर प्रतिच्छेद करती हैं।
अत: PT और AT वृत्त की दो अभीष्ट स्पर्श रेखाएँ हैं
जो एक-दूसरे के साथ 60° का कोण बनाती हैं।
औचित्य (उपपत्ति) : माना वृत्त का केन्द्र O है।
PT और AT वृत्त की दो स्पर्श रेखाएँ हैं जिनके बीच का कोण 60° है अर्थात्
∠PTA = 60°
परन्तु ∠POA + ∠PTA = 180° होता है।
∴ ∠POA = 180° – ∠PTA
= 180° – 60°
= 120°
∴ ∠POB = 180° – ∠POA
= 180° – 60°
= 120°
अत: PT तथा AT वृत्त की स्पर्श रेखाएँ हैं जो 60° के कोण पर झुकी हैं।

प्रश्न 5.
8 सेमी लम्बा एक रेखाखण्ड AB खींचिए । A को केन्द्र मानकर 4 सेमी त्रिज्या का एक वृत्त तथा B को केन्द्र मानकर 3 सेमी त्रिज्या का एक अन्य वृत्त खींचिए । प्रत्येक वृत्त पर दूसरे वृत्त के केन्द्र से स्पर्श रेखाओं की रचना कीजिए ।
हल :
दिया है रेखाखण्ड AB = 8.0 सेमी केन्द्र A से 4 सेमी त्रिज्या का एक वृत्त खींचा गया है तथा केन्द्र B से 3 सेमी त्रिज्या का एक अन्य वृत्त खींचा गया है।
रचना के चरण :

1. रेखाखण्ड AB = 8 सेमी खींचा।
2. केन्द्र A से 4 सेमी त्रिज्या का एक वृत्त खींचा और केन्द्र B से 3 सेमी त्रिज्या का एक वृत्त खींचा।
3. AB का लम्ब समद्विभाजक खींचा जो AB को M बिन्दु पर काटता है।
4. बिन्दु M को केन्द्र मानकर MB त्रिज्या का एक वृत्त खींचा जो A केन्द्र वाले वृत्त को S₁ व S₂ पर काटता है तथा B केन्द्र वाले वृत्त को T₁ व T₂ बिन्दुओं पर काटता है।
5. S₁B व S₂B और T₁A व T₂ को मिलाया।

अत: S₁B और S₂B केन्द्र A वाले वृत्त की स्पर्श रेखाएँ हैं तथा T₁A और T₂A केन्द्र B वाले वृत्त की स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) :
∵ चूँकि हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
अतः ∠AS₁B = ∠AS₂B = 90°
तथा ∠BT₁A = ∠BT₂A = 90°
∵ केन्द्र बिन्दु M वाले वृत्त का व्यास AB है।
∴ ∠AS₁B = ∠AS₂B, ∠BT₁A व ∠BT₂A अर्द्धवृत्त में बने कोण हैं। अत: प्रत्येक कोण समकोण है।
∴ AS₁ ⊥BS₁; AS₂ ⊥ BS₂ और BT₁ ⊥ AT₁ व BT₂ ⊥ AT₂
∴ S₁B व S₂B केन्द्र A वाले वृत्त की स्पर्श रेखाएँ हैं और AT₁ व AT₂ केन्द्र B वाले वृत्त की स्पर्श रेखाएँ हैं।

प्रश्न 6.
माना ABC एक समकोण त्रिभुज है, जिसमें AB = 6 सेमी, BC = 8 सेमी तथा ∠B = 90° है। B से AC पर BD लम्ब है। बिन्दुओं B, C व D से होकर जाने वाला एक वृत्त खींचा गया है। A से इस वृत्त पर स्पर्श रेखा की रचना कीजिए।
हल :
दिया है : एक समकोण त्रिभुज ABC जिसमें ∠B = 90°, AB = 6 सेमी तथा BC = 8 सेमी है। शीर्ष B से भुजा AC पर BD लम्ब खींचा गया है।
रचना के चरण :
1. सर्वप्रथम रेखाखण्ड BC = 8 सेमी खींचा।
2. बिन्दु B पर 90° कोण बनाते हुए, AB रेखाखण्ड 6 सेमी खींचा।

