Author name: Prasanna

Haryana State Board HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Very Short-Answer Type Questions

Question 1.
What is the meaning of health?
Answer:
The proper shape of the entire organs and systems of the body, their location, and functioning is called as health. Health is the normal position of the ability of physical, mental, and social life.

Question 2.
What is called an ailment or disease?
Answer:
Any kind of disruption developing in good health is called a disease. Or in other words, occurring of any disorder in the physical system is called a disease.

Question 3.
Give the name of a factor that influences health.
Answer:
Personal hygiene.

Question 4.
Give a reason that affects the community’s cleanliness.
Answer:
Failure of disposing of garbage scientifically.

Question 5.
How does food affect our health ?
Answer:
Excess of deficiency of nutritious elements in diet both leave ill effects on our health.

Question 6.
How does our environment harm us ?
Answer:
Unwanted materials found in the environment proved to be the cause of diseases for us.

Question 7.
Is it called health that the body is disease-free ?
Answer:
‘No, even if we are disease-free, we cannot consider ourselves healthy.

Question 8.
How is a disease diagnosed ?
Answer:
From the symptoms of the disease.

Question 9.
What are the symptoms of diseases ?
Answer:
The signs that indicate any disorder in our body, are called as the symptoms of disease.

Question 10.
What do the physicians do to confirm a certain disease ?
Answer:
They refer the laboratory’ test,

Question 11.
How many types of diseases are there (on the basis of time duration) ?
Answer:
Two types: Acute diseases, Chronic diseases.

Question 12.
What are called as acute diseases ?
Answer:
Ailments that are short lived are called as acute diseases.

Question 13.
Write down names of two acute diseases.
Answer:
Cough-cold, headache.

Question 14.
What are called as chronic diseases?
Answer:
Chronic diseases-the sustain for long time or throughout life.

Question 15.
Write down the name of chronic disease.
Answer:
Filariasis (elephantiasis).

Question 16.
Can a person fall ill in the inadequacy of diet?
Answer:
Yes, in the inadequacy diet a person can fall ill.

Question 17.
The antibiotic penicillin blocks the processes that build in bacteria.
Answer:
cell wall.

Question 18.
What are called as communicable diseases ?
Answer:
Diseases that spread through contact are called as communicable diseases.

Question 19.
Write down the names of two communicable diseases.
Answer:
Cholera, jaundice.

Question 20.
What are called as non-communicable diseases ?
Answer:
Diseases that do not spread through infection or pathogens are called as non-communicable diseases.

Question 21.
Write down names of two non-communicable diseases.
Answer:
Goitre and Diabetes.

Question 22.
Write names of two diseases caused due to virus.
Answer:
Cough-cold, Influenza, Polio.

Question 23.
What type of disease is cancer ?
Answer:
Non-communicable disease.

Question 24.
Which diseases are caused due to bacteria ?
Answer:
Cholera and Tuberculosis.

Question 25.
Give two examples of the diseases caused due to fungi.
Answer:
Skin diseases (dermatosis)-Ringworm, eczema etc.

Question 26.
Write names of two diseases caused by protozo
Answer:
Malaria and kala-azar.

Question 27.
What are the names of diseases caused by a worm?
Answer:
Elephantiasis (Filariasis), ascariasis.

Question 28.
Write down the name of the factor of acne or pimple.
Answer:
Staphylococci bacteria.

Question 29.
What is the cause of sleeping sickness?
Answer:
Trypanosoma protozoa.

Question 30.
Which disease is caused due to lachmania protozoan?
Answer:
Kala-azar.

Short-Answer Type Questions

Question 1.
What is health ? Write the advantages of good health.
Answer:
Health is a state of being well enough for a person to function well physically, mentally and socially. The structure and functions of all the body parts of a healthy person are normal. He is free from all types of psychological, mental and social tensions. A person free from only physical diseases is not frilly healthy.
Advantages: Following are the advantages of good health:
(1) The personal life of a healthy person is full of happiness.
(2) A healthy person utilises his potential fully as an individual, in the family, in society and in the country and does not become a burden on anyone.

Question 2.
What are the conditions necessary to maintain good health?
Answer:
The following conditions are necessary to maintain good health:
(1) Personal and family hygiene.
(2) Proper exercise and relaxation.
(3) Balanced diet.
(4) Community hygiene.
(5) Good habits and abstinence from intoxicants/drugs.

Question 3.
How does the environment affect health?
Answer:
The health of all persons is based on the surrounding environment. Physical factors constitute the environment. A person can fulfil his basic requirements only from the environment. This is possible only in a hygienic environment. A polluted environment has a bad effect on our health.

Question 4.
What things should be kept in mind while cleaning OUT homes 2
Answer:
Apart from dusting, cleaning cobwebs and disposing the wastes, we should also regularly keep a check on insects like cockroaches, mosquitoes, houseflies, spiders etc. Apart from this, it is necessary to white wash the house to keep it clean and beautiful. We should also keep our immediate surroundings neat and clean so that we don’t catch a disease.

Question 5.
What is a disease? How does disease occur?
Answer:
Any variation in the normal structure or function of a tissue or organ of a body. Or a change in psychological condition is called a disease. A diseased person does not feel comfortable.

Causes of disease:
The disease can occur because of any of the following three causes:
(1) Diseases caused due to deficiency or accumulation of some nutritive components e.g. Beri-beri, Obesity etc.
(2) Through infection (communicable) e.g., cholera, tuberculosis etc.
(3) Non-communicable diseases, like heart diseases, stones in the kidney or gall bladder etc.

Question 6.
What are non-communicable diseases ? Name any two.
Answer:
The diseases which are not caused by infection (and remain confined to the bearer only) are called non-communicable diseases, e.g. heart diseases, non-functioning of kidney, gall bladder or kidney stones etc.

Question 7.
What are hereditary (congenita!) diseases ? Give examples.
Answer:
The diseases which are carried by the parents to the offspring are called hereditary diseases. One such disease is haemophilia, in which the patient’s blood does not clot easily. Generally, the time taken for the clotting of blood increases from 2-8 minutes to 30 minutes-24 hours. A haemophilic person can die because of a continuous flow of blood even from a minor injury. Another such type of disease in sickle cell anaemia. The patient’s RBCs (Red Blood Cells) have defective haemoglobin, which reduces the oxygen-carrying capacity of the RBCs and the person goes on to become weak.

Question 8.
What are antibiotics? Why is their excessive use harmful?
Answer:
Antibiotics are the drugs which are derived from bacteria and are fatal for other bacteria. They do not harm the protoplasm of humans For example; penicillin, Terramycin, streptomycin etc. These save us by stopping the growth of pathogens or by. destroying them. If we use them excessively, then the useful micro-organisms in our alimentary canal will start getting destroyed which will affect our digestive system. .

Question 9.
What are the simple and easy precautions to remain free from the diseases caused by microorganisms ?
Answer:
The simple and easy precautions are:
(1) To keep the infected person away from a healthy one.
(2) By vaccination.
(3) To boil water before drinking.
(4) By keeping our surroundings neat and clean.
(5) By keeping check (on the growth) of mosquitoes and houseflies.
(6) By wearing properly washed clothes.
(7) By dumping the waste products in covered places. .
(8) By spraying insecticides so as to keep the surroundings free from harmful insects and pathogens.

Question 10.
What is the difference between bacteria and virus ?
Answer:
Given is the difference between bacteria and virus:
Bacteria:
1. They are living organisms.
2. They have a cellular structure.
3. They can be seen under a microscope.
4. Bacteria are bom from the same type of bacteria and multiply on their own.
5. Bacteria can be harmful as well as useful.

Virus:
1. They possess the features of both living as well as non-living.
2. Their structure is not cellular.
3. They can be seen only with the help of an electron microscope.
4. Viruses can be multiplied only inside a jost cell.
5. Viruses are only harmful.

Question 11.
How do antibiotics act against bacterial infection? Why are they not effective against virus? Explain with reason.
Answer:
Bacteria create a cell wall for their protection. But the antibiotic substances hinder the process of creation of the cell soil due to which bacteria cannot protect themselves and the antibiotics destroy them. Pathogens are destroyed by the same process. Because virus do not create any cell wall so, that antibiotics are not able to effect virus.

Question 12.
How does infection through air take place ?
Answer:
When an infected person exhales, sneezes or coughs, the pathogens spread in the air. When another person inhales the same air, these pathogens enter his body. Common cold, pneumonia and tuberculosis spread through this mode only. The occurence of common cold in crowded places is again due to this.

Question 13.
How does infection take place through water ?
Answer:
When the pathogens from the infected person enter the drinking water, then a healthy person also gets infected when he drinks this water. There is more incidence of water-borne diseases in rainy season. Cholera Spreads through this method.

Question 14.
Why are V.D. and gonorrhoea considered a social evil ?
Answer:
They are both sexual diseases. Their pathogens enter a healthy person’s body through cuts, wounds or genitals. In such a condition, these diseases are spread from a woman to a man or from a man to a woman during sexual contact. As these diseases also spread through sexual activity, so these diseases are considered a social evil because the incidence of these diseases is more in people who establish illegitimate sexual relations. Prostitution is the main cause of this. Social awareness can reduce the incidence of these diseases.

Question 15.
What is the sign inflammation ?
Answer:
During infection, immune system recruits many cells to the affected tissue to kill the disease causing microbes, this process is called imflammation. As a part of this process, there are local effects such as swelling, pain and general effects such as fever.

Question 16.
Why is vaccination of children and expectant mothers done ?
Answer:
Polio, whooping cough, measles diphtheria, typhoid and tetanus are some infective diseases common in the children. To prevent these diseases newly bom babies and children must be got vaccinated or immunised. To immunise all the children and expectant mothers under a plan, sufficient quantity of vaccines have been manufactured and distributed in India. If the expectant mothers are treated with the tetanus vaccine before she delivers the baby, the newly bom baby will have the ability of immunity. Therefore, lacs of lives of children can be saved by vaccination at proper time.

Essay Type Questions

Question 1.
How do pathogens spread diseases ?
Answer:
Those pathogens, that spread diseases are called as pathogens. With the entrance of pathogens into the human body and their reaction a man falls ill.

