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NCERT Solutions for Class 9th: Ch 1 The French Revolution History

The NCERT Solutions For Class 9th History (Social Science) are given below. Students Should also check NCERT Solutions Class 9 for other subjects to.

NCERT Book Solutions For Class 9 History Chapter 1

Page No: 24

Questions

1. Describe the circumstances leading to the outbreak of revolutionary protest in France.

Answer

The circumstances leading to the outbreak of revolutionary protest in France were:

→ Social Inequality: French society in the eighteenth century was divided into three estates namely The Clergy, The nobility and third estates. First two estates, that is, the clergy and the nobility enjoyed certain privileges by birth. They were exempt from paying taxes. The Third estate comprises of businessmen, merchants, Peasants and artisans, labours had to pay taxes to the state.

→ Political Causes: Long years of war had drained the financial resources of France. France had a debt of more than 2 billion livres. To meet its regular expenses, such as the cost of maintaining an army, the court, running government offices or universities, the state was forced to increase taxes which angered the people.

→ Economic Problems: The population of France also increased from 23 million in 1715 to 28 million in 1789. Food grains were now in great demand. The price of bread shot up. Wages did not keep pace with rising prices. This led to subsistence crisis.

→ Strong Middle Class: A new middle class emerged educated and wealthy during the eighteenth century. They believed that no group in society should be given privileges by birth. Ideas of equality and freedom were put forward by philosophers. The ideas of these philosophers were discussed intensively in salons and coffee houses and spread among people.

→ Immediate Causes: On 5 may, 1789, Louis XVI called together an assembly of Estates General to pass proposals for new taxes. Third estates protested against this proposal but as each estate have one vote, the king rejected this appeal. They walked out of the assembly.

2. Which groups of French society benefited from the revolution? Which groups were forced to relinquish power? Which sections of society would have been disappointed with the outcome of the revolution?

Answer

It was the richer members of the third estate who mostly benefited from the French Revolution.
The clergy and the nobility were forced to relinquish (surrender) their power.
The poor class of third estate and women would have been disappointed with the outcome of the revolution as the promise of equality, discussed during the revolution was not given. The poorer classes had no right to vote.

3. Describe the legacy of the French Revolution for the peoples of the world during the nineteenth and twentieth centuries.

Answer
The ideas of liberty and democratic rights were the most important legacy of the French Revolution. These spread from France to the rest of Europe during the nineteenth century, where feudal systems were abolished. It inspired the Germans, Italians, and Austrians to overthrow their oppressive regimes. The French Revolution inspired the struggling nations of Asia and Africa who were groaning under the oppression of European colonialism. Tipu Sultan and Rajaram Mohan Roy are two examples of individuals who responded to ideas coming from French revolution.

4. Draw up a list of democratic rights we enjoy today whose origins could be traced to the French Revolution.

Answer

We can trace the origin of the following democratic rights we enjoy today to the French revolution:
→ Right to Equality before law
→ Freedom of Speech and expression
→ Right against exploitation
→ Right to justice

5. Would you agree with the view that the message of universal rights was beset with contradictions? Explain.

Answer

Yes, the message of universal rights was beset with contradictions:

→ Law is the expression of the general will. All citizens have the right to participate in its formation, personally or through their representatives. – In this line, it is stated that every citizen has the right to participate in the law however only men above 25 years of age who paid taxes equal to at least 3 days of a labourer’s wage were given voting right. The remaining men and all women were classed as passive citizens and were deprived of voting rights.

Hence, the message of universal rights was not very clear. The Constitution is only available for the rich. Women were totally neglected in decision making.

6. How would you explain the rise of Napoleon?

Answer

After Robespierre’s rule came to an end a directory was formed to avoid concentration of power in one individual. Members of the directory often fought among themselves leading to total chaos and political instability. This created a political vacuum in France. This was a conducive situation and Napoleon Bonaparte took the reign of power as a military dictator.

Napoleon saw his role as a moderniser of Europe. He introduced many laws such as the protection of private property and a uniform system of weights and measures provided by the decimal system.

Important Links

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Most Important Questions – Ch 1 The French Revolution 

Important MCQs – Ch 1 The French Revolution 

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NCERT SOLUTIONS FOR CLASS 9 | Ch 15 Improvement in Food Resources Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 15 Improvement in Food Resources Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 15 Improvement in Food Resources Science

In text Questions

Page No: 204

1. What do we get from cereals, pulses, fruits and vegetables?

Answer

→ Cereals provide us with carbohydrates. Also, they are a rich source of energy.
→ Pulses give us proteins.
→ Fruits and vegetables are a rich source of vitamins and minerals. A small amount of proteins, carbohydrates, and fats are also present in them.

Page No: 205

1. How do biotic and abiotic factors affect crop production?

Answer

Biotic factors such as pests, insects and diseases reduce the crop production. A pest causes damage to our crops by feeding. Weeds also reduce crop productivity by competing with the main crop for nutrients and light.
Similarly, abiotic factors such as temperature, wind, rain etc. affect the net crop production. For example droughts and floods have a great impact on crops sometimes, destroying the entire crop.
 


2. What are the desirable agronomic characteristics for crop improvements?

Answer

The desirable agronomic characteristics for crop improvements are:
→ Tallness and profuse branching in any fodder crop.
→ Dwarfness in cereals.
 
Page No: 206

1. What are macro-nutrients and why are they called macro-nutrients?

Answer

Macro-nutrients are those nutrients which are required in large quantities for growth and development of plants. Since they are required in large quantities, they are known as macro-nutrient. The six macro-nutrients required by plants are nitrogen, phosphorus, potassium, calcium, magnesium, and sulphur.

2. How do plants get nutrients?

Answer

Plants get nutrients from air, water, and soil. Soil is the major source of nutrients. Thirteen of these nutrients are available from soil. The remaining three nutrients (carbon, oxygen, and hydrogen) are obtained from air and water.
 
Page No: 207

1. Compare the use of manure and fertilizers in maintaining soil fertility.
 
Answer

Manures increase soil fertility by enriching the soil with organic matter and nutrients as it is prepared by the decomposition of animal excreta and plant wastes. On the other hand, fertilizers are mostly inorganic compounds whose excessive use is harmful to the symbiotic micro-organisms living in soil. Their excessive use also reduces soil fertility. Hence, fertilizers are considered good for only short term use.
 
Page No: 208
 
1. Which of the following conditions will give the most benefits? Why?
(a) Farmers use high-quality seeds, do not adopt irrigation or use fertilizers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilizer.
(c) Farmers use quality seeds, adopt irrigation, use fertilizer and use crop protection measures.
 
Answer
 
(c)Farmers using good quality seeds, adopting irrigation, using fertilizers, and using crop protection measures will derive most benefits.
→ The use of good quality seeds increases the total crop production. If a farmer is using good quality seeds, then a majority of the seeds will germinate properly, and will grow into a healthy plant.
 

→ Proper irrigation methods improve the water availability to crops.
→ Fertilizers ensure healthy growth and development in plants by providing the essential nutrients such as nitrogen, phosphorus, potassium, etc.
→ Crop protection measures include various methods to control weeds, pests, and infectious agents. If all these necessary measures are taken by a farmer, then the overall production of crops will increase.

Page No: 209

1. Why should preventive measures and biological control methods be preferred for protecting crops?

Answer

Preventive measures and biological control methods should be preferred for protecting crops because excessive use of chemicals leads to environmental problems. Biological methods cause harm neither to crop nor to the environment.

2. What factors may be responsible for losses of grains during storage?

Answer

Factors causing loss of grains during storage:
→ Biotic Factors: Insects, rodents, fungi and bacteria etc.
→ Abiotic Factors: Moisture Content and Temprature etc.

Page No: 210

1. Which method is commonly used for improving cattle breeds and why?

Answer

Cross Breeding is commonly used for improving cattle breeds.Cross breeding between two good varieties of cattle will produce a new improved variety. For example, the cross between foreign breeds such as Jersey Brown, Swiss and Indian breeds such as Red Sindhi, Sahiwal produces a new variety having qualities of both breeds.

Page No: 211

1. Discuss the implications of the following statement:
“It is interesting to note that poultry is India’s most efficient converter of low fibre food stuff (which is unfit for human consumption) into highly nutritious animal protein food.”

Answer

Poultry in India is the most efficient converter of low fibre food stuff into highly nutritious animal protein food. In poultry farming, domestic fowls are raised to produce eggs and chicken. For this, the fowls are given animal feeds in the form of roughage, which mainly consists of fibres. Thus, by feeding animals a fibre rich diet, the poultry gives highly nutritious food in the form of eggs and chicken.

1. What management practices are common in dairy and poultry farming?

Answer

Common management practices in dairy and poultry farming are:
→ Proper shelter facilities and their regular cleaning.
→ Some basic hygienic conditions such as clean water, nutritious food, etc.
→ Animals are kept in spacious, airy, and ventilated place.
→ Prevention and cure of diseases at the right time is ensured.

2. What are the differences between broilers and layers and in their management?

Answer
Layers are meant for egg production, whereas broilers are meant for poultry meat. Nutritional, environmental, and housing conditions required by broilers are different from those required by egg layers. A broiler chicken, for their proper growth, requires vitamin rich supplements especially vitamin A and K. Also, their diet includes protein rich food and enough fat. They also require extra care and maintenance to increase their survival rate in comparison to egg layers.

Page No: 213

1. How are fish obtained?

Answer

Fish can be obtained by two ways:
→ Capture fishing: It is the process of obtaining fish from natural resources.
→ Culture fishery: It is the practice of farming fishes. Farming can be done in both freshwater ecosystem (which includes river water, pond water) and marine ecosystem.