3. AC को मिलाया। इस प्रकार समकोण ΔABC प्राप्त हुआ।
4. बिन्दु B से AC पर BD लम्ब खींचा जो AC को D बिन्दु पर काटता है।
5. अब ΔBCD की भुजाओं BD और CD के लम्ब समद्विभा जक किए जो परस्पर O बिन्दु पर काटते हैं।
6. O को केन्द्र मानकर OB त्रिज्या का एक वृत्त खींचा जो बिन्दुओं B, C D से होकर गुजरता है।
7. चूँकि AB स्वयं स्पर्श रेखा है। इसलिए A को केन्द्र मानकर AB त्रिज्या का चाप खींचा जो वृत्त को P बिन्दु पर काटता है। AP को मिलाया।
अतः AP अभीष्ट स्पर्श रेखा है।
औचित्य (उपपत्ति) :
∵ ∠ABC = 90° (रचना से)
स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∴ ∠ABC = 90°
अत: AB रेखा स्पर्श रेखा है।
∵ AP = PB
अत: रेखा AB बिन्दु A पर खींची गई दूसरी स्पर्श रेखा है।

प्रश्न 7.
किसी चूड़ी की सहायता से वृत्त खींचिए। वृत्त के बाहर एक बिन्दु लीजिए। इस बिन्दु से वृत्त पर स्पर्श रेखाओं की रचना कीजिए।
हल :
किसी चूड़ी की सहायता से एक वृत्त खींचने का अर्थ है कि वृत्त का केन्द्र अज्ञात है। सर्वप्रथम हम केन्द्र ज्ञात करेंगे।
रचना के चरण :
1. चूड़ी की सहायता से वृत्त खींचा। इस वृत्त का केन्द्रबिन्दु ज्ञात करने के लिए कोई दो जीवाएँ AB और CD खाँचते हैं।

2. जीवा AB और CD के लम्ब समद्विभाजक किए जो परस्पर बिन्दु पर काटते हैं।
बिन्दु O वृत्त का केन्द्र होगा।
[∵ OA = OB = OC = OD]
(वृत की त्रिज्याएँ हैं)
3. वृत्त के बाहर कोई बिन्दु P लिया।
4. OP को मिलाकर लम्ब समद्विभाजक किया जो OP को बिन्दु M पर काटता है।
5. बिन्दु M को केन्द्र मानकर MP त्रिज्या का वृत्त खींचते हैं जो O केन्द्र वाले वृत्त को T₁ व T₂ पर काटता है।
6. PT₁ और PT₂ को मिलाया।
अत: PT₁ व PT₂ अभीष्ट स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) : OT₁ को मिलाया।
हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∵ ∠OT₁P = ∠OT₂P = 90°
∵ OP वृत्त का व्यास है, ∠OT₁P तथा ∠OT₂P अर्द्धवृत्त में बने कोण हैं।
∴ ∠OT₁P = 90° तथा ∠OT₂P = 90°
अत: PT₁ और PT₂ वृत्त की स्पर्श रेखाएँ हैं।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically:
1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
2. 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
1. Let the number of boys be x and the number of girls be y.
Then, the equations formed as follows:
x + y = 10 ……. (1)
and y = x + 4,
i.e., y – x = 4 …… (2)
To draw the graphs of these equations,
we find two solutions for each equation.
For equation (1), x + y = 10 gives y = 10 – x.

x	0	5
y	10	5

For equation (2), y – x = 4 gives y = x + 4.

x	0	2
y	4	6

Two lines intersect at point (3, 7). Hence, x = 3 and y = 7 is the required solution of the pair of linear equations.
Thus, 3 boys and 7 girls took part in the quiz.
Verification : x = 3 and y = 7 satisfy both the equations x + y = 10 and y – x = 4.