Pathogens can spread diseases in the following ways:
1. Through Contact: Some pathogens spread diseases through direct contact with the patient or the thing used by the patient; for example, if a healthy person puts on the clothes of a patient or uses his bed, the pathogens get entered into his body and make him ill, like pathogens causing skin diseases.

2. Through eatables: Pathogens of typhoid, cholera etc. the remaining food of the patient enter into our body and spread diseases.

3. Through air: Pathogens of common cold, influenza etc. enter the body of a healthy person through breathing.

4. Through vectors: Houseflies, mosquitoes, mice, bugs etc. are some insects that spread diseases; like malaria spreads through female anopheles mosquito bite. Plague is spread by mice.

5. Through water: Pathogen on mixing into drinking water reach to the healthy person and diseases
like cholera, diarrhoea, typhoid etc. surround the human beings. Therefore, specially during rainy season it is suggested to boil drinking water and then cool it down to drink there after, so that the pathogens might destroy.

Question 2.
Give a brief introduction of the infectious diseases.
Answer:
Infectious diseases are caused due to micro-organisms like:
1. Cholera: The cause of qholera is vibriocoli. It occurs due to drinking dirty water or infected eatables. Pathogens of cholera are spread through housefly. The patient suffers from painless water loss of due to acute vomitting and diarrhoea.

2. Tuberculosis: It’s cause is micro bacterium tuberculosis. This disease affects the lungs. The patient suffers from constant fever, blood in sputum, bodyache, diarrhoea and the neck ribs get swollow.

3. Typhoid: It spreads through salmonela bacterium typall. Its main symptoms are fever, headache, bleeding nose, black circles on the skin, reddish blisters on the chest.

4. Cough and Cold: It is caused by rhino works. Constant flowing of liquid through nostrils, redness in the eyes, irritation and headache are its main symptoms.

5. Chickenpox: It is caused by the pathogen berisola joster. Here, big circles on the patients body
get formed, severe backache, high fever and headache are its other symptoms. Its a effect lasts long for 14 days. .

6. Polio: It is an infectious disease among the children. It spreads through the pathogen of polio mylitis.
Its symptoms are fever, vomiting, bodyache, stiffness and difficulty in controlling muscles. The children develop paralysis. .

7. Rabies: It is caused by the Rabies pathogen. It is caused due to the bite of dog, cat, bat, jackal, fox etc. Patient suffering from Rabies scares water (hydrophobia). It affects the nervous system. Due to this the patient meets a painful death.

8. AIDS: It comes under the category of sexually transmitted diseases which is caused due to virus. A person suffering from HIV-AIDS, Immunisation system becomes weaker, with the result the body fails to face even the minor infection. The patient of HIV runs with pneumonia even with minor cough and cold. Presently, AIDS is an incurabled disease.

Question 3.
Which programmes have been adopted the Govt, of control communicable diseases ?
Answer:
Following programme have been adopted by the Govt, of India to control communicable diseases :

1. National Malaria Eradiction Programme:
This started as a control measure in 1953. In 1975, it was converted from control of the disease to total eradication programme. The changed programmed was started in 1977 which was effectively enforced. The programme include inspection of areas affected by malaria educated the people using suitable medicines, spraying of insecticides to kill mosquitoes, identifying the malaria patients, distribution of antimalarial drugs to keep their accounts etc.

Proper Vaccination List:

2. National Tuberculosis Eradication Programme:

Following stages are there in this programme:
(1) To identify the person suffering from tuberculosis in the very first stage and giving him treatment.
(2) To prevent the diseases by inoculating the children at an early age by BCG vaccine.
(3) Many TB clinics, district centres and education centres started in country.

3. National Cholera Eradication Programme: It includes cleaning the cholera effected area, making fresh drinking water availability and proper disposal of urban faecol waste. To educate people and for immediate treatment, mobile treatment units have been established for it. To inoculate people with cholera injections has been arranged in these.

4. National Leprosy Eradication Programme : A survey was conducted for thistdiseases in India and many treatment centres have been setup. Houses have been constructed for rehabilitation of the infected persons. Tamilnadu Govt, has opened a Leprosy Education and Research Centres, where there is an advisory committee for this diseases. Sulpha drugs are used for the treatment of the disease.

Practical Work

Experiment 1.
To study the effect of possible calamities on the health of people.

Procedure:
There are two effects of natural calamities on the health. Prepare its table with the help of science-teacher.

Experiment 2.
To study the provisions made by local municipal corporation for the supply of clear drinking water.

Procedure:
Prepare a list of provisions made by the local municipal corporation for the supply of clear drinking water.
(1) The cholorination of drinking water is done.
(2) Water storage tank is cleaned regulary.
(3) Tubewells are used to get underground drinking water.
(4) Water treatment plant is setup for canal water.
(5) Water supplies by pipes to avoid pollution.

Experiment 3.
To study the provisions made by local authority to manage garbage.

Procedure:
Prepare a list of provisions related to it from officer related to local conservator.
(1) The garbage is cleaned by landfill.
(2) Organic garbage is treated by roasting.
(3) Covered vehicles are used to throw garbage.
(4) Composte prepared from organic garbage is selling to farmers.
(5) Local water is planned to set up solid waste treatment plant.

Quick Review of the Chapter

1. Cholera is a ………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(A) bacterial

2. Malaria spreads by:
(A) mouse
(B) fly
(C) mosquito
(D) bug
Answer:
(C) mosquito

3. Cholera spreads by :
(A) fly
(B) mouse
(C) mosquito
(D) none of these
Answer:
(A) fly

4. Respiratory diseases spread by :
(A) food
(B) water
(C) air
(D) animal
Answer:
(C) air

5. Communicable disease is :
(A) goitre
(B) beri-beri
(C) obesity
(D) tuberculosis
Answer:
(D) tuberculosis

6. Example of chronic disease is :
(A) cough
(B) cold
(C) elephantiasis
(D) headache
Answer:
(C) elephantiasis

7. Acute disease is:
(A) lungs tuberculosis
(B) Dengue fever
(C) blindness
(D) disentary
Answer:
(D) disentary

8. Which of the following disease is caused by virus ?
(A) Anthrax
(B) Dengue fever
(C) Malaria
(D) Cholera
Answer:
(B) Dengue fever

9. The reason of AIDS is :
(A) bacteria
(B) protozoa
(C) virus
(D) fungi
Answer:
(C) virus.

10. The prostitution in society is the reason which of the following disease
(A) pneumonia
(B) gonorrhoea
(C) giardiasis
(D) influenza
Answer:
(B) gonorrhoea

11. AIDS can spread by:
(A) kiss
(B) embracing
(C) combined used blade of shave
(D) hug
Answer:
(C) combined used blade of shave

12. AIDS cannot spread by:
(A) sexual relations
(B) blood transfusion
(C) from placenta of mother
(D) by shaking hands
Answer:
(D) by shaking hands

13. Which disease can be caused by attack on lever?
(A) diabetes
(B) malaria
(C) jaundice
(D) goitre
Answer:
(C) jaundice

14. Which disease can be caused by fungi?
(A) kala-azar
(B) cholera
(C) malaria
(D) skin disease
Answer:
(D) skin disease

15. Vomiting is related to :
(A) mind
(B) liver
(C) lungs
(D) kidney
Answer:
(A) mind

16. Which fight against microbes :
(A) RBC
(B) WBC
(C) platelets
(D) lymphocytes
Answer:
(B) WBC

17. Which disease happens first and last times ?
(A) rabies
(B) polio
(C) chicken pox
(D) small pox
Answer:
(D) small pox

18. The main symptom of AIDS is:
(A) the loss in resistance power
(B) fever
(C) arthritus
(D) high cough ‘
Answer:
(A) the loss in resistance power

19. Rabies is a …………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(B) viral

20. Before going to fair, there is vaccination of:
(A) tuberculosis
(B) hepatitis
(C) typhoid
(D) cholera
Answer:
(D) cholera

21. Deficiency disease is not:
(A) diarrohea
(B) scurvy
(C) pellagra
(D) xeropthalmia
Answer:
(A) diarrohea

22. Communicable disease is :
(A) intluenza
(B) hemophilia
(C) goitre
(D) anaemia
Answer:
(A) influenza

23. The disease in mother’s placenta is:
(A) polio
(B) AIDS
(C) goitre
(D) scurvy
Answer:
(B) AIDS

24. Vaccination of which of the following is not available?
(A) whooping cough
(B) polio
(C) diphtheria
(D) AIDS
Answer:
(D) AIDS

25. Which disease can be caused by throwing solid waste in open?
(A) elephantiasis
(B) goitre
(C) cholera
(D) AIDS
Answer:
(C) cholera

26. Match the following:

(A) a-iv, b-i,c-ii,d-iii
(B) a-i,b-ii,c-iii,d-iv
(C) a-ii,, b-ic—iii,d-ii,
(D) a-iii. b-Ii, c-i, d’-iv
Answer:
(A) a-iv b-i, c-ii d-III

27. Which disease is caused by viruses?
(A) Common cold
(B) Dengue fever
(C) Influenza
(D) All of these
Answer:
(D) All of these

Haryana State Board HBSE 9th Class Science Important Questions Chapter 12 Sound Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 12 Sound

Very Short-Answer Type Questions

Question 1.
What is sound?
Answer:
Sound is a kind of energy that is produced by vibration.

Question 2.
What is the state of the body that is producing sound?
Answer:
Vibrating state.

Question 3.
How does sound travel in air?
Answer:
Sound travels in compressions and rarefactions.

Question 4.
In which the speed of sound is more-solids, liquids or gases ?
Answer:
In solids.

Question 5.
What is the speed of sound in air at 0°C ?
Answer:
331 m/s.

Question 6.
What is the speed of sound in air at 22°C ?
Answer:
344 m/s.

Question 7.
What is the speed of sound in distilled water at 25°C ?
Answer:
1498 m/s.

Question 8.
In which the speed of sound is 5100 m/s at 20°C ?
(i) steel
(ii) aluminium
(iii) alcohol
(iv) mercury.
Answer:
(ii) aluminium.

Question 9.
What is the velocity of light ?
Answer:
3 x 10⁸ m/s.

Question 10.
Name two forms of energy.
Answer:
(i) Heat energy
(ii) Light energy.