2. What are the desirable characters of bee varieties suitable for honey production?

Answer

Bee varieties having the following desirable characters are suitable for honey production:
→They should yield high quantity of honey.
→ They should not sting much.
→ They should stay in the beehive for long durations.
→ They should breed very well.

1. What are the advantages of composite fish culture?

Answer

The advantages of composite fish culture are: → Fish can be grown in crop fields especially paddy.
→ Intensive Fish farming is possible because plenty of water is available during crop seasons.
→ In this system both local and imported fish species can be cultivated.

2. What is pasturage and how is it related to honey production?

Answer

Pasturage is the availability of flowers from which bees collect nectar and pollen. It is related to the production of honey as it determines the taste and quantity of honey.

Excercise

1. Explain any one method of crop production which ensures high yield.

Answer

Inter cropping is a method of crop production which ensures high yield. It is a practice of growing two or more crops simultaneously. in the same field in rows. In inter cropping definite row patterns are followed such as one row of main crop of is followed by two row of intercrop.
In inter cropping there is greater utilisation of the interspaced area, light, nutrients, water and air. As a result productivity per unit area increased.

2. Why are manures and fertilizers used in fields?

Answer

Manures and fertilizers are used in fields to enrich the soil with the required nutrients. Manure helps in enriching the soil with organic matter and nutrients. This improves the fertility and structure of the soil. On the other hand, fertilizers ensure a healthy growth and development in plants. They are a good source of nitrogen, phosphorus, and potassium. To get an optimum yield, it is instructed to use a balanced combination of manures and fertilizers in the soil.

3. What are the advantages of inter-cropping and crop rotation?

Answer

Inter-cropping and crop rotation both are used to get maximum benefit on limited land. Inter-cropping helps in preventing pests and diseases to spread throughout the field. It also increases soil fertility, whereas crop rotation prevents soil depletion, increases soil fertility, and reduces soil erosion. Both these methods reduce the need for fertilizers. It also helps in controlling weeds and controls the growth of pathogens and pests in crops.

4. What is genetic manipulation? How is it useful in agricultural practices?

Answer

Genetic manipulationis a process where the gene for a particular character is introduced inside the chromosome of a cell. When the gene for a particular character is introduced in a plant cell, a transgenic plant is produced. These transgenic plants exhibit characters governed by the newly introduced gene.
Genetic manipulation is useful in developing varities with Higher yield, Good Quality, Biotic and Abiotic resistance, short maturity period, wider adaptability and desirable agronomic characteristics

5. How do storage grain losses occur?

Answer

There are various biotic and abiotic factors that act on stored grains and result in degradation, poor germinability, discolouration, etc.

Biotic factors include insects or pests that cause direct damage by feeding on seeds. They also deteriorate and contaminate the grain, making it unfit for further consumption.
Abiotic factors such as temperature, light, moisture, etc., also affect the seed. They decrease the germinating ability of the seeds and make them unfit for future use by farmers. Unpredictable occurrence of natural calamities such as droughts and floods also causes destruction of crops.
 

6. How do good animal husbandry practices benefit farmers?

Answer

Cattle farming is one of the methods of animal husbandry that is most beneficial for farmers. Using this method, better breeds of draught animals can be produced. Such draught animals are engaged in agricultural fields for labour work such as carting, irrigation, tilling, etc.

7. What are the benefits of cattle farming?

Answer

Benefits of cattle farming:
→ Good quality and quantity of milk can be produced.
→ Draught labour animals can be produced for agricultural work.
→ New variety that are resistant to diseases can be produced by crossing two varieties with the desired traits.

8. For increasing production, what is common in poultry, fisheries and bee-keeping?

Answer

The common factor for increasing production in poultry, fisheries, and bee keeping is the proper management techniques that are to be followed. Regular cleaning of farms is of utmost importance. Maintenance of temperature and prevention and cure of diseases is also required to increase the number of animals.

9. How do you differentiate between capture fishing, mariculture and aquaculture?

Answer

→ Capture fishing the method of obtaining fishes from natural resources

→ Mariculture is the culture of marine fishes for commercial use.

→ Aquaculture involves the production of aquatic animals that are of high economic value such as prawns, lobsters, fishes, crabs, etc.

Important Link

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Important MCQs- Ch 15 Improvement in Food Resources

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NCERT SOLUTIONS FOR CLASS 9 | Ch 14 Natural Resources Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 14 Natural Resources Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 14 Natural Resources Science

In text Questions

Page No: 193

1. How is our atmosphere different from the atmospheres on Venus and Mars?

Answer

Earth’s atmosphere is a mixture of nitrogen (79%), oxygen (20%), and a small fraction of carbon dioxide, water vapours and other gases. This makes the existence of life possible on Earth. However, the atmospheres on Venus and Mars mainly consist of carbon dioxide. The amount of carbon dioxide on these planets can range from 95% to 97%.

2. How does the atmosphere act as a blanket?

Answer

The atmosphere acts as a blanket by performing the following functions:
→ It keeps the average temperature of the Earth fairly constant during day time and even during the course of whole year.
→ It prevents a sudden increase in the temperature during day time.
→ It slows down the escape of heat from the surface of the Earth into outer space during night time.

3. What causes winds?

Answer

An uneven heating of the Earth’s surface causes winds. On being heated, air becomes lighter and rises up. As a result, a region of low pressure is created. Then, air from a high pressure region moves to a low pressure region, causing wind.

4. How are clouds formed?

Answer

During day time, on being heated, a large amount of water evaporates from various water bodies and goes into the air. A part of this water vapour also reaches the atmosphere through biological activities such as transpiration and respiration. This causes the air in the atmosphere to heat up. When this heated air rises, it expands and cools, which results in the condensation of water vapour forming water droplets. The presence of dust and other suspended particles in air also facilitates the process of condensation. The formation of water droplets leads to the formation of clouds.

5. List any three human activities that you think would lead to air pollution.

Answer

Three human activities leading to air pollution are:
→ Smoke from industries e.t.c.
→ Burning of fossil fuels like coal and petroleum→ Deforestation.

Page No: 194

1. Why do organisms need water?

Answer

The organisms need water:
→ For different cellular process.
→ For transportation of substance from one place to another inside the body.

2. What is the major source of fresh water in the city/town/village where you live?

Answer

Rivers.

3. Do you know of any activity which may be polluting this water source?

Answer

The discharge of waste water from homes, industries, hospitals, etc. into the river pollutes this fresh water source.

Page No: 196

1. How is soil formed?

Answer

Soil is formed by breaking down of rocks at or near the surface of the Earth through various physical, chemical, and biological processes by various factors such as the sun, water, wind, and living organisms.→ Sun: During day time, the rocks are heated up by solar rays. This causes the rocks to expand. During night time, these rocks cool down and contracts, thus the cracks devlop in the rock and they break down.
→ Water: It helps in breaking of rocks in two ways:
(i) It goes into the cracks and crevices formed in the rocks. When this water freezes, its volume increases. As a result, the size of the cracks also increases. This helps in the weathering of rocks.
(ii) Water moving in fast speed carries big and small particles of rock downstream. These rocks rub against each other, resulting in breaking down of rocks.These smaller particles are carried away by running water and deposited down its path.→ Wind: Strong winds carry away rocks, which causes rubbing of rocks. This results in the breaking down of rocks into smaller and smaller particles.
→ Living Organism: Some living organisms like lichens help in the formation of soil. Lichens grow on rock surfaces and converts them into powdery form and make soil layer. In the same way, the plants like moss also help in the making of fine soil particles.

2. What is soil erosion?

Answer

The blowing away or washing away of land surface bywind or water is known as soil erosion.

3. What are the methods of preventing or reducing soil erosion?

Answer

The methods of preventing or reducing soil erosion are:
→ Plantation of tress and plants→ Prevention of deforestation
→ Prevent excessive grazing

Page No: 201

1. What are the different states in which water is found during the water cycle?

Answer

Water is found in three different states during the water cycle:
→ Solid (Snow, Ice)
→ Liquid Water (ground water, river water, etc.)
→ Gaseous State (Water vapours)

2. Name two biologically important compounds that contain both oxygen and nitrogen.

Answer

Two biologically important compounds that contain both oxygen and nitrogen are:
(i) Amino acids
(ii) Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA)
3. List any three human activities which would lead to an increase in the carbon dioxide content of air.

Answer

Three Human activities are:
→ Burning of fuels in various processes like heating, cooking, transportation, and industry.
→ Human induced forest fires
→ The process of deforestation includes the cutting down of trees. This decreases the uptake of carbon dioxide for photosynthesis. Eventually, the content of carbon dioxide increases.

4. What is the greenhouse effect?

Answer

Some gases like carbon dioxide, methane, nitrous oxide prevent the escape of heat from the Earth’s surface by trapping it. This increases the average temperature of the Earth. This is called the green house effect.

5. What are the two forms of oxygen found in the atmosphere?

Answer

The two forms of oxygen found in the atmosphere are:
→ Diatomic molecular form with chemical formula O2.
→ Triatomic molecular form with chemical formula O3known as ozone.

Excercise

1. Why is the atmosphere essential for life?

Answer

The atmosphere is essential for life because it maintains an appropriate climate for the sustenance of life by carrying out the following activities:
→ Atmosphere keeps the average temperatureof the Earth fairly constant during day time.
→ It prevents a sudden increase in temperature during day time.
→ It also slows down the escape of heat from the surface of the Earth into outer space during night time.

