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These NCERT Solutions for Class 9 Science Chapter 13 Why Do We Fall Ill Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Why Do We Fall Ill NCERT Solutions for Class 9 Science Chapter 13

Class 9 Science Chapter 13 Why Do We Fall Ill InText Questions and Answers

Question 1.
State any two conditions essential for good health.
Answer:
The two conditions essential for good health are:

1. An individual must have better health facilities and more professionals to deal with health problems.
2. All basic necessary conditions to prevent diseases must be present. For example, proper garbage collection and disposal, clearing of drains, supply of healthy drinking water, etc.

Question 2.
State any two conditions essential for being free of disease.
Answer:
The two conditions essential for being free of diseases are:

1. Personal hygiene and cleanliness are necessary to stay away from diseases.
2. Individuals should take a balanced diet that contains carbohydrates, fats, proteins, vitamins, fibres, and proper quantity of water.

Question 3.
Are the answers to the above questions necessarily the same or different? Why?
Answer:
No. The answers to the above questions may not necessarily be the same. This is because a disease free state is not the same as being healthy. Good health is the ability of an individual to realise his or her full potential. Individuals can have poor health without having any identifiable disease. Also, health is related to society and community, whereas having a disease is about an individual sick person. Hence, the conditions for good health and for being disease free can be same or even different.

Question 4.
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:
Symptoms such as a headache, stomach pain, nausea, vomiting, fever, etc., make us feel that we are sick and must visit a doctor. These symptoms basically indicate that there might be a disease, but we cannot predict the kind of disease. Therefore, it becomes necessary to visit a doctor so that the disease can be identified and can be treated with proper medication.

However, 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.

Question 5.
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.

Why?
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.

Question 6.
Why are we normally advised to take bland and nourishing food when we are sick?
Answer:
We are normally advised to consume bland and nourishing food when we are sick so that we can get the nutrients and energy quickly to fight off the foreign disease-causing agents.

Question 7.
What are the different means by which infectious diseases are spread?
Answer:
Diseases can be spread through various means such as air, water, sexual contact, blood, and vector.

• 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.
• Sometimes causal micro-organisms get mixed with drinking water and spread water borne diseases. Cholera for example is waterborne disease.
• Sexual act between two people can lead to the transfer of diseases such as syphilis, gonorrhoea, AIDS, etc.
• Certain diseases such as AIDS can spread via blood-to-blood contact during blood transfusion or pregnancy.
• Certain diseases spread by animals called vectors. For example mosquitoes spread malaria.

Question 8.
What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:
Precautions to reduce incidence of infectious diseases are:

• Stay away from the diseased person.
• Cover your mouth or nose while coughing or sneezing to prevent the spread of disease.
• Drink safe water.
• Keep the environment clean to prevent mosquitoes from breeding.

Question 9.
What is immunization?
Answer:
Immunizations defined as protection of the body from communicable diseases by administration of some agent that mimics the microbe. This suspension of killed microbes that mimics the disease-causing microbes is known as vaccines.

Question 10.
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.

Class 9 Science Chapter 13 Why Do We Fall Ill Textbook Questions and Answers

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 illnesses.
Answer:
This varies from person to person. Some people fall ill several times in a year, while others do not fall ill at all. A person’s immune system and hygiene-related habits play a major role in determining the person’s health.

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:
The following precautions must be taken by a doctor/ nurse/ health-worker:

• Wearing a mask when in contact with a diseased person.
• Keeping yourself covered while moving around an infected place.
• Drinking safe water.
• Eating healthy and nutritious food.
• Ensuring proper cleanliness and personal hygiene.

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:
Three most common diseases are:

1. Tuberculosis
2. Typhoid
3. Jaundice

Steps to be taken to bring down the incidence of these diseases are:

1. Proper disposal of sewage.,
2. Ensuring supply of safe drinking water.
3. Providing a clean environment and preventing mosquitoes from breeding.

Question 4.
A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out
(a) that the baby is sick?
(b) what is the sickness?
Answer:
(a) The baby is sick can be determined by his/ her behavioural changes such as constant crying of baby, improper intake of food, frequent mood changes, etc.
(b) The sickness is determined by symptoms or indications that can be seen in the baby. The symptoms include vomiting, fever, loose motion, paleness in the toe body, etc.

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 chickenpox.
(c) when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chickenpox.
Why?
Answer:
(c) A person is more likely to fall sick when she is cm a four day fast after recovering from malaria and is taking care of someone who is suffering from chickenpox. This is because she is fasting during recovery, and her immune system is so weak that it is not able to protect its own body from any foreign infection. If she is taking care of someone suffering from chickenpox, then she has more chances of getting infected from chickenpox virus and will get sick again with this disease.

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 today.
(c) when your friend is suffering from measles.
Why?
Answer:
(c) You are more likely to fall sick when your friend is suffering from measles. This is because measles is highly contagious and can easily spread through respiration i.e., through air. Thus, if your friend is suffering from measles, stay away from him otherwise you might easily get infected with the disease.

Class 9 Science Chapter 13 Why Do We Fall Ill Additional Important Questions and Answers

Choose the correct option:

Question 1.
Which one of the following is not a viral disease?
(a) Dengue
(b) AIDS
(c) Typhoid
(d) Influenza
Answer:
(c) Typhoid

Question 2.
Which one of the following is not a bacterial disease?
(a) Cholera
(b) Tuberculosis
(c) Anthrax
(d) Influenza
Answer:
(d) Influenza

Question 3.
Which one of the following diseases is not transmitted by mosquitoes?
(a) Brain fever
(b) Malaria
(c) Typhoid
(d) Dengue
Answer:
(c) Typhoid

Question 4.
Which one of the following disease is caused by bacteria?
(a) Typhoid
(b) Anthrax
(c) Tuberculosis
(d) Malaria
Answer:
(d) Malaria

Question 5.
Which one of the following diseases is caused by protozoans?
(a) Malaria
(b) Influenza
(c) AIDS
(d) Cholera
Answer:
(a) Malaria

Question 6.
Which one of the following has a long-term effect on the health of an individual?
(a) Common cold
(b) Chickenpox
(c) Chewing tobacco
(d) Stress
Answer:
(c) Chewing tobacco

Question 7.
Which of the following can make you ill if you come in contact with an infected person?
(a) High blood pressure
(b) Genetic abnormalities
(c) Sneezing
(d) Blood cancer
Answer:
(c) Sneezing

Question 8.
AIDS cannot be transmitted by
(a) sexual contact
(b) hugs
(c) breastfeeding
(d) blood transfusion
Answer:
(b) hugs

Question 9.
Making anti-viral drugs is more difficult than making anti-bacterial medicines because
(a) viruses make use of host machinery
(b) viruses are on the borderline of living and non-living
(c) viruses have very few biochemical mechanisms of their own
(d) viruses have a protein coat
Answer:
(c) viruses have very few biochemical mechanisms of their own

Question 10.
Which one of the following causes kala-azar?
(a) Ascaris
(b) Trypanosoma
(c) Leishmania
(d) Bacteria
Answer:
(c) Leishmania

Question 11.
If you live in a overcrowded and poorly ventilated house, it is possible that you may suffer from which of the following diseases
(a) Cancer
(b) AIDS
(c) Airborne diseases
(d) Cholera
Answer:
(c) Airborne diseases

Question 12.
Which disease is not transmitted by mosquitoes?
(a) Dengue
(b) Malaria
(c) Brain fever or encephalitis
(d) Pneumonia
Answer:
(d) Pneumonia

Question 13.
Which one of the following is not important for individual health?
(a) Living in clean space
(b) Good economic condition
(c) Social equality and harmony
(d) Living in a large and well furnished house
Answer:
(d) Living in a large and well furnished house

Question 14.
Choose the wrong statement
(a) High blood pressure is caused by excessive weight and lack of exercise.
(b) Cancers can be caused by genetic abnormalities
(c) Peptic ulcers are caused by eating acidic food
(d) Acne in not caused by staphylococci
Answer:
(c) Peptic ulcers are caused by eating acidic food

Question 15.
We should not allow mosquitoes to breed in our surroundings because they
(a) multiply very fast and cause pollution
(b) are vectors for many diseases
(c) bite and cause skin diseases
(d) are not important insects
Answer:
(b) are vectors for many diseases

Very Short Answer Questions

Question 1.
Write the full form of AIDS.
Answer:
AIDS stand for Acquired immunodeficiency syndrome.

Question 2.
Name the three diseases for which a child is vaccinated with DPT?
Answer:
DPT is vaccine for diphtheria, whooping cough and tetanus.

Question 3.
Which body cells contribute to the immunity against the infection?
Answer:
WBC, particularly the phagocytes and lymphocytes.

Question 4.
Which diseases are more fatal, acute or chronic? Why?
Answer:
Acute diseases are more fatal than chronic because of their quick manifestations.

Question 5.
Give three examples of each the followings:
(a) Waterborne disease
(b) Airborne disease
(c) STD
Answer:
(a) Typhoid, cholera and jaundice.
(b) Tuberculosis, influenza and common cold
(c) AIDS, syphilis and gonorrhea

Question 6.
Give three examples of each of the followings:
(a) Viral disease
(b) Bacterial diseases
(c) Protozoan diseases
Answer:
Viral diseases: Chickenpox, polio and measles Bacterial disease: Typhoid, cholera and tuberculosis Protozoan diseases : Malaria, sleeping sickness and kala-azar

Question 7.
Give three examples of each of the followings:
(a) Disease transmitted by water
(b) Diseases transmitted by air
(c) Diseases transmitted by vector
Answer:
(a) Cholera, typhoid and jaundice.
(b) Tuberculosis, meningitis and polio.
(c) Malaria, Sleeping sickness and dengue.

Question 8.
Name the nutrients which contribute to the body’s immunity.
Answer:
Vitamins and minerals

Question 9.
Which disease you expect a person to suffer from if he joins an overcrowded place?
Answer:
Airborne disease

Question 10.
Which is the commonly found element in the most of antibiotics?
Answer:
Sulphur

Short Answer Type Questions

Question 1.
What do you mean by health?
Answer:
Health is a state of being well enough to function well physically, mentally and socially and not merely the absence of disease. In order to keep good health, we need to take food full of nutrients, maintain good personal and domestic hygiene and regular exercise, sleep and relaxation. Our social environment and public cleanliness is also necessary for good individual health.

Question 2.
What do you mean by a disease?
Answer:
The word disease mean disturbed ease or not at ease. Disease literally means being uncomfortable. When we talk of a disease, we find a specific and particular cause for discomfort.

Question 3.
Is it possible to be in poor health without actually suffering from a particular disease?
Answer:
Yes, it is possible for a person to be in a poor health without actually suffering from a particular disease. Simply not being diseased is not same as being healthy. In order to keep healthy, we need to be happy if we mistreat each other or are afraid of each other, we can not be healthy. Good health for a dancer means he should be able to stretch his body into different but graceful positions. If the dancer is not suffering from any disease but he is not able to stretch his body according to the dance, he will not be considered in a good health.

Question 4.
How do we know that there is a disease?
Or
How do we know that there is something wrong with the body?
Answer:
When there is a disease, either the functioning or the appearance of one or more system of the body will change for the worse. These changes give rise to symptoms and sign of disease. So, a headache, cough, loose motion and wound with pus are all symptoms. These indicate that there may be a disease.

Signs of disease are what physicians will look for on the basis of symptoms. Signs will give a little indication of presence of a disease. Physician will also get laboratory tests done to pinpoint the disease further.

Question 5.
What are acute and chronic diseases? Give examples.
Answer:
Acute diseases: The diseases which last for a short periods of time, are called acute diseases. example : Common cold, influenza, typhoid fever, cholera etc. These diseases if not treated well in time can be fatal. Chronic diseases: Some diseases which last for a long time i.e. one year or more or in’ some cases even as much as lifetime. Examples : Tuberculosis, elephantitis, AIDS, cancers etc. Chronic disease causes prolonged general poor health.

Question 6.
A baby is suffering from loose motions. What are the immediate and contributory causes?
Answer:

• The immediate cause of loose motions is an infection with a virus or bacteria.
• The lack of good nourishment or poor nourishment and genetic difference are the other contributory causes of the disease. Because without virus, the genetic difference and poor nourishment alone would not lead to loose motions.
• The poor public services such as supply of uncleaned drinking water in the area where the family lives, is the third possible cause of disease.

Question 7.
What are infectious diseases and non infectious diseases? Give examples?
Answer:
Infectious diseases. Diseases where microorganism or microbes are the immediate causes are called inflectious diseases. This is because these microbes can spread in the community, and the diseases they cause will spread with them. Example: common cold, influenza, dengue fever, cholera, tuberculosis and anthrax etc.

Non-infectious diseases: These are diseases that are neither caused by infectious agents like microbes nor can spread in the community. Instead, the cause of disease are mostly internal i. e. malfunctioning of body organs.
Example : High blood pressure, Cancer, Stones in different body organs, etc.

Question 8.
Name the different diseases caused by the following organisms.
Answer:

Question 9.
How does cholera spread from an infected person to a healthy person?
Answer:
Cholera spread from one person to another through water. This occurs if tire excreta from one person suffering from cholera, get mixed with drinking water used by the other healthy people living nearby. The cholera causing microbe Vibrio cholerae will enter new host through the water, they drink and cause disease in them.

Question 10.
How is the AIDS transmitted from one person to another?
Answer:
AIDS is transmitted from a person to another by the following ways,

• By sexual contact from an infected partner to another.
• The AIDS virus can also spread through blood-to-blood contact with infected person.
• It can also spread from an infected mother to her baby during pregnancy or through breastfeeding.

Question 11.
Give some ways in which AIDS does not spread from one person to another.
Answer:
AIDS does not spread from one person to another in the following ways.
AIDS is not transmitted by handshakes or hugs or sports like wrestling, or by any of the other ways, we touch each other socially.

Question 12.
How does malaria spread from one person to another person?
Answer:
Malaria spread from infected person to healthy person by the bite of a female anopheles mosquito. These insects carry the infecting agent plasmodium from sick person to another healthy person and transfer disease from person to person.

Question 13.
What are air borne diseases? Give examples.
Answer:
The diseases which spread from infected person to healthy person through air are called air borne diseases. For example. Common cold, pneumonia and tuberculosis.

Question 14.
How do the sign and symptoms of a disease depend upon the organ affected by microbes? Explain with example.
Answer:
The signs and symptoms depends upon the tissues or organ which the microbe targets. For example if the lungs are targeted, then symptoms will be cough and breathlessness. If the liver is targeted, there will be jaundice. If the brain is the target, we will observe headache, vomiting, fits or unconsciousness.

Question 15.
What is AIDS? What is die cause of this disease? What are effects of HIV-AIDS?
Answer:
AIDS (Acquired Immuno Deficiency Syndrome) is a chronic fatal disease in which immune system of the body is seriously affected. The cause of this disease is a virus which is known as HIV (Human Immuno Deficiency Virus). Many effects of HIV-AIDS are because the body can no longer fight off the many minor infections that we face everyday. Instead, every small cold can become pneumonia. Similarly minor gut infections can produce major diarrhoea, with blood loss. Ultimately, it is these other infections that kill person suffering from HIV-AIDS.

Question 16.
How does the severity of disease manifestations depend upon number of microbes in the body?
Answer:
The severity of disease depends upon the number of microbes in the body. If the number of microbes is very small, the disease manifestations may be minor, or may go unnoticed. But if the number of microbes is very large, the disease can be severe enough to be life threatening.

Question 17.
Explain why is it harder to make antiviral medicines than antibacteria medicines.
Answer:
One reason why making antiviral medicines is harder than making anti bacterial medicines is that viruses have few biochemical mechanisms of their own. They enter our cells and use our machinery for their life processes. This means there are few virus specific targets to aim at. But bacterial synthesis pathways are different from that used by our cells. So we can easily find a drug that blocks the multiplication of bacteria without affecting our body. This is achieved by antibiotics.

Question 18.
Why is prevention of diseases better than their cure? Give three reason.
Answer:
Prevention of diseases is better than their cure because

1. If someone has a disease, their body functions are damaged and may never recover completely.
2. The treatment of disease will take time, which means that person suffering from a disease is likely tube Wed ridden for sometime even if we can give proper treatment.
3. The disease, from which a person is suffering, may spread to other person.

Question 19.
What type of drugs should be given to a patient suffering from diseases caused by bacteria and why?
Answer:
Antibiotics should be given to a patient suffering from diseases caused by bacteria because they commonly blocks the biochemical pathways important for bacteria. Bacteria make cell-wall to protect themselves. The antibiotic penicillin blocks the bacterial processes that build the cell-wall. As a result growing bacteria became unable to make cell-walls and die easily.

Question 20.
What type of diseases are caused by the following.
(a) If lungs are infected by bacteria.
(b) If liver is infected by virus.
(c) If brain is infected by the virus.
Answer:
(a) Tuberculosis.
(b) Jaundice.
(c) Japanese encephalitis or meningitis.

Question 21.
What are the principles of treatment?
Answer:
The principles of treatment are as follows:

1. First the doctor should know the cause of the diseases by examining the patient.
2. The doctor should provide treatment that reduce the effect of the disease.
3. The doctor should prescribe drugs which can kill the microbes or stop the proliferation of disease causing microbes.

Question 22.
What are two general ways to prevent diseases.?
Answer:
General ways to prevent diseases are as follows:

1. By preventing exposure to infectious microbes.
2. The second basic principle of prevention of infection disease is the availability of proper and sufficient food for everyone, so that the functioning of the immune system remain good.

Question 23.
What is the specific way of preventing infectious diseases?
Answer:
The specific way of prevention of infectious disease is to strengthen the immune system of the body by giving some vaccines. Many vaccines are now available for preventing a whole range of inflections diseases and provide a disease specific means of preventions. There are vaccines against tetanus, diphtheria, measles, polio and many others.

Question 24.
There is a vaccine for hepatitis A, in the market. But the majority of children in many parts of India are already immune to hepatitis A by the time they are five years old? Why is it so?
Answer:
Some hepatitis viruses, which cause jaundice are transmitted through water. The majority of the children in many parts of India are exposed to the hepatitis A virus through water the time they are five years old. So, they become immune to hepatitis A.

Question 25.
If you had chickenpox once, there is no chance of suffering from it again. Why?
Answer:
This happens because when the immune system first sees an infectious agent or microbe, it responds against it and then remembers it specifically. So, the next time that particular microbe, or its close relatives enter the body, the immune system responds with even greater vigour. This eliminates the infection even more quickly then the first time around. So, having the disease once is a means of preventing subsequent attacks of the same disease.

Question 26.
Name the physician who discovered the vaccine against smallpox and cowpox?
Answer:
English physician named Edward Jenner, discovered the vaccine against smallpox. Dr. Jenner realised that milkmaids who had cowpox did not catch the smallpox even during epidemics. Cowpox is a very mild disease. Jenner tried deliberately giving cowpox virus to the people and found that they were now resistant to smallpox. This was because smallpox virus is closely related to cowpox virus. In this way Dr. Jenner discovered the vaccine against smallpox.

Long Answer Type Questions

Question 1.
Why do we fall ill? What does disease look like.
Answer:
There are many tissues in the body. These tissues make up organ systems that carry out body functions. Each of the organ systems has specific organs as its parts, and it has particular functions. For example digestive system has the stomach and intestine, and it helps to digest food taken in from outside the body. The musculoskeletal system, which is made up of bones and muscles, holds the body parts together and helps the body move.