Question 11.
What is medium ?
Answer:
The liquid or material through which sound propagates, is called medium.

Question 12.
What moves forward in sound propagation-particle of medium or disturbance ?
Answer:
Disturbance moves forward.

Question 13.
Why are sound waves called mechanical waves ?
Answer:
Because sound waves are characterise by the particles of medium. Therefore, the sound waves are called mechanical waves.

Question 14.
What is compression ?
Answer:
The region of high pressure in medium is called compression (C).

Question 15.
What is rarefaction ?
Answer:
The region of low pressure in medium is called rarefaction (R).

Question 16.
What is the meant of 256 written on tuning fork ?
Answer:
It means its frequency is 256 Hz.

Question 17.
What are electromagnetic waves ?
Answer:
Those waves which can travel without any medium are called electromagnetic waves.

Question 18.
Can the sound waves travel in vacuum ?
Answer:
No, sound waves cannot travel in vacuum.

Question 19.
Which type of waves are used by astronauts to talk on moon ?
Ans:
Radio waves.

Question 20.
Define wave velocity.
Answer:
The distance covered by the wave per unit time is called wave velocity.

Question 21.
What is the relation between frequency and time period ?
Answer:
Frequency is the reciprocal of time period. That is v = \(\frac {1}{T}\) .

Question 22.
What is the relation among velocity of wave, wavelength and frequency ?
Answer:
The velocity of wave is equal to the product of frequency and wavelength. That v = v X

Question 23.
How many type of waves can slinky produce ?
Answer:
Slinky can produce three types of waves :
(1) Transverse waves.
(2) Longitudinal waves.
(3) Periodic waves.

Question 24.
Which physical quantity is propagated by waves ?
Answer:
Energy.

Question 25.
What is the SI unit of frequency ?
Answer:
Hertz.

Question 26.
What is the SI unit of wavelength ?
Answer:
Metre.

Question 27.
What is the SI unit of wave velocity ?
Answer:
Metre/second (m/s)

Question 28.
Define pulse.
Answer:
Waves of shorter duration is called pulse.

Question 29.
What is the effect on the particles of the medium when a wave moves ?
Answer:
When a wave moves, the particles of the medium execute only pendulum motion, and not moves with sound.

Question 30.
What are longitudinal waves ?
Answer:
When the individual particles of the medium move in a direction parallel to the direction of propagation of disturbance are called
longitudinal waves.

Question 31.
How many types are there of waves ?
Answer:
There are three types of waves:
(i) Longitudinal waves
(ii) Transverse waves
(iii) Periodic waves.

Question 32.
Which are longitudinal waves among these ?
(i) waves on pond water
(ii) sound waves
(iii) light waves.
Answer:
(ii) Sound waves.

Question 33.
What are transverse waves ?
Answer:
When the individual particles of the medium moves in a direction perpendicular to the direction of wave propagation, are called
transverse waves.

Question 34.
The propagation of sound is the example of which wave ?
Answer:
Longitudinal wave.

Question 35.
What types of waves are produced on pond water ?
Answer:
Transverse waves.

Question 36.
How do the longitudinal waves travel ?
Answer:
The longitudinal waves travel in compressions and rarefactions.

Question 37.
What is a trough ?
Answer:
The maximum displacement in negative direction on Y-axis in the graph of periodic waves is called a trough.

Question 38.
Define the pitch.
Answer:
How the brain interprets the frequency of an emitted sound is called the pitch.

Question 39.
Define note.
Answer:
A sound which is produced due to a mixture of several frequencies is called a note.

Question 40.
What are the causes of variations in sounds ?
Answer:
The main cause of variations in sounds are loudness of sound, quality, pitch of sound and frequency.

Question 41.
What is the hearing range of our ears ?
Answer:
20 Hz to 20,000 Hz.

Question 42.
Define ultrasonic sound.
Answer:
Frequencies higher than 20 kHz are called ultrasonic sound or ultrasound.

Question 43.
What is the frequency of the waves produced during earthquake and valcano eruption ?
Answer:
Frequencies less than 20 Hz.

Question 44.
Write the name of two creatures which produce sounds of frequencies less than 20 Hz.
Answer:
(i) Whale
(ii) Elephant.

Question 45.
Write the name of two creatures which produce ultrasonic sound.
Answer:
(i) Bats
(ii) Dogs.

Question 46.
Which sound helps the bat to locate their way ?
Answer:
Ultrasound.

Question 47.
What is reverberation ?
Answer:
The repeated reflection that results in the persistence of sound is called reverberation.

Question 48.
What is meant by supersonic speed ?
Answer:
When the speed of any object exceeds the speed of sound it is said to be travelling at supersonic speed.

Question 49.
Which types of aircraft produce shock waves ?
Answer:
Supersonic aircraft.

Question 50.
What is the sound called produced by motion of supersonic aircrafts ?
Answer:
Sonic boom.

Question 51.
What is meant by reflection of sound ?
Answer:
When sound strikes on a surface, it changes its direction, it is known as reflection of sound.

Question 52.
When sound produces in deep wells, huge buildings and mountains, we experience a sound after some time besides mean sound, what is that ?
Answer:
Echo.

Question 53.
On which principle the megaphone works ?
Answer:
Megaphone works on the principle of reflection of sound.

Question 54.
What should be the minimum time to have echo between mean sound and echo ?
Answer:
0.1 second or \(\frac {1}{10}\) second.

Question 55.
How much minimum distance should be their to listen echo from a reflecting object ?
Answer:
17.2 metres.

Question 56.
What will be the minimum distance covered by the sound to hear echo ?
Answer:
34.4 metres.

Question 57.
Define sonar.
Answer:
Sonar means Sound Navigation And Ranging. It is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.

Question 58.
Where is SONAR used ?
Answer:
It is used to determine the depth of the sea and to locate underwater hill, valleys, submarine, icebergs and sunken ships, etc.

Question 59.
Which type of waves are used in SONAR system ?
Answer:
Ultrasonic waves.

Question 60.
Upto what frequency, the bat can produce ultrasonic waves ?
Answer:
Upto 120 kHz.

Question 61.
By emitting which waves to the painful part of body, we feel very like rubing (malish) ?
Answer:
Ultrasound wave.

Question 62.
Why is it beneficial to use ultrasound instead of X-rays to take the photograph of inner body ?
Answer:
Because ultrasound does not affect tissues and cells.

Question 63.
What is ultrasonography ?
Answer:
The technique which is used to examine the foetus during pregnancy to detect congenial defects and growth abnormalities is called ultrasonography.

Question 64.
Which type of waves are used to locate the eye tumor and to cure that ?
Answer:
Ultrasonic waves.

Question 65.
Which characterstics of ultrasound is used to develop photographic film ?
Answer:
Mixing of immiscible liquids by ultrasound.

Question 66.
Which type of waves are used in breadth therapy ?
Answer:
Ultrasonic waves.

Short Answer Type Questions

Question 1.
Prove that sound is a form of energy.
Answer:
When we speak for long time or loudly, we get tired. We spend our energy during talking. This energy produces vibration in throat. It proves that sound is a form of energy.

Question 2.
How is sound produced ? Justify with example.
Answer:
It is compulsory to vibrate for the production of sound. For example; when we speak, the vocal cord of throat vibrate. When trumphets or guitar play, its wires vibrate. When bell is rung, there is also vibrations. The vibrations in bell can be felt by touching it by hand. In this way, the sound is produced by vibrations.

Question 3.
Define compression and rarefaction. Draw a graph to show the density of air molecules due to sound waves.
Answer:
Compression: A region of high pressure is called compression.
Rarefaction: A region of low pressure is called rarefaction.
The graph between the density of air in unit interval and distance can be drawn.

In fig. the distribution of air atoms produced by sound propagation of sound wave is shown. The graph between density and distance is also shown.

Question 4.
Justify with an example that how the compression and rarefaction formed in vibrating body move ?
Answer:
Experiment:
Take a spiral spring as shown in fig. Press one end of spring slowly by hand and then leave. We see that the compressions and rarefactions are formed in the spring and they propagate from one end to another end. The propagating motion of compressions and rarefactions is called wave motion. In the same way, the compressions and rarefactions produced in air propagate from one layer to another and in this way, the sound travels from one place to another.

Question 5.
Justify the sound system in air by tuning fork.
Answer:
The prongs of tuning fork start vibrating when hit with a rubber pad. The vibrating prong exert pressure of air particles and displaces them. The air particles come in motion from rest and come in original postition after displacing the nearby particles. The compression of high pressure is formed at the region where the air particles come close, and after detaching pressure of vibration, the particles come in their original position. This region of low pressure is called rarefaction. The particles of medium do not move forward itself but disturbance of particle moves forward by compression and rarefaction. This is called sound wave. Air is the most common medium for the propagation of sound.

Question 6.
Can sound make a light spot dance ? Justify it.
Answer:
Take an empty tin with both ends open. Tie stretched membrane of balloon on one end. Take one piece of plane mirror and stick it to the membrane of balloon in such a way that its reflecting surface points upward and the rays coming from a source should fall on the mirror and after reflection the light spot should be on the wall. According to figure, make a loud sound from the open end of tin. You will see the light will dance.

Cause:
Due to vibration of sound, the membrane and the mirror on it vibrate, due to this the image formed on wall seems to be shaking. Stop the sound. The spot will be on fixed place because the vibration of membrane is stopped.

Question 7.
Define longitudinal and transverse waves.
Answer:
(i) Longitudinal wave : When the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance is called longitudinal wave.
(ii) Transverse wave : When the individual particles of the medium move in a direction perpendicular to the direction of wave propagation, is called transverse wave.

Question 8.
Define crest, trough and amplitude and show them for a transverse wave by graph.
Answer:
The crest, trough and amplitude of a transverse wave are shown in fig.
Crest: Peaks formed during the motion of transverse wave is called crest.
Trough: The depression formed during the motion of transverse wave is called trough.
Amplitude: The maximum displacement of medium particles from their mean position during transmission of waves is called amplitude. The unit of amplitude is metre.