2. Why is water essential for life?

Answer

Water is essential for life because of the following reasons:
→ Most biological reactions occur when substances are dissolved in water. Thus, all cellular processes need water as a medium to take place.
→ Transportation of biological substances needs water as a medium.

3. How are living organisms dependent on the soil? Are organisms that live in water totally independent of soil as a resource?

Answer

Almost all living organisms are dependent on soil. Some depend directly, while some depend indirectly.
Plants need soil for getting support as well as nutrients to prepare their food.
On the other hand, organisms depend on plants for food and other substances that are essential for life. Herbivores depend directly upon plants, and carnivores depend upon animals, which in turn depend upon plants for food. This makes them depend on soil indirectly.
Organisms that live in water are not totally independent of soil as a resource. These organisms depend on aquatic plants for food and other substances. These aquatic plants in turn require minerals for their sustenance. These minerals are carried to water bodies from soil by rivers, rain water, etc. Without the supply of minerals from the soil to the water bodies, it is impossible to imagine aquatic life.

4. You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?

Answer

The meteorological department of the government collects data on the elements of weather such as maximum and minimum temperatures, maximum and minimum humidity, rainfall, wind speed, etc. They are able to study these elements using various instruments. The maximum and minimum temperature of a day is measured by a thermometer known as the maximum-minimum thermometer. Rain fall is measured by an instrument known as the rain gauge. Wind speed is measured by anemometers. There are various instruments used to measure humidity.


Page No: 202

5. Write a note on how forests influence the quality of our air, soil and water resources.

Answer

Forests influence the quality of our air, soil, and water resources in various ways. Some of them are:
→ Forests balance the percentages of carbon dioxide and oxygen in the atmosphere. The increasing amount of carbon dioxide caused by human activities is balanced by a larger intake of carbon dioxide by plants during the process of photosynthesis. Simultaneously, a large amount of oxygen is released.
→ Forests prevent soil erosion. Roots of plants bind the soil tightly in a way that the surface of the soil cannot be eroded away by wind, water, etc.
→ Forests help in the replenishment of water resources. During the process of transpiration, a huge amount of water vapour goes into the air and condenses to form clouds. These clouds cause rainfall that recharge water bodies.

6. We know that many human activities lead to increasing levels of pollution of the air, water-bodies and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution?

Answer

Yes. Isolating human activities to specific areas would help in reducing levels of pollution. For example, setting up of industries in isolated regions will control pollution to some extent. The pollution caused by these industries will not contaminate water resources, agriculture land, fertile land, etc.

Important Link

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Most Important Questions- Ch 14 Natural Resources

Important MCQs- Ch 14 Natural Resources

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NCERT SOLUTIONS FOR CLASS 9 | Ch 13 Why Do We Fall ill Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 13 Why Do We Fall Ill Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 13 Why Do We Fall ill Science

In-text Questions

Page No: 178

1. State any two conditions essential for good health.

Answer

Two conditions that are essential for good health are:
→ Proper nutrition and a blanced diet
→ Good Social environment.
 
2. State any two conditions essential for being free of disease.

Answer

Two conditions essential for being disease-free are:
→ Person should take balance diet.
→ Personal and community hygiene.

3. Are the answers to the above questions necessarily the same or different? Why?
 
Answer

To some extent they are the same, because if the conditions that are essential for good health are maintained, then automatically the chances of getting a disease will be minimized. But at the same time, we can say that they are different because being health or good health means physical, mental and social well-being while being disease-free means not suffering from a particular disease.

Page No: 180

1. List any three reasons why you would think that you are sick and ought to see a doctor. If only one of these symptoms were present, would you still go to the doctor? Why or why not?

Answer

Common symptoms which indicate sickness are:
→ Headache
→ Cough
→ Dysentery

If only one of these symptoms is present, we usually do not visit a doctor. This is because such symptoms do not have much effect on our general health and ability to work. However, if a person is experiencing these symptoms for quite sometime, then he needs to visit a doctor for proper treatment.

2. In which of the following case do you think the long-term effects on your health are likely to be most unpleasant?
• If you get jaundice,
• if you get lice,
• If you get acne.

Answer

Jaundice is a disease that can cause long-term effects on our health. It is a chronic disease that lasts for a long period of time. Jaundice does not spread rapidly, but it develops slowly over a period of time.
 
Page No: 187
 
1. Why we are normally advised to take bland and nourishing food when we are sick?

Answer

When we are sick the normal body functions get disturbed. In such situation food that is easily digestible and contains adequate nutrients are required for the speedy recovery. Thus bland and nourishing food is given during sickness.
 
2. What are the different means by which infectious diseases are spread?

Answer

The different modes of transmission of infectious diseases are:
→ Through Air: Certain disease-causing micro-organisms are expelled in air by coughing, sneezing, talking, etc. These micro-organisms can travel through dust particles or water droplets in air to reach other people. For example, tuberculosis, pneumonia, etc. spread through air.
→ Through Water: Sometimes causal micro-organisms get mixed with drinking water and spread water borne diseases. Cholera for example is water borne disease.
→ Through Sexual Contact: Sexual act between two people can lead to the transfer of diseases such as syphilis, gonorrhoea, AIDS, etc.
→ Through Vectors: Certain diseases spread by animals called vectors. For example mosquitoes spread malaria.

3. What precautions can you take in your school to reduce the incidence of infectious diseases?

Answer

Precautions to reduce incidence of infectious diseases are:
→ Staying away from the infected person.
→ Covering mouth or nose while coughing or sneezing to prevent the spread of disease.
→ Drinking safe water.
→ Keeping the school environment clean to prevent multiplication vectors.

4. What is immunization?

Answer

Immunizationis defined as protection of the body from communicable diseases by administration of some agent that mimics the microbe.

5. What are the immunization programmes available at the nearest health centre in your locality? Which of these diseases are the major health problems in your area?

Answer

The immunization programmes available at the nearest health centre are DPT (Diphtheria, Pertusis, and Tetanus), polio vaccine, hepatitis B, MMR (Measles, Mumps, and Rubella), jaundice, typhoid, etc.
Of all these diseases, jaundice and typhoid are major health problems

Exercise

 

Question 1. How many times did you fall ill in the last one year? What were the illnesses?
(a) Think of one change you could make in your habits in order to avoid any of/ most of the above illnesses.
(b) Think of one change you would wish for in your surroundings in order to avoid any of/most of the above illness.


Answer: The illness was 2-3 times, common-cold, occurred in a year.
(a) One change I would make in my habits in order to avoid the above illness is that I would take proper diet rich in vitamin C and would avoid too cold food.
(b) The surroundings should be neat, and clean,

 

Question 2. A doctor/nurse/health worker is exposed to more sick people than others in the community. Find out how she/he avoids getting sick herself/himself?


Answer: A doctor/nurse/health worker when exposed to sick people they keep their nose and mouth covered, take care of hygiene, wash hands with soap before drinking water or eating food. They use mask, gloves, etc to avoid the direct contact with the person suffering from infectious diseases.

 

Question 3. Conduct a survey in your neighbourhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to . bring down the incidence of these diseases.


Answer:

Common-diseases In neighbourhood Steps to bring down the diseases spread
1. Malaria

 

2. Typhoid

3. Cough and cold

1. Clean surrounding

2. Clean drinking water

3. Childhood immunisation

 

Question 4. A baby is not able to tell bis/her.caretakers that she/he is sick. What would help us to find out
(a) that the baby is sick?
(a) what is the sickness?


Answer:
(a) The symptoms like body temperature, fever, cough, cold, loose-motions, non-stop crying improper or no food intake etc. would help up to find that the baby is sick.
(b) The symptoms could help us to find out the sickness of the body.

Question 5. Under which of the following conditions is a person most likely to fall sick?
(a) When she is recovering from malaria.
(b) When she has recovered from malaria and is taking care of someone suffering from chicken-pox.
(c) When she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken-pox.
Why?


Answer: (c) When she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken-pox.
As the person is not taking proper diet which is required for her proper health and healing of body.
Her chances of getting chicken-pox also high as her body’s immunity has lowered.

 

Question 6. Under which of the following conditions are you most likely to fall sick?
(a) When you are taking examinations.
(b) When you have travelled by bus and train for two days.
(c) When your friend is suffering from measles.
Why?


Answer: (c) When your friend is suffering from measles, as it is an infectious disease.

Important Link

Quick Revision Notes- Ch 13 Why Do We Fall ill

Most Important Questions- Ch 13 Why Do We Fall ill

Important MCQs- Ch 13 Why Do We Fall ill

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NCERT SOLUTIONS FOR CLASS 9 | Ch 12 Sound Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 12 Sound Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 12 Sound Science

In Text Questions

Page No: 162

1. How does the sound produced by a vibrating object in a medium reach your ear?

 
Answer
 

When an object vibrates, it sets the particles of the medium around it vibrating. The particles in the medium in contact with the vibrating object displace from its equilibrium position. It then exerts force on the adjacent particles. After displacing the adjacent particle the first particle of medium comes back in its original position. This process continues in the medium till the sound reaches your ear.

Page No: 163

1. Explain how sound is produced by your school bell.

Answer

When the bell continues to move forward and backward, it creates a series of compressions and rarefactions making production of sound.

2. Why are sound waves called mechanical waves?

Answer

Sound waves needs material medium to propagate therefore, they are called mechnical waves. Sound waves propagate through a medium because of theinteraction of the particles present in that medium.

3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Answer

No, because sound waves needs a medium through which they can propagate. Since there is no material medium on the moon due to absence of atmosphere, you cannot hear any sound on the moon.