2. Let the cost of each pencil be ₹ x and the cost of each pen be ₹ y.
Then, from the given information, we receive the following equations:
5x + 7y = 50 ……… (1)
7x + 5y = 46 ……….. (2)
To draw the graphs of these equations, we find two solutions for each equation.
For equation (1), 5x + 7y = 50 gives
y = \(\frac{50-5 x}{7}\)

x	3	10
y	5	0

For equation (2), 7x + 5y = 46 gives
y = \(\frac{46-7 x}{5}\)

x	3	8
y	5	-2

Two lines intersect at point (3, 5). Hence, x = 3 and y = 5 is the required solution of the pair of linear equations.
Thus, the cost of each pencil is ₹ 3 and the cost of each pen is ₹ 5.
Verification : x = 3 and y = 5 satisfy both the equations 5x + 7y = 50 and 7x + 5y = 46.

Question 2.
On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
1. 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
3. 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
5x – 4y + 8 = 0; 7x + 6y – 9 = 0
For the given pair of linear equations,
a₁ = 5, b₁ = -4, c₁ = 8, a₂ = 7, b₂ = 6 and c₂ = -9
Now, \(\frac{a_1}{a_2}=\frac{5}{7}, \quad \frac{b_1}{b_2}=\frac{-4}{6}=-\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{8}{-9}=-\frac{8}{9}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the lines representing the given pair of linear equations intersect at a point.

2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
For the given pair of linear equations, a₁ = 9, b₁ = 3, c₁ = 12, a₂ = 18, b₂ = 16 and c₂ = 24.
Now, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the lines representing the given pair of linear equations are coincident lines.

3. 6x – 3y + 10 = 0; 2x – y + 9 = 0
For the given pair of linear equations, a₁ = 6, b₁ = -3, c₁ = 10, a₂ = 2, b₂ = -1 and c₂ = 9.
Now, \(\frac{a_1}{a_2}=\frac{6}{2}=3, \quad \frac{b_1}{b_2}=\frac{-3}{-1}=3\)
and \(\frac{c_1}{c_2}=\frac{10}{9}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the lines representing the given pair of linear equations are parallel lines.

Question 3.
On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent or inconsistent:
1. 3x + 2y = 5; 2x – 3y = 7
2. 2x – 3y = 8; 4x – 6y = 9
3. \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
4. 5x – 3y = 11; -10x + 6y = -22
5. \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
Solution:
1. 3x + 2y = 5; 2x – 3y = 7
For the given pair of linear equations, a₁ = 3, b₁ = 2, c₁ = -5, a₂ = 2, b₂ = -3 and c₂ = -7.
Now, \(\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{-3}=-\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.

2. 2x – 3y = 8: 4x – 6y = 9
For the given pair of linear equations,
a₁ = 2, b₁ = -3, c₁ = -8, a₂ = 4, b₂ =-6 and c₂ = -9.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

3. \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
Multiplying the first equation by 6 and expressing both the equations in the standard form, we get following equations:
9x + 10y – 42 = 0; 9x – 10y – 14 = 0
For the given pair of linear equations, a₁ = 9, b₁ = 10, c₁ = -42, a₂ = 9, b₂ = -10 and c₂ = -14.
Now, \(\frac{a_1}{a_2}=\frac{9}{9}=1, \frac{b_1}{b_2}=\frac{10}{-10}=-1\)
and \(\frac{c_1}{c_2}=\frac{-42}{-14}=3\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.

4. 5x – 3y = 11; -10x + 6y = -22
For the given pair of linear equations, a₁ = 5, b₁ =-3, c₁ = -11, a₂ = -10, b₂ = 6 and c₂ = 22.
Now, \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-11}{22}=-\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.

5. \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
For the given pair of linear equations, a₁ = \(\frac{4}{3}\), b₁ = 2, c₁ = -8, a₂ = 2, b₂ = 3 and c₂ = -12.
Now, \(\frac{a_1}{a_2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}, \quad \frac{b_1}{b_2}=\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
1. x + y = 5; 2x + 2y = 10
2. x – y = 8; 3x – 3y = 16
3. 2x + y – 6 = 0; 4x – 2y – 4 = 0
4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
1. x + y = 5; 2x + 2y = 10
For the given pair of linear equations,
a₁ = 1, b₁ = 1, c₁ = -5, a₂ = 2, b₂ = 2 and c₂ = -10.
Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.
Now, we draw the graphs of both the equations.
x + y = 5 gives y = 5 – x.

x	0	5
y	5	0

2x + 2y = 10 gives y = \(\frac{10-2 x}{2}\)

x	1	3
y	4	2

Here, lines representing both the equations. are coincident. Hence, any point on the line gives a solution. In general, y = 5 – x, where x is any real number is a solution of the given pair of linear equations.