Question 9.
Distinguish between longitudinal and transverse waves.
Answer:
The differences between longitudinal and transverse waves are as follows

Question 10.
What do you know about these :
(i) Loudness of sound
(ii) Pitch
(iii) Frequency.
Or
Write two differences between the loudness and pitch of sound.
Answer:
(i) Loudness of sound:
The level of the sensation of the sound is called the loudness of sound. This level of sensation depends upon the sharp or light vibrations of source. When vibrations are more, the loudness of sound will be more and when vibrations are light, the loudness of sound will be low.

(ii) Pitch:
That property’of sound by which we can differentiate between the louder and softer sound. This property depends upon the frequency of the source of sound. When the frequency of the source of sound is more, the pitch is also more.

(iii) Frequency:
The number of vibrations completed by any object in one second is called frequency.

Question 11.
What is the difference between the sound of mosquito and lion ?
Answer:
The loudness of the sound of mosquito is less but frequency is more. On the other side, the loudness of the sound of lion is more but frequency is less. So the sound of mosquito is more sharp than that of lion.

Question 12.
Show through graph that how the variations occur in density and pressure when sound wave travels in a medium.
Answer:

On a certain time, both the density’ and pressure of the medium vary up and down from their average value. In fig. a region of high pressure is compression and low pressure is rarefaction. The peak is called crest and valley is called trough of wave.

Question 13.
What is the difference between the sound of violin and flute of an orchestra ?
Answer:
Both the violin and flute travel in same medium and reach to our ear in same time, the speed of both is same, but both sounds are obtained from different characterstics. That is pitch. How the brain interprets the frequency of an emitted sound is called the pitch. Therefore, the sound of violin and flute are heard differently.

Question 14.
Differentiate between low and high pitch through graph.
Answer:

Question 15.
What is the main difference between soft and loud sound ? Clarify it by graph.
Answer:
The softness and loudness of sound depends upon the amplitude. For instance, knocking the door slowly produces a sound of’low amplitude and on loudly it produces a sound of high amplitude. Less energy is emerged in low amplitude and more energy is emerged in high amplitude. That is why the sound of low amplitude covers less distance while of high amplitude covers more distance

Question 16.
Show by an experiment that speed of sound is more in liquids than in air.
Answer:
Take a balloon full of water and rub its surface slowly. You will hear a sound. Now keep the balloon near your ear and rub it from other side, we will hear more and clear sound. It proves that the speed of sound is more in liquids than in air.

Question 17.
How can we hear the sound of train coming from very far when ear is kept on track ?
Answer:
We know that the speed of sound in iron on normal temperature is 5950 m/s. Therefore, we can hear the sound of train when it is so far by keeping the ear on line which is made of iron girders.

Question 18.
Describe an experiment to show the reflection of sound.
Answer:
Take two identical wide tubes of plastic or card. Also take one sheet of plywood or metal. Arrange these according to fig. Keep a clock near the end of one of the tubes. Keep your ear near the other end of tube. Put a thick membrane between two tubes so that sound of watch cannot reach to your ears. Arrange the angle of second tube in such a way that you should be able to hear the sound of clock. You will see that in this situation both the tubes make equal angle to normal. This proves that sound is reflected.

Question 19.
What is meant by reflection of sound ? How is reflection of sound used to measure the depth of sea ?
Answer:
When sound changes its direction after striking a surface, it is called reflection of sound. For example, when the sound is produced near the mouth of well, the wave of mean sound produced enter in the well and after some time they came back after reflecting from the surface of water and heard again like mean sound as echo.

To measure the depth of sea by reflection of sound:
The depth of sea is measured by sonar device. From this instrument, the sound signal is sent to the bottom of sea from the ship. This signal comes back to the instrument after the reflection from the sea bottom. In this way, the time taken to reach the sea bottom and reflection is noted. The depth of sea is measured by the speed of sound in water. Now the depth of sea is calculated by the time and speed of sound in water.

Question 20.
What is the difference between ultrasound and echo ?
Answer:
Ultrasound:
The frequencies more than 20 kHz are called ultrasonic sound. We cannot hear this sound. Some animals like dogs have the capacity to hear ultrasound. The bats have the capacity to produce and listen ultrasound.

Echo:
When sound is produced in deep well, big buildings, mountains, etc., the same sound is heard as mean sound after some time. This is known as echo. We can hear echo only when there is a minimum difference of 1/10 s between the mean sound and reflected sound. The minimum distance covered by the sound should be 34.4 metres to hear the echo. Thus, the object should be at a distance of 17.2 metres or more from the source of sound because the speed of sound at 22°C is 344 m/s. Time taken by sound to reach the obstacle and returning back should be 1/10 seconds. The sensation of sound in our brain lasts for 1/10 seconds.

Question 21.
What is the audible range of different animals ?
Answer:
The audible range of different animals is different. For example,
(1) Man – 20 Hz to 20,000 Hz.
(2) Dog and babies less than 5 yrs – upto 25 kHz.
(3) Rhino – 5 Hz.
(4) Whale and elephant – Infra sound.
(5) Dolphins, bats and porpoises – sounds of frequency more than 20 kHz.
(6) Moths – The moths can hear the high frequency squeaks of the bat.

Question 22.
What is the reason that we cannot hear the waves produced by earthquake whiie bats or dogs can?
Answer:
Our ear can hear the waves of frequencies from 20 Hz to 20 kHz. This range of frequency is known as audible range. The frequency of the waves produced in earthquake does not lie in this limit. Therefore, we cannot hear these waves, but the bats and dogs are sensitive to these waves and can hear these waves.

Question 23.
Define infrasound waves, audible range and ultrasound.
Answer:
1. Infrasound: Those waves whose frequency is less than 20 Hz are known as infrasound waves. We cannot hear them. i.e. earthquake wave.
2. Audible range : The waves of frequency between 20 Hz and 20 kHz is called audible range.
3. Ultrasound waves : Those waves whose frequency is more than 20 kHz is known as ultrasound waves. We cannot hear them also. Dogs and bats can hear these waves.

Question 24.
Write the uses of ultrasound.
Answer:
The uses of ultrasound are following :
(1) Radar works on this principle.
(2) The milk exposed to them does not get spoilt for long.
(3) Sprouted is more in grain when affected by it.
(4) The cracks can be found in metal sheets melted by it.
(5) The bat find its prey by using this sound.
(6) The depth of sea can be measured by using it.
(7) It is used to wash clothes.
(8) It is used to find the cancer in body parts.
(9) It is used to cut hard substances and to join aluminium.

Question 25.
A person has been kept in a closed room whose doors and windows are made of glass. When he makes a sound in the room, nobody hear that; but when he knocks the window, the outsider persons hear the knocking. Why ?
Answer:
When a person in a room makes a sound, the waves produced by him is reflected from the glass. This produces weak vibrations in the glass, which is incapable to vibrate the air outside the room. As a result of it sound is not heard outside. Opposite to it, when the glass of window is knocked, the waves produced, reach the outer part of the glass and vibrate the air in its contact. Therefore, the sound of knocking is heard by persons.

Question 26.
What is sonic boom ? How is it harmful ?
Answer:
Jet aircrafts produce shock waves in air. These shocking waves travel with supersonic speed. These waves carry a large amount of energy. The air pressure variation associated with this type of shock waves produces a very sharp and loud sound called the “sonic boom”. The shock waves produced by a supersonic aircraft have enough energy to shatter glass and even damage buildings. Sometime sonic boom produces such a loud noise which is beyond our intolerable limits. This can create trouble in our ears. Remember it, when supersonic aircraft flies, it continuously produces sonic boom around it.

Question 27.
Write a short note on whispering galleries.
Answer:
The whispering gallery of dome of St. Paul’s Cathedral in London is very famous and serves an interesting example of reflection of sound. The gallery is running round the base in the inside of the dome in a circular form. When we whisper near any point of walls, the sound produced is reflected at every place. This can be clearly heard near the walls. But at some distance towards centre, let at point A, this sound can never be heard. One such gallery is in Gol Gumbaz at Bijapur in India.

Question 28.
What is SONAR ? What are its uses ?
Answer:
SONAR: The instrument which produces sound waves and measures the small interval of time of reflected sound waves is known as SONAR.
Uses:
(1) To measure the depth of sea.
(2) To find the distance of groups of fishes.
(3) To locate the submarines and enemy torpedo.
(4) To detect cracks and flaws in metal blocks or sheets.

Question 29.
How can the bat fly freely in dark by using the principle of sonar ?
Answer:
Bats use the principle of SONAR naturally. Bat can produce ultrasound of frequency 120 kHz or 1.2 x 10⁵ Hz and can hear them. The wavelength of such high frequencies is very less, approximately 0.29 cm. Bat is blind in reality, but in intense dark, it can fly without any difficult while crossing all the obstacles. It uses ultrasound to find its prey. When it flies, emits ultrasound signals of high frequency for small interval of time. These signals strike the surrounding objects and reflect. Thus, the bat calculates the time taken by ultrasonic waves after reflecting from any object. In this way the bat can fly freely in dark.

Question 30.
On what factors the speed of sound depends ?
Answer:
The speed of sound depends on following factors:
1. Density of medium: The speed of sound is proportional to the density of medium. This is, the speed increases as the density increases and decreases with the decrease in density.
2. Temperature: Speed of sound increases with increase in temperature.
3. Wavelength: The speed of sound is proportional to wavelength.
4. State of medium: The speed of sound is maximum in solids, less in liquids and minimum in gases.

Question 31.
What is the audible range ? Write the name and uses of the frequencies more than this range.
Answer:
Audible range:
A person can hear waves of frequency from 20 Hz to 20 kHz. This limit is called audible range.

The sound of frequency more than 20 kHz is called ultrasound. Its applications are as follows :
(1) Radar is made on this principle.
(2) It is used to find the cancer in body parts.

Question 32.
How can we measure the depth of sea ?
Answer:
An instrument is fitted on the ship to measure the depth of sea which sends ultrasound in sea level. These waves return back after reflecting from sea level and the receiver on ship receives these reflected rays. The time taken (t) by these ultrasonic waves to reach the sea level and return back and the speed of sound waves (v) help to calculate the distance (d) between the ship and sea level.
We know that,
Velocity =
or Distance = velocity x time.
= 2 d = v x t
d = \(\frac{\mathrm{v} \times \mathrm{t}}{2}\)
This will be the depth of sea at a certain place.