When there is any disease either the functioning or the appearance of one or more systems of the body will change for the worse. These changes give rise to symptoms and sign of diseases. Symptoms of disease are the things we feel as being wrong, i.e, if our digestive system is not good, we have loose motion or constipation or pain in the stomach or some gastric trouble.

If our lungs are infected than we loose our weight, become short of breath, feel tired all the time, have fever, cough. These symptoms indicate the tuberculosis of lungs. If we have headache it means just examination stress for any other type of stress or, very rarely, it may mean meningitides or any of the other disease on the basis of certain laboratory test.

Only symptoms do not indicate what the disease is. On the basis of symptoms, the physician get laboratory test i.e. blood sugar, blood pressure, x-ray of a certain organ etc. to pinpoint the disease further.

Question 2.
What do you mean by the (a) immediate cause and (b) contributory cause of a particular disease ? Explain with the help of an example.
Answer:
There are many causes of a particular disease. These causes inlude:
(a) Immediate cause. Immediate cause of a particular disease may be a infectious agents mostly microorganism such as virus, bacteria, worms or protozoans etc. The disease where microbes are the immediate cause are called infectious diseases.

In some diseases immediate cause are not infectious agent such as some cancer. These are caused by genetic abnormalities or heart diseases caused by malfunctioning of heart.

(b) Contributory causes. The microbes can spread in the community due to unhygienic condition or poor nourishment. Thus, unhygienic conditions, poor nourishment are the contributory causes.

Example. If there is baby suffering from loose motions, we can say that the cause of the loose motions is an infection with a virus. So, the immediate cause of the disease is a virus. We find that this virus came through unclean drinking water. But many babies must have had this unclean water. So, why is it that only one baby developed loose motion while the other babies did not.

One reason for the infection is that baby is not healthy because he is not well nourished and does not get balanced diet. So, genetic difference, lack of good nourishment and unclean drinking water are contributory causes of loose motion. Without the virus, the poor nourishment or genetic difference alone may not lead to loose motion.

Question 3.
What are the different types of infectious micro-organisms which cause infectious diseases? Also give the diagrams and the diseases caused by some of these organisms.
Answer:
Organism that can cause disease are found in a wide range of categories of classification. Some of these are viruses, bacteria, fungi, single-celled animals or protozoans and worms of different kind.

(i) Virus: All virus live inside host cells and multiply very quickly. The diseases caused by viruses are common cold, influenza, dengue fever SARS and AIDS.
The picture of the SARS virus coming out of the surface of an infected cell is shown below.

(ii) Bacteria: Bacteria are organism which are closely related to each other which means many important life process are similar in the bacteria group. Diseases like typhoid fever, cholera, tuberculosis and anthrax are caused by bacteria.

The picture of staphylococci, the bacteria which can cause acne, is shown below.

(iii) Protozoans: Protozoans are single celled animals. These microbes causes many familiar diseases such as malaria and kala-azar. Some protozoa also cause sleeping sickness. The protozoa, leishmania is shown in the figure causes kalaazar. The organism is oval shaped and each has one long whip-like structure.

(iv) Worms: Some worms like roundworm, tapeworm live in the intestine. Ascariasis and Taeniasis diseases are caused by the roundworm and tapeworm respectively. The disease elephantiasis is also caused by a species of worms.
The pictures below shows the roundworm. This type of roundworm live in small intestine.

Question 4.
How do infectious diseases spread from infected person to healthy person? Give any four ways with examples.
Answer:
Many microbial agents can commonly move from an affected person to someone else (healthy person) in different ways. These diseases can be communicated from one person to another, so these diseases are called communicable diseases. These communicable diseases can spread by the following ways.

(i) By air. Common cold, preumonia and tuberculosis spread through air from one person to another. This occurs through the little droplets thrown out by an infected person who sneezes or coughs. Some one standing close by can breathe in these droplets and the microbes get a chance to start new infection. Obviously, the more crowded our living conditions are, the more likely it is that such air bone diseases will spread.

(ii) By water. The diseases like cholera, jaundice spread through water. This occurs if the excreta from some one suffering from an infectious disease, such as cholera, get mixed with the drinking water used by the people living nearby. The cholera causing microbes will enter new hosts through the water they drink and cause diseases in them.

(iii) By sexual contact. Some diseases like HIV- AIDS are transmitted by sexual contact from one partner to the other. However, such sexually transmitted diseases do not spread by physical contact such as hand shakes or hugs or sports like wrestlings.

(iv) By Blood to Blood contact: The AIDS virus can also spread through blood-to-blood contact from an infected people. It can also spread from an infected mother to her baby through breast feeding or during pregenancy.

(v) By other creature or animals. Many diseases are transmitted from infected people to the other healthy people through other animals. These animals carry the infecting agents from a sick person to another potential host or healthy person. These animals are called vectors.

The commonest vectors are mosquitoes. The female mosquitoes need highly nutritious food in the form of blood in order to be able to lay mature eggs. Thus, the malaria causing microbes entering through a mosquito bite, will go to the liver and then to the red blood cells. The virus causing Japanese encephalitis or brain fever also enter through mosquito bite.

Question 5.
What are the principles of treatment of infectious diseases? Explain with the help of examples.
Answer:
There are two ways to treat an infectious diseases:
(i) One would be to reduce the effect of the disease; we can provide the treatment that will reduce the symptoms. The symptoms are usually because of inflammation. For example, we can take medicine that bring down the fever, reduce pain or loose motions. We can take bed rest so that we can conserve our energy. But this kind of symptom-directed treatment by itself will not make the infecting microbes go away and the disease will not be cured.

(ii) The second way is to use medicine that kill the microbes which are main cause of disease. Microbe are classified in different categories. As each groups of microbes have some essential biochemical life process which is peculiar to that group and not shared with the other group. These pathways or processes are not used by us either.

For example : Our cells may make new substances by a mechanism different from that used by the bacteria or virus. We have to find a drug that blocks the bacterial synthesis pathway without affecting our own. The antibiotic drugs blocks the synthesis pathways related to the bacteria. Similarly, there are drugs which kill the protozoa such as malarial parasite. There are antiviral drugs also which can keep HIV infection under control.

Question 6.
What are the principles of preventing infectious diseases?
Answer:
The principles of preventing infectious diseases are as follows:
1. The general way of preventing infectious diseases mostly relate to the preventing of exposure to the infectious microbes.

For Example: In case of airborne microbes, we can prevent ourselves by providing living condition that are not over crowded. For water home microbes, we can prevent exposure by providing safe drinking water. The. drinking water should be filtered and free from any microbial contamination.

For vector-borne diseases, we require clean environment or we can say that the public hygiene is the basic key to prevention of infectious diseases. Clean environ-ment would not allow any mosquito breeding. We can prevent the spreading of disease like malaria and Japanese fever etc.

2. The second basic principle of prevention of infectious diseases is the availability of proper and sufficient food for everyone. The functioning of our immune system in our body will not be good if proper and sufficient nourishment food is not available. The immune cell manage to kill the microbes. If the number of infecting microbes are controlled, the manifestation of disease will be minor.

3. The infectious diseases can be prevented by immunisation. In general, we befool the immune systems to develop a memory for a particular infection by putting something, that mimic the microbes we want to vaccinate against, into the body. This does not actually cause the disease but this would prevent any subsequent exposure to the infecting microbe from turning into actual disease.

Many such vaccines e.g. vaccines against tetanus, diptheria, measles, polio etc. are available. These provide a disease-specific means of prevention.

These NCERT Solutions for Class 9 Science Chapter 12 Sound Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Sound NCERT Solutions for Class 9 Science Chapter 12

Class 9 Science Chapter 12 Sound InText Questions and Answers

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way, vibrations produced by an object are transferred from one particle to another till it reaches the ear.

Question 2.
Explain how sound is produced by your school bell.
Answer:
When the school bell vibrates, it forces the adjacent particles in air to vibrate. This disturbance gives rise to a wave and when the bell moves forward, it pushes the air in front of it. This creates a region of high pressures known as compression. When the bell moves backwards, it creates a region of low pressure know as rarefaction. As the bell continues to move forward and backward, it produces a series of compressions and rarefactions. This makes the sound of a bell propagate through air.

Question 3.
Why are sound waves called mechanical waves?
Answer:
Sound waves force the medium particles to vibrate. Hence, these waves are known as mechanical waves. Sound waves propagate through a medium because of the interaction of the particles present in that medium.

Question 4.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
Sound needs a medium to propagate. Since the moon is devoid of any atmosphere, you cannot hear any sound on the moon.

Question 5.
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.

Question 6.
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) × Frequency (u)
v = λ × u

Question 7.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Ans. Frequency of the sound wave, υ = 220 Hz
Speed of the sound wave, v = 440 m s^-1
For a sound wave,
Speed = Wavelength × Frequency
v = λ × u
∴ λ = \(\frac{v}{u}=\frac{440}{220}\) = 2 m
Hence, the wavelength of the sound wave is 2m.

Question 8.
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 = \(\frac{1}{\text { Frequency }}=\frac{1}{500}\) = 0.002s

Question 9.
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.

Question 10.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Answer:
Speed of sound, v = 342 ms^-1
Echo returns in time, t = 3s
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 = \(\frac{1026}{2}\) m = 513 m

Question 11.
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.

Question 12.
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. Humans cannot hear sounds having frequencies less than 20 Hz and greater than 20,000 Hz.

Question 13.
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.

Question 14.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in saltwater 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 saltwater, v = 1531 ms^-1
Total distance covered by the sonar pulse = Speed of sound × Time taken
Total distance covered by the sonar pulse
= 1.02 × 1531 = 1561.62 …..(i)
Let d be the distance of the cliff from the submarine.
Total distance covered by the sonar pulse = 2d
⇒ 2d = 1561.62 [From(i)]
⇒ d = 780.81 m

Class 9 Science Chapter 12 Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound is produced by vibration. When a body vibrates, it forces the neighbouring particles of the medium to vibrate. This creates a disturbance in the medium, which travels in the form of waves. This disturbance, when reaches the ear, produces sound.

Question 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 creates a 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 (as shown in the following figure).

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Answer:
Take an electric bell and hang this bell inside an empty bell-jar fitted with a vacuum pump (as shown in the following figure).

Initially, one can hear the sound of the ringing bell. Now, pump out some air from the bell-jar using the vacuum pump. It will be observed that the sound of the ringing bell decreases. If one keeps on pumping the air out of the bell-jar, then at one point, the glass-jar will be devoid of any air. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. When there is no air present inside, we can say that a vacuum is produced. Sound cannot travel through vacuum. This shows that sound needs a material medium for its propagation.

Question 4.
Why is sound wave called a longitudinal wave?
Answer:
The vibration of the medium that travels along or parallel to the direction of the wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Hence, a sound wave is called a longitudinal wave.

Question 5.
Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
Quality of sound is that characteristic which helps us identify a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.

Question 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 × 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.

Question 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 ms^-1.
Answer:
For a sound wave,
Speed = Wavelength × Frequency
v = λ × υ
Given that the speed of sound in air = 344 m/s
(j) For, υ = 20.Hz
λ1 = \(\frac{v}{v_{1}}=\frac{344}{20}\) = 17.2m
(ii) For, υ₂ = 20,000 Hz
λ₂ = \(\frac{v}{v_{2}}=\frac{344}{20,000}\) = 0.0172 m
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Question 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:
Let the length of the aluminium rod be d.
Speed of sound wave in aluminium at 25°C,
v_Al = 6420 ms^-1
Therefore, time taken by the sound wave to reach the other end,
\(t_{\mathrm{Al}}=\frac{d}{v_{\mathrm{Al}}}=\frac{d}{6420}\)
Speed of sound wave in air at 25°C, vAir = 346 ms^-1 .
Therefore, time taken by sound wave to reach the other end,
\(t_{\text {Air }}=\frac{d}{v_{\text {Air }}}=\frac{d}{346}\)
The ratio of time taken by the sound wave in air and aluminium: \(\frac{t_{\text {Air }}}{t_{\mathrm{Al}}}=\frac{\frac{d}{346}}{\frac{d}{6420}}=\frac{6420}{346}=18.55\)

Question 9.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency is defined as the number of oscillations per second. It is given by the relation:
Frequency = \(\frac{\text { Number of oscillations }}{\text { Total Time }}\)
Number of oscillations = Frequency × Total time
Given, Frequency of sound = 100 Hz
Total time = 1 min = 60 s
Number of oscillations / Vibrations = 100 × 60 = 6000
Hence, the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.

Question 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.

Question 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 interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Hence, on a hotter day, the time interval between the original sound and the reflected sound will decrease. Therefore, an echo can be heard only if the time interval between the original sound and the reflected sound is greater than 0.1 s.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:

• 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.

Question 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 ms^-2and speed of sound = 340 ms^-1.
Answer:
Height of the tower, s = 500 m
Velocity of sound, v = 340 ms^-1
Acceleration due to gravity, g = 10 ms^-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:
s = ut1 + \(\frac {1}{2}\)gt₁²
500 = 0 × t1 + \(\frac {1}{2}\) × 10 × t₁²
r₁² = 100
t₁ = 10 s
Now, time taken by the sound to reach the top from the base of the tower, t₂ = \(\frac {500}{340}\) = 1.47S
Therefore, the.splash is heard at the top after time, t
Where, t = t₁ + t₂ = 10 + 1.47 = 11.47 s

Question 14.
A sound wave travels at a speed of 339 ms^-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 ms^-1
Wavelength of sound, λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × υ
∴ υ = \(\frac{v}{\lambda}=\frac{339}{0.015}\) = 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.

Question 15.
What is reverberation? How can it be reduced?
Answer:
Persistence of sound (after the source stops producing sound) due to repeated reflection is known as reverberation. As the source produces sound, it starts travelling in all directions. Once it reaches the wall of a room, it is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Due to this, sound can be heard even after the source has ceased to produce sound.

To reduce reverberations, sound must be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibre board, rough plastic, heavy curtains, and cushioned seats can be used. to reduce reverberation.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
A loud sound has high energy. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Question 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.

Question 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.

Question 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 underwater objects such as submarines and shipwrecks with the help of ultrasounds. It is also used to measure the depth of seas and oceans.

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 seawater. 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 (f) 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’.

Question 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 = 5s
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 sovmd in water,
v = \(\frac{2 d}{t}=\frac{2 \times 3625}{5}\) = 1450 ms^-1

Question 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.

Question 22.
Explain how the human ear works.
Answer:
Different sounds produced in our surroundings are collected by pinna that sends these sounds to the eardrum via the ear canal. The eardrum starts vibrating back and forth rapidly when the sound waves fall on it. The vibrating eardrum sets the small bone hammer into vibration. The vibrations are, passed from the hammer to the second bone anvil, and finally to the third bone stirrup. The vibrating stirrup strikes on the membrane of the oval window and passes its vibration to the liquid in the cochlea. This produces electrical impulses in nerve cells. The auditory nerve carries these electrical impulses to the brain. These electrical impulses are interpreted by the brain as sound and we get a sensation of hearing.

Class 9 Science Chapter 12 Sound Additional Important Questions and Answers

Multiple choice Question
Choose the correct option:

Question 1.
Note is a sound
(a) of mixture of several f.requencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) a ways unpleasant to listen
Answer:
(c) of a single frequency

Question 2.
A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case
(a) sound will be louder but pitch will not be different
(b) sound will be louder and pitch will also be higher
(c) sound will be louder but pitch will be lower
(d) both loudness and pitch will remain unaffected
Answer:
(a) sound will be louder but pitch will not be different

Question 3.
In SONAR, we use
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves
(d) audible sound waves
Answer:
(a) ultrasonic waves

Question 4.
Sound travels in air if
(a) particles of medium travel from one place to another
(b) there is no moisture in the atmosphere
(c) disturbance moves
(d) both particles as well as disturbance travel from one place to another.
Answer:
(c) disturbance moves

Question 5.
When we change feeble sound to loud sound we increase its
(a) frequency
(b) amplitude
(c) velocity
(d) wavelength
Answer:
(b) amplitude

Question 6.
In the curve (Fig.12.1) half the wavelength is

Answer:
(b)

Question 7.
Earthquake produces which kind of sound before the main shock wave begins
(a) ultrasound
(b) infrasound
(c) audible sound
(d) none of the above
Answer:
(b) infrasound

Question 8.
Infrasound can be heard by
(a) dog
(b) bat
(c) rhinoceros
(d) human beings
Answer:
(c) rhinoceros

Question 9.
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting
(a) intensity of sound only
(b) amplitude of sound only
(c) frequency of the sitar string with the frequency of other musical instruments
(d) loudness of sound
Answer:
(c) frequency of the sitar string with the frequency of other musical instruments

Very Short Answer Questions

Question 1.
What is sound?
Answer:
It is a form of energy that enables us to hear.

Question 2.
What is necessary for a body to produce sound?
Answer:
The vibrating bodies can only produce sound.

Question 3.
Which three parameters are used to describe sound?
Answer:
The three parameter used to describe sound are amplitude, frequency and quality.

Question 4.
What determines the loudness of a sound wave?
Answer:
Amplitude determines the loudness of a sound.

Question 5.
What determines the pitch of a sound.
Answer:
The frequency of sound wave determines the pitch.

Question 6.
What are two types of mechanical wave motions?
Answer:
The two types of mechanical wave motions ate transverse waves and longitudinal waves.

Question 7.
Name the wave motion in which the wave propagates in the direction of motion.
Answer:
In longitudinal wave motion, the waves propagate in the direction of motion.

Question 8.
Why is sound wave called a mechanical wave?
Answer:
Sound cannot travel through vacuum, it requires the presence of a medium for its propagation.

Question 9.
In which physical medium, the sound travels the fastest?
Answer:
Sound travels the fastest in solids.

Question 10.
What are curtains and furniture in a house, good reflectors or absorbers of sound?
Answer:
They are good absorbers of the sound.

Question 11.
What do you understand by the reflection of sound?
Answer:
It refers to the bouncing back of the sound in the same medium after striking a non-absorptive solid surface.

Question 12.
Which part of human body helps produce the sound?
Answer:
The vocal cord also called Adam’s apple help produce the sound.

Question 13.
In which medium will the reflection of sound be faster, air or water?
Answer:
In water, the density of water is more than that of air and hence, the speed of sound in water is more than in air.

Question 14.
Name the property of sound, human ears respond to after hearing the sound.
Answer:
Human ears are receptive to the loudness therefore, they only respond to loudness.

Question 15.
Which property is used to distinguish two sounds from different sources but having the same amplitude and pitch ?
Answer:
The property used to distinguish the sound waves having same pitch and frequency is called quality.

Question 16.
In which form the sound waves travels across a medium.
Answer:
The sound waves travel in the form of. compression and rarefaction.

Question 17.
How is an ultrasonic wave different from the infrasonic wave?
Answer:
Infrasonic waves have the frequency less than 20Hz while the ultrasonic waves have the frequency of more than 20KHz.

Question 18.
If the speed of sound incident on a surface is doubled, will there be any effect on the angle of reflection of sound ?
Answer:
No, there will be no effect on the angle of reflection of sound because it is dependent on the angle of incidence, not on the speed of sound.

Question 19.
What will happen, if a sound wave is made incident at 90° on a solid surface?
Answer:
No reflection will occur but the development of resonance Will take place.

Question 20.
Can sound waves be transformed into » electrical impulses or vice versa?
Answer:
Yes, the sound waves can be transformed into electrical impulses or vice versa as it happens in telephone.

Question 21.
What is the law of conservation of energy?
Answer:
The law states that energy can neither be created nor be destroyed but one form of energy can be transformed into another form of energy.