Page No: 166

1. Which wave property determines (a) loudness, (b) pitch?

Answer

(a) Amplitude

(b) Frequency


2. Guess which sound has a higher pitch: guitar or car horn?

Answer

Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has high frequency than that of car horn. High the frequency higher is the pitch.

 
1. What are wavelength, frequency, time period and amplitude of a sound wave?

Answer

→ Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is known as the wavelength. Its SI unit is metre (m).

→ Frequency: The number of complete oscillations per second is known as the frequency of a sound wave. It is measured in hertz (Hz).

→ Amplitude: The maximum height reached by the crest or trough of a sound wave is called its amplitude.
 

2. How are the wavelength and frequency of a sound wave related to its speed?

Answer

Speed, wavelength, and frequency of a sound wave are related by the following equation:
Speed (v) = Wavelength (λ) x Frequency (ν)
v = λ x ν

3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.

Answer

Frequency of the sound wave, ν= 220 Hz
Speed of the sound wave, v = 440 m s-1
For a sound wave,
Speed = Wavelength x Frequencyv = λ x ν
∴ λ= v / ν = 440 / 220 = 2m
Hence, the wavelength of the sound wave is 2 m.

4. A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?

Answer

The time interval between two successive compressions is equal to the time period of the wave. This time period is reciprocal of the frequency of the wave and is given by the relation:T= 1 / Frequency = 1/ 500 = 0.002 s

1. Distinguish between loudness and intensity of sound.

Answer

Intensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. Loudness is a measure of the response of the ear to the sound. The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.

Page No: 167

1. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?

Answer

The speed of sound depends on the nature of the medium. Sound travels the fastest in solids. Its speed decreases in liquids and it is the slowest in gases. Therefore, for a given temperature, sound travels fastest in iron.

Page No: 168

1. An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m s−1?

Answer

Speed of sound, v = 342 m s−1
Echo returns in time, t = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.

 


Hence, the distance of the reflecting surface from the source= 1026 / 2 m = 513 m.

Page No: 169

1. Why are the ceilings of concert halls curved?

Answer

Ceilings of concert halls are curved so that sound after reflection (from the walls) spreads uniformly in all directions.

Page No: 170

1. What is the audible range of the average human ear?

Answer

The audible range of an average human ear lies between 20 Hz to 20,000 Hz.

2. What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?

Answer

(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

Page No: 172

1. A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?

 
Answer
 

Time taken by the sonar pulse to return, t = 1.02 s
Speed of sound in salt water, v = 1531 m s – 1
Distance of the cliff from the submarine = Speed of sound x Time taken
Distance of the cliff from the submarine = 1.02 x 1531 = 1561.62 m

 

Distance travelled by the sonar pulse during its transmission and reception in water = 2 x Actual distance = 2d

Actual Distance, d= Distance of the cliff from the submarine/2
= 1561/2
= 780.31 m

Excercise
 
1. What is sound and how is it produced?

Answer

Sound is a form of eneergy which gives the sensation of hearing. It is produced by the vibrations caused in air by vibrating objects.

2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

Answer

When a vibrating body moves forward, it createsa region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction. As the body continues to move forward and backwards, it produces a series of compressions and rarefactions. This is shown in below figure.

Series of compressions and rarefactions
 
3. Cite an experiment to show that sound needs a material medium for its propagation.
 
Answer
 
Take an electric bell and an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside the jar, and press the switch of the bell. You will be able to hear the bell ring. Now pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out.This shows that sound needs a material medium to travel.

Experiment Figure
 
4. Why is sound wave called a longitudinal wave?
 
Answer
 
Sound wave is called longitudinal wave because it is produced by compressions and rarefactions in the air. The air particles vibrates parallel to the direction of propagation.
 
5. Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer

The quality or timber of sound enables us to identify our friend by his voice.

6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Answer

The speed of sound (344 m/s) is less than the speed of light(3 x 108 m/s). Sound of thunder takes more time to reach the Earth as compared to light. Hence, a flash is seen before we hear a thunder.
7. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s−1.

Answer

For a sound wave,
Speed = Wavelength x Frequencyv = λ x ν
Speed of sound in air = 344 m/s (Given)
(i) For, ν= 20 Hz
λ1= v/ν = 344/20 = 17.2 m

(ii) For, ν= 20000 Hz
λ2= v/ν = 344/20000 = 0.172 m

Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

8. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Answer

Velocity of sound in air= 346 m/s
Velocity of sound wwave in aluminium= 6420 m/s
Let length of rode be 1

Time taken for sound wave in air, t1= 1 / Velocity in air
Time taken for sound wave in Aluminium, t2= 1 / Velocity in aluminium
 

Therefore, t1 / t2 = Velocity in aluminium / Velocity in air = 6420 / 346 = 18.55 : 1

9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Answer

Frequency = 100 Hz (given)
This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds = 100 x 60 = 6000 times.

10. Does sound follow the same laws of reflection as light does? Explain.

Answer

Sound follows the same laws of reflection as light does. The incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.

11. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?

Answer

An echo is heard when the time for the reflected sound is heard after 0.1 s
Time Taken= Total Distance / Velocity
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1 sec it will not be heard.

12. Give two practical applications of reflection of sound waves.

Answer

Two practical applications of reflection of sound waves are:
→ Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.
 

→ Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection of sound.

13. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s−2 and speed of sound = 340 m s−1.

Answer

Height of the tower, s = 500 m
Velocity of sound, v = 340 m s−1
Acceleration due to gravity, g = 10 m s−2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, t1
According to the second equation of motion:


Now, time taken by the sound to reach the top from the base of the tower, t2= 500 / 340 = 1.47 s
Therefore, the splash is heard at the top after time, t
Where, t= t1 + t2 = 10 + 1.47 = 11.47 s.

14. A sound wave travels at a speed of 339 m s−1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer

Speed of sound, v= 339 m s – 1
Wavelength of sound, λ= 1.5 cm = 0.015 m
Speed of sound = Wavelength x Frequencyv= λ x v
∴ v= v / λ = 339 / 0.015 = 22600 Hz
The frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not audible.
 
Page No: 175

15. What is reverberation? How can it be reduced?

Answer

The repeated multiple reflections of sound in any big enclosed space is known as reverberation.
The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.
 


16. What is loudness of sound? What factors does it depend on?

Answer

The effect produced in the brain by the sound of different frequencies is called loudness of sound.
Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

17. Explain how bats use ultrasound to catch a prey.

Answer

Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are reflected by objects such as preys and returned to the bat’s ear. This allows a bat to know the distance of his prey.

18. How is ultrasound used for cleaning?

Answer

Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

19. Explain the working and application of a sonar.

Answer

SONAR is an acronym for Sound Navigation And Ranging. It is an acoustic device used to measure the depth, direction, and speed of under-water objects such as submarines and ship wrecks with the help of ultrasounds. It is also used to measure the depth of seas and oceans.

Sonar Working
 
 
A beam of ultrasonic sound is produced and transmitted by the transducer (it is a device that produces ultrasonic sound) of the SONAR, which travels through sea water. The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the under-water object is calculated from the time (t) taken by the echo to return with speed (v) is given by 2d = v × t. This method of measuring distance is also known as ‘echo-ranging’.

20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.

Answer

Time taken to hear the echo, t= 5 s
Distance of the object from the submarine, d= 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water= 2d
Velocity of sound in water, v= 2d / t = 2 x 3625 / 5 = 1450 ms-1.

21. Explain how defects in a metal block can be detected using ultrasound.

Answer

Defects in metal blocks do not allow ultrasound to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.

Ultrasound Waves helpful in detecting metal defects
 
22. Explain how the human ear works.
 
Answer
 
The human ear consists of three parts – the outer ear, middle ear and inner ear. 
→ Outer ear: This is also called ‘pinna’. It collects the sound from the surrounding and directs it towards auditory canal.
→ Middle ear: The sound reaches the end of the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones- hammer, anvil and stirrup. 
→ Inner ear: These vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.

Working of human ear

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NCERT SOLUTIONS FOR CLASS 9 | Ch 11 Work and Energy Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 11 Work and Energy Science Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 11 Work and Energy Science

In-Text Questions

Page No: 148

1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Fig 11.3
Answer

We know that if force F acting on an object to displace it through a distance S in one direction, then the work done W on the body by the force is given by:

Work done = Force × Displacement
W = F × S
Where,
F = 7 N
S = 8 m
Therefore, work done, W = 7 × 8
= 56 Nm
= 56 J

Page No: 149

1. When do we say that work is done?

Answer

Work is done whenever the given conditions are satisfied:
→ A force acts on the body.
→ There is a displacement of the body caused by the applied force along the direction of the applied force.

2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Answer

When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression:
Work done = Force × Displacement
W = F × s

3. Define 1 J of work.

Answer

1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.

4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Answer

Work done by the bullocks is given by the expression:
Work done = Force x Displacement
W= F ×d
Where,
Applied force, F = 140 N
Displacement, d = 15 m
W= 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the field.

Page No: 152

1. What is the kinetic energy of an object?

Answer

The energy possessed by a body by the virtue of its motion is called kinetic energy. Every moving object possesses kinetic energy. A body uses kinetic energy to do work. Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of air is used to run wind mills, etc.

2. Write an expression for the kinetic energy of an object.

Answer

If a body of mass mis moving with a velocity v, then its kinetic energy Ekis given by the expression,
Ek= 1/2 mv2
Its SI unit is Joule (J).