Essay Type Questions

Question 1.
What is sound ? Prove it by experiment that sound is produced by vibration.
Answer:
Sound:
Sound is a form of energy which produces a sensation of hearing in our ears. It is produced by vibrating bodies. It can be proved by following experiments.

Experiment 1:
Take a tong. Hold it from the curved end and strike it against a surface or a stone. We will see the arms of the tong vibrating. Due to these vibrations a sound is heard. Repeat this process and bring the tong near to the ear. The sound of the tong is heard clearly.

Experiment 2:
Fix two nails on a wooden plank and tie a thin wire tightly between the two nails. Put a triangular piece of wood below the wire. Now pluck the wire and release it. A sound is heard. If we put a small piece ofpaper on the wire and again pluck it and release, the paper wil I fall dowm proving thatthe sound is produced due to vibrations.

Experiment 3.
Take a tuning fork. Strike it against a rubber pad, and make it stand straight on a table. A sound is produced which we can hear. Now bring a ball near to the prongs of the vibrating tuning fork. The prong strikes again and again with the ball. This proves the vibration of the tuning fork. If we bring our finger near to it, we can feel the vibrations. When we touch it the sound coming from the tuning fork will stop. This again proves that the sound is produced due to vibrations.

Question 2.
Describe two types of wave motions.
Answer:
There are two types of wave motions:
(a) Transverse waves:
A wave in which individual particles of medium vibrate about their mean position in a direction perpendicular to the direction of propagation of the wave through the medium, is called transverse wave.

Examples:
(1) Waves produced in a slinky
(2) Waves produced on the surface of the water.

(1) The waves produced in a slinky: Take a slinky, Tie its one end with a book to the wall. Hold its other end and give a jerk. The disturbance produced in the slinky moves from hand to the slinky and the particles of the slinky vibrate up and down from their mean position. This is an example of transverse wave.

(2) The wave produced on the surface of the water:
Throw a stone from a certain height in the still water of a pond then ripples are produced on the surface of the water which moves in all the directions. If we put a leaf on the ripples, the ripples will make the leaf move up and down at the same point. The raised upper part of the wave is called the ‘crest’ and the lower part is called the ‘trough’.

(b) Longitudinal waves:
A wave in which the individual particles of medium vibrate back and forth from their mean position in direction parallel to the direction of propagation of the wave through the medium is called longitudinal wave.

Examples:
(1) Waves produced in a slinky
(2) Waves propagating in air.

(1) Waves produced in a slinky : Longitudinal waves can also be produced with slinky. Tie the one end of slinky with a hook on wall and stretch other end with your hand. Now press some turns in the small part of slinky between your fingers. The compression will reach the other end of slinky.

(2) The waves propagating in air : Take a tuning fork, suspend it forcefully with rubber pad. Its prongs will vibrate left-right. When one of its vibrating prongs moves in right, it press the air molecules due to this the air molecules compressed and the distance between them decreases. This is known as compression when the prong of vibrating fork moves in left, the vacuum is created in right side. To fill this, the air molecules expand to right side, and the distance among these air molecules increases which is known as rarefaction.

Question 3.
What are the characterstics of periodic waves ? Establish relationship among wave velocity, frequency and wavelength.
Answer:
The periodic wave has following characterstics:
1. Amplitude: The magnitude of maximum displacement of vibrating particles from mean position is called amplitude of pendulum.
2. Time period: Time taken by a particle to complete one oscillation is called time period.
3. Frequency: The number of vibrations in a second by a vibrating particle is called frequency. It is denoted by Greek Letter (v). Its unit is Hertz (Hz).
4. Wavelength: The time in which the particles of medium complete one vibration, the distance covered by the wave in that period is called wavelength. It is denoted by X .
5. Velocity: The distance moved by the wave in unit time is called velocity of the wave. It is denoted by v.

The relation among frequency, wavelength and velocity of wave:
According to definition:

The distance covered by the wave T second = λ
The distance covered by the wave in 1 second = \(\frac {λ}{T}\)
But, the distance covered in 1 second = Velocity (v)
∴ v = \(\frac {λ}{T}\) ………….(i)
If time taken in one vibration is T,
Then number of vibrations in T seconds = 1
Numbers of vibrations in 1 second = \(\frac {1}{T}\)
But, number of vibrations in 1 second is called frequency ‘υ’.
υ = \(\frac {1}{T}\)
Putting the value of y in equation (i)
v = \(\frac {λ}{T}\) = λ \(\frac {1}{T}\) [∴ \(\frac {1}{T}\) = v]
y = λv
Thus, Velocity = wavelength x frequency.

Question 4.
What is meant by reflection of sound ? Write three uses of it.
Answer:
Reflection of Sound: When the sound strikes any surface it changes its direction which is called reflection of sound.
Uses of reflection of sound: The practical uses of reflection of sound are following :

1. Megaphones:
You have seen the use of megaphones to address the crowd in fairs or tourists places. Megaphone is simply a horn shaped instrument. When a person speaks in the megaphone, the sound waves are prevented from spreading out due to repeated and successive reflections from the walls of megaphone tube. For this reason, loudspeakers have horn shapes openings.

2. Ear trumpet or Hearing aid:
People with hard of hearing may need a hearing aid. Hearing aid is an electronic battery operated device. The hearing aid receives sound through a microphone. The microphone converts the sound waves to electrical signals. These electrical signals are amplified by an amplifier. The amplifier electric signals are given to a speaker of the hearing aid. The speaker converts the amplified electrical signal to sound and sends to the ear for clear hearing.

3. Sound drum or boards : The reflection of sound is also possible from curved surfaces. This fact is used in bigger halls to spread the sound in whole hall. Sound drums are used for this purposes. The speaker S is focussed at S sound drum. Sometimes a curved sunboard may be placed behind the stage so that the sound after reflecting from the sound board, spreads evenly across the width of the hall as shown in fig.

Question 5.
Describe the different parts of human ear.
Answer:
There are three main parts of human ear:
(a) External ear : It is that part which is seen from outside. It consists of projecting part known as pinna and the passage auditory canal which is connected to a thin membrane called the eardrum or thin panic membrane.
1. Pinna: It consists of cartilage and muscles. Its lower part is soft and elastic. Pinna collects the sound waves and gives direction to them. It is joined with auditory canal.

2. Auditory Canal: It is a slant about 3-4 cm long. Its skin contains hair and micro glands and liquid comes out of them. This is called ear wax. It prevents the ears from dust particles and insects.

3. Eardrum: There is a thin membrane in the end of auditory canal which is called eardrum. It sends the sound waves to middle ear.

(b) Middle ear: The middle ear is like a air filled room. Middle ear is attached to throat through a canal which is called ‘eustachian canal’. This canal maintains the air pressure in two sides of eardrum. The middle ear contains three tiny bones –
(i) hammer
(ii) anvil
(iii) stirrup.

These three bone are together called ear ossicles.
(c) The Inner Ear: It contains
(i) cochlea
(ii) auditory nerves
(iii) semi-circular canals.

(i) Cochlea: The shape of cochlea is spiral and has two half turns. It contains a liquid which is called ‘andolimph’. It helps in hearing.
(ii) Auditory nerve: It is a small chamber which is connected with cochlea from front and from back with semi circular canals.
(iii) Semi-circular canals: These are three in number and arranged at right angles to-each other in three different planes. The ends of auditory canals swim in ‘andolimph.’

Important Table

Numerical Questions

Question 1.
The frequency and wavelength of a sound wave is 2 kHz and 25 cm respectively. How much time will it take to move a distance of 1.5 km ?
Solution:
Here, Frequency v = 2 kHz = 2000 Hz
Wavelength λ = 25cm = 0.25 m
We know that, velocity of wave v= wavelength x frequency v = λv
v = 0.25m x 2000 Hz = 500m/s
Time taken by wave to move a distance of 1.5 km,
t = \(\frac{\mathrm{d}}{\mathrm{v}}=\frac{1.5 \times 1000 \mathrm{~m}}{500 \mathrm{~ms}^{-1}}=\frac{15}{5}\)s = 3s
Thus, sound will take 3 s to travel on distance of 1.5 km.

Question 2.
A person clapped his hands near a cliff and heard the echo after 6 s. What is the distance of the cliff from the person if the speed of the sound, v is taken as 346 ms^-1 ?
Solution:
Given, Speed of sound, (v) = 346 ms^-1
Time taken for hearing the echo, (t) = 6s
Distance travelled by the sound = v x t = 346 ms^-1 x 6s = 2076 m
In 5 s sound has to travel twice the distance between the cliff and the person. Hence, the distance between the cliff and the person = 2076 m/2= 1038 m.

Question 3.
A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is 1531 m/s, what is the distance of the seabed from the ship ?
Solution:
Given, time between transmission and detection, (t) = 3.42 s.
Speed of ultrasound in sea water, (v) =1531 m/s
Distance travelled by the ultrasound = 2 x depth of the sea = 2d
Where d is the depth of the sea.
2 d = speed of sound x time = 1531 m/s x 3.42 s = 5236 m
d = 5236 m/s = 2618 m
Thus, the distance of the seabed from the ship is 2618 m or 2.62 km.