Question 22.
State the energy transformations which take place when you clap your hands.
Answer:
When dapping, the muscular energy is transformed into sound and heat energy.

Question 23.
How is a light wave different from a sound wave?
Answer:
Sound wave requires a medium for its propagation but not light wave. It can even pass through a vacuum.

Question 24.
The distance in between two adjoining crest and trough is ‘d’. What is the wavelength of the wave?
Answer:
The wavelength is the distance in between two successive troughs or crests, therefore the wavelength would be 2d.

Question 25.
What do you mean by supersonic speed?
Answer:
When the speed of an object/body exceeds the speed of sound, then the speed is called supersonic speed.

Short Answer Type Questions

Question 1.
What is a wave? Can sound be visualized as a wave?
Answer:
Wave is a disturbance that moves through a medium due to repeated oscillatory motion of the particles of the medium about their mean position. This oscillatory motion is passed over from one particle to another progressively. Hence, a wave only involves the transfer of energy not particles of the material medium.

Sound is visualized as a wave because the disturbance set by the sound in the medium travel through the medium in form of energy instead of the particle of the medium.

Question 2.
What are longitudinal waves? Give an example.
Answer:
The waves in which the particles of the medium vibrate in the direction of the propagation of the wave are called longitudinal waves. During propagation of these waves, the particles of the medium only exhibit the vibratory motion along their mean position.

Sound waves propagate in the form of longitudinal waves.

Question 3.
What are transverse waves? Give an example.
Answer:
The waves in which the particles of the medium vibrate in the direction perpendicular to the direction of the propagation of wave is called transverse wave. The positive displacement from the mean position makes up the crest while the negative displacement from the mean position makes up the trough.

When a pebble is dropped in stagnant water, water ripples formed on the surface water represents the transverse waves.

Question 4.
What are mechanical and non¬mechanical waves? Give examples.
Answer:
The waves which require the presence of a material medium for their propagation are called mechanical or elastic waves e.g. sound waves. These waves can propagate through all the three states of matter but not through a vacuum.

The waves which do not require the presence of a material medium for their propagation are called non-mechanical waves or non-elastic waves e.g. light waves, microwaves or radio waves. These waves can easily propagate across the vacuum.

Question 5.
How is the loudness or softness of a sound determined? Give the wave shape of a loud and soft sound of same frequency.
Answer:
The loudness or softness of a sound depends upon its amplitude. A loud sound has higher amplitude than a soft sound when the amplitude describes the maximum displacement of the particle from its mean position.

Question 6.
The sound produced by a vehicle and a flute travels through the same medium, air and arrive at the ear at same time. Are the two sounds different? If yes, give reasons.
Answer:
No, the two sounds are different. The pitch of the sound is one of the characteristics responsible for the difference. The sound from the flute has higher frequency and has pleasing effect but the sound from vehicle has low pitch. It is cracking sound and causes irritation.

Question 7.
Different musical instruments produce different sounds. Why ?
Answer:
The different musical instruments have different modes of vibrations. These instruments with different sizes under different conditions vibrate at different frequency producing sounds of different pitches and hence, they are different from one another.

Question 8.
Give a graphical representation of for a sound wave having low pitch and high pitch.
Answer:
The pitch of a sound is determined by its frequency. Therefore, a sound with low pitch has low frequency i.e. number of the waves produced per second and sound with higher pitch has higher frequency. (Fig. 12.6).

Question 9.
In humans, whose voice is sharp, a male or female? Why?
Answer:
In humans, the voice of a female is sharper than the voice of a male because the sound produced by a female has higher frequency i.e. pitch than the sound produced by a male.

Question 10.
A person standing on the railway platform of a rural area could neither see nor hear the sound of the train. With none to help him, how can he assess the arrival time of the train?
Answer:
The person can assess the arrival time of the train by keeping his ear close to the rail line and sense the vibrations of the incoming train.

In steel, the speed of sound is 5960 m/s as compared to the speed of sound in air which is only 340 m/s. Hence, he can easily estimate the possible arrival time of the train. The fast vibration would indicate the early arrival while the slow vibrations would indicate the late arrival.

Question 11.
A person standing 1000 m away from a siren hears the sound. When will he hear the sound earlier and why?
(i) On a hot day or a calm day?
(ii) On a dry day or cloudy day having same temperature?
Answer:
(i) The person will hear the sound earlier on a hot day than on a calm day because with increasing temperature, the speed of the sound increases. At 273 K, the speed of sound is 331 m/s while at 295 k, it is 344 ms.

(ii) The person will hear the sound earlier on a cloudy day than on a dry day because the speed of the sound depends upon the density of the medium of propagation. On a cloudy day because of the presence of the moisture in air, its density is more than on a dry day.

Question 12.
What is the audible range for the human beings?
Answer:
The human ears respond to the sound waves having frequency of 20 Hz to 20 kHz. Therefore, this range of sound waves is called audible range. However, children can hear the sound up to the frequency of 25 kHz while with increasing age the range declines as old people become less sensitive to higher frequency of sound, their maximum frequency range goes up to 15,000 Hz.

Question 13.
What are infrasound and ultrasound? Give examples.
Answer:
The classification of sound waves into infrasound and ultrasound is based on the human audible range of 20 Hz to 20 kHz. The sound waves having frequency less than 20 Hz are called infra sounds and the sound waves having frequency greater than 20 kHz tire called ultrasounds.

Rhinoceros, whales and elephants can produce and respond to the infrasound waves.

Bats and rats can produce and receive ultrasound waves. Rats when playing use ultrasound waves while bats use ultrasound waves for flying in dark and capture their prey.

Question 14.
What is sonic boom? Is it harmful, if yes then give reason?
Answer:
When sound producing source moves with the speed of sound or above such as some fighter aircrafts, it produces shock wave in air. These shock waves contain a large amount of energy. The air pressure variations associated with this type of shock wave sharp and loud sound called sonic boom.

The shock waves of sonic boom possess lot of energy. They can cause the shattering of glass windows and damage the buildings.

Question 15.
What do you mean by the reflection of sound? What are the laws of reflection of sound?
Answer:
The reflection of the sound refers to the bouncing back of sound waves after being incident on a hard polished surface in the same medium.

According to the laws of the reflection of sound:

• The incident sound wave, reflected sound wave and the normal drawn at the point of incidence are in the same plane.
• The direction in which the sound is incident and is reflected make equal angle with the normal at the point of incidence.

Question 16.
What are megaphones? Why the loudness of sound is increased by megaphones?
Answer:
A megaphone is a simple horn shape tube followed by a conical opening.
In a megaphone, the sound waves from a source are reflected successively from the conical surface of the tube and directed towards audience without spreading the sound. Moreover, the amplitude of the sound waves adds up increasing the loudness of sound.

Question 17.
What is an echo? State the conditions necessary for echo formation.
Answer:
An echo refers to the reflected sound that is reheard by a speaker/listener himself. For the echo formation to occur

• there have to be a good reflective surface to reflect the sound.
• the minimum distance between the source of sound and the reflective surface has to be 17.2 m.
• there should not be any obstruction in the path of speaker and the reflective surface.

Question 18.
What is the minimum distance between the listener and the reflecting surface for hearing the distinct echo? Why?
Answer:
In humans, the time period of the persistence of sound is 0.1 second. The speed of sound in air at 295 k (22°C) is 344 m/s.

For a listener to hear a distinct echo, the reflected sound should reach his ears after a period of 0.1s.

Hence, the total distance sound of echo has to cover is = 344 × 0.1 = 34.4 m.
Therefore, the minimum distance of the reflecting surface has to be 344/2 = 17.2 m

However, this difference is temperature dependent because with the increasing temperature, the speed of the sound increases.

Question 19.
Explain why the walls and roof of the auditorium are covered with sound absorbing material?
Answer:
The walls and roof of a good auditorium are covered with sound absorbing material such as fibre boards and draperies to prevent the reverberation i.e., the repeated reflection of the sound waves. The reverberation makes the sound blurred and difficult to understand and interpret.

Question 20.
Why is the flash of lightning seen much before the thunder is heard on a rainy day, although both occur simultaneously?
Answer:
Although both lightning and thunder formation occur simultaneously at the same height in air yet the lightning flash is seen much before the hearing of thunder sound because of the difference in the speed of the sound and light The sound traveling with speed of 342 m/s takes a longer time than light traveling at the speed of 3 × 10⁸ m/s to cover the same distance.

Question 21.
State some of the practical applications of echo.
Answer:
An echo which is a reflected sound has its own applications such as

• In echo ranging or sonar, it is used to determine the depth of the oceans.
• The submarines floating in water not only measure the depth of the ocean but also the obstruction in their path in front if any.
• Bats use the echo or sound reflection in finding an obstruction free path for flying and capturing their prey.
• In medical sciences the ultrasound waves are used widely in diagnosis of structural disorders of body parts.

Question 22.
What is ultrasonography? State its use.
Answer:
Ultrasonography is a technique in which the ultrasonic waves are used to asses the structure of a body part or tissue. The ultrasound waves are made to travel through tire body tissues. Where ever there is a change in tissue density, these waves get reflected, then the reflected waves are transformed into electrical signals to generate the picture of the tissue.

The technique is widely used in examination of the growing foetus in womb of a pregnant mother and in analysis of stones in different body organs such as gall bladder, liver, etc.

Question 23.
State some of the advantages of the ultrasounds in medical sciences.
Answer:
Ultrasound consists of sound waves having frequency greater than 20 kHz. These waves are used in

• Echo-cardiography, a technique used to diagnose the blockage of heart valves or arteries.
• Creating pictures of different body organs to detect the presence of stone or any other structure like tumor.
• In breaking the small stones formed in kidney for their easy removal.

Question 24.
What is a stethoscope? On what principle does it work?
Answer:
A stethoscope is a medical instrument used by a doctor to hear the heart sounds, ‘Lub- Dub’. It’s working is based on the repeated and multiple reflection of the sound waves, received by the broad round receiver.

Question 25.
How is pressure variation in a sound wave amplified in human ear?
Answer:
The pressure vibrations produced by a vibrating object in form of compression and rarefaction reach inside the external ear and make the eardrum to vibrate. The three bones of the middle year: hammer, anvil and stirrup being solid causes the amplification of the waves by several times.

Question 26.
A longitudinal wave is produced on a toy slinky. The wave travels at the speed of 30 cm/s. If the frequency of the wave is 20 Hz. What is the minimum distance between the two consecutive compressions?
Answer:
Velocity of wave, v = 30 cm/s = 0.3 m/s
Frequency of wave, n = 20 Hz
Distance between two consecutive compressions = λ.
We know that v = λn
λ = \(\frac{v}{n}\)
\(\frac {0.30}{20}\) = 0.015 m = 15 cm

Question 27.
A child hears an echo from a cliff 4 seconds later after the sound from a powerful cracker is produced. How far away is the cliff from the child?
Answer:
Let the distance between the child and cliff = x
Total distance travelled by sound = 2x
Velocity of sound = 344 m/s
Velocity = \(\frac{\text { Distance travelled }}{\text { Time }}\)
344 = \(\frac{2 x}{4}\)
Or 2x = 688m

Question 28.
A sound wave have frequency of 2 kg Hz and wavelength of 35 cm. How long will it take to travel 1.5 km?
Answer:
Frequency, v = 2k Hz = 2000 Hz
Wavelength λ = 35 cm = 0.35 m
We know that, velocity of wave = frequency × wavelength
v = v × A.
v = 20 × 0.35 = 700 m/s
Let time taken by wave = 5 km = 1500 m
t = \(\frac{\text { Distanec travedlled }}{\text { velocity of wave }}\)
\(\frac {1500}{700}\) = 2.14 second

Question 29.
A person clapped hands near a mountain and heard the echo after 5s. What is the distance of the mountain from the person if the speed of sound at a given temperature is 346 m/s?
Answer:
Given
Speed of sound, v = 346 m/s
Time taken for hearing the echo, t = s
Let the distance between the person and mountain = x
We know that
Velocity = \(\frac{\text { Distanec travedlled }}{\text { time }}\)
346 = \(\frac{2 x}{5}\)
2x = 346 × 5 = 1730
x = \(\frac {1730}{2}\) = 865 m
Hence, distance between the person and mountain = 865 m Ans.

Question 30.
A ship sends out ultrasound produced by the transmitter that return from the sea bed and detected after 3.42s. If the speed of ultrasound through seawater is 1531 m/s. What is the distance of sea bed from the ship?
Answer:
Given
Time between transmission and detection, t = 3.42 s
Speed of the ultrasound in sea water, v = 1531 m/s
Let the distance of sea bed from ship = d
Distance travelled by ultrasound = 2 × d
We know that
v = \(\frac{\text { distance wavelength }}{\text { time taken }}\)
v = \(\frac{2 d}{t}\)
1531 = \(\frac{2 \times d}{3.42}\)
Hence, the distance of sea bed from the ship = 2618 m.

Question 31.
The frequency of a sound wave is 550 Hz. What is its wavelength? Sound travels with the a speed of 330 m/s. Calculate the time period of the wave also.
Answer:
Given
Frequency of sound wave, υ = 550 Hz
Velocity of sound wave, v = 330 m/s
Wavelength, λ = ?
λ = \(\frac{v}{υ}=\frac{330}{550}=\frac{3}{5} m\) = 0.6m
Time period = \(\frac{1}{υ}=\frac{1}{550}\)
= 0.0018 second.

Question 32.
If the velocity of sound, in a medium is 1400 m/s and its wavelength is 100m. What is its frequency? Can you hear this sound?
Answer:
Velocity of sound, v = 1400 m/s
Wavelength, λ = 100m
Frequency υ = ?
We know that, v = υ × λ
υ = \(\frac{v}{\lambda}=\frac{1400}{100}\) = 14 Hz
We can hot hear this sound, because we can hear the sound having frequency range from 20 Hz to 20000 Hz.

Question 33.
A longitudinal wave travels in a coiled spring or slinky at the rate of 4m/s. The distance between two consecutive compression is 20 cm. Find (i) wavelength of longitudinal wave (ii) frequency of longitudinal wave.
Answer:
Velocity of the longitudinal wave, v = 4 m/s
Distance between two consecutive compressions, wavelength, λ = 20 cm
= 0.2m
Frequency υ = ?
We know that
v = υ × λ.
υ = \(\frac{v}{\lambda}=\frac{4}{0.2}\) = 20 Hz

Question 34.
A body vibrating with a time period of 2 milli seconds produces a wave travelling with a velocity of 1250 m/s. What is the frequency of vibrating body? (ii) What is the wavelength of the travelling wave?
Ans.wer:
Given, Time period, T = 2 miliseconds
= 2 × 10^-3 S
Frequency, υ = \(\frac{1}{\mathrm{~T}}=\frac{1}{2 \times 10^{-3}}\) = 500Hz
Wavelenght of the wave, λ = ?
Velocity of the Wave, v = 1250 m/s
We know that
v = \(\frac{v}{υ}=\frac{1250}{500}\) = 2.5m

Question 35.
A Sonar echo takes 2.2 s to return from a whale. How far away is the whale? (Take speed of ultrasound to be 1531 m/s in seawater).
Answer:
Time taken by ultra sound between Transmission and reception, t = 2.2s
Speed of ultra sound wave, v = 1531 m/s
Depth of the whale from sea level = h
Distance travelled by ultrasound = 2 × h
We know that,
distance travelled = speed × time
2h = 1531 × 2.2
h = \(\frac{1531 \times 2.2}{2}\)
h = \(\frac{1531 \times 2.2}{2}\) = 1684.1 m
Hence, whale is 1684.1 away from the sea level

Long Answer Qestions

Question 1.
Give a simple activity to show that sound is produced by vibration.
Answer:
Suspend a smallplastic ball by a thread from a support. (Fig. 12.7) Take a tuning fork just touch, the ball with prong of the turning fork without setting it into vibration. There is no displacement in the ball. Now set the tuning fork into vibration by striking its prong with a rubber pad. Now touch the suspended ball with the prong of tuning fork. The ball get displaced from its position. Now, bring the tuning fork near your ear. You will hear sound. The ball get displaced due to thee vibration of the prong of the tuning fork. This activity shows that sound is produced due to vibration.

Fig.: Vibrating tuning fork just touching the suspended ball

Question 2.
Define speed, frequency and wavelength of a wave and derive the relation between them.
Answer:
Speed: The speed of a sound wave is defined as foe distance travelled by a point on a wave such as compression or a rarefaction in a unit time.
Speed = \(\frac{\text { distance }}{\text { time }}\)
Frequency: The number of compressions or rarefactions or crest and trough produced per unit time is known as foe frequency. It is denoted by

Wave length: The distance two consecutive compressions or rarefactions is called wavelength. It is denoted by X.
Or the distance between two consecutive crests or troughs is called wavelength.

Relation between speed, frequency and wavelength: We know that foe speed of sound wave is given by
Speed = \(\frac{\text { distance travelled by wave }}{\text { time taken }}\)
Now we known that foe wavelength (λ) is equal to foe distance travelled in one complete oscillation and foe time taken to complete one oscilaltion is called time period and is denoted by T.

Now distance travelled in time T = λ
Therefore, Speed = \(\frac{\lambda}{\mathrm{T}}\)
or v = \(\frac{\lambda}{\mathrm{T}}\) ………(i)
But we know that the frequency is the number of oscillation per unit time.
or frequency = \(\frac{1}{\text { Time Period }}\)
v = \(\frac{1}{\mathrm{~T}}\) ……….(ii)
Now putting foe value of \(\frac{1}{\mathrm{~T}}\) in equation (i) we get v = λ × \(\frac{1}{\mathrm{~T}}\)
v = λ × υ (∵ υ = \(\frac{1}{\mathrm{~T}}\))
Hence, Speed = wavelength × frequncy.

Question 3.
(i) What is reflection of sound? Prove that the reflection of sound follows the same law of reflection of light.
(ii) What are the uses of multiple reflection of sound?
Answer:
(i) Reflection of sound: When a sound wave strikes a solid or liquid surface, it is reflected back according to the laws of reflection i.e. the direction of incident and reflected sound makes equal angle with the normal to the reflecting surface and the three are in the same plane. These laws can be proved by the following activity.

Activity: Take two identical pipes as shown in the fig. 12.8. The length of the pipe should be sufficiently long (about 75 cm). Arrange them on a table near wall. Keep a clock near the open end of one of the pipes and try to hear the sound of the clock through the other pipe. Now adjust the pipes so that you can best hear the sound of the clock.

Now, measure the angle between the incident sound and the normal and the reflected sound and the normal to the reflecting surface. Now repeat the activity by changing the angles of the pipes. We find that the angle of indicent of sound wave is always equal to the angle of reflection of sound wave.

Now, becasue both the tubes are placed on the table, it also proves that incidient sound, normal and relfected sound lie th one plane.

(ii) Uses of multiple reflection of sound : (1) Megaphones or loudspeaker, horns, musical instrument such as trumpets and shehanais all designed in such a way to send sound in a particular direction. In these instruments the conical opening reflects sound successively and the amplitude of the sound wave adds up and the loudness of sound increases.
(ii) In stethoscope, the sound of heartbeat and lungs reaches the doctor’s ears by multiple reflection.
(iii) The ceilling of the conference hall and cinema halls are made curved so that sound after reflection reaches to all the corners of the halls.
(iv) Sometimes curved sound board is placed behind the stage so that sound after reflection from the sound board spread evenly across the width of the hall.