3. The kinetic energy of an object of mass, m moving with a velocity of 5 m s−1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Answer

K.E. of the object= 25 J
Velocity of the object, v= 5 m/s
∵ K.E.= 1/2 mv2
⇒ m = 2 × K.E./v2
⇒ m = 2 × 25/25 = 2 kg

If velocity is double, v = 2 × 5= 10 m/s
∴ K.E. (for v = 10 m/s) = 1/2 mv2 = 1/2 × 2 × 100= 100 J

If velocity is tripled, v = 3 × 5 = 15 m/s
∴ K.E. (for v = 10 m/s) = 1/2 mv2 1/2 × 2 × 225 = 225 J

Page No: 156

1. What is power?

Answer

Power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,
Power= Work/Time = Energy/Time
P = W/T
It is expressed in watt (W).

 

2. Define 1 watt of power.

Answer

A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s, i.e.,1 W= 1J / 1s

3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Answer

Power = Work Done/Time
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
Power= 1000 / 10 = 100 Js-1=100 W

4. Define average power.

Answer

The average Power of an agent may be defined as the total work done by it in the total time taken. Average Power = Total Work Done/Total time taken

Excercise
 
1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.

Answer

Work is done whenever the given two conditions are satisfied:
→ A force acts on the body.
→ There is a displacement of the body by the application of force in or opposite to the direction of force.

(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Seema while swimming.

(b) While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since, displacement is perpendicular to force, the work done is zero.

(c) A wind mill works against the gravitational force to lift water. Hence, work is done by the wind mill in lifting water from the well.

(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.

(e) An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.

(f) Food grains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in the Sun.

(g)Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Answer

Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions/heights of the object, which is zero.
Work done by gravity is given by the expression,
W= mgh
Where,
h= Vertical displacement = 0
W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joule.

3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer

When a bulb is connected to a battery, then the chemical energy of the battery is transferred into electrical energy. When the bulb receives this electrical energy, then it converts it into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:
Chemical Energy → Electrical Energy → Light Energy + Heat Energy

4. Certain force acting on a 20 kg mass changes its velocity from 5 m s−1 to 2 m s−1. Calculate the work done by the force.

Answer

Kinetic energy is given by the expression,(Ek)v= 1/2 mv2
Where,
Ek= Kinetic energy of the object moving with a velocity, v
(i) Kinetic energy when the object was moving with a velocity 5 m s-1
(Ek)5= 1/2 × 20 × (5)2 = 250 J
Kinetic energy when the object was moving with a velocity 2 m s-1
(Ek)2= 1/2 × 20 × (2)2 = 40 J.

5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

Answer

Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression,
W= mgh
Where,
Vertical displacement, h = 0
∴W= mg x 0 = 0
Hence, the work done by gravity on the body is zero.

6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer
No. The process does not violate the law of conservation of energy. This is because when the body falls from a height, then its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equal to an increase in the kinetic energy of the body. During the process, total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is not violated.

7. What are the various energy transformations that occur when you are riding a bicycle?

Answer

While riding a bicycle, the muscular energy of the rider gets transferred into heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular Energy → Kinetic Energy + Heat Energy
During the transformation, the total energy remains conserved.
 
8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Answer

When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred into heat energy, which causes our body to become hot.

9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Answer

1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1 kWh
1 kWh = 3.6 × 106 J
Therefore, 250 units of energy = 250 x 3.6 × 106 = 9 × 108 J.

10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

Answer

Gravitational potential energy is given by the expression,
W = mgh
Where,
h = Vertical displacement = 5 m
m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 m s−2
∴ W = 40 × 5 × 9.8 = 1960 J.
At half-way down, the potential energy of the object will be 1960/2 = 980 J.
At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be 980 J.

11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Answer

Work is done whenever the given two conditions are satisfied:
→ A force acts on the body.
→ There is a displacement of the body by the application of force in or opposite to the direction of force.

If the direction of force is perpendicular to displacement, then the work done is zero.

When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, the work done on the satellite by the Earth is zero.

12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

Answer

Yes. For a uniformly moving object
Suppose an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

Page no: 159

13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

Answer

Work is done whenever the given two conditions are satisfied:
→ A force acts on the body.
→ There is a displacement of the body by the application of force in or opposite to the direction of force.

When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. Although, force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.

14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer

Energy consumed by an electric heater can be obtained with the help ofthe expression,
P= W / T
Where,
Power rating of the heater, P = 1500 W = 1.5 kW
Timefor which the heater has operated, T= 10 h
Work done = Energy consumed by the heater
Therefore, energy consumed = Power × Time
= 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.

15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer

The law of conservation of energy states that energy can be neither created nor destroyed. It can only be converted from one form to another.
Consider the case of an oscillating pendulum.
Oscilating Pendulum

When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

The bob does not oscillate forever. It comes to rest because air resistance resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after some time.

The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remain conserved.

16. An object of mass, mis moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

Answer

Kinetic energy of an object of mass, moving with a velocity, v is given by the expression,
Ek= 1/2 mv2
To bring the object to rest, 1/2 mv2 amount of work is required to be done on the object.

17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer

Kinetic energy, Ek = 1/2 mv2
Where,
Mass of car, m = 1500 kg
Velocity of car, v = 60 km/h = 60 × 5/18 ms-1
Hence, 20.8 x 104 J of work is required to stop the car.
 
18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Answer
 
Case I
In this case, the direction of force acting on the block is perpendicular to the displacement. Therefore, work done by force on the block will be zero.
 
Case II
In this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.

Case III
In this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done by force on the block will be negative.

19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Answer

Acceleration in an object could be zero even when several forces are acting on it. This happens when all the forces cancel out each other i.e., the net force acting on the object is zero. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.

Answer

Energy consumed by an electric device can be obtained with the help of the expression for power,

P = W/T
Where,
Power rating of the device, P = 500 W = 0.50 kW
Time for which the device runs, T = 10 h
Work done = Energy consumed by the device
Therefore, energy consumed = Power × Time
= 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 h will be 4 × 5 kWh = 20 kWh = 20 Units.

21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer

When an object falls freely towards the ground, its potential energy decreases and kinetic energy increases. As the object touches the ground, all its potential energy gets converted into kinetic energy. As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. It can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.

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NCERT SOLUTIONS FOR CLASS 9 | Ch 10 Gravitation Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 10 Gravitation Science Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 10 Gravitation Science

In Text Questions

Page No: 134

1. State the universal law of gravitation

Answer

The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Masses of two objects of = m1and m2
Distance between the two objects = d
Force of attraction acting between them = F
It will be given by the universal law of gravitation.
F = Gm1m2/d2
where, G is the universal constant.
G = 6.67×10-11 Nm2kg-2

2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

 
Answer
 

Mass of the Earth = ME
Mass of an object on its surface = m.
Radius of the Earth = R
According to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:
F = Gm1m2/r2
Thus, the formula will be, F = GMEm/R2

Page No: 136

1. What do you mean by free fall?

Answer

Gravity of earth attracts every object towards its center. When an object is dropped from a certain height , it begins to fall towards Earth’s surface under the influence of gravitational force. Such a motion of object is called free fall.

2. What do you mean by acceleration due to gravity?

Answer

When an object falls freely towards the surface of earth from a certain height, then its velocity changes. This change in velocity produces acceleration in the object which is known as acceleration due to gravity denoted bu letter g . The value of acceleration due to gravity is g= 9.8 m/s2.

Page No: 138

1. What are the differences between the mass of an object and its weight?

Answer

Mass
Weight
Mass is the quantity of matter contained in the body. Weight is the force of gravity acting on the body.
It is the measure of inertia of the body. It is the measure of gravity.
Mass is a constant quantity. Weight is not a constant quantity. It is different at different places.
It only has magnitude. It has magnitude as well as direction.
Its SI unit is kilogram (kg). Its SI unit is the same as the SI unit of force, i.e., Newton (N).


2.  Why is the weight of an object on the moon 1/6th its weight on the earth?

Answer

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitional attraction on the moon is about one sixth when compared to earth. Hence, the the weight of an object on the moon 1/6th its weight on the earth.

Page No: 141

1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Answer

It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.


2. What do you mean by buoyancy?

Answer

The upward force exerted by a liquid on an object that is immersed in it is known as buoyancy.

3. Why does an object float or sink when placed on the surface of water?


Answer

→ An object sink in water if its density is greater than that of water.
→ An object floats in water if its density is less than that of water.

Page No: 142

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Answer

When we weigh our body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

Answer

The cotton bag is heavier than the iron bar. The cotton bag experiences larger up thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

Excercises

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer

According to Universal Law of gravitation, the gravitational force of attraction between any two objects of mass M and m is proportional to the product of their masses and inversely proportional to the square of distance r between them
So, force F is given by

Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Answer

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m).

Answer

Given that,
Mass of the body, m= 1 kg
Mass of the Earth, M= 6×1024 kg
Radius of the earth, R= 6.4×106 m
Now, magnitude of the gravitational force (F) between the earth and the body can be calculated.


4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Answer

According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth.

  1. If the moon attracts the earth, why does the earth not move towards the moon?


Answer

The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon.

Page No: 144

6. What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?

Answer

From Universal law of gravitation, force exerted on an object of mass m by earth is given by
So as the mass of any one of the object is doubled the force is also doubled.

(ii) The force F is inversely proportional to the distance between the objects. So if the distance between two objects is doubled then the gravitational force of attraction between them is reduced to one fourth(1/4) of its original value.
Similarly, if the distance between two objects is tripled, then the gravitational force of attraction becomes one ninth(1/9) of its original value.

(iii) Force F is directly proportional to the product of both the masses. So, if both the masses are doubled then the gravitational force of attraction becomes four times the original value.