Question 4.
Two children are standing at two opposite ends of an iron pipe. One boy hit one end of pipe with stone. Find the ratio of the time taken by the sound to travel through iron and air (Given, speed of sound in air is 344 m/s and in iron 5130 m/s).
Solution:
Here,
Let length of the pipe = L m
Velocity of sound in air (v₁) = 344 m/s
Velocity of sound in iron (v₂) = 5130 m/s
∴ Time taken by sound in the air (T₁) = \(\frac{\mathrm{L}}{\mathrm{v}_1}=\frac{\mathrm{L}}{344}\)s
Time taken by sound in iron (T₁) = \(\frac{\mathrm{L}}{\mathrm{v}_2}=\frac{\mathrm{L}}{5130^{\mathrm{s}}}\)
The ratio between two time = \(\frac{T_1}{T_2}=\frac{\frac{\mathrm{L}}{344}}{\frac{\mathrm{L}}{5130}}\)
= \(\frac{\mathrm{L}}{344} \times \frac{5130}{\mathrm{~L}}\)
= \(\frac{14.9}{1}\)
T₁ : T₂ = 14.9 : 1

Question 5.
A boat at anchor is rocked by waves whose crests are 100 m apart and whose velocity is 20 m/s. How often do the crest reach the boat?
Solution:
Here,
The distance between two consecutive crests of wave = 100 m
Velocity of wave (v) = 20 m/s
Frequency of wave (v) = \(\frac{v}{\lambda}=\frac{20}{100}\) Hz = 0.2 Hz

Question 6.
A source of wave produces 40 crests and 40 troughs in 0.4 s. Find the frequency of the wave.
Solution:
Here,
Number of wave crests = Number of wave troughs = 40
time = 0.4 s
Number of crests per second Number of troughs per second =\(\frac{40}{0.4}\) = 100
Thus, the frequency of wave = 100 Hz

Question 7.
Using the SONAR, sound pulses are emitted at the surface. The pulses after bdng reflected from the bottom are detected. If the time interval from the emission to the detection of the sound pulse is 2s, find the depth of the wster. IVelocity of sound ¡n water is 1498 rn/sI
Solution:
The velocity of sound in water (y) = 1498 m/s
Time taken in coming and going pulse = 2 s
Thus, time taken in going pulse = \(\frac{2}{2}\) = 1 s
Depth of water = velocity x time = 1498 x 1 = 1498 metres

Question 8.
A child hears an echo from a cliff 4 seconds after the sound from a powerful cracker is produced. How far away is the cliff from the child?
Solution:
Total time taken by sound in going and returning back to the child = 4 s
The time taken by sound in going to the child = \(\frac{4}{2}\) = 2s
The velocity of sound in air 344 = m/s
Thus, the distance between the child and the cliff = velocity x time = 344 x 2 = 688 m

Question 9.
A stone is dropped into a well 44.1 metres deep. The sound of the splash is heard 3.13 s after the stone is dropped. Find the velocity of sound in air.
Solution:
Here,
Depth of the well (s) = 44.1 m
Suppose time taken by the stone to reach the well = t₁ s
And the time taken by the sound to reach on water = t₂ s
According to the question t₁ + t₂ = 3.13
second u = 0, g =9.8 m/s²

We know that s = u t + \(\frac{1}{2}\) gt²
44.1 = 0 x t₁ + gt₁² (For t₁)
or 441 = \(\frac{1}{2}\) x 9.8 x t₁²
or t₁² = \(\frac{44.1 \times 2}{9.8}\) = 9
or t₁ = 3s
t₂ = 3.13 – t₁ = 3.13 – 3 = 0.13 s
Velocity of sound = = \(\frac{44.1}{0.13}\) = \(\frac{441 \times 100}{10 \times 13}\) = 339.2 m/s

Question 10.
A tuning fork produces waves of frequency 512 Hz by vibration, which moves with a velocity of 330 m/s. Find its wavelength.
Solution:
Here,
Frequency (υ) = 512 Hz
Wave velocity (v) = 330 mis
Wavelength (λ) = ?
We know that,
v = υ λ
Or λ = \(\frac{v}{v}=\frac{330}{512}\) m = 0.645 m

Question 11.
What is the speed of that water wave whose wavelength ¡s 20cm and frequency is 20 Hz.
Solution:
Here,
Wavelength (λ) = 20 cm = \(\frac{20}{100}\) m = 0.2m
Frequency (υ) = 20 Hz
Velocity of wave (v) = ?
We know that, v = υ λ = 20 x 0.2 = 4 m/s

Question 12.
Find the frequency of that wave whose time period is 0.02 s.
Solution:
Here,
Time period (T) = 0.02 s
Frequency (υ) = ?
We know that
υ = \(\frac{\mathrm{I}}{\mathrm{T}}=\frac{1}{0.02}\) = \(\frac{100}{2}\) = 50 Hz

Question 13.
If the frequency of a tuning fork is 400 Hz and the speed of sound in air is 320 m/s, find the distance covered by the sound when tuning fork completes 20 vibrations.
Solution:
Here,
Wave velocity (v) = 320 mIs
Frequency (υ) = 400 Hz
We know that v = υ λ
or λ = \(\frac{v}{v}=\frac{320}{400}\) = 0.8 m
Thus, the distance covered in I vibration = 0.8 m
The distance covered in 20 vibrations = 20 x 0.8 = 16 m

Question 14.
An object completes 6000 vibrations in one minute. If the speed of sound in air is 330 m/s, find the (a) frequency of vibrations (b) wavelength of produced waves.
Solution:
(a) 1 minute = Number of vibrations in 60 s = 6000
Number of vibrations in I second = \(\frac{6000}{60}\) = 100
Thus, frequency (υ) = 100 Hz

(b) Here, velocity of wave (v) = 330m1s
Frequency (υ) loo Hz
Wavelength (λ) = ?
We know that v = υλ
or λ = \(\frac{\mathrm{v}}{v}=\frac{330}{100}\) = 3.3 m

Question 15.
The velocity of light wave is 3 x 10⁵ km/s and wavelength is 6 x 10⁷ m. Find its frequency.
Solution:
Here,
Velocity of wave (v) = 3 x 10⁵ km/s
3 x 10⁵ x 1000.m/s
3 x 10⁸ m/s
Wavelength (λ) = 6 x 10^-7 m
Frequency (υ) = ?
We know that v = υλ
υ = \(\frac{v}{\lambda}=\frac{3 \times 10^8}{6 \times 10^{-7}}\) = 0.5 x 10¹⁵ Hz
Thus, frequency is 0.5 x 10¹⁵ Hz.

Question 16.
An observer is standing at a distance of 200 metres from a cliff and whistles. If the speed of sound is 332 mIs, when the echo of whistle will be heard?
Solution:
Distance of observer from cliff = 200 m
To hear the echo,
Total distance covered = 200 + 200 400 m
Speed of sound = 332 m/s
Time =
\(\frac{400}{332}\) = 12 s

Question 17.
A Sound signal is send from a boat. That signal return after 0.8s by reflecting from the sea bottom. Find the velocity of sound in water if the depth of sea is 600m.
Solution:
Here,
Depth of sea = 600 m
Distance travelled by the signal = 600 + 600 = 1200 m
Time = 0.8s
Velocity of Sound in water =
= \(\frac{1200}{0.8}\) m/s
1500 m/s

Question 18.
A sound signal is send from a boat. That signal return after 0.8 s by reflecting from the sea bottom. If the velocity of sound is 1500 mIs, then find the depth of sea.
Solution:
Here,
Velocity of sound in water = 1500 mis
Time taken = 0.8 s
Distance travelled by sound = Velocity of sound x time
1500 x 0.8 = 1200m
Thus Depth of sea latex]\frac{1200}{2}[/latex] = 600 m

Practical Work

Experiment 1.
To study the vibrations produced by tuning fork.

Procedure:
(i) Suspend a small plastic ball by a thread from a support.
(ii) Bring the non-vibrating tuning fork near the ball and observe the situation.
(iii) Now bring the vibrating tuning fork near the ball and observe the situation.
In first situation, the ball does not move far from tuning fork while in second situation, the ball moves far as the tuning fork is taken near to it, because the prong of tuning fork applies the force on ball and pushes it backward.

Experiment 2.
To study the vibrating parts of musical instruments.

Procedure:
Go into the music room of your school and know about the system of producing sound in musical instruments from music teacher.

For example:
(i) In drum: by the vibration of membrane.
(ii) Sitar: by vibration in wire.
(iii) Ektara: by vibration in wire.
(iv) In flute: by air column.
(v) In water wave: by water.

Experiment 3.
To produce longitudinal wave by slinky.

Procedure:
(i) Take a slinky. Ask your friend to hold one end and you hold the other end.
(ii) Give it a sharp push towards your friend. A special motion will be seen in slinky while repeated.
Mark a dot on the slinky and repeat the process again. The dot on the slinky will move back and forth parallel to the direction of the propagation of the disturbance. This is longitudinal wave.

Experiment 4.
To understand the reflection of sound.

Procedure:
Take a hallow cylinder of cardboard opened from both ends. Keep a small clock near the open end of cylinder and keep the wooden block in such a way that the person standing on the other end of the cylinder can hear the sound of tik-tik of clock. Now keep the other wooden block far from the other open end of cylinder. At this time sound reaches the ears after striking from surface. The change in the direction of sound after striking from any surface is called reflection of sound.

Quick Review of the Chapter

1. The sound of a bell is
(A) mechanical energy
(B) sound energy
(C) heat energy
(D) light energy
Answer:
(B) sound energy

2. Sound is produced due to
(A) striking
(B) vibration
(C) beating
(D) rubbing
Answer:
(B) vibration

3. Sound does not travel/transmit in:
(A) solids
(B) liquids
(C) gases
(D) vacuum
Answer:
(D) vacuum

4. Vave is a …………
(A) disturbance
(B) compression
(C) rarefaction
(D) vibration
Answer:
(A) disturbance

5. A region of high pressure is called :
(A) wave
(B) compression
(C) rarefaction
(D) disturbance
Answer:
(B) compression

6. In which of the following the speed of sound maximum?
(A) Dry air
(B) steel
(C) wet air
(D) water
Answer:
(B) steel

7. For sound propagation, which is the most common medium?
(A) water
(B) earth
(C) air
(D) space
Answer:
(C) air

8. Which are waves produced in water?
(A) longitudinal
(B) transverse
(C) partially both (A) and (B)
(D) sometime (A) and sometime (B)
Answer:
(B) transverse

9. The SI unit of wave velocity is:
(A) m
(B) mis
(C) km
(D) kmls
Answer:
(B) m/s

10. Which of the following is represented by λ (lambda)?
(A) Wavelength
(B) Wave velocity
(C) Frequency
(D) Amplitude
Answer:
(A) Wavelength

11. The transfer of takes place in sound propogation.
(A) mass
(B) both (A) and (C)
(C) energy
(D) none of these
Answer:
(C) energy

12. The longitudinal waves are produced in :
(A) solids
(B) liquids
(C) gases
(D) all three
Answer:
(D) all three

13. The transverse waves are produced ¡n :
(A) solid and liquids
(B) liquids and gases
(C) gases and solids
(D) all three
Answer:
(A) solid and liquids