Question 4.
What are the application of ultrasound?
Answer:
The uses of ultrasound are as follows:
1. Ultrasound can be used to detect cracks and flaws in metal blocks. Ultrasonic waves are allowed to pass through the metal blocks and detectors are used to detect the transmitted wave. If there is a defect, the ultrasound gets reflected back indicating the cracks or flaws in the metal block.

2. Ultrasound scanner uses ultrasound wave for getting images of internal organs of human body such as liver, gall bladler, uterus, kidney etc. It helps the doctor for detection of stones in gall bladder and kidney and tumor in different organs. This technique is called ultrasonography. It is also used for examination of the foetus during pregnancy to detect congenial and growth abnormalities if any.

3. Ultrasound may be employed to break small stones formed in the kidneys into fine grains. These grains get flushed out with urine.

4. Ultrasound waves are made to reflect from the various parts of heart and form image of heart. This technique is called ‘Echocardiography’.

5. Ultrasound is also used to clean parts I coated in hard to reach places i.e. spiral tube, odd shapes parts of machines.

These NCERT Solutions for Class 9 Science Chapter 11 Work and Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Work and Energy NCERT Solutions for Class 9 Science Chapter 11

Class 9 Science Chapter 11 Work and Energy InText Questions and Answers

Question 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?

Answer:
When a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by:
Work done = Force x Displacement
W = F × S
Where,
F = 7N
S = 8m
Therefore, work done, W = 7 × 8
= 56 Nm
= 56 J

Question 2.
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.

Question 3.
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

Question 4.
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.

Question 5.
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 × Displacement
W = F × d
Where,
Applied force, F = 140 N
Displacement, d = 15m
W = 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the field.

Question 6.
What is the kinetic energy of an object?
Answer:
Kinetic energy is the energy possessed by a body by the virtue of its motion. 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 windmills, etc.

Question 7.
Write an expression for the kinetic energy of an object.
Answer:
If a body of mass ‘m’ is moving with a velocity v, then its kinetic energy E_k is given by the expression,
E_k = \(\frac {1}{2}\)mv²
Its SI unit is Joule (J j.

Question 7.
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:
Expression for kinetic energy is
E_K = \(\frac {1}{2}\)mv²
m = Mass of the object
v = Velocity of the object = 5ms^-1
Given that kinetic energy, E_K = 25 J
(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 ms^-1.
Therefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.
(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 9 = 225 J.

Question 8.
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 = \(\frac{\text { Work }}{\text { Time }}=\frac{\text { Energy }}{\text { Time }}\)
P = \(\frac{\mathrm{W}}{\mathrm{T}}\)
It is expressed in watt (W).

Question 9.
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.,
1W = \(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}\)

Question 10.
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
Power is given by the expression,
Work done = \(\frac{\text { Workdone }}{\text { Time }}\)
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
Power = \(\frac{1000}{10}\) = 100J S^-1 = 100 W

Question 11.
Define average power.
Answer:
A body can do different amount of work in different time intervals. Hence, it is better to define average power. Average power is obtained by dividing the total amount of work done in the total time taken to do this work.
Average Power = \(\frac{\text { Total work done }}{\text { Total time taken }}\)

Class 9 Science Chapter 11 Work and Energy Textbook Questions and Answers

Question 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 windmill 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:
(i) A force acts on the body.
(ii) 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 windmill works against the gravitational force to lift water. Hence, work is done by the windmill 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) Foodgrains 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.

Question 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.

Question 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

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.
Answer:
Kinetic energy is given by the expression,
(Ek)v = \(\frac {1}{2}\)mv2
Where,
Ek = Kinetic energy of the object moving with a velocity, v
m = Mass of the object
(i) Kinetic energy when the object was moving with a velocity 5 m s-1
(Ek)5 = \(\frac {1}{2}\) × 20 × (5)2 = 250J
ii. Kinetic energy when the object was moving with a velocity 2 ms-1
(Ek)2= \(\frac {1}{2}\) × 20 × (2)2 = 40J
Work done by force is equal to the change in kinetic energy.
Therefore, work done by force = (Ek)2 – (Ek)5
= 40 – 250 = -210 J
The negative sign indicates that the force is acting in the direction opposite to the motion of the object.

Question 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 × 0 = 0
Hence, the work done by gravity on the body is zero.

Question 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.

Question 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.

Question 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.

Question 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 × 3.6 × 10⁶ = 9 × 10⁸ J

Question 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 ms^-2
∴ W = 40 × 5 × 9.8 = 1960 J.
At half-way down, the potential energy of the object will be \(\frac{1960}{2}\) = 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.

Question 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.

Question 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.

Question 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.

Question 14.
An electric heater is rated 1500 W. How much energy does if use in 10 hours?
Answer:
Energy consumed by an electric heater can be obtained with the help of the expression,
P = \(\frac{W}{T}\) Where,
Power rating of the heater, P = 1500 W = 1.5 kW
Time for 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.

Question 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.

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.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done oh the object in order to bring the object to rest?
Answer:
Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,
E_k = \(\frac {1}{2}\)mv²
To bring the object to rest, \(\frac {1}{2}\)mv² amount of work is required to be done on the object.

Question 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 = \(\frac {1}{2}\)mv²
Where,
Mass of car, m = 1500 kg
Velocity of car, v = 60 km/h = 60 × \(\frac {5}{18}\)ms-1
∴ Ex = \(\frac {1}{2}\) × 1500 × \(\left(60 \times \frac{5}{18}\right)^{2}\) =20.8 × 104J
Hence, 20.8 × 104 J of work is required to stop the car.

Question 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:
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.

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.

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.

Question 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.

Question 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 = \(\frac{W}{T}\)
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.

Question 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.

Class 9 Science Chapter 11 Work and Energy Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
When a body falls freely towards the earth, then its total energy
(a) increases
(b) decreases
(c) remains constant
(d) first increases and then decreases
Answer:
(c) remains constant

Question 2.
A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial
Answer:
(a) does not change

Question 3.
In case of negative work the angle between the force and displacement is
(a) 00
(b) 450
(c) 900
(d) 1800
Answer:
(d) 1800

Question 4.
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy
Answer:
(a) acceleration

Question 5.
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 ms^-2)
(a) 6 × 103 J
(b) 6J
(c) 0.6 J
(d) zero
Answer:
(d) zero

Question 6.
Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour
Answer:
(c) kilowatt

Question 7.
The work done on an object does not depend upon the
(a) displacement
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object
Answer:
(d) initial velocity of the object

Question 8.
Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy
Answer:
(d) potential energy

Question 9.
A body is falling from a height h. After it has fallen a height h/2, it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy
Answer:
(c) half potential and half kinetic energy

Very Short Answer Type Questions

Question 1.
A boy climbs up the staircase to reach to second, floor from first floor. Has he done any work?
Answer:
Yes, he has done work against the gravitational force by climbing up.

Question 2.
In the following activities, has work been done or not? Why?
(i) A girl pulls a trolly and trolley moves.
(ii) A book is lifted through a height ‘h’.
Answer:
(i) When a girl pulls a trolley, work is done because trolley experiences a displacement from its mean position.
(ii) When a book is lifted through a height ‘h’ because book has gained displacement in its height from the ground level.

Question 3.
Is work a scalar or vector quantity?
Answer:
Although work done = Force × Displacement, yet it is taken as scalar quantity.

Question 4.
What is the work done in lifting an object of mass ‘m’ through a height ‘h’ from the ground?
Answer:
Force acting on body, F = m × g
Displacement = h.
Therefore, workdone = Force × Displacement
= mg × h = mgh

Question 5.
What is the angle between the line of force and the direction of displacement for maximum work?
Answer:
With W = F.s. cos θ, θ = 0 for the maximum work.

Question 6.
Define energy. What is the S.I. unit?
Answer:
Energy is the capacity to do work. Its S.I. unit is joule.

Question 7.
Which physical quantity has the unit Js^-1 ?
Answer:
Power.

Question 8.
Which of the two out of work, power and energy have same unit?
Answer:
Work and energy.

Question 9.
What are the two forms of mechanical energy?
Answer:
The two forms of mechanical energy are kinetic energy and potential energy.

Question 10.
State the relation in kinetic energy of a body in motion with respect to its (i) mass and (ii) velocity.
Answer:

• Kinetic energy of a body is directly proportional to its mass.
• Kinetic energy of a body is directly proportional to the square of its velocity.

Question 11.
State the relation in potential energy of a body at rest with respect to its mass and height.
Answer:
Potential energy of a body is directly proportional to both the mass and height.

Question 12.
What is the relationship between Watt and Horsepower?
Answer:
1 Horsepower = 746 Watt.

Question 13.
The work done by the heart for each beat is 1J. Calculate the power of the heart if it beats 72 times in a minute.
Answer:
Total work done in minute = 1J × 72 = 72J
Time = 1 minute = 1 × 60 = 6 seconds
Power = \(\frac{\text { Work done }}{\text { Time }}\) = \(\frac {72}{60}\) =1.2 Watt

Question 14.
What type of energy transformation takes place in (i) dynamo and (ii) Electric motor?
Answer:
(i) Dymamo—Mechanical energy into electrical energy,
(ii) Electric motor—Electrical energy into mechanical energy

Question 15.
Which type of energy is possessed by stretched rubber band?
Answer:
Elastic potential energy

Question 16.
A rubber ball and a leather cricket ball are moving with same velocity. Which will have greater kinetic energy? Why ?
Answer:
The leather cricket ball will have greater kinetic energy because of its large mass as compared to rubber ball.

Question 17.
Which will require more force to be pushed forward, a truck in rest or a slowly rolling truck?
Answer:
A truck in state of rest would require more force because of greater static friction than rolling friction.

Question 18.
What happens to kinetic energy of a body in motion when it comes to rest?
Answer:
Kinetic energy of body-in motion gets transformed into its potential energy when brought to rest while some is lost in overcoming friction as heat.

Short Answer Type Questions

Question 1.
A student when preparing for examinations draws diagram, read hooks and discuss problems with his friends. In the process does he do any work?
Answer:
No work is done by the student in reading or discussing problems with his friends as long as he remains sitting at one position to avoid any displacement.

Question 2.
What is work, a scalar or vector quantity? why?
Answer:
Work is a scalar quantity because on the application of force, the work done is in the direction of force. Hence, there is no change in the direction for work to be called a vector quantity.

Question 3.
Derive an expression for the work done on an object when the force is acting on the body in a direction making an angle ‘θ’ with the direction of displacement of the object.
Answer:
Let a force F acts on a body of mass ‘m’ by making an angle ‘θ’ with the direction of the displacement of the object.

The component of the force in direction of the displacement = F cos θ
Let displacement of body = s
Work done on the object = (F cos θ) × s = F × s cos θ

Question 4.
What is the workdone against force of gravity if a person carries a 30 kg load on his head?
Answer:
Work done against force of gravity W = F. s. cos θ
With θ = 90° and cos θ = 0
W = 0 × s = 0
Hence, total work done will be zero.

Question 5.
What do you mean by negative work and positive work? Give example.
Answer:
When a force acts on a body in a direction opposite to the direction of displacement
Work done = – F cos θ × s
The negative sign indicates that work done is negative.

When a force acts on a body in a direction of the displacement of body, the work done is positive.
When a bullock pulls a cart, the work done is called positive while when a player tries to stop a moving ball, the work done is called negative.

Question 6.
If a ball tied to a string is whirled in a circular orbit with centre being the hand holding the free end of string, then what is the work done on the ball?
Answer:
The force acting on the ball is along the radius of the circular path i.e. the direction of the force is perpendicular to the direction of motion of the ball.
Therefore, work done,
W = F. cos θ × s
= O × s = 0

Hence, work done on the ball is zero.

Question 7.
Take a toy car, wind it using key. Place the car on the ground and answer the following questions:
(i) Did the car move? Why?
(ii) Does the energy acquired by the car depends upon the number of times you turn the key?
(iii) From where did it get energy?
Answer:
(i) The car would move because the potential energy acquired by the car, due to widing of the spring of car got transformed into its kinetic energy.

(ii) The potential energy gained by the car depends upon the number of turns of the widing of the key with increasing number of turns, the greater potential energy is acquired by the car. This acquired potential energy later changes into car’s kinetic energy.

(iii) Car got the energy to move freon the spring inside it.

Question 8.
Two bodies having equal masses are kept at a height of 20m and 30m respectively. What is the ratio of their potential energy?
Answer:
Ratio of potential energies = mgh₁ : mgh₂
mg × 20 : mg × 30
= 20 : 30
= 2 : 3.

Question 9.
State the type of energy changes taking place in the following:
(i) A body is thrown up in air.
(ii) In a green leaf during photosynthesis.
(iii) In an oscillation of a pendulum.
Answer:
(i) When a body is thrown in upward direction, the muscular energy which is chemical potential energy is transformed into the kinetic energy of the sail. As the ball keeps rising, kinetic energy continues to change into potential energy. When at the maximum height, the body has no kinetic energy but the maximum potential energy.

(ii) In a leaf during photosynthesis, solar energy changes into chemical potential energy.

(iii) An oscilating pendulum has the maximum potential energy at its peak position where it temporarily come to state of rest while at its mean position, it has maximum kinetic energy.

Question 10.
Suppose a hammer which falls purely on a nail placed on a wood has a mass of 1 kg. If it falls from a height of 1m, how much kinetic energy will it have just before hitting the nail?
Answer:
Mass of the hammer = 1 kg
Height of the hammer, h = 1m
Acceleration due to gravity, g = 10 ms^-2
Potential energy at the highest point = mgh = 1 × 10 × 1 = 10 J
When the hammer falls, its whole potential energy would be converted into kinetic energy.
Kinetic energy just before hitting the nail 10 J.

Question 11.
A man whose mass is 50 kg climbs up 30 stairs in 30 seconds. If each step is 20 cm high. Calculate the power used in climbing the stairs.
Answer:
Mass of the man, m = 50 kg
Number of the stairs = 30
Height of one stair = 20 cm = 0.20 m
Total height of stairs = 30 × 20 = 6 m
Workdone in climbing the stairs=m × g × h
=50 × 10 × 6 = 3000 J
Tune taken by man to climb up 30 stairs =30
Power used inclimbing stairs = \(\frac {3000}{30}\) = 100 w

Question 12.
A porter lifts a luggage of 15 kg from the ground and put it on his head 2 m above the ground. Calculate the work done by him on the luggage Tig = 10 ms^-2)
Answer:
Mass of luggage, m = 15 kg
Height or displacement, h = 2m
Work done by porter = m.g × h
= 15 × 10 × 2 = 300 J

Question 13.
A force of 10 N acts on an object in the direction of displacement. If the displacement of the object is 2 m. Calculate the work done by the force.
Answer:
Force applied, F = 10 N.
Displacement, s = 2 m
Work done = F × S = 10 × 2 = 20 N.

Question 14.
An object of mass 15 kg is moving with a constant velocity of 4 ms^-1. What is the kinetic energy possessed by the body?
Answer:
Mass of the object, m = 15 kg.
Velocity of the object, v = 4 ms^-1
Kinetic energy possessed by the body = ½ mv².
= ½ × 15 × (4)² = ½ × 15 × 16= 120 J

Question 15.
What is the work done to increase the velocity of a car from 30 km h^-1 to 60 km h^-1 if the mass of the car is 1500 kg ?
Answer:
Mass of the car, m = 1500 kg

Work done = change in kinetic energy
= ½ m (v² – u²)

Question 16.
Find the energy present in any object of mass 10 kg when it is at a height of 6m above the ground. (Take g = 10 ms^-2).
Answer:
Mass of the object, m = 10 kg
Height, h = 6 m
g = 10 ms^-2
Potential energy possessed by the object = mg × h
= 10 × 10 × 6 = 600 J

Question 17.
An object of mass 12 kg is at a certain height above the ground. If the gravitational potential energy of the object is 480 J. Find the height at which the object is with respect to the ground, (Give g =10 ms^-1).
Answer:
Mass of the object, m = 12 kg
Potential energy, P.E. = 480 J
Let height of the object from ground = h
Then P.E. = m × g × h
h = \(\frac{\mathrm{P.E}}{m \times g}=\frac{480}{12 \times 10}=4 m\)

Question 18.
A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10s. Calculate the power used by her. (Take g = 10 ms^-2).
Answer:
Mass of the bucket, m = 5 kg
Height, h = 10 m
g = 10 ms^-2
Energy consumed in pulling the bucket = m × g × h
= 5 × 10 × 10 = 500 J
Time taken = 10s

Question 19.
Two girls A and B each of weight 400 N climb up a rope to a height of 8m. Girl A takes 20 seconds while girl B takes 50 seconds to accomplish their task. What is the power spent by each girl?
Answer:
(i) Weight of the girl ‘A’ = 400 N.
Displacement (height) h = 8m
Work done by girl A = mg × h
=400 × 8 = 3200 J
Time taken by girl A, t = 20 seconds
Power spent by girl A,
= \(\frac{\text { Workdone }}{\text { Time taken }}=\frac{3200}{20}\) = 160 W

(ii) Weight of the girl B, = 400 N
Displacement (height), h = 8 m
= 400 × 8 = 3200 J
Time taken, t = 50 seconds
Power spent by girl B
= \(\frac{\text { Workdone }}{\text { Time taken }}=\frac{3200}{20}\) = 64 W

Question 20.
A boy of 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step is 15 cm. Calculate the power of the boy.
Answer:
Mass of the boy, m = 50 kg
Height of each step = 15 cm = 0.15 m
Number of steps in stair case = 45
Total height of stair case = 45 × 0.15 = 6.75 m
Time taken to climb, t = 9 seconds

Question 21.
A boy pulls a toy car with a force of SON through a string which makes an angle of 30° with the horizontal so as to move the toy 2 m horizontally. Calculate the work done by the boy on the toy car.
Answer:
Force applied by the boy = F = SON
Angle of direction of force with the direction of displacement, θ = 30°
Displacement, S = 2m
Work done = F × s cos θ
Work done = 50 × 2 cos 30°
= 50 × 2 × \(\frac{\sqrt{3}}{2}\) = 50\(\sqrt{3}\)J
= 50 × 1.732 = 86.5 J

Question 22.
Calculate the velocity of an object of mass 100 g moving with a kinetic energy of 200 J.
Answer:
Mass of the object = 100g = 0.1 kg
Kinetic Energy (K.E.) = 20 J
Let velocity of the object = v
K.E. = ½ mu²
20 = ½ × 0.1 × v²
or v² = 400
v = 20 ms^-1

Question 23.
Calculate the power of a pump which can lift 200 kg of water to store it in a tank at height of 19m in 25 seconds. (Take g = 10 ms^-2).
Answer:
Mass of water, m =200 kg
Height, h = 19 m
Time taken to lift the water, t = 25 s
Power of the pump = \(\frac{\text { Workdone }}{\text { Time taken }}\)
= \(\frac{m \times g \times h}{t}=\frac{200 \times 10 \times 19}{25}\)
= 1520 W

Question 24.
Calculate the power of a pump in horse power (H.P.) which can lift 600 kg of water into the water tank at a height of 40m in 10 minutes.
Answer:
Mass of water m =600 kg
Height, h =40 m
g = 10ms^-2
Time taken by pump, t = 10 min = 600 s
Power of the pump = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{m g}{600}\)
= \(\frac{600 \times 10 \times 40}{600}\) = 400 W
or = \(\frac {400}{746}\)H.P = 0.52H.P. (∵ 1H.P. = 746W)

Question 25.
An object of mass 2 kg is thrown vertically upward with an initial velocity of 20m/ s. What will-be the potential energy at the highest point (Take g= 10 ms^-2).
Answer:
Given u =20 m/s
g = -10 ms^-2
v = 0
Let maximum height of the object = h
We know that
v² – u² = 2gh
h = \(\frac{v^{2}-u^{2}}{2 g}=\frac{0-(20)^{2}}{2 \times(-10)}=\frac{-400}{-20}\) = 20 m
Potential energy of the object = mgh.
= 2 × 10 × 20 = 400 J

Long Answer Type Questions

Question 1.
Define kinetic energy and potential energy ? Give some examples in each case.
Answer:
Kinetic energy: The energy possessed by an object by virtue of its motion is known as its kinetic energy.
Kinetic energy (E_k) = ½ mv².
Kinetic energy is directly proportional to it is mass and its is directly proportional to the square of its velocity.