7. What is the importance of universal law of gravitation?


Answer

Universal law of Gravitation is important because it it tells us about:
→ the force that is responsible for binding us to Earth.
→ the motion of moon around the earth
→ the motion of planets around the sun
→ the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both sun and moon on the earth.

  1. What is the acceleration of free fall?

    Answer

    Acceleration of free fall is the acceleration produced when a body falls under the influence of the force of gravitation of the earth alone. It is denoted bygand its value on the surface of the earth is 9.8 ms-2.

    9. What do we call the gravitational force between the Earth and an object?

    Answer

    Gravitational force between the earth and an object is known as the weight of the object.

    10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of gis greater at the poles than at the equator].

    Answer

    Weight of a body on the Earth is given by:
    W = mg
    Where,
    m = Mass of the body
    g = Acceleration due to gravity
    The value of gis greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

    11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

    Answer

    When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

  2. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?

    Answer

    Weight of an object on the moon = 1/6 ×Weight of an object on the Earth
    Also,
    Weight = Mass × Acceleration
    Acceleration due to gravity, g = 9.8 m/s2
    Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N
    And, weight of the same object on the moon= 1.6 × 9.8 = 16.3 N.

  3. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.

Answer

According to the equation of motion under gravity:
v2 – u2 = 2gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v= 0
u = 49 m/s
During upward motion, g = -9.8 m s-2
Let h be the maximum height attained by the ball.
Hence,
0 – 492 = 2×9.8×h
⇒ h = 492/(2×9.8)
⇒ h = 2401/19.6 = 122.5

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
v = u + gt
We get,
0 = 49 + t x (- 9.8)
9.8t = 49
t = 49 / 9.8 = 5s
Also,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Answer

According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s-2
∴ v2 − 02 = 2 × 9.8 × 19.6
v2 = 2 × 9.8 × 19.6 = (19.6)2
v = 19.6 m s-1
Hence, the velocity of the stone just before touching the ground is 19.6 m s-1.

15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer

According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = −10 m s−2
Let h be the maximum height attained by the stone.
Therefore,
0 – (40)2 = 2×h×(-10)
h = (40×40)/20 = 80 m

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (−80) = 0

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.

Answer

According to question, 
MS = Mass of the Sun = 2×1030 kg
ME = Mass of the Earth = 6×1024 kg
R = Average distance between the Earth and the Sun = 1.5×1011 m
From Universal law of gravitation,

  1. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

    Answer

Let t be the point at which two stones meet and let h be their height from the ground.
Height of the tower is H = 100 m (Given)

It is clear from the question that we need to calculate time when the two stones met. After calculating time, we will also be able to calculate the distance. 

Now, first consider the stone which falls from the top of the tower.
Initial velocity (u) = 0
So, distance covered by this stone at time t can be calculated using equation of motion

18. A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.

Answer

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = −9.8 ms−2
Equation of motion, v = u + gt will give,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms− 1
Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s−2
From the equation of motion, s= ut + 1/2 at2
h= 29.4 × 3 + 1/2 × -9.8 × (3)2 = 44.1 m


(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.
Equation of motion, s= ut + 1/2 gt2 will give,
s = 0 × t + 1/2 × 9.8 × 12 = 4.9 m
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.

Page No: 145

19. In what direction does the buoyant force on an object immersed in a liquid act?

Answer

An object immersed in a liquid experiences buoyant force in the upward direction.

20. Why does a block of plastic released under water come up to the surface of water?

Answer

For an object immersed in water two force acts on it
→ gravitational force which tends to pull object in downward direction
→ buoyant force that pushes the object in upward direction
here in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

21. The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm−3, will the substance float or sink?

Answer

If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
Here, density of the substance = Mass of the substance / Volume of the substance
= 50/20
= 2.5 g cm-3
The density of the substance is more than the density of water (1 g cm−3). Hence, the substance will sink in water.

22. The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm−3? What will be the mass of the water displaced by this packet?

Answer

Density of the 500 g sealed packet= Mass of the Packet / Volume of the Packet
= 500 / 350
= 1.428 g cm−3
The density of the substance is more than the density of water (1 g cm−3). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

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NCERT SOLUTIONS FOR CLASS 9 | Ch 9 Force and Laws of Motion Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 9 Force and Laws of Motion Science Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 9 Force and Laws of Motion Science

In Text Questions

Page No: 118

1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin?
 
Answer
 
Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.

(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.

(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.

(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.

2. In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.

Answer

The velocity of football changes four times.
First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team.

Agent supplying the force:
→ First case – First player
→ Second case – Second player
→ Third case – Goalkeeper
→ Fourth case – Goalkeeper

3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer

Some leaves of a tree get detached when we shake its branches vigorously because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer

In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain to inertia of motion. As a result, a forward force is exerted on him.
Similarly, the passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.
 
Page No: 126

1. If action is always equal to the reaction, explain how a horse can pull a cart.
 
Answer

A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman
decreases. Hence, it is difficult for him to remain stable while holding the hose.

3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−1. Calculate the initial recoil velocity of the rifle.

Answer

Mass of the rifle, m1= 4 kg
Mass of the bullet, m2= 50g= 0.05 kg
Recoil velocity of the rifle= v1
Bullet is fired with an initial velocity, v2= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m1+m2)v= 0
Total momentum of the rifle and bullet system after firing:
= m1v1 + m2v2 = 0.05 × 35 = 4v1 + 1.75

According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing  4v1 + 1.75= 0
v1 = -1.75/4 = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

Page No: 127

4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms−1 and 1 ms−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms−1. Determine the velocity of the second object.

Answer

Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other object, m2 = 200 g = 0.2 kg
Velocity of m1 before collision, v1= 2 m/s
Velocity of m2 before collision, v2= 1 m/s

Velocity of m1 after collision, v3= 1.67 m/s

Velocity of m2 after collision= v4


According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision

Therefore, m1v1 + m2v2 = m1v3 + m2v4
                  2(0.1) + 1(0.2) = 1.67(0.1) + v4(0.2)
                  0.4 = 0.167 + 0.2v4
                  v4= 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Excercises

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer

Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.

2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer

When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.

3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer

When a bust starts suddenly, the lower part of the luggage kept on the roof being in contact with the bus begins to move forward with the speed of bus, but the upper part tends to remain at rest due to inertia of rest. Therefore, the upper part is left behind and hence luggage falls backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.

4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Answer

The ball slows down and comes to rest due to opposing forces of air resistance and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the motion is correct.

5.  A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg).

Answer

Initial velocity, u = 0 
Distance travelled, s = 400 m
Time taken, t = 20 s

We know, s = ut + ½ at2

Or, 400 = 0 + ½ a (20)2
Or, a = 2 ms–2
Now, m = 7 metric tonne = 7000 kg, a = 2 ms–2
Or, F = ma = 7000 × 2 = 14000 N Ans.

6. A stone of 1 kg is thrown with a velocity of 20 m s−1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer

Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance covered by the stone, s = 50 m
Since, v2 – u2 = 2as,
Or, 0 – 202 = 2a × 50,
Or, a = –4 ms-2
Force of friction, F = ma = – 4N

7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c)the force of wagon 1 on wagon 2.

Answer

(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F − Ff = 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.

(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon x Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass, M = m  = 10000 kg
From Newton’s second law of motion:
Fa= Ma
a = Fa/M  = 35000 10000 = 3.5 ms-2
Hence, the acceleration of the wagons and the train is 3.5 m/s2.

(c) Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s2
Thus, force exerted on all the wagons except wagon 1
= 8000 × 3.5 = 28000 N
Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 ​ N.
Hence, the force exerted by wagon 1 on wagon 2 is 28000 ​ N.

8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s−2?

Answer

Mass of the automobile vehicle, m= 1500 kg
Final velocity, v= 0 (finally the automobile stops)
Acceleration of the automobile, a= −1.7 ms−2
From Newton’s second law of motion:
Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N
Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.

9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv

Answer

(d) mv
Mass of the object = m
Velocity = v
Momentum = Mass x Velocity
Momentum = mv

10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer

The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.
11. Two objects, each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Answer

Mass of one of the objects, m1 = 1.5 kg Mass of the other object, m2 = 1.5 kg
Velocity of m1 before collision, u1 = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, u2 = −2.5 m/s
Let v be the velocity of the combined object after collision. By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m1 + m2) v = m1u1 + m2u2
Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms–1

Page No: 129

12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer

The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.

13. A hockey ball of mass 200 g travelling at 10 m s−1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s−1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer

Mass of the hockey ball, m = 200 g = 0.2 kg Hockey ball travels with velocity, v1 = 10 m/s
Initial momentum = mv1
Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s
Final momentum = mv2
Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg ms−1
Hence, the change in momentum of the hockey ball is 3 kg ms−1.
 
14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s−1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer

Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)a = -150/0.03 = -5000 m/s2
 
(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m

Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s−1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer

Mass of the object, m1 = 1 kg
Velocity of the object before collision, v1 = 10 m/s
Mass of the stationary wooden block, m2 = 5 kg
Velocity of the wooden block before collision, v2 = 0 m/s
∴ Total momentum before collision = m1 v1 + m2 v2
= 1 (10) + 5 (0) = 10 kg ms−1

It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m1 + m2
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1 v1 + m2 v2 = (m1 + m2)v
⇒ 1 (10) + 5 (0) = (1 + 5)v
⇒ v = 10/6
= 5/3

The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg ms−1
Total momentum just after the impact = (m1 + m2)v = 6 × 5/3 = 10 kg ms-1
Hence, velocity of the combined object after collision = 5/3 ms-1
 

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s−1 to 8 m
s−1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer

Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms−1
Final momentum = mv = 100 × 8 = 800 kg ms−1
Force exerted on the object, F = (mv – mu)/ t
= m (v-u)/t
= 800 – 500
= 300/6
= 50 N

Initial momentum of the object is 500 kg ms−1.
Final momentum of the object is 800 kg ms−1.
Force exerted on the object is 50 N.

17. Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Answer

The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong. 
The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motor car, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momenta are also the same. Only the signs of changes in momenta are opposite, i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same.

18. How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s−2.

Answer

Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v2 = u2 + 2as
⇒ v2 = 0 + 2 (10) 0.8
⇒ v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv = 10 × 4 = 40 kgms−1

Page No: 130


Additional Excercises
 
1. The following is the distance-time table of an object in motion:

Time in seconds
Distance in metres
0
0
1
1
2
8
3
27
4
64
5
125
6
216
7
343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b)What do you infer about the forces acting on the object?

Answer

(a) There is an unequal change of distance in an equal interval of time.
Thus, the given object is having a non-uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.

(b) The object is in accelerated condition. According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say unbalanced force is acting on the object.
 
2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s−2. With what force does each person push the motorcar?
(Assume that all persons push the motorcar with the same muscular effort)

Answer

Mass of the motor car = 1200 kg
Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person,
a = 0.2 m/s2
Let the force applied by the third person be F.
From Newton’s second law of motion:
Force = Mass × Acceleration
F = 1200 × 0.2 = 240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motor car.

3. A hammer of mass 500g, moving at 50 ms−1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer

Mass of the hammer, m = 500 g = 0.5 kg
Initial velocity of the hammer, u = 50 m/s
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, f = m(v-u)/t
= 0.5(0-50)/0.01
= -2500 N

The hammer strikes the nail with a force of −2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.

4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Answer

Mass of the motor car, m = 1200 kg
Initial velocity of the motor car, u = 90 km/h = 25 m/s
Final velocity of the motor car, v = 18 km/h = 5 m/s
Time taken, t = 4 s
According to the first equation of motion:
v = u + at
⇒ 5 = 25 + a (4)
⇒ a = − 5 m/s2
Negative sign indicates that its a retarding motion i.e. velocity is decreasing.
Change in momentum = mv − mu = m (v−u)
= 1200 (5 − 25) = −24000 kg m s−1
∵ Force = Mass × Acceleration
= 1200 × −5 = −6000 N
Acceleration of the motor car = −5 m/s2
Change in momentum of the motor car = −24000 kg m s−1
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)

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NCERT SOLUTIONS FOR CLASS 9 | Ch 8 Motion Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 8 Motion Science Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 8 Motion Science

In Text Question

Page No: 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer

Yes,an object can have zero displacement even when it has moved through a distance.This happens when final position of the object coincides with its initial position. For example,if a person moves around park and stands on place from where he started then here displacement will be zero.

2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?



Answer


Given, Side of the square field= 10m
Therefore, perimeter = 10 m × 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 × 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m

Therefore, in 1s distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter
= 140 m /40 m  = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position.

  1. Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer

None of the statement is true for displacement. First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.

Page No: 102

1. Distinguish between speed and velocity.

Answer

Speed

Velocity
Speed is the distance travelled by an object in a given interval of time.  Velocity is the displacement of an object in a given interval of time.
Speed = distance / time Velocity = displacement / time
Speed is scalar quantity i.e. it has only magnitude. Velocity is vector quantity i.e. it has both magnitude as well as direction.
  1. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

    Answer

    The magnitude of average velocity of an object is equal to its average speed, only when an object is moving in a straight line.

    3. What does the odometer of an automobile measure?

    Answer

    The odometer of an automobile measures the distance covered by an automobile.

    4. What does the path of an object look like when it is in uniform motion?

    Answer

    An object having uniform motion has a straight line path.

    5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108m s−1.

    Answer

    Speed= 3 × 108ms−1
    Time= 5 min = 5 × 60 = 300 secs.

Distance= Speed × Time
Distance= 3 × 108 ms−1 × 300 secs. = 9 × 1010 m

Page No: 103

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer

(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant.

  1. A bus decreases its speed from 80 km h−1to 60 km h−1in 5 s. Find the acceleration of the bus.

    Answer

 

  1. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1in 10 minutes. Find its acceleration.

    Answer



    Page No: 107

    1. What is the nature of the distance – ‘time graphs for uniform and non-uniform motion of an object?


Answer

When the motion is uniform,the distance time graph is a straight line with a slope.

Graph of uniform motion

When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.

Graph of non uniform motion

  1. What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis?

Answer

If distance time graph is a straight line parallel to the time axis, the body is at rest.

Distance time graph showing body is at rest

  1. What can you say about the motion of an object if its speed – ‘time graph is a straight line parallel to the time axis?

Answer


If speed time graph is a straight line parallel to the time axis, the object is moving uniformly.


  1. What is the quantity which is measured by the area occupied below the velocity -time graph?

Answer

The area below velocity-time graph gives the distance covered by the object.

Page No: 109

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer

Initial speed of the bus, u= 0

Acceleration, a = 0.1 m/s2
Time taken, t = 2 minutes = 120 s

(a) v= u + at
v= 0 + 0.1 × 120
v= 12 ms–1

(b) According to the third equation of motion:
v2 – u2= 2as
Where, s is the distance covered by the bus
(12)2 – (0)2= 2(0.1) s
s = 720 m

Speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.

Page No: 110

2. A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.

Answer

Initial speed of the train, u= 90 km/h = 25 m/s
Final speed of the train, v = 0 (finally the train comes to rest)
Acceleration = – 0.5 m s-2
According to third equation of motion:
v2= u2+ 2 as
(0)2= (25)2+ 2 ( – 0.5) s

Where, s is the distance covered by the train


The train will cover a distance of 625 m before it comes to rest.


  1. A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after the start?

    Answer

    Initial Velocity of trolley, u= 0 cms-1
    Acceleration, a= 2 cm s-2
    Time, t= 3 s
    We know that final velocity, v= u + at = 0 + 2 x 3 cms-1
    Therefore, The velocity of train after 3 seconds = 6 cms-1

  2. A racing car has a uniform acceleration of 4 m s – ‘2. What distance will it cover in 10 s after start?

    Answer

    Initial Velocity of the car, u=0 ms-1
    Acceleration, a= 4 m s-2
    Time, t= 10 s
    We know Distance, s= ut + (1/2)at2
    Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 102
    = 0 + (1/2) × 4 × 10 × 10 m
    = (1/2) × 400 m
    = 200 m

    5. A stone is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 m s−2in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

    Answer

    Given Initial velocity of stone, u=5 m s-1

Downward of negative Acceleration, a= 10 m s-2
We know that 2 as= v2– u2

Excercise

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer

Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 × ( 22 / 7 ) × 100
Speed of the athlete (v) = Distance / Time
= (2 × 2200) / (7 × 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 × 40) × (2 × 60 + 20)
= 4400 / (7 × 40) × 140
= 4400 × 140 /7 × 40
= 2200 m

Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 1/2
After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure

 
Hence, Displacement of the athlete with respect to initial position at x= xy

= Diameter of circular track
= 200 m

2.  Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Answer

Total Distance covered from AB = 300 m
Total time taken = 2 × 60 + 30 s
=150 s

Therefore, Average Speed from AB = Total Distance / Total Time
=300 / 150 m s-1
=2 m s-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1
=2 m s-1
Total Distance covered from AC =AB + BC
=300 + 200 m


Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 × 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s-1
= 1.904 m s-1

Displacement (S) from A to C = AB – BC
= 300-100 m
= 200 m
 
Time (t) taken for displacement from AC = 210 s

Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s-1
= 0.952 m s-1


3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 40 km h−1. What is the average speed for Abdul’s trip?

Answer

The distance Abdul commutes while driving from Home to School = S
Let us assume time taken by Abdul to commutes this distance = t1

Distance Abdul commutes while driving from School to Home = S

Let us assume time taken by Abdul to commutes this distance = t2
Average speed from home to school v1av = 20 km h-1
Average speed from school to home v2av = 30 km h-1
Also we know Time taken form Home to School t1 =S / v1av
Similarly Time taken form School to Home t2 =S/v2av
Total distance from home to school and backward = 2 S
Total time taken from home to school and backward (T) = S/20+ S/30
Therefore, Average speed (Vav) for covering total distance (2S) = Total Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600]
= 1200S / 50S
= 24 kmh-1

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s−2 for 8.0 s. How far does the boat travel during this time?

Answer

Given Initial velocity of motorboat,  u = 0
Acceleration of motorboat,  a = 3.0 m s-2
Time under consideration,  t = 8.0 s
We know that Distance, s = ut + (1/2)at2
Therefore, The distance travel by motorboat = 0 ×8 + (1/2)3.0 × 82
= (1/2) × 3 × 8 × 8 m
= 96 m
5. A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer

As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh−1 and 3 kmh−1 respectively.

Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) × OR × OP
= (1/2) × 5s × 52 kmh−1
= (1/2) × 5 × (52 × 1000) / 3600) m
= (1/2) × 5 × (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) × OQ × OS
= (1/2) × 10 s × 3 kmh−1
= (1/2) × 10 × (3 × 1000) / 3600) m
= (1/2) × 10 x (5/6) m
= 5 × (5/6) m
= 25/6 m
= 4.16 m

 

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?
(d)How far has B travelled by the time it passes C?


Answer

(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km

Therefore, Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.