14. Light is wave.
(A) transverse
(B) mechanical
(C) longitudinal
(D) electromagnetic
Answer:
(A) transverse

15. The frequency of a source of sound is loo Hz. How many times does it vibrate in Half an hour?
(A) 1,80,000
(B) 90,000
(C) 60,000 (D) I ,00,000
Answer:
(A) 1,80,000

16. The distance between is called wavelength.
(A) two consecutive compressions
(B) two consecutive rarefactions
(C) a compression and a rarefaction
(D) both (A) and (B)
Answer:
(D) both (A) and (B)

17. The SI unit of wavelength is.
(A) metre
(B) joule
(C) erg
(D) newton
Answer:
(A) metre

18. What is the unit of frequency of a sound wave?
(A) Hertz
(B) Metre
(C) erg
(D) Joule
Answer:
(A) Hertz

19. The SI unit of time period is:
(A) Hertz
(B) Metre
(C) Second
(D) Nm
Answer:
(C) Second

20. The magnitude of the maximum distrubance in the medium on either side of the mean is called
(A) frequency
(B) amplitude
(C) velocity
(D) none of the above
Answer:
(B) amplitude

21. What is represented by AR?
(A) frequency
(B) velocity
(C) amplitude
(D) time period
Answer:
(C) amplitude

22. The sound of single frequency is called
(A) note
(B) tone
(C) noise
(D) none of these
Answer:
(B) tone

23. A sound which is produced due to mixture of several frequencies is called
(A) note
(B) tone
(C) noise
(D) lyrics
Answer:
(A) note

24. The amount of sound passing each second through unit area is called . of sound.
(A) loudness
(B) intensity
(C) pitch
(D) frequency
Answer:
(B) intensity

25. In which medium the speed of sound is maximum at 25°C?
(A) glass
(B) steel
(C) aluminium
(D) nickel
Answer:
(C) aluminium

26. In which liquid the speed of sound is maximum at 25°C?
(A) sea water
(B) distilled water
(C) ethanol
(D) methanol
Answer:
(A) sea water

27. To listen the echo clearly the minimum distance of reflector should be:
(A) 172m
(B) 17.2m
(C) 344 m
(D) 344 m
Answer:
(B) 17.2 m

28. In which gas the speed of sound is maximum at 25°C?
(A) oxygen
(B) air
(C) hydrogen
(D) helium
Answer:
(C) hydrogen

29. The sensation of sound can remain about second in our brain.
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
Answer:
(A) 0.1

30. Sound obsorber substances are installed walls and roofs of building to reduce
(A) echo
(B) sound reflection
(C) reverberation
(D) supersonic sound
Answer:
(C) reverberation

31. The principle of reflection of sound is used:
(A) in megaphone and horns
(B) stethoscope
(C) in concert hail, cinema halls
(D) in all aboe mentioned
Answer:
(D) in all above mentioned

32. What is hearing range of sound in man?
(A) 10-15 Hz
(B) 20-20.000Hz
(C) 30.000 Hz-40,000 Hz
(D) 45,000 Hz-50,000 Hz
Answer:
(B) 20-20,000 Hz

33. How much Hz sound can the children and dogs of age below S hear?
(A) to25kHz
(B) to30kHz
(C) to 35 k Hz
(D) to 45 k Hz
Answer:
(A) to 25 k Hz

34. Sounds of frequencies below are called infrasonic sound.
(A) 20 Hz
(B) 200 Hz
(C) 2000 Hz
(D) 20000 Hz
Answer:
(A) 20 Hz

35. Frequencies higher than is called ultrasonic sound.
(A) 10kHz
(B) 20kHz
(C) 25 k Hz
(D) 30 k Hz
Answer:
(B) 20 k Hz

36. Ultrasonic sound is used :
(A) for cleaning dust, oilness and dirt
(B) to detect cracks in bodies of metal
(C) in ECG and ultrasound
(D) in all above mentioned
Answer:
(D) in all above mentioned

37. SONAR is used to measure the of bodies under sea water.
(A) distance
(B) direction
(C) speed
(D) all above mentioned
Answer:
(D) all above mentioned

38. Which part is lamella of an ear?
(A) outer part
(B) midd’e ear
(C) inner ear
(D) auditory nerve
Answer:
(A) outer part

39. How are the waetength and frequency of a sound wave related to its speed ?
(A) λ= vυ
(B) v = \(\frac {υ}{λ}\)
(C) υ = λ υ²
(D) v = υλ
Answer:
(D) v= vA

40. The number of complete oscillations per unit time is called
(A) Amplitude
(B) Wavelength
(C) Frequency
(D) Time period
Answer:
(C) Frequency

Haryana State Board HBSE 9th Class Science Important Questions Chapter 11 Work and Energy Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 11 Work and Energy

Very Short-Answer Type Questions

Question 1.
What is work ?
Answer:
When force is applied on an object and it moves in the direction of force, it is said to be work done.

Question 2.
On which the quantity of work depends ?
Answer:
The quantity of work depends upon the applied force & distance moved.

Question 3.
What is the formula of work ?
Answer:
Work = Force x Displacement.

Question 4.
What is the unit of work ?
Answer:
Newton-metre (Joule).

Question 5.
What is energy ?
Answer:
The capacity of doing work is called energy.

Question 6.
What are the types of mechanical energy ?
Answer:
Mechanical energy is of two types:
(1) Kinetic energy,
(2) Potential energy.

Question 7.
What is potential energy ?
Answer:
The energy possessed by a body due to its position or change in shape is called potential energy.

Question 8.
Which type of energies are in aircraft flying on height ?
Answer:
Both type of energies-kinetic and potential.

Question 9.
Which type of energy is there in a stone lying on a hill ?
Answer:
Potential energy.

Question 10.
What is mechanical Energy ?
Answer:
The sum of kinetic energy and potential energy of a body is called mechanical energy.

Question 11.
Which type of energy is there when a bullet is fired from a gun ?
Answer:
Kinetic energy.

Question 12.
Write the formula for kinetic energy.
Answer:
Kinetic energy = \(\frac {1}{2}\) mv², where m → mass, v → velocity.

Question 13.
Write the formula for potential energy.
Answer:
Potential energy = mgh, where m → mass, g → acceleration due to gravity and h → height.

Question 14.
What is the law of conservation of energy ?
Answer:
According to this law, energy can neither be created nor be destroyed but can only be transferred from one form to another.

Question 15.
What is the effect on energy when work is done ?
Answer:
When work is done, energy decrease in some extent.

Question 16.
What is the work done by a man when he lifts a luggage of mass 10 kg on his head and moves 50 metres on horizontal road ?
Answer:
zero.

Question 17.
From where, our body gets energy ?
Answer:
From food taken by us.

Question 18.
Which type of energy is used by plants ?
Answer:
Solar energy.

Question 19.
Write an important use of wind energy.
Answer:
Wind energy is used to grind the flour by wind mills.

Question 20.
Write the name of any two natural sources of energy.
Answer:
(i) Sun
(ii) Water.

Question 21.
Write any four types of energy.
Answer:
(i) Mechanical Energy
(ii) Chemical Energy
(iii) Light Energy
(iv) Sound Energy.

Question 22.
What is gravitational potential energy ?
Answer:
The energy stored in a body when work is done against force of gravity is called gravitational potential energy.

Question 23.
What is mean by elastic potential energy ?
Answer:
The energy possessed by an object due to change in its shape is called elastic potential energy.

Question 24.
Why spark is produces when two stones collides with each other ?
Answer:
Because kinetic energy of stone is converted into light energy.

Question 25.
How arrow gets its kinetic energy ?
Answer:
The arrow converts the potential energy from bow into kinetic energy.

Question 26.
Out of energy and acceleration, which is vector ?
Answer:
Acceleration is a vector quantity.

Question 27.
The kinetic energy goes on decreasing when it is thrown vertically upward. What will be the motion at highest point ?
Answer:
Zero.

Question 28.
Write the name of that device which converts electric energy into mechanical energy.
Answer:
Electric motor.

Question 29.
Which energy is converted in other in a hydroelectric power centre ?
Answer:
The kinetic energy of water is converted into electric energy.

Question 30.
What is power ? What is its unit ?
Answer:
The rate of doing work is power. Its unit is watt.

Question 31.
If the velocity of an object is increased three times, what will be its kinetic energy ?
Answer:
The kinetic energy will increase nine times.

Question 32.
Which type of energy is there in a expanded spring ?
Answer:
Potential energy.

Question 33.
Which type of energy is there in a dam ?
Answer:
Potential energy.

Question 34.
Which type of energy is there in the spring of watch ?
Answer:
Potential energy which gives the kinetic energy to the needles of watch. ,

Question 35.
Which type of energy is there in running train ?
Answer:
Kinetic energy.

Question 36.
What will be potential energy of that body whose height is doubled ?
Answer:
Its potential energy will be doubled.

Question 37.
What is the relation between 1 kilowatt and watt ?
Answer:
1 kilowatt = 1000 watt.

Question 38.
Why the palms become hot when rubbed ?
Answer:
Because kinetic energy of palms is converted into heat energy.

Question 39.
How much energy is required to lift a stone of mass 1 kg to a height of 1 metre ?
Answer:
9.8 Joules [∵ W = mgh = 1 x 9.8 x 1 = 9.8 joules]

Question 40.
A person and his son are running with uniform velocity. If the mass of person become two times more than of his son, then what will be the ratio between kinetic energy of both ?
Answer:
2 : 1 [ ∵ K.E. = \(\frac {1}{2}\) mv² and v is equal]

Question 41.
Which type of energy is required to move for a sailboat ?
Answer:
The sailboat moves by the kinetic energy of wind.

Question 42.
What will be the energy a body has a capacity to do a work of 10 joules ?
Answer:
10 Joules.

Question 43.
Write the name of two units of power used in daily life.
Answer:
Kilowatt and megawatt.

Question 44.
How many watts are there in one horse power ?
Answer:
1 horse power = 746 watt.

Question 45.
What is meant by 1 kilowatt hour energy ?
Answer:
The energy consumed by a source of 1 kilowatt power in 1 hour.