Example:

1. The energy possessed by a moving wind can move the blade of a wind-mill due to its kinetic energy.
2. The flowing water also have kinetic energy which can make the water-mill to work.
3. The speeding car, moving bullet, rotating wheel all have kinetic energy and can do work.

Potential Energy: The energy posessed by a an object by virute of its position or change in configuration is known as its potential energy.
Potential energy =m × g × h
where m = mass of the object, h = height, g = acceleration due to gravity.

Example:

1. When a body is lifted to a certain height against the force of gravitation, the work done on the body is stored in the body in the form of potential energy.
2. The water stored at a height in a dam possesses potential energy
3. The energy possessed b a stretched rubber or stretched spring or compressed spring is also an example of potential energy.

Question 2.
What is gravitational potential energy and elastic potential energy? Give examples.
Answer:
(i) Gravitational potential energy : When a body is lifted to a certain height then die work done in raising the object from the ground to that point against gravity is stored in the body and is called gravitational potential energy.
The graviational potential energy of the object = m.g.h

The work done by the gravity depends upon the difference in height of initial position and final position and not on the path along which object is moved.

Example: Any object lifted from the ground level or any other reference level possess potential energy.

(ii) Elastic potential energy : The energy possesed by an object due to change in its shape is known as elastic potential energy It is mainly associated with the compression or extension of an object.

Example :

1. A stretched spring or rubber possesses elastic potential energy.
2. A compressed spring also possesses elastic kinetic energy
3. Derive an expression for the kinetic energy of an object of mass ‘m’ moving with the velocity V.

Or

Derive the expression of kinetic energy (Ek = ½ mv², where m is the mass of the object and v is the constant velocity with which the body is moving.
Answer:
The kinetic energy of an object is the energy possessed by the object by virtue of its motion. The kinetic energ of a body moving with a vertain velocity is equal to the work done on it to make it acquire that velocity from rest.

Consider an object of mass ‘tri at rest. Letitbe displaced through a distance ‘S’ when a constant force ‘F acts on its in the direction of displacement. The work done is given by
W = F × S ……..(i)
The work done on the object will cause a change in its velocity. Let velocity of the object changed from zero (position of rest) to v. Let ‘a’ be the acceleration produced in the body, then

According to Newtons second law of motion F = m × a …….(ii)
But we know that
v² – u² = 2as
or s = \(\frac{v^{2}-u^{2}}{2 a}\)
Object starts moving from rest, therefore u = 0
or s = \(\frac{v^{2}}{2 a}\) …….(iii)
Now substituting the value of ‘F’ and ‘s’ from equation (ii) and (iii) in equation (i) we get
W = m × a\(\left(\frac{v^{2}}{2 a}\right)\)
= ½ mv²
Now, the work done is stored in the form of kinetic energy of the object.
Hence, kinetic energy (E_k) = ½ mv²

Question 4.
What is the law of conservation of energy ? Prove it when an object falls freely from a certain height under the force of gravity.
Answer:
Law of conservation of energy: Energy can neither be created nor destroyed but it can be changed from’one form into another form. The total energy after and before the transformation always remains constant i.e. total energy of a closed system remains constant.

Let an object of mass ‘m’ is dropped freely from ascertain height ‘h’ from the ground.
(i) The potential energy of the body at point A = mgh
Kinetic energy at A = 0 (Because velocity is zero at A).
Total energy at point A (highest point)
= mgh + 0 mgh

(ii) After a certain interval of time it reaches at B. It moves through a distance V due to force of gravitation in downward direction.
The potential energy at B = m.g. (h – x)

Let the velocity of the body at B = v
We know that v² – u² = 2 gs
v² = 2g(x)(u = 0)
Kinetic energ at B = ½ mv²
= ½ m(2gx)
= mgx

Total energy at B = Potential energ + kinetic energy
= m.g. (h – x) + mgx
= m.g.h. – mgx + mgx
= mgh.

(iii) As the fall continues, the potential energy goes on changing into kinetic energy. The potential energu would decrease while kinetic energy would increase, when the object is about the reach the ground then, h – 0.
The potential energy at C = mg × h = mg × 0 = 0.
Let v be the velocity of the object just before reaching the ground.
u = 0, s = 0
v² – u² = 2gh
or v² – 0 = 2gh
Kinetic energy of the object = ½ mv^-1
= ½ m (2gh) – mgh
Total energy = mgh + 0 = mgh.

The total energy i.e. the sum of potential energy and kinetic energy at all the three point is same. Similarly, the total energy at any point will be the same. This proves the law of conservation of energy.

These NCERT Solutions for Class 9 Science Chapter 10 Gravitation Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Gravitation NCERT Solutions for Class 9 Science Chapter 10

Class 9 Science Chapter 10 Gravitation InText Questions and Answers

Question 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.

For two objects of masses ml and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:
F = \(\frac{G m_{1} m_{2}}{r^{2}}\)

Where, G is the universal gravitation constant given by:
G = 6.67 × 10^-11 Nm² kg^-2

Question 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:
Let ME be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:
F = \(\frac{G m_{1} m_{2}}{r^{2}}\)

Question 3.
What do you mean by free fall?
Answer:
Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall.

Question 4.
What do you mean by acceleration due to gravity?
Answer:
When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s².

Question 5.
What are the differences between the mass of an object and its weight?
Answer:
Mass:

1. It refer to quantity contained in an object
2. It is a scalar quantity.
3. It’s S.I. unit is Kg.
4. It remains constant at places.
5. It is measured by physical or beam balance

Weight:

1. It refers to force with which a body is attracted towards the earth.
2. It is a vector quantity
3. It’s S.I. unit is Newton.
4. It varies depending upon the value of g.
5. It is measured by spring balance.

Question 6.
Why is the weight of an object on the moon \(\frac {1}{6}\)th its weight on the earth?
Answer:
Let ME be the mass of the Earth and m be an object on the surface of the Earth. Let RE be the radius of the Earth. According to the universal law of gravitation, weight W_E of the object on the surface of the Earth is given by,
W_E = \(\frac{\mathrm{GM}_{\mathrm{E}} m}{\mathrm{R}_{\mathrm{E}}{ }^{2}}\)
Let M_M and R_M be the mass and radius of the moon.
Then, according to the universal law of gravitation, weight W_M of the object on the surface of the moon is given by,
W_M = \(\frac{\mathrm{GM}_{\mathrm{E}} m}{\mathrm{R}_{\mathrm{M}}{ }^{2}}\)
\(\frac{W_{M}}{W_{\mathrm{E}}}=\frac{M_{\mathrm{M}} R_{\mathrm{E}}{ }^{2}}{\mathrm{~W}_{\mathrm{E}} \mathrm{R}_{\mathrm{M}}{ }^{2}}\)
Where, ME = 5.98 × 10²⁴ kg, MM = 7.36 × 10²² kg
R_E = 6.4 × 10⁶ m, R_M = 1.74 × 10⁶ m
∴ \(\frac{W_{M}}{W_{\mathrm{E}}}=\frac{7.36 \times 10^{22} \times\left(6.37 \times 10^{6}\right)^{2}}{5.98 \times 10^{24} \times\left(1.74 \times 10^{6}\right)^{2}}=0.165 \approx \frac{1}{6}\)
Therefore, weight of an object on the moon is \(\frac {1}{6}\) of its weight on the Earth.

Question 7.
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.

Question 8.
What do you mean by buoyancy?
Answer:
The upward force exerted by a liquid on an object immersed in it is known as buoyancy. When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water.

Question 9.
Why does an object float or sink when placed on the surface of water?
Answer:
If the density of an object is more than the density of the liquid, then it sinks in the liquid.

This is because the buoyant force acting on the object is less than the force of gravity. On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity.

Question 10.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
When you weigh your 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.

Question 11.
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 bag of cotton is heavier than the iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag heavier than its actual value. For this reason, the iron bar and the bag of cotton show the same mass on the weighing machine, but actually the mass of cotton bag is more than that of the iron bar.

Class 9 Science Chapter 10 Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r) between them, i.e.,
F ∝ \(\frac{1}{r^{2}}\)
If distance r becomes r/2, then the gravitational force will be proportional to
\(\frac{1}{\left(\frac{r}{2}\right)^{2}}=\frac{4}{r^{2}}\)
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 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 resistance). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 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:
According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:
F = \(\frac{\mathrm{GMm}}{r^{2}}\)
Where,
Mass of Earth, M = 6 × 10²⁴ kg
Mass of object, m = 1 kg
Universal gravitational constant, G = 6.7 × 10^-11 Nm² kg^-2
Since the object is on the surface of the Earth, r = radius of the Earth (R)
r = R = 6.4 × 10⁶ m
Gravitational force, F = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
= \(\frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^{6}\right)^{2}}\) = 9.8 N

Question 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.

Question 5.
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

Question 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:
(i) Doubled
(ii) One-fourth and one-ninth
(iii) four times
According to the universal law of gravitation, the force of gravitation between two objects is given by:
F = \(\frac{\mathrm{G} m_{1} m_{2}}{r^{2}}\)
(i) F is directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.
Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

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

Question 7.
What is the importance of universal law of gravitation?
Answer:
The universal law of gravitation proves that every object in the universe attracts every other object.

Question 8.
What is the acceleration of free fall?
Answer:
When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms^-2, which is constant for all objects (irrespective of their masses).

Question 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.

Question 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 g is 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 g is 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.

Question 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.

Question 12.
Gravitational force on the surface of the moon is only \(\frac {1}{5}\) 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 = \(\frac {1}{6}\) × 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 = 98N
And, weight of the same object on the moon = \(\frac {1}{6}\) × 98 = 16.3 N

Question 13.
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:
(i) 122.5 m (ii) 10 s
According to the equation of motion under gravity: v² – u² = 2 gs
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-(49)² = 2 × (-9.8) × h
h = \(\frac{49 \times 49}{2 \times 9.8}\) = 122.5 m
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 × (-9.8)
9.8t = 49
t = \(\frac{49}{9.8}\) = 5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 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 ms^-1
Hence, the velocity of the stone just before touching the ground is 19.6 m s^-1.

Question 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 = \(\frac{40 \times 40}{20}\) = 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

Question 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 the universal law of gravitation, the force of attraction between the Earth and the Sim is given by:
F = \(\frac{\mathrm{GM}_{\text {Sun }} \mathrm{M}_{\text {Earth }}}{\mathrm{R}^{2}}\)
Where,
MSun = Mass of the Sun = 2 × 10³⁰ kg
MEarth = Mass of the Earth = 6 × 10²⁴ kg
R = Average distance between the Earth and the Sun = 1.5 × 10¹¹ m
G = Universal gravitational constant = 6.7 × 10^-11 Nm² kg^-2
F = \(\frac{6.7 \times 10^{-11} \times 2 \times 10^{36} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^{2}}\) = 3.57 × 10²² N

Question 17.
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 the two stones meet after a time t.
For the stone dropped from the tower:
Initial velocity, u = 0
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 m s^-2
From the equation of motion,
S = ut + \(\frac {1}{2}\)gt²
= 0 × t + \(\frac {1}{2}\) × 9.8 × t² ………. (i)

(ii) For the stone thrown upwards:
Initial velocity, u = 25 m s^-1
Let the displacement of the stone from the ground in time t be s’.
Acceleration due to gravity, g = -9.8 ms^-2
Equation of motion,
s’ = ut + \(\frac {1}{2}\)gt²
= 25f – \(\frac {1}{2}\) × 9.8t²
∴ s’ = 25t – 9t² ………. (2)
The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.
∴ s + s’ = 100
\(\frac {1}{2}\)gt² + 25t – \(\frac {1}{2}\)gt² = 100
∴ t = \(\frac {100}{25}\) = 4 s
In 4 s, the falling stone has covered a distance given by equation (1) as
s = \(\frac {1}{2}\) × 10 × 4² = 80m
Therefore, the stones will meet after 4 s at a height (100 – 80) = 20 m from the ground

Question 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) 29.4m/s
(b) 44.1 m
(c) 39.2 m above the ground

(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 ms^-1.

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

(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 4s -3s = 1s.
Equation of motion, s = ut + \(\frac {1}{2}\)gt² will give,
s = 0 × t + \(\frac {1}{2}\) × 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.

Question 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.

Question 20.
Why does a block of plastic released under water come up to the surface of water?
Answer:
Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within water. Due to this reason, a block of plastic released under water comes up to the surface of the water.

Question 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 = \(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}=\frac{50}{20} 2.5 \mathrm{~g} \mathrm{~cm}^{-3}\)
The density of the substance is more than the density of water (1 g cm^-3). Hence, the substance will sink in water.

Question 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
\(\frac{\text { Mass of the packet }}{\text { Volume of the packet }}=\frac{500}{350}=1.428 \mathrm{gcm}^{-3} \)
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.

Class 9 Science Chapter 10 Gravitation Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
Two objects of different masses falling freely near the surface of moon would
(a) have same velocities at any instant
(b) have different accelerations
(c) experience forces of same magnitude
(d) undergo a change in their inertia
Answer:
(a) have same velocities at any instant

Question 2.
The value of acceleration due to gravity
(a) is same on equator and poles
(b) is least on poles
(c) is least on equator
(d) increases from pole to equator
Answer:
(c) is least on equator

Question 3.
The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become
(a) F/4
(b) F/2
(c) F
(d) 2F
Answer:
(a) F/4

Question 4.
A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone
(a) will continue to move in the circular path
(b) will move along a straight line towards the centre of the circular path
(c) will move along a straight line tangential to the circular path
(d) will move along a straight line perpendicular to the circular path away from the boy
Answer:
(c) will move along a straight line tangential to the circular path

Question 5.
An object is put one by one in three liquids having different densities. The object floats with 1/9, 2/11, 3/7, and parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₁ > d₂ < d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃
Answer:
(d) d₁ < d₂ < d₃

Question 6.
In the relation F = GMm/d2, the quantity G
(a) depends on the value of g at the place of observation
(b) is used only when the earth is one of the two masses
(c) is greatest at the surface of the earth
(d) is universal constant of nature
Answer:
(d) is universal constant of nature

Question 7.
Law of gravitation gives the gravitational force between
(a) the earth and a point mass only
(b) the earth and Sim only
(c) any two bodies having some mass
(d) two charged bodies only
Answer:
(c) any two bodies having some mass

Question 8.
The value of quantity G in the law of gravitation
(a) depends on mass of earth only
(b) depends on radius of earth only
(c) depends on both mass and radius of earth
(d) is independent of mass and radius of the earth
Answer:
(d) is independent of mass and radius of the earth

Question 9.
Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be
(a) 1/4 times
(b) 4 times
(c) 1/2 times
(d) unchanged
Answer:
(b) 4 times

Question 10.
The atmosphere is held to the earth by
(a) gravity
(b) wind
(c) clouds
(d) earth’s magnetic field
Answer:
(a) gravity

Question 11.
The force of attraction between two unit point masses separated by a unit distance is called
(a) gravitational potential
(b) acceleration due to gravity
(c) gravitational field
(d) universal gravitational constant
Answer:
(d) universal gravitational constant

Question 12.
The weight of an object at the centre of the earth of radius R is
(a) zero
(b) infinite
(c) R times die weight at the surface of the earth
(d) 1/R₂ times the weight at surface of the earth
Answer:
(a) zero

Question 13.
An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N
Answer:
(a) 2 N

Question 14.
A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base
(b) maximum when breadth and width form the base
(c) maximum when width and length form the base
(d) the same in all the above three cases
Answer:
(b) maximum when breadth and width form the base

Very Short Answer Type Questions

Question 1.
State what causes the revolution of planets around the sun?
Answer:
Gravitational force of attraction between the sun and planets.

Question 2.
What is the direction of gravitation force acting between the two objects?
Answer:
The force of gravitation between two objects act along the line joining their centres of mass.

Question 3.
How does the gravitation force of attraction vary with the distance between them?
Answer:
The gravitation force of attraction is universely proportional to the square of distance between die centres of their mass.

Question 4.
State the effect on the gravitational force of attraction between two objects
(i) When distance is halved
(ii) When distance is doubled
Answer:
(i) When distance is halved, the force increases by four times.
(ii) When distance is doubled, the force decreases by four times.

Question 5.
How does the gravitation force of attraction between two objects depend upon their masses?
Answer:
The force of gravitation is directly proportional to the product of the masses of two objects.
F ∝ m₁.m₂

Question 6.
If the mass of a body is doubled and that of another body is tripled, what would be the effect of gravitational force?
Answer:
The gravitational force would increase by six times.

Question 7.
State any one physical quantity that does not affect the magnitude of acceleration due to gravity.
Answer:
MaSs of the body.

Question 8.
What is the magnitude of universal gravitational constant?
Answer:
6.67 × 10^-11 Nm²kg^-2.

Question 9.
Does the value of universal gravitational constant change in universe?
Answer:
No, its a constant value.

Question 10.
What is the Value of ‘g’ acceleration due to gravity on moon?
Answer:
The value of acceleration due to gravity on moon surface is 1/6 th of value of acceleration of gravity on earth i.e. 1.63 ms^-2.

Question 11.
If a body of mass ‘m’ revolves in a circle of radius ‘r’, then what would be the magnitude of the acceleration produced in the body?
Answer:
The magnitude of acceleration produced (a) = \(\frac{v^{2}}{r}\)

Question 12.
What is the relation between ‘g’ and ‘G’?
Answer:
g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
When M is mass of earth and R is the radius of earth.

Question 13.
Why do the nails and pins have pointed ends?
Answer:
The nails and pins have pointed ends to reduce area and increase the effect of applied force i.e. pressure.

Question 14.
What is the S.I. unit of density and relative density?
Answer:
Relative density has no unit but for density S.I. unit kg/m³.

Question 15.
State the principle that describes the working of hydrometer or lactometer.
Answer:
Archimedes principle.

Question 16.
What is lactometer?
Answer:
It is a device used to measure the purity of milk i.e. amount of water present in milk.

Question 17.
What is hydrometer?
Answer:
It is a device used to measure the density of a liquid.

Question 18.
State Archimedes principle.
Answer:
The principle states that when an object is immersed in fluid partially or completely, it experiences an upward thrust that is equal to the weight of the fluid displaced by it.

Question 19.
Where will a student find the mass of 1kg heavier, when weighted in air or in water? Why?
Answer:
In air because the buoyant force in air is less than in water.

Question 20.
If a student drops a feather and an iron block in vacuum from a particular height. Which of the two will land first? Why?
Answer:
Both feather and iron block would land at the same time because acceleration due gravity is independent of mass.

Short Answer Questions

Question 1.
What do you understand by gravitation and gravity?
Answer:
Gravitation refers to the force with which any two objects in universe attract each other irrespective of their masses and distance between them. Gravity refers to force with which earth attracts an object towards its centre.