2. x – y = 8; 3x – 3y = 16
For the given pair of linear equations, a₁ = 1, b₁ = -1, c₁ = -8, a₂ = 3, b₂ = -3 and c₂ = -16.
Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

3. 2x + y – 6 = 0; 4x – 2y – 4 = 0
For the given pair of linear equations, a₁ = 2, b₁ = 1, c₁ = -6, a₂ = 4, b₂ = -2 and c₂ = -4.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.
Now, we draw the graphs of both the equations.
2x + y – 6 = 0 gives y = 6 – 2x.

x	0	3
y	6	0

4x – 2y – 4 = 0 gives y = \(\frac{4 x-4}{2}\) = 2x – 2.

x	0	1
y	-2	0

Here, the lines intersect at point (2, 2). Hence, x = 2 and y = 2 is the unique solution of the given pair of linear equations.

4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
For the given pair of linear equations, a₁ = 2, b₁ = -2, c₁ = -2, a₂ = 4, b₂ = -4 and c₂ = -5.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the length and breadth of the rectangular garden be x m and y m respectively.
Then, from the given data, x = y + 4 and 36 = \(\frac{1}{2}\)[2(x + y)] i.e., x + y = 36 as the perimeter of a rectangle = 2 (length + breadth).
To draw the graphs, we find two solutions of each equation.
x = y + 4 gives y = x – 4

x	8	24
y	4	20

x + y = 36 gives y = 36 – x

x	12	24
y	24	12

Here, the lines intersect at point (20, 16). Hence, x = 20 and y = 16 is the unique solution of the pair of linear equations.
Thus, for the rectangular garden, length = 20 m and breadth = 16 m.

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
1. intersecting lines
2. parallel lines
3. coincident lines
Solution:
1. For intersecting lines, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 3x + 4y – 24 = 0. Here, \(\frac{a_1}{a_2}=\frac{2}{3}\) and \(\frac{b_1}{b_2}=\frac{3}{4}\) satisfying \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

2. For parallel lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 6x + 9y – 10 = 0.
Here, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{1}{3}\) and \(\frac{c_1}{c_2}=\frac{4}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).

3. For coincident lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0.
We can give another equation as 10x + 15y – 40 = 0. Here, \(\frac{a_1}{a_2}=\frac{1}{5}\), \(\frac{b_1}{b_2}=\frac{1}{5}\) and \(\frac{c_1}{c_2}=\frac{1}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Question 7.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
x – y + 1 = 0 gives y = x + 1

x	-1	2
y	0	3

3x + 2y – 12 = 0 gives y = \(\frac{12-3 x}{2}\)

x	0	4
y	6	0

The vertices of the triangle formed by the given lines and the x-axis are (-1, 0), (4, 0) and (2, 3).

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:
1. x + y = 5 and 2x – 3y = 4
2. 3x + 4y = 10 and 2x – 2y = 2
3. 3x – 5y – 4 = 0 and 9x = 2y + 7
4. \(\frac{x}{2}+\frac{2 y}{3}=-1\) and x – \(\frac{y}{3}\) = 3
Solution:
1. Elimination method:
x + y = 5 ………(1)
2x – 3y = 4 ……..(2)
We multiply equation (1) by 3 and equation (2) by 1 to get following equations:
3x + 3y = 15 ……(3)
2x – 3y = 4 ……(4)
Adding equations (3) and (4), we get
(3x + 3y) + (2x – 3y) = 15 + 4
∴ 5x = 19
∴ x = \(\frac{19}{5}\)
Substituting x = \(\frac{19}{5}\) in equation (1), we get
\(\frac{19}{5}\) + y = 5
∴ y = 5 – \(\frac{19}{5}\)
∴ y = \(\frac{6}{5}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{19}{5}\), y = \(\frac{6}{5}\)
Substitution method:
x + y = 5 ……(1)
2x – 3y = 4 ……(2)
From equation (1), we get y = 5 – x.
Substituting y = 5 – x in equation (2), we get
2x – 3(5 – x) = 4
∴ 2x – 15 + 3x = 4
∴ 5x = 19
∴ x = \(\frac{19}{5}\)
Substituting x = \(\frac{19}{5}\) in y = 5 – x, we get
y = 5 – \(\frac{19}{5}\)
∴ y = \(\frac{6}{5}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{19}{5}\), y = \(\frac{6}{5}\)