(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

(c)

7 square box = 4 km
∴ 1 square box = 4/7 km
C is 4 blocks away from origin therefore initial distance of C from origin = 16/7 km
Distance of C from origin when B passes A = 8 km
Thus, Distance travelled by C when B passes A = 8 – 16/7 = (56 – 16)/7 = 40/7 = 5.714 km

(d)

 
Distance travelled by B by the time it passes C = 9 square boxes
9×4/7 = 36/7 = 5.143 km


Page No: 113

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?

Answer

Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’
Initial Velocity of ball,  u =0
Distance or height of fall,  s =20 m
Downward acceleration,  a =10 m s-2
As we know, 2as =v2-u2
v2 = 2as+ u2
= 2 x 10 x 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds

8. The speed-time graph for a car is shown is Fig. 8.12.


(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?
 


Answer

(a)

The shaded area which is equal to 1/2 × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer

(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.

(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its direction.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer

Radius of the circular orbit, r = 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v = (2π r)/t
= [2× (22/7)×42250 × 1000] / (24 × 60 × 60)
= (2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1
= 3073.74 m s -1

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Important MCQs- Ch 8 Motion

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NCERT SOLUTIONS FOR CLASS 9 | Ch 7 Diversity in Living Organisms Science | EDUGROWN

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Ch 7 Diversity in Living Organisms Science Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 9 Ch 7 Diversity in Living Organisms Science

In Text Questions

Page No: 80

1. Why do we classify organisms?

Answer

There are millions of organisms on this earth. So, it is harder to study them one by one. Therefore, we look for similarities among them and classify them into different classes to study these different classes as a whole. Classification makes our study easier.

2. Give three examples of the range of variations that you see in life-forms around you.
 
Answer
 
Examples of range of variations observed in daily life are:
→ Organisms vary greatly in size-from microscopic bacteria to elephants, whales and large trees.
→ The colour of various animals is quite different. Some worms are even colourless or transparent. Various types of pigments are found in plants.
→ The life span of different organisms is also quite varied. For example, a crow lives for only 15 years, whereas a parrot lives for about 140 years.

Page No: 82

1. Which do you think is a more basic characteristic for classifying organisms?
(a) The place where they live.
(b) The kind of cells they are made of. Why?

Answer

The more basic characteristic for classifying organisms is the kind of cells they are made of because different organisms may share same habitat but may have entirely different form and structure. So, the place where they live cannot be a basis of classification.

2. What is the primary characteristic on which the first division of organisms is made?

Answer

The primary characteristic on which the first division of organisms is made is the nature of the cell – prokaryotic or eukaryotic cell.

3. On what basis are plants and animals put into different categories?

Answer

Plants and animals are put into different categories on the basis of Mode of nutrition.Plants are autotrophs. They can make their food own while animas are heterotrophs which are dependent on others for food. Also, locomotion, absence of chloroplasts etc. make them different.

Page No: 83

1. Which organisms are called primitive and how are they different from the so-called advanced organisms?

Answer

A primitive organism is the one which has a simple body structure and ancient body design or features that have not changed much over a period of time.As per the body design, the primitve organisms which hav simple structures are different from those so-called advanced organisms which have complex body structure and organization.

2. Will advanced organisms be the same as complex organisms? Why?

Answer

Yes, because the advanced organisms also were like the primitive ones once. They have acquired their complexity relatively recently. There is a possibility that these advanced or ‘younger’ organisms acquire more complex structures during evolutionary time to compete and survive in the changing environment.

Page No: 85

1. What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?

Answer

The organisms belonging to Kingdom Monera are unicellular and prokaryotic whereas the organisms belonging to Kingdom Protista are unicellular and eukaryotic.

2. In which kingdom will you place an organism which is single-celled, eukaryotic and photosynthetic?
► Kingdom Protista

3. In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?

Answer

In the hierarchy of classification,a species will have the smallest number of organisms with a maximum of characteristics in common, whereas the kingdom will have the largest number of organisms.

Page No: 88

1. Which division among plants has the simplest organisms?
► Division Thallophyta

2. How are pteridophytes different from the phanerogams?

Answer

Pteridophyta
Phanerogams
They have inconspicuous or less differentiated reproductive organs. They have well developed reproductive organs.
They produce naked embryos called spores. They produce seeds.
Ferns, Marsilea, Equisetum, etc. are examples of pteridophyta. Pinus, Cycas, fir, etc. are examples of phanerogams.

3. How do gymnosperms and angiosperms differ from each other?

Answer

Gymnosperm
Angiosperm
They are non-flowering plants. They are flowering plants.
Naked seeds not enclosed inside fruits are produced. Seeds are enclosed inside fruits.
Pinus, Cedar, fir, Cycas, etc. are some examples of gymnosperms. Coconut, palm, mango, etc. are some examples of angiosperms.

Page No: 94

1. How do poriferan animals differ from coelenterate animals?

Answer

Porifera
Coelenterate
They are mostly marine, non-motile, and found attached to rocks. They are exclusively marine animals that either live in colonies or have a solitary life-span.
They show cellular level of organisation. They show tissue level of organisation.
Spongilla, Euplectella, etc. are poriferans. Hydra, sea anemone, corals, etc. are coelenterates.
2. How do annelid animals differ from arthropods?

Answer

Annelids
Arthropods
The circulatory system of annelids is closed. Arthropods have an open circulatory system.
The body is divided into several identical segments. The body is divided into few specialized segments.

3. What are the differences between amphibians and reptiles?

Answer

Amphibian
Reptiles
They have a dual mode of life. They are completely terrestrial.
Scales are absent. Skin is covered with scales.
They lay eggs in water. They lay eggs on land.
It includes frogs, toads, and salamanders. It includes lizards, snakes, turtles, chameleons, etc.

4. What are the differences between animals belonging to the Aves group and those in the mammalia group?

Answer

Aves
Mammals
Most birds have feathers and they possess a beak. They do not have feathers and the beak is also absent.
They lay eggs. Hence, they are oviparous. Some of them lay eggs and some give birth to young ones. Hence, they are both oviparous and viviparous.

Excercise

1. What are the advantages of classifying organisms?

Answer

Following are the advantages of classifying organisms:
→ It makes us aware of and gives us information regarding the diversity of plants and animals.

→ It makes the study of different kinds of organisms much easier.
→ It tells us about the inter-relationship among the various organisms.
→ It helps us understanding the evolution of organisms.
→ It helps in the development of other life sciences easy.

→ It helps environmentalists to develop new methods of conservation of plants and animals.


2. How would you choose between two characteristics to be used for developing a hierarchy in classification?

Answer

We choose that characteristics which depends on the first characteristics and determines the rest variety.

3. Explain the basis for grouping organisms into five kingdoms.

Answer

The basis for grouping organisms into five kingdoms are:
→ Complexity of cell structure – There are two broad categories of cell structure: Prokaryotic and Eukaryotic. Thus, two broad groups can be formed, one having prokaryotic cell structure and the other having eukaryotic cell structure. Presence or absence of cell wall is another important characteristic.
→ Unicellular and multicellular organisms – This characteristic makes a very basic distinction in the body designs of organisms and helps in their broad categorizations.

→ Cell Wall: Presence and absence of cell wall leads into grouping.
→ Mode of nutrition -Organisms basically have two types of nutritions – autotrophic who can manufacture their own food and heterotrophic who obtain their food from external environment, i.e., from other organisms). Thus, organisms can be broadly classified into different groups on the basis of their mode of nutrition.


4. What are the major divisions in the Plantae? What is the basis for these divisions?

Answer

The major divisions in Kingdom Plantae are:
→Thallophyta
→ Bryophyta
→ Pteridophyta
→ Gymnosperms
→ Angiosperms

The following points constitute the basis of these divisions:
→ Whether the plant body has well differentiated, distinct components.
→ whether the differentiated plant body has special tissues for the transport of water and other substances.
→ The ability to bear seeds.
→ Whether the seeds are enclosed within fruits.

5. How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?

Answer

The characteristics used to classify plants is different from animals because the basic design are different, based on the need to make their own food (plants) or acquire food (animals).
Criteria for deciding divisions in plants are:
→ Differentiated/ Undifferentiated plant body
→ Presence/ absence of vascular tissues
→With/without seeds
→ Naked seeds/ seeds inside fruits

 
But the animals can’t be divided into groups on these criteria. It is because the basic designs of animals are very different from plants. They are divided on the basis of their body structure.


6. Explain how animals in Vertebrata are classified into further subgroups.

Answer

Animals in Vertebrata are classified into five classes:

(i) Class Pisces: This class includes fish such as Scoliodon, tuna, rohu, shark, etc. These animals mostly live in water. Hence, they have special adaptive features such as a streamlined body, presence of a tail for movement, gills, etc. to live in water.

(ii) Class Amphibia: It includes frogs, toads, and salamanders. These animals have a dual mode of life. In the larval stage, the respiratory organs are gills, but in the adult stage, respiration occurs through the lungs or skin. They lay eggs in water.

(iii) Class Reptilia: It includes reptiles such as lizards, snakes, turtles, etc. They usually creep or crawl on land. The body of a reptile is covered with dry and cornified skin to prevent water loss. They lay eggs on land.

(iv) Class Aves: It includes all birds such as sparrow, pigeon, crow, etc. Most of them have feathers. Their forelimbs are modified into wings for flight, while hind limbs are modified for walking and clasping. They lay eggs.

(v) Class Mammalia: It includes a variety of animals which have milk producing glands to nourish their young ones. Some lay eggs and some give birth to young ones. Their skin has hair as well as sweat glands to regulate their body temperature.

Important Link

Quick Revision Notes- Ch 7 Diversity in Living Organisms 

Most Important Questions- Ch 7 Diversity in Living Organisms 

Important MCQs- Ch 7 Diversity in Living Organisms 

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