Question 2.
Does earth attracts moon towards itself? If yes, why does the moon then not fall on the earth?
Answer:
Earth and moon have gravitational force that acts in between them. Moon revolving round the earth undergoes a change in its acceleration because of changing direction. Moreover, the motion of moon around earth is because of centripetal gravitational force which is balanced by moon’s centrifugal gravitational force to prevent it’s falling down on the earth.

Question 3.
It is seen that an apple is attracted towards the earth. Does the apple attract the earth? If yes, why does the earth not move towards the apple?
Answer:
According to universal law of gravitation apple also attracts the earth with the same force as the earth attracts the apple but in opposite direction.
Force acting on the apple due to earth’s attraction, F = mg.
Apple also attracts the earth with same force. Therefore, acceleration produced on the earth
(a) = \(\frac{\mathrm{F}}{\mathrm{M}_{\mathrm{e}}}=\frac{m g}{\mathrm{M}_{e}}\)

However, the mass of earth is 6.0 × 10²⁴ kg which is very large as compared to the mass of an apple. Therefore, the value of acceleration produced in the earth is very-very small or negligible for the earth to show any visible movement towards the apple.

Question 4.
State the factors which affect the gravitation force of attraction in between two objects?
Answer:
There are two factors which affect the force of attraction between two objects. These factors include:

1. Mass, because the force of gravitation is directly proportional to the product of the masses of two bodies.
2. Distance between the two bodies because of the force of attraction is inversely proportional to the square of the distance in between them.

Question 5.
What is gravitational constant ‘G’ ? What is its S.I. unit?
Answer:
The gravitational force of attraction between two bodies/objects is given by
F = \(\mathrm{G} \frac{m_{1} m_{2}}{d^{2}}\)
Where ‘G’ is universal gravitational constant.
G = \(\frac{\mathrm{F} d^{2}}{m_{1} m_{2}}\)
Putting the S.I. units of different physical quantities.
G = \(\frac{\mathrm{N} m^{2}}{\mathrm{~kg} . \mathrm{kg}}\) = nm²kg^-2
Hence, S.I. units of G is Mm² kg^-2.

Question 6.
What do you understand by acceleration due to gravity? Does it depend upon the mass?
Answer:
It refers to the acceleration gained by a free falling body towards the earth. It is represented by symbol ‘g’ and its value is 9.8 m/s² i.e. a free falling body gains an acceleration of 9.8 m/s² every second during tire course of its fall.
It does not depend upon the mass of the object.

Question 7.
What is the difference between ‘g’ and ‘G’ ?
Answer:

Question 8.
Calculate the value of ‘g’ on the surface of earth.
Answer:
We know that g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
M, mass of earth = 6.0 × 10²⁴ kg
R1 radius of earth = 6.4 × 10⁶ m.
G, Gravitational constant = 6.67 × 10^-11 Nm² kg^-2
g = \(\frac{6.67 \times 10^{-11} \mathrm{Nm} \mathrm{kg}^{-2} \times 6.0 \times 10^{24} \mathrm{~kg}}{\left(6.4 \times 10^{6}\right)^{2} \mathrm{~m}}\)
= \(\frac{400.38}{40.96}=9.8 \mathrm{~ms}^{-2}\)

Question 9.
What will happen to the weight of a body when it is taken down inside the earth?
Answer:
The weight of the body decreases because the effective mass of the earth with which it attracts the body towards it centre decreases. The decrease in weight due to decresing effecting mass of the earth overpowers the increase in weight due to decrease in distance.

Question 10.
Define mass and weight. What are their S.I. units?
Answer:
Mass is defined as the quantity of matter contained in a body that remains unchanged throughout the universe. It’s S.I. units is kilograms. Weight is defined as the force with which the earth attracts the object towards its centre.
Weight (W) = m × g
where m = mass of body
g = accelertion due to gravity
The S.I. unit of weight is Netwon(N).

Question 11.
Where do you expect the weight of a body to be more, at poles or at equator? Why?
Answer:
The earth is not a perfect spherical structure. It is compressed at poles therefore the radius of earth is greater at equator than at poles. With value of ‘g’ inversely proportional to square of distance from the centre of earth, the value of ‘g’ is comparitively more at poles than at equator. Therefore, the weight of an object is maximum at poles and minimum at equator.

Question 12.
Write the equation of motion for free falling bodies due to gravitational force of attraction on earth.
Answer:
When u = initial velocity, v = final velocity,
g = acceleration due to gravity, s = distance covered and time = t, then
V = u + gt
s = ut + \(\frac {1}{2}\)gt²
2gs = v² – u²

Question 13.
According to Cartesian convention what are the sign for acceleration due to gravity ‘g’. Velocity ‘v’ and distance covered ‘s’ ?
Answer:
(i) According to cartesian convention ‘g’ is always taken in negative since its acts in downward direction. It is always taken in negative whether a body goes up or comes down.

(ii) The value of velocity V is taken in positive when a body goes in upward direction and when it come down, it value is taken in negative. (Fig. 10.2)
(iii) Distance is also taken in positive when a body goes up and taken in negative when falls down on the earth.

Question 14.
What is thrust and pressure? How are they different?
Answer:

Question 15.
Why cutting tools like pair of scissors, knives or axes have sharp edges?
Answer:
The edges of pair of scissors, knives and axes are made sharp to increase the effect of the applied force on them. The pressure is universely proportional to area i.e. with the decreasing area, pressure increases.

Question 16.
What is bouyant force? How will you determine the bouyant force?
Answer:
Bouyant force is an upward force that is exerted on a body when it is partially or fully immersed in a liquid like water. However, the magnitude of bouyant force depends upon the density of liquid, with increasing density of fluid, the bouyant force also increases.

The magnitude of the force of bouyancy is determined by measuring the weight of the fluid displaced by the object when immersed in a liquid.

Question 17.
Does a body experience the same bouyant force in different fluids? If not, why?
Answer:
A body immersed in different fluids such as water, alcohol and cooking oil will experience different bouyant force because the bouyant force exerted on a body immersed in a fluid depends upon, the fluid’s density. The bouyant force increases with increasing density of the fluid.

Question 18.
What is law of floatation?
Answer:
The law of floatation describes, the conditions in which a body would float or sink. When immersed in a liquid, the law states that:

• If relative density of a body is more than one, it would sink in the water.
• If relative density is less than one, it will float on water.

Question 19.
Why a nail or a rupee coin sinks in water but a ship floats in water?
Answer:
The weight of water displaced by a nail or a rupee coin is less than the weight of the nail or rupee coin. Therefore, with less upward thrust applied on them, they sink in water.

A ship being large is so designed that the water displaced by the ship is more than the weight of the ship i.e. upward thrust is more on ship to keep it floating in water.

Question 20.
What do you understand by density and the relative density? What are their S.I. units?
Answer:
Density refers to the mass per unit volume of a substance.
Mass of substance = \(\frac{\text { Mass of substance }}{\text { Volume of substance }}\)
It’s S.I. unis is kg/m³.
Relative density refers to the ratio of the density of a substance to the density of pure water.
Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\)
Relative density being a ratio has no unit.

Question 21.
What is the force of gravitation in between two boys having mass equal to 50 kg and 40 kg respectively and separated by a distance of lm from each other?
Answer:
Mass of one boy m₁ = 50 kg, Mass of other boy, m₂ = 40 kg Distance between them, d = 1m.
Value of G = 6.67 × 10^-1 Nm² kg^-2
According to law of gravitation, force of attraction between them is given by
F = G \(\frac{m_{1} m_{2}}{d^{2}}\)
= \(\frac{6.67 \times 10^{-11} \times 50 \times 40}{(1)^{2}}\)
= 1.334 × 10^-7 N

Question 22.
Calculate the force exerted by earth on the moon. If the mass of the earth 6.0 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. The distance between the earth and the moon is 3.84 × 10^-11. (G = 6.67 × 10^-11 Nm² kg^-2).
Answer:
Mass of the earth, M = 6.0 × 10²⁴ kg.
Mass of the moon, m = 7.4 × 10²² kg.
The distance between moon and earth, r = 3.84 × 10²⁴ m.
The force exerted by earth on moon is given by

Question 23.
A car falls off a ledge and drops in % seconds. Let g = -10 ms^-2.
(i) What will be its speed on striking the ground?
(ii) What is its average speed during the % second?
(iii) How high is the ledge from the ground.
Answer:
Given: Initial velocity of car = 0 ms^-1
Acceleration due to gravity, g = -10 ms^-2
lime taken, t = 1/2 second.
(i) Let speed on striking the groun d,v = ?
v = u + gt
= 0 – 10 × \(\frac {1}{2}\) = -5 ms^-1
Negative sign indicates that car is moving downward.

(ii) Average speed of the car
\(\frac{u+v}{2}=\frac{0+(-5)}{2}=2.5 \mathrm{~ms}^{-1}\)
Average speed is 2.5 ms^-1 downward.

(iii) Let the height of ledge from the ground
h = ut + \(\frac {1}{2}\)gt²
= 0 × \(\frac {1}{2}\) + \(\frac {1}{2}\) (-10)(\(\frac {1}{2}\))2
= \(-\frac{5}{4}\) = -1.25 m.
The distance travelled has a negative sign this means that the car is moving in downward direction.

Question 24.
A stone is dropped from 180 m high tower. How long does the stone tak to reach the ground ? What is the velocity when it touches the ground ? (Take g = -10 ms^-2)
Answer:
Height of the tower, h = – 180m
Initial velocity, u = 0
Let time taken to reach the ground, t = ?
Let velocity with which it touches the ground, v = ?
We know that
h = ut + \(\frac {1}{2}\)gt²
-180 = 0 × t + \(\frac {1}{2}\) (-10)t²
180 = 5t²
t2 = \(\frac {180}{5}\) = 36 or t = 6
(ii) u = u+gt = 0 – 15 × 6
= -60 m/s
Negative sign indicates that the body is moving in downward direction.

Question 25.
A body is thrown vertically upward rises to a height of 10m. Calculate.
(i) The velocity with which body was thrown upward.
(ii) The time taken by the body to reach the highest point.
Answer:
(i) Maximum height reached by the body, h = 10m.
Let initial velocity =?
Final velocity, v =0
Acceleration due to gravity g = – 9.8 ms^-2
We know that v² – u² = 2 (-9.8) × 10
u² = 196
u = \(\sqrt{196}\) = 14ms^-1

(ii) Initial velocity of the body = 14 ms^-1
v = u + gt
0 = 14 – 9.8 × t
t = \(\frac {14}{9.8}\) = 1.43 second
Time taken by the body to reach maximum height = 1.43 s.

Question 26.
The mass of an object is 50 kg on earth. What is its weight on the earth and what will be its weight on the moon?
Answer:
Mass of the object = 50 kg.
Acceleration due to gravity at earth, g = 9.8 ms^-2
Weight on earth = m × g = 50 × 9.8 = 490 N
Acceleration due to gravity, g on moon = \(\frac{9.8}{6} m \mathrm{~s}^{-2}\)
Weight on moon = m × g
= \(\frac{50 \times 9.8}{6}\) = 81.67 N

Question 27.
Find the mass of an object whose weight on the earth is 49 N. What is its mass on the surface of moon?
Answer:
Weight on the earth, W = 49 N.
Acceleration due to gravity on earth, g = 9.8 ms^-2.
We know that W = m.g.
m = \(\frac{W}{g}=\frac{49}{9.8}\) = 5 kg.
Mass of the body on earth = 5 kg.
Mass of the body on moon = 5 kg
(Because the mass remains constant at every place).

Question 28.
Calculate the mass of the moon. If the radius of the moon ‘R_m‘ is 1.60 × 10⁶ m. The acceleration due to gravity at moon is 1/6th of the gravity at earth.
Answer:
Let mass of the moon = M_m
Mass of the earth = m_e
We know that acceleration due to gravity ‘g’ at each is given by g = \(\frac{\mathrm{GM}_{e}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}\)
Similarly, gravity at moon, g will be given by

Question 29.
A block of wood is a kept on tabletop. The mass of the wooden block is 5kg and its dimension are 40 cm × 20 cm × 10cm. Find the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop with its side& of dimension (i) 20m × 10cm. (ii) 40 cm × 20 cm.
Answer:
The mass of the wooden block = 5 kg.
The force exerted by wooden block = mg = 5 × 9.8 = 49 N
(i) When wooden block lie with its 20 cm × 10 cm side on the tabletop.
Area of block in contact with table top = 20 × 10 = 200 cm² = 0.02 m²
Force = \(\frac{\text { Force }}{\text { Area }}=\frac{49}{0.02}\) = 2450 Nm²

(ii) When wooden block lie with its side 40 × 20 cm on the table top:
Area in contact with table = 40 × 20 = 800 cm² = 0.08 m².
Pressure = \(\frac{\text { Force }}{\text { Area }}=\frac{49}{0.08}\) = 612.5 Nm^-2.

Question 31.
The relative density of silver is 10.8. The density of water is 1 kg m^-3. What is the density of silver?
Answer:
Relative density of silver = 10.8 kgm^-3
Density of water = 10³kg m^-3
Relative density = \(\frac{\text { Density of silver }}{\text { Density of water }}\)
Density of silver = Relative density × Density of water
= 10.8 × 10³ kgm^-3

Question 32.
A cubical block of mass 5 kg with each side 5 cm is lying on a table top. Calculate the pressure exerted by the block.
Answer:
Mass of the block = 5 kg
Thrust exerted by the block =m × g
= 5 × 9.8 = 49.0 N
Side of cubical block = 5 cm = 0.05 m
Area of the surface in contact with table top
= (0.05)² = 0.0025 m² = 2.5 × 10^-3 m²
Pressure exerted = \(\frac{\text { Thrust }}{\text { Area }}\)
= \(\frac{49.0}{2.5 \times 10^{-3}}\) = 1.86 × 10⁴ Nm^-2 or Pa.

Long Answer Questions

Question 1.
State the Kepler’s laws that describe the motion of the planets around the earth.
Answer:
The motion of the planets are governed by the following Kepler’s laws.
(i) The orbit of a planet is elliptical with the Sun at one of the focus, as shown in the figure 10.3.
(ii) The line joining the planet and the sun sweeps equal area in equal intervals of time. Thus, in the given figure if the time of interval of moving from A to B is the same as that from C to D, then the area OAB and OCD are equal.

(iii) The cube of the mean distance of a planet from the Sun is proportional to the square of its orbital period, i.e. time taken by the planet to complete its orbit T:
r³ ∝ T²
or \(\frac{r^{3}}{\mathrm{~T}^{2}}\) = constant
where r = Mean distance of the planet from sun.
T = Orbital period.

Question 2.
What is inverse square rule? How did Newton guess or arrived the inverse square rule?
Answer:
Inverse square rule: When a planet revolves around the sun, the force of attraciton between sun and planet is inversely proportional to the square of radius of the orbit.
F ∝ \(\frac{1}{r^{2}}\)
F = Force of attraction between sun and planet.
r = Radius of the orbit of a planet.

Newton showed that the cause of planetary motion is the gravitational force of attraction that sun exert on planets. Newton used the third law of Kepler to guess the inverse square rule.

Suppose a planet is moving in an orbit. We can assume Planetary orbits as circular orbit. Let the radius of the circular orbit is ‘r’ and planet is moving with a linear velocity ‘v’.
Then, the force ‘F’ acting on the planet of mass ‘m’ is given by
F = \(\frac{m v^{2}}{r}\)
or F ∝ \(\frac{v 2}{\bar{r}}\) ………(i)
(m = mass of planet is contant)
If let T is the taken by the planet to complete one orbit, then,

Therefore, we have, v² ∝ \(\frac{1}{r}\)
combining (i) and (ii) we have
F ∝ \(\frac{1}{r^{2}}\)

Question 3.
What do you mean by acceleration due to gravity? How will you prove that acceleration due to gravity is independent of the mass and nature of the body?
Answer:
Acceleration due to gravity: When body falls freely due to the gravitational force of attraction of the earth, there is no change in its direction of motion but its velocity changes by equal amounts in each second. So the acceleration produced in the body due to earth’s gravitation force of attraction is known as the acceleration due to gravity. It is denoted by.

Let a body of mass ‘m’ is falling downward due to force of gravity and acceleration produced in the body is ‘g’. Then according the Newton’s second law of motion, gravitation force acting on the body is given by
F = mg ……..(i)
But the gravitation force of attraction between the body of mass ‘m’ and earth is given by
F = \(\frac{\mathrm{GM}_{m}}{d^{2}}\)
where M = mass of the earth
d = distance between the body and the centre of the earth from (i) and (ii)
mg = \(\frac{\mathrm{GM}_{m}}{d^{2}}\)
g = \(\frac{\mathrm{GM}}{d^{2}}\)
The expression of ‘g’ is independent of the mass of the object. Hence, acceleration due to gravity of the earth does not depend upon the mass and nature of the object.

Question 4.
Show that the weight of an object on the moon’s surface is 1/6th that of the earth.
Answer:
Let the mass of an object on moon’s surface is ‘m’.
Let weight of the object on moon’s surface = W
Let, Mass of moon =M
Radius of moon = R
Now the force of attraction by which moon attracts an object is equal to the weight of the object.
Therefore, Wm = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\) ………(1)
Let the weight of the same object on earth = We
The force with which earth attracts the object is equal to the weight of the object on earth.
Therefore We = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{e}^{2}} \) ……..(2)
Where Me = mass of the earth and Re = Radius of earth.

But Mass of the earth is 100 times that of the moon and radius of the earth is 4 times the radius of moon.
Therefore, M_e = 100 M
and R_e = 4R
Putting these values in equation (2) we have
We = \(\frac{G(100 M) \times m}{(4 R)^{2}}\)
= \(\mathrm{G} \frac{100 \mathrm{Mm}}{16 \mathrm{R}^{2}}\) ……..(3)
Dividing equation (1) by (3)

or Weight on the moon = \(\frac {1}{6}\) × weight on the earth.

Question 5.
Show with the help of an activity that whenever a body is immersed in a liquid, it loses same weight due to force acting on the body in the upward direction (bouyant force).
Or
Show with the help of an activity that whenever an object is immersed in a fluid, a force acts on the object in upward direction.
Answer:
When an object is immersed in liquid, a force acts on the body in upward direction. ‘It can be shown by the following activity. (Fig. 10.4)

Activity: Take a piece of stone and tie it to one end of a spring balance. Suspend the stone by holding the string as shown in the figure. Note the extension produced in the reading in the spring balance due to the weight of the stone. Now, slowly dip the stone in water in a container.

As the stone is dipped in water, it loses the reading on spring balance scale. It continues declining till it gets completely immersed in water.

The decrease in extension of the spring shows that a force is acting on the stone in the upward direction. This upward thrust causes or brings a loss in the weight.

(a) Observe the elongation of the rubber string due to the weight of a piece of stone suspended form it in air.
(b) The elongation decreases as the stone, is immersed in water.

Question 6.
What is Archimedes principle? State some of its applications.
Answer:
Archimedes principle states that when an object is immersed in a fluid partially or fully, it experiences an upward thrust or force that is equal to the weight of the fluid displaced. The principle has helped.

• In designing ships and submarines.
• In designing lactometer to assess the purity of milk.
• In desgining hydrometer to measure the density of liquids.
• In assessing the purity of a given substance.

These NCERT Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

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

Class 9 Science Chapter 7 Diversity in Living Organisms InText Questions and Answers

Question 1.
Why do we classify organisms?
Answer:
There are a wide range of life forms (about 10 million – 13 million species) around us. These life forms have existed and evolved on the Earth over millions of years ago. The huge range of these life forms makes it very difficult 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. Thus, classification makes our study easier.