2. Elimination method:
3x + 4y = 10 ……(1)
2x – 2y = 2 ……(2)
We multiply equation (1) by 1 and equation (2) by 2 to get following equations:
3x + 4y = 10 ……(3)
4x – 4y = 4 ……(4)
Adding equations (3) and (4), we get
(3x + 4y) + (4x – 4y) = 10 +4
∴ 7x = 14
∴ x = 2
Substituting x = 2 in equation (1), we get
3(2) + 4y = 10
∴ 4y = 10 – 6
∴ 4y = 4
∴ y = 1
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

Substitution method:
3x + 4y = 10 ……(1)
2x – 2y = 2 ……(2)
From equation (2), we get 2x = 2y + 2.
i.e., x = y + 1.
Substituting x = y + 1 in equation (1), we get
3(y + 1) + 4y = 10
∴ 3y + 3 + 4y = 10
∴ 7y = 7
∴ y = 1
Substituting y = 1 in x = y + 1, we get
x = 1 + 1
x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

3. Elimination method:
3x – 5y – 4 = 0 ……(1)
9x = 2y + 7 ……(2)
i.e., 3x – 5y = 4 ……(3)
9x – 2y = 7 ……(4)
We multiply equation (3) by 3 and equation (4) by 1 to get following equations:
9x – 15y = 12 ……(5)
9x – 2y = 7 ……(6)
Subtracting equation (5) from equation (6),
we get
(9x – 2y) – (9x – 15y) = 7 – 12
∴ 9x – 2y – 9x + 15y = -5
∴ 13y = -5
∴ y = –\(\frac{5}{13}\)
Substituting y = –\(\frac{5}{13}\) in equation (5), we get
9x – 15(-\(\frac{5}{13}\)) = 12
∴ 9x + \(\frac{75}{13}\) = 12
∴ 9x = 12 – \(\frac{75}{13}\)
∴ 9x = \(\frac{75}{13}\)
∴ x = \(\frac{9}{13}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\)

Substitution method:
3x – 5y – 4 = 0 ………(1)
9x = 2y + 7 ………(2)
From equation (2), we get x = \(\frac{2 y+7}{9}\).
Substituting x = \(\frac{2 y+7}{9}\) in equation (1).
we get
3(\(\frac{2 y+7}{9}\)) – 5y – 4 = 0
∴ \(\frac{2 y+7}{9}\) – 5y – 4 = 0
∴ 2y + 7 – 15y – 12 = 0 (Multiplying by 3)
∴ -13y – 5 = 0
∴ -13y = 5
∴ y = –\(\frac{5}{13}\)
Substituting y = –\(\frac{5}{13}\) in x = \(\frac{2 y+7}{9}\)
we get,

Thus, the solution of the given pair of linear equations is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\).

4. Elimination method:
\(\frac{x}{2}+\frac{2 y}{3}=-1\) ………(1)
\(x-\frac{y}{3}=3\) ………(2)
We multiply equation (1) by 3 and equation (2) by 6 to get following equations:
\(\frac{3}{2}\)x + 2y = -3 ………(3)
6x – 2y = 18 ………(4)
Adding equations (3) and (4), we get
(\(\frac{3}{2}\)x + 2y) + (6x – 2y) = -3 + 18
∴ \(\frac{15}{2}\)x = 15
∴ x = 2
Substituting x = 2 in equation (3), we get
\(\frac{3}{2}\)(2) + 2y = -3
∴ 3 + 2y = -3
∴ 2y = -6
∴ y = -3
Thus, the solution of the given pair of linear equations is x = 2, y = -3.