Question 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:

• Variety of living organisms in terms of size ranges from microscopic bacteria to tall trees of 100 metres.
• The colour, shape, and size of snakes are. completely different from those of lizards.
• 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.

Question 3.
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 kind of cells that living organisms
are made up of is a more basic characteristic for classifying organisms, than on the basis of their habitat. This is because on the basis of the kind of cells, we can classify all living organisms into eukaryotes and prokaryotes. On the other hand, a habitat or the place where an organism lives is a very broad characteristic to be used as the basis for classifying organisms. For example, animals that live on land include earthworms, mosquitoes, butterfly, rats, elephants, tigers, etc.

These animals do not resemble each other except for the fact that they share a common habitat. Therefore, the nature or kind of a cell is considered to be a fundamental characteristic for the classification of living organisms.

Question 4.
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. It is considered to be the fundamental characteristic for classifying all living organisms. Nature of the cell includes the presence or absence of membrane-bound organelles.

Therefore, on the basis of this fundamental characteristic, we can classify all living organisms into two broad categories of eukaryotes and prokaryotes. Then, further classification is made on the basis of cellularity or modes of nutrition.

Question 5.
On what basis are plants and animals put into different categories?
Answer:
Plants and animals differ in many features such as the absence of chloroplasts, presence of cell wall, etc. But, locomotion is considered as the characteristic feature that separates animals from plants.

This is because the absence of locomotion in plants gave rise to many structural changes such as the presence of a cell wall (for protection), the presence of chloroplasts (for photosynthesis) etc. Hence, locomotion is considered to be the basic characteristic as further differences arose because of this characteristic feature.

Question 6.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
A primitive organism or lower 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. An advanced organism or higher organism has a complex body structure and organization. For example, an Amoeba is more primitive as compared to a starfish. Amoeba has a simple body structure and primitive features as compared to a starfish. Hence, an Amoeba is considered more primitive than a starfish.

Question 7.
Will advanced organisms be the same as complex organisms? Why?
Answer:
It is not always true that an advanced organism will have a complex body structure. But, there is a possibility that over the evolutionary time, complexity in body design will increase. Therefore, at times, advanced organisms can be the same as complex organisms.

Question 8.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
The criterion for the classification of organisms belonging to kingdom Monera or Protista is the presence or absence of a well-defined nucleus or membrane-bound organelles. Kingdom Monera includes organisms that do not have a well-defined nucleus or membrane-bound organelles and these are known as prokaryotes.

Kingdom Protista, on the other hand, includes organisms with a well-defined nucleus and membrane-bound organelles and these organisms are called eukaryotes.

Question 9.
In which kingdom will you place an organism which is single-celled, eukaryotic and photosynthetic?
Answer:
Kingdom Protista includes single celled, eukaryotic, and photosynthetic organisms.

Question 10.
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.

Question 11.
Which division among plants has the simplest organisms?
Answer:
Thallophyta is the division of plants that has the simplest organisms. This group includes plants, which do not contain a well differentiated plant body. Their body is not differentiated into roots, stems, and leaves. They are commonly known as algae.

Question 12.
How are pteridophytes different from the phanerogams?
Answer:

Question 13.
How do gymnosperms and angiosperms differ from each other?
Answer:

Question 14.
How do poriferan animals differ from coelenterate animals?
Answer:

Question 15.
How do annelid animals differ from arthropods?
Answer:

Question 16.
What are the differences between amphibians and reptiles?
Answer:

Question 17.
What are the differences between animals belonging to the Aves group and those in the mammalia group?
Answer:

Class 9 Science Chapter 7 Diversity in Living Organisms Textbook Questions and Answers

Question 1.
What are the advantages of classifying organisms?
Answer:
There are a wide range of life forms (about 10 million-13 million species) around us. These life forms have existed and evolved on the Earth over millions of years ago. The huge range of these life forms makes it very difficult to study them one by one. Therefore, we look for similarities among them and classify them into different classes so that we can study these different classes as a whole. This makes our study easier.

Therefore, classification serves the following advantages:

• It determines the methods of organising the diversity of life on Earth.
• It helps in understanding millions of life forms in detail.
• It also helps in predicting the line of evolution.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
For developing a hierarchy of classification, we choose the fundamental characteristic among several other characteristics. For example, plants differ from animals in the absence of locomotion, chloroplasts, cell wall, etc. But, only locomotion is considered as the basic or fundamental feature that is used to distinguish between plants and animals. This is because the absence of locomotion in plants gave rise to many structural changes such as the presence of a cell wall for protection, and the presence of chloroplast for photosynthesis (as they cannot move around in search of food like animals).

Thus, all these features are a result of locomotion. Therefore, locomotion is considered to be a fundamental characteristic. By choosing the basic or fundamental characteristic, we can make broad divisions in living organisms as the next level of characteristic is dependent on these. This goes on to form a hierarchy of characteristics.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
RE Whittaker proposed a five kingdom classification of living organisms on the basis of Linnaeus system of classification. The five kingdoms proposed by Whittaker are Monera, Protista, Fungi, Plantae, and Animalia.

The basis for grouping organisms into five kingdoms is as follows:

(i) On the basis of the presence or absence of membrane-bound organelles, all living or
ganisms are divided into two broad categories of eukaryotes and prokaryotes. This division lead to the formation of kingdom Monera, which includes all prokaryotes.

(ii) Then, eukaryotes are divided a unicellular and multicellular, on the basis of cellularity. Unicellular eukaryotes form kingdom Protista, and multicellular eukaryotes form kingdom Fungi, Plantae, and Animalia.

(iii) Animals are then separated on the basis of the absence of a cell wall.

(iv) Since fungi and plants both contain a cell wall, they are separated into different kingdoms on the basis of their modes of nutrition. Fungi have saprophytic mode of nutrition, whereas plants have autotrophic mode of nutrition. This results in the formation of the five kingdoms.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
The kingdom Plantae is divided into five main divisions: Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, and Angiosperms.

The classification depends on the following criteria:

• Differentiated/ Undifferentiated plant body
• Presence /absence of vascular tissues
• With/without seeds
• Naked seeds/ seeds inside fruits

(i) The first level of classification depends on whether a plant body is well differentiated or not. A group of plants that do not have a well differentiated plant body are known as Thallophyta.

(ii) Plants that have well differentiated body parts are further divided on the basis of the presence or absence of vascular tissues. Plants without specialised vascular tissues are included in division Bryophyta, whereas plants with vascular tissues are known as Tracheophyta.

(iii) Tracheophyta is again sub-divided into division Pteridophyta, on the basis of the absence of seed formation.

(iv) The other group of plants having well developed reproductive organs that finally develop seeds are called Phanerogams. This group is further sub-divided on the basis of whether the seeds are naked or enclosed in fruits. This classifies them into gymnosperms and angiosperms. Gymnosperms are seed bearing, non-flowering plants, whereas angiosperms are flowering plants in which the seeds are enclosed inside the fruit.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
The criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals because

• plants are fixed at a specific position while the animals are motile.
• plants are autotrophic while animals are heterotrophic therefore, their requirements and the circulation of the necessary substances in body is different.
• plants lack nervous system but animals from the phylum cinidaria onwards posses nervous system.

Question 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.

Class 9 Science Chapter 7 Diversity in Living Organisms Additional Important Questions and Answers

Question 1.
Find out incorrect sentence
(a) Protista includes unicellular eukaryotic organisms
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms
(c) Both Monera and Protista may be autotrophic and heterotrophic
(d) Monerans have well defined nucleus
Answer:
(d) Monerans have well defined nucleus

Question 2.
Which among the following has specialised tissue for conduction of water?
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(c) (iii) and (iv)

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(d) Gymnosperms

Question 4.
Which one is a true fish?
(a) Jellyfish
(b) Starfish
(c) Dogfish
(d) Silverfish
Answer:
(c) Dogfish

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c)Mollusca
(d) Pisces
Answer:
(b) Echinodermata

Question 6.
Which among the following have open circulatory system?
(i) Arthropoda
(ii) Mollusca
(iii) Annelida
(iv) Coelenterata
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (ii)

Question 7.
In which group of animals, coelom is filled with blood?
(a) Arthropoda
(b) Annelida
(c) Nematoda
(d) Echinodermata
Answer:
(a) Arthropoda

Question 8.
Elephantiasis is caused by
(a) Wuchereria
(b) Pinworm
(c) Planarians
(d) Liver flukes
Answer:
(a) Wuchereria

Question 9.
Which one is the most striking or (common) character of the vertebrates?
(a) Presence of notochord
(b) Presence of triploblastic condition
(c) Presence of gill pouches
(d) Presence of coelom
Answer:
(a) Presence of notochord

Question 10.
Which among the following have scales?
(i) Xmphibians
(ii) Pisces
(iii) Reptiles
(iv) Mammals
(a) (i) and (iii)
(b) (iii) and (iv)
(c) (ii) and (iii)
(d) (i) and (ii)
Answer:
(c) (ii) and (iii)

Question 11.
Find out the false statement
(a) Aves are warm blooded, egg laying and have four chambered heart
(b) Aves have feather covered body, fore limbs are modified as wing and breathe through lungs
(c) Most of the mammals are viviparous
(d) Fishes, amphibians and reptiles are oviparous
Answer:
(d) Fishes, amphibians and reptiles are oviparous

Question 12.
Pteridophyta do not have
(a) root
(b) stem
(c) flowers
(d) leaves
Answer:
(c) flowers

Question 13.
Identify a member of Porifera
(a) Spongilla
(b) Euglena
(c) Penicillium
(d) Hydra
Answer:
(a) Spongilla

Question 14.
Which is not an aquatic animal?
(a) Hydra
(b) Jellyfish
(c) Corals
(d) Filaria
Answer:
(d) Filaria

Question 15.
Amphibians do not have the following
(a) Three chambered heart
(b) Gills or lungs
(c) Scales
(d) Mucus glands
Answer:
(c) Scales

Question 16.
Organisms without nucleus and cell organelles belong to
(i) fungi
(ii) protista
(iii) cyano bacteria
(iv) archae bacteria
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(b) (iii) and (iv)

Question 17.
Which of the following is not a criterion for classification of living organisms?
(a) Body design of the organism
(b) Ability to produce one’s own food
(c) Membrane bound nucleus and cell organelles
(d) Height of the plant
Answer:
(d) Height of the plant

Question 18.
The feature that is not a characteristic of protochordata?
(a) Presence of notochord
(b) Bilateral symmetry and coelom
(c) Jointed legs
(d) Presence of circulatory system
Answer:
(c) Jointed legs

Question 19.
The locomotory organs of Echinoderms are
(a) tube feet
(b) muscular feet
(c) jointed legs
(d) parapodia
Answer:
(a) tube feet

Question 20.
Corals are
(a) Poriferans attached to some solid support
(b) Cnidarians, that are solitary living
(c) Poriferans present at the sea bed
(d) Cnidarians that live in colonies
Answer:
(d) Cnidarians that live in colonies

Very Short Answer Questions

Question 1.
Who is called as the father of taxonomy?
Answer:
Carious Linneaus

Question 2.
Which division of plants is also considered as amphibian of plant kingdom?
Answer:
Bryophytes having motile sperms are considered as amphibian of plant kingdom.

Question 3.
Which is the first division of plants to bear true shoot and root?
Answer:
Pteridophytes are the first to bear true root and shoot.

Question 4.
What are sporophylls in ferns?
Answer:
The leaves bearing the sporangia on their lower surface are called the sporophylls.

Question 5.
What are circinate leaves in a dryopteris?
Answer:
Circinate leaves are young leaves with the top margins folded inwardly.

Question 6.
Which part of pinus plant bears naked seeds?
Answer:
The female cone in pinus plant bears the seeds.

Question 7.
Which is the lateral appendage of shoot in angiosperms that acts as the reproductive organ?
Answer:
Flower is the lateral appendage of shoot that acts as the reproductive organ.

Question 8.
Which phase is dominant in a Funaria plant, gametophytic i sporophytic?
Answer:
In funaria, the gametophytic phase is dominant over the sporophytic phase.

Question 9.
Which type of spores are produced by the Funaria sporophyte?
Answer:
The sporophyte always produces the haploid spores that give rise gametophytic thallus.

Question 10.
In phanerogams, which plants bear the tap root system, gymnosperms or angiosperms?
Answer:
In phanerogams, the gymnosperms bear the tap roots.

Question 11.
In which plants the algae and fungi are present in the symbiotic relationship?
Answer:
Lichens consist of algae and fungi in the symbiotic relationship.

Question 12.
Which type of plants do you expect to observe in the artic regions?
Answer:
Bryophytes

Question 13.
Which division of plants consists mostly of ornamental plants?
Answer:
Pteridophytes, mostly ferns.

Question 14.
Name an aquatic pteridophyte that is capable of nitrogen fixation?
Answer:
Azolla

Question 15.
Name any two products of economical value that are obtained from algae.
Answer:
Iodine from kelps and silica from diatoms.

Question 16.
Which term is used to describe the network of hyphae in Rhizopus?
Answer:
Mycelium

Question 17.
Which nutrient medium is mostly used in laboratory for culturing microorganisms?
Answer:
Agar-agar, a product obtained from an algae is widely used as the culture medium.

Question 18.
What are the male and female reproductive organs in a moss or fern?
Answer:
In a moss or fern, the male and female reproductive organs are multicellular. They are called antheridum and archegonium respectively.

Question 19.
Name the division of plants whose members show a clearly distinct phases of alternation of generation.
Answer:
The members of bryophytes shows a clearly distinct alternation of generation with gametophytic phase being dominant over the sporophytic phase.

Question 20.
How is a protozoan different from a metazoan?
Answer:
A protozoan is unicellular but a metazoan is multicellular.

Question 21.
What is reproductive nature of an earthworm, unisexual or bisexual?
Answer:
An earthworm is bisexual, although the ovaries and testes mature at the different rimes.

Question 22.
Which cockroach has the larger antennae, male or female?
Answer:
The male cockroach has larger antennae than the female cockroach to perceive the odour in the surroundings.

Question 23.
Which phylum in animal kingdom has the largest and the second largest number of animals?
Answer:
The phylum arthropods has the largest and phylum mollusca has the second largest number of animals.

Question 24.
Which animal phylum is characterized by the presence of water canal system and water vascular system?
Answer:
Animals belonging to phylum Echinodermata have water vascular system, while the animals belonging to phylum Porifera have water canal system.

Question 25.
Which is the common physical feature in a bird and a fish?
Answer:
A bird and a fish both have streamlined body.

Question 26.
Name the animal that resembles both the reptiles and mammals.
Answer:
Duck billed platypus as it lays eggs like reptiles.

Question 27.
Name an aerial and an aquatic mammal.
Answer:
The only aerial mammal is bat and whale is an aquatic mammal.

Question 28.
Give an example of a parasitic annelida?
Answer:
Leech ( Hinidinaria) is the parasitic annelida. It is parasitic on cattle fronl whom it sucks the blood.

Question 29.
Name any three animals that show the generation of their body parts.
Answer:
Star fish regenerates it lost arm, frog regenerates its limb and a hous lizard regenerates it lost tail.

Question 30.
The animals of which phylum are characterized by the presence of protective hard shell?
Answer:
In the phylum Mollusca, all animals posses protective hard shell.

Question 31.
Sea horse is a vertebrate. In which class has it been placed li classification?
Answer:
Sea horse has been placed in the classes of Pisces under Oestiochythe

Question 32.
Name the three layers present in the embryo of a triploblastic animal.
Answer:
The three layers are ectoderm, mesoderm and endoderm.

Short Answer Questions

Question 1.
Who proposed binomial nomenclature? State its significance.
Answer:
The concept of binomial nomenclature was proposed by Carolus Linnaeus It was meant to give the scientific name to an organism for

• easy identification of an organism because the first name refers to the gen while the second name is the specific epithet.
• maintaining the uniformity of the names worldwide.

Question 2.
Draw a labelled diagram of a typical bacterium.
Answer:

Question 3.
How are prokaryotes different from eukaryotes?
Answer:
The organisms with the prokaryotic cell are characterized by the abs of well defined nucleus and membrane bound cell organelles while those with t eukaryotic cell, plant or animal cell are called eukaryotes.

In five kingdom system, prokaryotes have been placed in kingdom Me while the eukaryotic organisms have been placed in the rest of the four kingdoms.

Question 4.
Why are viruses considered to be at the border line of living and non-livings?
Answer:
Viruses are the obligate parasites, they are living only when in body of a living host otherwise, they act as non-living. They can be filtered and crystallized like chemicals.

Question 5.
What are protozoa? Give examples.
Answer:
Protozoa are the eukaryotic unicellular animals which belongs to the kingdom ‘Protista’. These animals are either free living or parasitic. The free living protozoa include amoeba and paramecium while the parasitic protozoa which are the cause of infectious disease to mankind include

• Plasmodium, the causative agent of malaria.
• Trypanosoma, the causative agent of sleeping sickness.
• Entameoba histolytica, the causative agent of amoebiasis.

Question 6.
Why are the saprophytes called the natural cleaners of the environment?
Answer:
The saprophytes are called the natural cleaners of the environment because they decompose the dead organic matter of living organisms such as the agricultural wastes and animal excreta into simple inorganic form to allow the recycling of the different nutrients.

Question 7.
Name the unicellular fungi with its importance.
Answer:
One of the unicellular fungi is yeast. It is commercially an important gradient as it lays the foundation of brewery and bakery industry because of its ability of anaerobically breaking down of sugar, producing the alcohol, ethanol and carbon dioxide. The ethanol is the product used in brewery industries and produced carbon dioxide is used in bakery industry to make the bread or cake soft and spongy.

Question 8.
Differentiate in the followings:
(a) Monera and Protista
(b) Bryophytes and pteridophytes
(c) Monocots and dicots
(d) Mollusca and echinodermata
(e) Chordata and non-chordata
(f) Cartilaginous and bony fishes
(g) Platyhelminthes and nematoda
Answer:

(b) Bryophytes and pteridophytes

(c) Monocots and dicots

(d) Mollusca and echinodermata

(e) Chordata and non-chordata

(f) Cartilaginous and bony fishes

(g) Platyhelminthes and Nematoda

Question 9.
List some of the important characteristics of protochordates.
Answer:
The protochordates are primitive to chordates.

• They are triploblastic and bilaterally symmetrical with a coelom.
• They are exclusively marine animals.
• They posses a notochord at some stage of their life that separates nervous tissue with gut.
• They have muscles attached to notochord for easy movement.
  Example: Herdmania, Amphixous and Balanglossus.

Question 10.
Write the characteristic adaptive features in the followings:
(a) Fishes for swimming in water.
(b) Birds for flying in air.
Answer:
(a) Fishes are well adapted for swimming in water because they have

• streamlined body to reduce friction when swimming in water.
• scales to prevent the rotting action of water on body.
• have fins to swim and tail fin to steer their motion in water.
• have gills for the gaseous exchange.
• have swim bladder to change their buoyancy i.e. the depth in water.

(b) Birds are well adapted for flying because of the presence of

• streamlined body with kneeled chest to reduce the air friction in air.
• forelimbs modified in form of wings to help bird fly in air.
• pneumatic bones to keep the body weight light.
• air bladder to change their height in air.