Substitution method:
\(\frac{x}{2}+\frac{2 y}{3}=-1\) ………(1)
\(x-\frac{y}{3}=3\) ………(2)
From equation (2), we get x = \(\frac{y}{3}+3\)
Substituting x = \(\frac{y}{3}+3\) in equation (1).
we get

Substituting y = -3 in x = \(\frac{y}{3}+3\), we get
x = \(\frac{-3}{3}+3\)
∴ x = -1 + 3
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = -3.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:

Question 1.
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
Solution:
Let the numerator of the required fraction be x and the denominator be y.
Then, the required fraction is \(\frac{x}{y}\)
From the first condition given, we get
\(\frac{x+1}{y-1}=1\)
∴ x + 1 = y – 1
∴ x – y = -2 …..(1)
From the second condition, we get
\(\frac{x}{y+1}=\frac{1}{2}\)
∴ 2x = y + 1
∴ 2x – y = 1 ………(2)
Now, subtracting equation (1) from equation (2), we get
(2x – y) – (x – y) = 1 – (-2)
∴ 2x – y – x + y = 3
∴ x = 3
Substituting x = 3 in equation (1), we get
3 – y = -2
∴ -y = – 2 – 3
∴ -y = -5
∴ y = 5
So, the fraction \(\frac{x}{y}=\frac{3}{5}\)
Thus, the pair of linear equations formed is x – y = -2 and 2x – y = 1; and the required fraction is \(\frac{3}{5}\)

Question 2.
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
Solution:
Let the present age of Nuri be x years. and the present age of Sonu be y years.
Five years ago, the age of Nuri was (x – 5) years and the age of Sonu was (y – 5) years.
Then, from the first condition given, we get
(x – 5) = 3(y – 5)
∴ x – 5 = 3y – 15
∴ x – 3y = -10 ……….(1)
Ten years later the ages of Nuri and Sonu will be (x + 10) years and (y + 10) years respectively.
Then, from the second condition given, we get
(x + 10) = 2(y + 10)
∴ x + 10 = 2y + 20
∴ x – 2y = 10 ………..(2)
Subtracting equation (1) from equation (2).
we get
(x – 2y) – (x – 3y) = 10 – (-10)
∴ x – 2y – x + 3y = 10 + 10
∴ y = 20
Substituting y = 20 in equation (2).
we get
x – 2 (20) = 10
∴ x – 40 = 10
∴ x = 50
So, the present ages of Nuri and Sonu are 50 years and 10 years.
Thus, the pair of linear equations formed is x – 3y = -10 and x – 2y = 10 and the present ages of Nuri and Sonu are 50 years and 20 years respectively.

Question 3.
The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let the digit at tens place be x and the digit at units place be y in the original number.
Then, the original number = 10x + y.
From the first condition given, we get x + y = 9 ………..(1)
If the digits are reversed, in the new number the digit at tens place is y and the digit at units place is x.
Then, the new number = 10y + x.
From the second condition given, we get
9(10x + y) = 2(10y + x)
∴ 90x + 9y = 20y + 2x
∴ 88x – 11y = 0
∴ 8x – y = 0 (Dividing by 11) ……….(2)
Adding equations (1) and (2), we get
(x + y) + (8x − y) = 9 + 0
∴ 9x = 9
∴ x = 1
Substituting x = 1 in equation (1), we get
1 + y = 9
∴ y = 8
So, the original number = 10x + y
= 10(1) + 8 = 18
Thus, the pair of linear equations formed is x + y = 9 and 8x – y = 0; and the original number is 18.

Question 4.
Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
Solution:
Suppose Meena received x notes of ₹ 50 and y notes of ₹ 100.
So, the total amount received by her = ₹ (50x + 100y)
From the first condition given, the total amount is ₹ 2000. So, we get
50x + 100y = 2000
∴ x + 2y = 40 (Dividing by 50) ……(1)
From the second condition given, we get
x + y = 25 ……(2)
Subtracting equation (2) from equation (1),
we get
(x + 2y) – (x + y) = 40 – 25
∴ y = 15
Substituting y = 15 in equation (2), we get
x + 15 = 25
∴ x = 10
Hence, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
Thus, the pair of linear equations formed is x + 2y = 40 and x + y = 25, and Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Question 5.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
Let the fixed charge for first three days be ₹ x and the additional charge for each day exceeding the first three days be ₹ y. Saritha kept the book for 7 days.
So, she has to pay the fixed charge plus the additional charge for 4(7 – 3) days. Hence, we get the following equation for Saritha:
x + 4y = 27 …………(1)
Similarly, Susy has to pay the fixed charge plus the addition charge for 2 (5 – 3) days.
Hence, we get the following equation for Susy:
x + 2y = 21 ……….(2)
Subtracting equation (2) from equation (1),
we get
(x + 4y) – (x + 2y) = 27 – 21
∴ 2y = 6
∴ y = 3
Substituting y = 3 in equation (1), we get
x + 4(3) = 27
∴ x + 12 = 27
∴ x = 15
Hence, the fixed charge for first three days is ₹ 15 and the addition charge for each day thereafter is ₹ 3.
Thus, the pair of linear equations formed is x + 4y = 27 and x + 2y = 21; and the fixed charge and the additional charge per day are ₹ 15 and ₹ 3 respectively.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1.
Divide the polynomial p (x) by the polynomial g(x) and find the quotient and remainder in each of the following:
1. p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
2. p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
3. p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Solution:
1.

Quotient x – 3, Remainder = 7x – 9

2. p(x) = x⁴ – 3x² + 4x + 5
= x⁴ + 0x³ – 3x² + 4x + 5
and g(x) = x² + 1 – x = x² – x + 1

Quotient = x² + x – 3, Remainder = 8

3. p(x) = x⁴ – 5x + 6
= x⁴ + 0x³ + 0x² – 5x + 6
and g(x) = 2 – x² = -x² + 2

Quotient = -x² – 2, Remainder = – 5x + 10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
1. t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
2. x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
3. x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution:
1.

As the remainder is 0, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

2.

As the remainder is 0, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

3.

As the remainder is not 0, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)
Solution:
Since \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are given two zeroes of the polynomial, \(\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=x^2-\frac{5}{3}\) is a factor of the given polynomial. Then, to obtain the other zeroes of the polynomial, we divide it by x² – \(\frac{5}{3}\)

Now,
3x² + 6x + 3 = 3(x² + 2x + 1)
= 3(x + 1)²
= 3(x + 1)(x + 1)
Hence, the two zeroes of 3x² + 6x + 3 are 1 and – 1.
Hence, the two zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5 other than the given zeroes are 1 and -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
Here, dividend p(x) = x³ – 3x² + x + 2, quotient q(x) = (x – 2) and remainder r(x) = (-2x + 4).
Now, p(x) = g(x) × q(x) + r(x)
∴ x³ – 3x² + x + 2 = g(x) × (x – 2) + (-2x + 4)
∴ (x³ – 3x² + x + 2) – (-2x + 4) = g(x) × (x – 2)
∴ x³ – 3x² + 3x – 2 = g(x) × (x – 2)
∴ g(x) = \(\frac{x^3-3 x^2+3 x-2}{x-2}\)

Hence, g(x) = x² – x + 1.

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
1. deg p(x) = deg q(x)
2. deg q(x) = deg r(x)
3. deg r(x) = 0.
Solution:
1. deg p (x) = deg q (x) implies that deg g (x) = 0. i.e., g(x) is a non-zero constant. One such example can be given as p (x) = 3x² + 15x + 13, g(x) = 3, q(x) = x² + 5x + 4 and r(x) = 1, which satisfies the division algorithm as:
3x² + 15x + 13 = 3(x² + 5x + 4) + 1.

2. deg q(x) = deg r (x) implies that deg g (x) > deg q(x), because
deg g (x) > degr (x). One such example can be given as p (x) = x³ + 5x² + 2x + 7, g(x) = x² + 1, q(x) = x + 5 and r(x) = x + 2, which satisfies the division algorithm as:
x³ + 5x² + 2x + 7 = (x² + 1)(x + 5) + (x + 2).

3. deg r(x) = 0 implies that the remainder is a constant. One such example can be given as p (x) = x³ + 4x² + 5x + 9, g(x) = x + 3, q(x) = x² + x + 2 and r(x) = 3, which satisfies the division algorithm as:
x³ + 4x² + 5x + 9 = (x + 3)(x² + x + 2) + 3.
Note: There can be several examples in each of the above cases.