Question 11.
Do all birds fly? If not, give examples.
Answer:
No, all birds do not fly. They are few non-flying birds such as

• Ostrich of Africa,
• Kiwi of Newzealand
• Penguin of Antarctica region

Long Answer Questions

Question 1.
How have the organisms being classified on the basis of their mode of nutrition?
Answer:
The organisms on the basis of their mode of nutrition have been basically classified into two types, autotrophs and heterotrophs. Autotrophs can synthesize their own food like green plants but heterotrophs are organisms that are dependent on others for their nutrition. These organisms based on that how they obtain their food have been further categorized into different types such as:

Saprophytes: Organisms who obtain their nourishment from other living organisms called host. They cause serious damage to the host as they obtain their nourishment such as malaria parasite, leech, ticks, etc.

Symbiotic: Organisms which while living together help each other in gaining the nourishment such as lichen. They are composed of algae and fungi, the fungi absorbs water and minerals while green algae synthesize the food that is equally shared by the both algae and fungi.

Question 2.
State the different characteristic features of monera.
Answer:
All members of Monera have the following characteristics:

• They lack multicellular body design.
• The body cell lack well defined nucleus and membrane bound cell organelles.
• Some posses cell wall like bacteria and cyanobacteria while others like mycoplasm lack it.
• They can be autotrophic, heterotrophic or symbiotic in their mode of nutrition.
• Example: Bacteria, mycoplasma and cyanobacteria like nostoc, anabena and oscillatoria.

Question 3.
State the different characteristic features of protista.
Answer:
The kingdom Protista include number of eukaryotic unicellular organisms.

• They vary in their structure, some have locomotory organs like cila or flagella while others don’t like amoeba.
• They vary in their mode of nutrition, some are autotrophic while others are heterotrophic from holozoic to parasitic. Euglena is the only Protista that shows both autotrophic and heterotrophic mode of nutrition.

Question 4.
State the chief characteristics of the division Thallophyta with examples.
Answer:
The chief characteristics of thallophytes include:

• The plant body is in form of individual thallus ire. not differentiated between stem, root and leaf.
• They lack vascular system and tissues for conduction of water and food.
• They are predominantly aquatic.
• Their reproductive organs are single celled.
• They are autotrophic like algae, symbiotic like lichens or saprophytic like fungi in their mode of nutrition.

Question 5.
Which conventions are followed when writing the scientific name of organism?
Answer:
When writing the scientific name of an organism, the sign conventions followed are:

• The name of genus is written first followed by the name of species.
• The name of genus starts with capital alphabet while that of species starts with small alphabet.
• When printed, it is always written in italics and if written by hand, it is underlined.

These NCERT Solutions for Class 9 Science Chapter 6 Tissues Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Tissues NCERT Solutions for Class 9 Science Chapter 6

Class 9 Science Chapter 6 Tissues InText Questions and Answers

Question 1.
What is a tissue?
Answer:
Tissue is a group of cells that are similar in structure and are organised together to perform a specific task.

Question 2.
What is the utility of tissues in multi-cellular organisms?
Answer:
In unicellular organisms, a single cell performs all the basic functions such as respiration, movement, excretion, digestion, etc. But in multicellular organisms, cells are grouped to form tissues. These tissues are specialised to carry out a particular function at a definite place in the body. For example, the muscle cells form muscular tissues which helps in movement, nerve cells form the nervous tissue which helps in transmission of messages. This is known as division of labour in multicellular organisms. It is because of this division of labour that multicellular organisms are able to perform all functions efficiently.

Question 3.
Name types of simple tissues.
Answer:
Simple permanent tissues are of three types: Parenchyma, Collenchyma, and Sclerenchyma. Parenchyma tissue is of further two types – aerenchyma and chlorenchyma.

Question 4.
Where is apical meristem found?
Answer:
Apical meristem is present at the growing tips of stems and roots. Their main function is to initiate growth in new cells of seedlings, at the tip of roots, and shoots.

Question 5.
Which tissue makes up the husk of coconut?
Answer:
The husk of a coconut is made up of sclerenchyma tissue.

Question 6.
What are the constituents of phloem?
Answer:
Phloem is the food conducting tissue in plants. It is made up of four components:

• Sieve tubes
• Companion cells
• Phloem parenchyma
• Phloem fibres

Question 7.
Name the tissue responsible for movement in our body.
Answer:
The muscular tissue is responsible for movement in our body.

Question 8.
What does a neuron look like?
Answer:
A neuron consists of a cell body with a nucleus and cytoplasm. It has two important extensions known as the axon and dendrites. An axon is a long thread-like extension of nerve cells that transmits impulses away from the cell body. Dendrites, on the other hand, are thread-like extensions of cell body that receive nerve impulses. Thus, the axon transmits impulses away from the cell body, whereas the dendrite receives nerve impulses. This coordinated function helps in transmitting impulses very quickly.

Question 9.
What are the functions of areolar tissue?
Answer:
Functions of areolar tissue:

• It helps in supporting internal organs.
• It helps in repairing the tissues of the skin and muscles.

Class 9 Science Chapter 6 Tissues Textbook Questions and Answers

Question 1.
Define the term “tissue”.
Answer:
Tissue is a group of cells that are similar in structure and are organized together to perform a specific task.

Question 2.
How many types of elements together make up the xylem tissue? Name them.
Answer:
There are four different types of cells that make up the xylem tissue. They are:

• Tracheids
• Vessels
• Xylem parenchyma
• Xylem fibres

Question 3.
How are simple tissue? different from complex tissues in plants?
Answer:
Simple tissue:

1. It consists of the cells that similar in both their structure and functions.
2. Parenchyma, collenchyma and sclerenchyma.

Complex tissue:

1. It consists of the cells that are neither similar in structure nor in function but collectively help to perform a specific function.
2. Xylem and phloem

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall.
Answer:

Question 5.
What are the functions of the stomata?
Answer:
Functions of the stomata:

• They allow the exchange of gases (CO₂ and O₂) with the atmosphere.
• Evaporation of water from the leaf surface occurs through the stomata. Thus, the stomata help in the process of transpiration.

Question 6.
Diagrammatically show the difference between the three types of muscle fibres.
Answer:
The three types of muscle fibres are: Striated muscles, smooth muscles (unstriated muscle fibre), and cardiac muscles.

Question 7.
What is the specific function of the cardiac muscle?
Answer:
The specific function of the cardiac muscle is to control the contraction and relaxation of the heart.

Question 8.
Draw a labelled diagram of a neuron.
Answer:

Question 9.
Name the following:
(a) Tissue that forms the inner lining of our mouth.
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Tissue that forms the inner lining of our mouth ? Epithelial tissue
(b) Tissue that connects muscle to bone in humans? Dense regular connective tissue (tendons)
(c) Tissue that transports food in plants ? Phloem
(d) Tissue that stores fat in our body ? Adipose tissue
(e) Connective tissue with a fluid matrix ? Blood
(f) Tissue present in the brain ? Nervous tissue

Question 10.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:
Skin: Stratified squamous epithelial tissue
Bark of tree: Simple permanent tissue
Bone: Connective tissue
Lining of kidney tubule: Cuboidal epithelial tissue
Vascular bundle: Complex permanent tissue

Question 11.
Name the regions in which parenchyma tissue is present.
Answer:
Leaves, fruits, and flowers are the regions where the parenchyma tissue is present.

Question 12.
What is the role of epidermis in plants?
Answer:
Epidermis is present on the outer surface of the entire plant body. The cells of the epidermal tissue form a continuous layer without any intercellular space. It performs the following important functions:

• It is a protective tissue of the plant body
• It protects the plant against mechanical injury
• It allows exchange of gases through the stomata

Question 13.
How does the cork act as a protective tissue?
Answer:
The outer protective layer or bark of a tree is known as the cork. It is made up of dead cells. Therefore, it protects the plant against mechanical injury, temperature extremes, etc. It also prevents the loss of water by evaporation.

Question 14.
Complete the following chart:

Answer:

Class 9 Science Chapter 6 Tissues Additional Important Questions and Answers

Question 1.
Which of the following tissues has dead cells?
(a) Parenchyma
(b) Scierenchyma
(c) Collenchyma
(d) Epithelia tissue
Answer:
(b) Scierenchyma

Question 2.
Find out incorrect sentence
(a) Parenchymatous tissues have intercellular spaces
(b) Collenchymatous tissues are irregularly thickened at corners
(c) Apical and intercalary meristems are permanent tissues
(d) Meristeniatic tissues, in its early stage, lack vacuoles
Answer:
(c) Apical and intercalary meristems are permanent tissues

Question 3.
Girth of stem increases due to
(a) apical meristem
(b) lateral meristem
(c) intercalary meristem.
(d) vertical meristem
Answer:
(b) lateral meristem

Question 4.
Which cell does not have perforated cell wall?
(a) Tracheids
(b) Companion cells
(c) Sieve tubes
(d) Vessels
Answer:
(b) Companion cells

Question 5.
Intestine absorb the digested food materials. What type of epithelial cells are responsible for that?
(a) Stratified squamous epithelium
(b) Columnar epithelium
(c) Spindle fibres
(d) Cuboidal epithelium
Answer:
(b) Columnar epithelium

Question 6.
A person met with an accident in which two long bones of hand were dislocated. Which among the following may be the possible reason?
(a) Tendon break
(b) Break of skeletal muscle
(c) Ligament break
(d) Areolar tissue break
Answer:
(c) Ligament break

Question 7.
While doing work and running, you move your organs like hands, legs etc. Which among the following is correct?
(a) Smooth muscles contract and pull the ligament to move the bones
(b) Smooth muscles contract and pull the tendons to move the bones
(c) Skeletal muscles contract and pull the ligament to move the bones
(d) Skeletal muscles contract and pull the tendon to move the bones
Answer:
(d) Skeletal muscles contract and pull the tendon to move the bones

Question 8.
Which muscles act involuntarily?
(i) Striated muscles
(ii) Smooth muscles
(iii) Cardiac muscles
(iv) Skeletal muslces
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) (ii) and (iii)

Question 9.
Meristematic tissues in plants are
(a) localised and permanent
(b) not limited to certain regions
(c) localised and dividing cells
(d) growing in volume
Answer:
(c) localised and dividing cells

Question 10.
Which is not a function of epidermis?
(a) Protection from adverse condition
(b) Gaseous exchange
(c) Conduction of water
(d) Transpiration
Answer:
(c) Conduction of water

Question 11.
Select the incorrect sentence
(a) Blood has matrix containing proteins, salts and hormones
(b) Two bones are connected with ligament
(c) Tendons are non-fibrous tissue and fragile
(d) Cartilage is a form of connective tissue
Answer:
(c) Tendons are non-fibrous tissue and fragile

Question 12.
Cartilage is not found in
(a) nose
(b) ear
(c) kidney
(d) larynx
Answer:
(c) kidney

Question 13.
Fats are stored in human body as
(a) cuboidal epithelium
(b) adipose tissue
(c) bones
(d) cartilage
Answer:
(b) adipose tissue

Question 14.
Bone matrix is rich in
(a) fluoride and calcium
(b) calcium and phosphorus
(c) calcium and potassium
(d) phosphorus and potassium
Answer:
(b) calcium and phosphorus

Question 15.
Contractile proteins are found in
(a) bones
(b) blood
(c) muscles
(d) cartilage
Answer:
(c) muscles

Very Short Answer Questions

Question 1.
Which tissue in plants has cells mostly found in state of division?
Answer:
Meristematic tissue present at the root and shoot tip has cells found in the state of division.

Question 2.
Which permanent tissue in plants consists of cells with thin cell wall?
Answer:
Parenchyma tissue

Question 3.
What is chlorenchyma?
Answer:
It’s a type of parenchyma whose cells possess the chloroplast containing chlorophyll.

Question 4.
Name the three types of simple permanent tissues found in plants.
Answer:
Parenchyma, collenchyma and sclerenchyma

Question 5.
Which simple permanent plant tissue have cell walls thickened at the comers?
Answer:
Collenchyma has somewhat more elongated cells with thickened walls corner.

Question 6.
Which are the two types of complex permanent tissues found in plants.
Answer:
Xylem and phloem

Question 7.
Which plant tissue cells are found in the apical part of stem and root?
Answer:
Meristematic tissue consists of parenchymatous cells with active protoplasm.

Question 8.
Name the only living component present in xylem?
Answer:
Xylem parenchyma

Question 9.
Name the only non-living component present in phloem.
Answer:
Phloem fibres

Question 10.
Name the four types of animal tissues?
Answer:
The lour types of animal tissues are epithelial tissue, connective tissues, musclar tissues and nervous tissue.

Question 11.
Which tissue forms the outer human skin?
Answer:
Epithelial tissue forms the outer human skin consisting of dead cells.

Question 12.
Which animal tissue you expect to observe when watching a thin slice Jther?
Answer:
Epithelial tissue as it makes, the skin from which the leather is made after tanning?

Question 13.
Bone is a type of tissue. In which tissue type will you classify the bones?
Answer:
Bone is a hard skeletal tissue, a type of connective tissue.

Short Answer Questions

Question 1.
How is a simple permanent tissue in plants different from compound permanent tissue?
Answer:
A simple permanent tissue consists of cells that are similar both in their structure and functions but the compound permanent tissue consists of cells that are different in their structure but are associated in performing a specific common function.

Question 2.
How is a cell from the meristematic tissue different from the cell from a simple permanent tissue?
Answer:
The meristematic tissue cells are thin walled with active protoplasm and rrninent nucleus. They also have reduced vacuole as compared to the living is of the permanent tissues.

Question 3.
What are the characteristics of the parenchymatous tissue?
Answer:
The chief characteristics of parenchymatous tissue include:

• Cells are mostly round or polygonal but can also be elongated.
• Cells have thin cell wall with uniform thickness.
• Cells are live with nucleus and large vacuoles.
• Cells may or may not have the intracellular, spaces.

Question 4.
What is specific about the cells of collenchymatous tissue in plants?
Answer:
The cells of collenchymatous tissue have thin cells walls but these walls are thickened at the corners because of the deposition of extra cellulose and pectin.

These cells also lack intercellular spaces being closely packed to each other.

Question 5.
State the functions of collenchymatous tissues with their locations in a plant.
Answer:
Collenchymatous tissue mostly found located under the epidermis helps

• providing mechanical support and elasticity to plant parts.
• providing flexibility to plant parts like leaves and branches to prevent them from breaking or tearing in blowing wind.
• manufacture of sugar and starch if they contain chloroplasts.

Question 6.
Describe the structure and function of sclerenchyma as a simple plant tissue.
Answer:
Sclerenchyma as a tissue consists of dead cells because of the excessive deposition of cutin, pectin and lignin in their cell walls. These chemicals act as cement hence, makes the cell wall hard. The cells are long, narrow and fine

Question 7.
Differentiate in the followings:
(a) Bone and cartilage
(b) Tendon and ligament
(c) Meristematic and permanent tissue
(d) Cardiac muscles and striated muscles
(e) Blood and lymph
Answer:
(a) Bone and Cartilage

(b) Tendon and ligament

(c) Meristematic and permanent tissue

(d) Cardiac muscles and striated muscles

(e) Blood and lymph

Question 8.
What do you understand by apical dominance? Why is it exercised?
Answer:
Apical dominance refers to the dominance of apical meristem over other meristems. This in most plants delays the branching of stem. Therefore, to promote: ranching in hedge plants, a gardener cuts off the apical meristem.

Long Answer Questions

Question 1.
What are meristematic tissues? State their characteristics with locations.
Answer:
Meristematic tissues are tissues found at the meristematic regions i.e. growing regions in plants. Their cells have active protoplasm, with small or no vacuole, no chloroplasts but a large active nucleus. The cells are compactly packed with less or no intercellular spaces. The cells are mostly involved in division to form new cells which later undergo differentiation to from different cells to add to plant growth.

Meristems on basis of its location has been classified into three types:
Apical meristem, present at the growing tips of stem and root. It contributes to vertical plant growth.
Intercalary meristem, present at the base of leaves and internodes where they contribute to the branching of stem.

Lateral meristem, present on the lateral side of stem and root in form of cambium. These meristems contribute to the growth in thickness i.e. increase in thickness or girth of stem and root.

Question 2.
State the functions of the different parenchymatous tissue.
Answer:
Parenchymatous tissue performs the following functions:

• They mostly store and assimilate food material.
• When in form of chlorenchyma in leaves and other green plant parts, they help in photosynthesis.
• In aquatic plants, their cells bear large air cavities and hence, help provide the necessary buoyancy to the plant parts for floating in water.
• They provide mechanical support also particularly the xylem parenchyma and pholem parenchyma.

Question 3.
What are vascular bundles? State their functions.
Answer:
Vascular bundles are the bundles comprising of xylem and pholem.

Xylem consist of four components: tracheids, vessel cells, xylem fibres and xylem parenchyma. Of these components, only xylem parenchyma is living the rest are the dead structures. Tracheids and vessels cells are tubular structures with a lumen to allow the transport of water and minerals from root to leaves. Their thick walls have pits too, for the lateral transport of water.

Phloem consists of four components: sieve cells, companion cells, phloem parenchyma and phloem fibres. In these components, phloem fibres are dead and are meant to provide, mechanical strength to the tissue, while the rest are living structures. However, the sieve cells are live but lack nucleus and hence, they are always present along with companion cells. These cells perform cellular activities for sieve cells. The sieve cells or sieve tubes allow the bidirectional flow of food.

Question 4.
Describe briefly the different types of epithelial tissues with their functions.
Answer:
The different types of epithelial tissues include:
(i) Squamous epithelial tissue: Their cells are thin, flattened, polygonal like the tiles of floor. They form the outer layer of skin, tongue, esophagus and the inner lining of mouth.

(ii) Cuboidal epithelial tissue: Their cells are mostly cuboidal in shape. They form the lining of kidney tubules and ducts of salivary glands. These cells provide mechanical support and if ciliated, they help in pushing the content forward such as in oviduct.

(iii) Columnar epithelial tissue: Their cells are long column or pillar like. They form the inner lining of the intestine and if ciliated, they help in movement, such as in respiratory tract, the ciliated columnar epithelial cells help in pushing the mucus upward from lungs to mouth.

(iv) Glandular epithelial tissue: Their cells are also columnar or cuboidal shape and involved in secretion of mucus and digestive enzymes and therefore are found in different body glands.

Question 5.
Name the different types of connective tissues with their functions.
Answer:
Areolar tissue: ft is present in between the skin and underlying muscles contains different types of cells that are responsible for the formation of different fibre types i.e. collagen fibres and elastin fibres. Some cells present also synthesize heparin, an anticoagulant and histamine.

Tendon: These are white fibrous tissues which attaches the muscles to bones.

Ligament: These are yellow fibrous tissues which attaches the two bones together at the point of their joint.

Adipose tissue: This tissue underlying the skin stores excess of fats and provide insulation to the body.

Blood is fluid connective tissue that helps in circulation of nutrients, metabolic waste, gases, etc in body along with providing the immunity or defense against any infection.

Question 6.
What are the different components of blood? What are their functions?
Answer:
Blood is the fluid connective tissue that consists of blood cells found immersed in blood plasma. Blood plasma is the fluid part of blood. In blood three different cell types are found: RBC, WBC and blood platelets.

Red blood cells(RBC) are carrier of haemoglobin and therefore responsible for the transport of oxygen from lungs to body parts.

White blood cells(WBC) are either amoeboid shaped called phagocytes or large oval cells with large nucleus called lymphocytes. Phagocytes provide defense t engulfing the germs and lymphocytes provide defense by secreting antibodies: fight to invading germs.

Blood platelets are small fragments like cells that are responsible for blood coagulation after the injury.