Question 46.
What is the relation between kilowatt hour and joule ?
Answer:
1 kilowatt hour = 36,00,000 joules.

Question 47.
Which type of energy is used when a person climbs stairs ?
Answer:
Muscular energy.

Question 48.
Which type of energy is there in our body ?
Answer:
Chemical energy.

Question 49.
The person requires the least amount of energy in which of the following activity ?
(i) In swimming
(ii) In movement
(iii) In sleeping
(iv) In running foot.
Answer:
(iii) In sleeping.

Question 50.
What is simple pendulum ?
Answer:
Simple pendulum is a metalic bob which is hang by weightless string.

Question 51.
Write the name of energy when simple pendulum is at its extreme point.
Answer:
Potential energy.

Question 52.
Write the name of energy when pendulum is in middle position.
Answer:
Kinetic energy.

Short-Answer Type Questions

Question 1.
What is meant by work ? How is it measured ? What is its unit ?
Answer:
Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the
applied force.
Thus, Work done = Force x The distance moved in the direction of force
W = F x s
Work is a vector quantity. Unit of work is newton-metre or joule (J).

Question 2.
What is energy ? Write its unit and definition. What are its various types ?
Answer:
The capacity of doing work is energy. Its unit is also joule.
Joule: Work done is said to be one joule when a force of 1 newton acting on an object and the object is displaced through 1 m in the direction of force.

Various types of energy:
(i) Mechanical energy
(ii) Heat energy
(iii) Chemical energy
(iv) Electric energy
(v) Nuclear energy’
(vi) Light energy
(vii) Atomic energy
(viii) Magnetic energy
(ix) Sound energy,
(x) Solar energy.

Question 3.
Justify with example that moving bodies have capacity to do work ?
Answer:
The examples are following :
(i) A moving marble has a capacity to displace that marble which is in rest.
(ii) The blowing wind works by rotating the wheels of windmill.
(iii) The moving water also takes the objects from one place to another.
(iv) The blowing wind is helpful for sailing boat.
Thus, we can say that moving bodies have capacity to do work.

Question 4.
What is meant by potential energy and kinetic energy ? Give two examples for each.
Answer:
Potential energy’: The energy possessed by a body due to its change in position or shape is called the potential energy.
Examples:
(1) The book lying on the table has potential energy due to its position.
(2) A binded watch has potential energy.

Kinetic energy:
The energy possessed by a body due to its motion is called the kinetic energy.

Examples:
(1) Moving air has kinetic energy.
(2) The running water has kinetic energy.

Question 5.
Is there energy in following substances ? If yes, then what is the type-potential energy or kinetic energy or both ?
(i) The switched off ceiling fan.
(ii) A man climbing up mountain.
(iii) Flying bird.
(iv) The water in a dam.
(v) The stretched bow more than its shape.
(vi) The rubber locket on a table.
(vii) Stretched locket of rubber.
Answer:
Yes, all substances have energy’.
(i) A switched off ceiling fan has potential energy only.
(ii) A man climbing up mountain has both potential and kinetic energy.
(iii) A flying bird has both kinetic and potential energy.
(iv) The water in the dam has potential energy.
(v) A stretched bow has potential energy.
(vi) The rubber locket on the table has potential energy.
(vii) Stretched locket of rubber has potential energy.

Question 6.
Write the difference between potential energy and kinetic energy.
Answer:
The differences between potential energy and kinetic energy are as follows:
Potential Energy (P.E.):
1. The energy possessed by an object due to its position is called potential energy.
2. Potential energy = mgh
3. The potential energy of an object depends upon the height or depth from the surface of earth.

Kinetic Energy (K.E.):
1. The energy possessed by an object due to its motion is called kinetic energy.
2. Kinetic energy = \(\frac {1}{2}\) mv²
3. Kinetic energy depends upon the motion of the object.

Question 7.
Which type of energy is in following ?
(1) Coiled spring of watch.
(2) Running buffallo.
(3) The arrow released from bow.
(4) A table lying on roof.
(5) Compressed air.
(6) A mango on tree.
Answer:
(1) Potential energy
(2) Kinetic energy
(3) Kinetic energy
(4) Potential energy
(5) Potential energy
(6) Potential energy.

Question 8.
When the capacity of doing work by human body decreases ?
Answer:
The capacity of doing work by human body decreases in following circumstances :
(1) When a person is ill.
(2) When a person grows old.
(3) When a person takes food less than required, his muscular energy decreases.

Question 9.
What is the relation between work and energy ? Write their units.
Answer:
The capacity of doing work is called energy. The magnitude of energy is always equal to that work which has a capacity to do that. Energy is measured in joule and the unit of work is also joule.

For example:
To lift a piece of stone of mass 1 kg upto a height of 1 metre requires 9.8 joules energy. Opposite to it, when stone is dropped from one metre, the stone does the same amount of work (9.8 J). Thus, if an object has an energy of 100 joules. It has a capacity of doing work of 100 joules.

Question 10.
How the energy is transferred in key toy ?
Answer:
There is a spring in key toy or key watch which contracts while binding and stored potential energy in it. After some moments, the spring loses slowly and the toy comes in motion. Thus, potential energy is transferred in kinetic energy.

Question 11.
A ball is thrown upward and it returns back. What is the change in kinetic and potential energy of the ball ? Explain.
Answer:
When a ball is thrown upward its kinetic energy changes continuously into potential energy. At highest point its whole kinetic energy changes into potential energy. When the ball returns back its potential energy changes continuously into kinetic energy and whole of its potential energy changes into kinetic energy near the earth but on reaching the earth, whole of its kinetic energy changes into potential energy.

Question 12.
A boy is standing by holding some weight in his expanded hand. In this stage, there seems no change in the position of both boy and weight. Is he doing any work ? If yes, then justify.
Answer:
Yet, there is no change in the position of boy and weight externally, but internally the boy is working. The muscles of boy are stretched due to weight on palm. Thus, their shape get changes. The heart pumps more blood into muscles. The energy of boy wastes in all these chemical activities and he tired soon. If the boy stand more enough in same position, the external change will be observed as sweating.

Question 13.
How much work is done when an artificial satellite revolve round the earth ? Explain in brief. –
Answer:
The artificial satellite revolves round the earth in almost circular track. The gravitational force between the earth and the satellite works along the line joining the earth and satellite and the satallite moves perpendicular to the direction of force of gravity. Thus, the magnitude of displacement of satellite in the direction of force will be zero and the work done by the earth on the satellite will also be zero.

Question 14.
How the kinetic energy of wind is utilised ? Write its important uses.
Answer:
The electric energy can be produced by the energy in the blowing air. It has much waste. Thus, wind mill is used in grinding flour and drawing out water from ponds. A small amount of electric energy is produced from wind mill. Those places where the wind blows throughout the years, wind mill are used to grind flour and drawing out water.
Important uses of mechanical energy of wind :
(1) The fields can be irrigated by drawing water from ponds.
(2) It can be used in grinding the flour.
(3) The mechanical energy of wind can be converted into electric energy by joining the various wind wheels together.

Question 15.
What is water energy ? Write its two important uses.
Answer:
The running water has a capacity to do work due to its kinetic energy, which is known as water energy.

Uses of water energy: Following are the uses of water energy :
(1) Water energy is converted into electric energy and it is used in houses and factories.
(2) Water energy is used to grind flour by windmills.
(3) The heavier wooden pieces is taken from one place to another due to water energy.

Question 16.
How water energy is used to produce electricity ?
Answer:
The kinetic energy of running water is stored as potential energy in dam. The stored water is dropped on the larger blades of water turbines. The shaft attached with running turbine rotates the generator or dyanmo quickly. The rotating coil of dyanmo produces electricity which is called ‘hydroelectric power’. The electricity is produced on Bhakhra dam by this method. This is also known as the principle of hydroelectricity.

Question 17.
Why the nail becomes hot when it is hammered in the sheet of wood ?
Answer:
The nail becomes hot when it is hammered in the sheet of wood because the lifted hammer has potential energy due to its position. When the hammer is dropped on the nail, it comes in rest position, but transfers, its whole energy in the nail. As a result of this, the nail gets some kinetic energy and moves inside the wood. When the nail enters completely and we continuous hammering the nail becomes hot because the mechanical energy of the hammer is transferred in nail and changes into heat energy.

Question 18.
What are the benefits of the solar energy ?
Answer:
The benefits of the solar energy are as follows :
(1) The solar energy is used in solar cooker to prepare food,
(2) Solar energy is used to warm the water by solar heater in hotels, hospitals and industries.
(3) Solar energy is used in the production of electricity.
(4) It is used to melt the metals in solar furnances.

Question 19.
Differentiate among work, energy and power. Write the SI unit for each. Or What is Power? Write its S.I. unit.
Answer:
Work:
When force is applied on a body and it moves in the direction of force, it is said to be work done.
Thus, Work (W) = Force (F) x Distance (S)
SI unit of work is joule.

Energy: t he capacity ot doing work is called energy. SI unit of energy is joule.

Power: The rate of doing work is called power.
Thus,

SI unit of power is watt.

Question 20.
What do you mean by transfer of energy ? Write two examples of transfer of energy.
Answer:
According to the law of conservation of the energy, The energy can neither be created nor destroyed. It can be transferred from one form to another. It is called the transfer of energy. Thus, the total magnitude of energy is never changed.

Examples:
(1) Ice on mountains melts and becomes water. This is the transfer of potential energy into kinetic energy.
(2) The electric energy in bulb changes into light energy and glows.

Numerical Questions

Question 1.
A boy of mass 40 kg is running with a ve’ocity of 0.5 ms. What is his kinetic energy ?
Solution:
Here,
m = 40 k
v = 0.5 ms^-1
We know that,
Kinetic Energy (K.E.) = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 40 x 0.5 x 0.5 = 5 Joule Answer:

Question 2.
An object of mass 15 kg is moving with a uniform velocity (if 4 ms. What is tue kincth energy possessed by the object?
Solution:
Here
Mass of the body (m) = 1 5 kg
Velocity of the body (v) = 4 rn/s
We know that
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 15 x 4(2)²
= \(\frac {1}{2}\) x 15 x 16
= 120 Jule

Question 3.
A stone of mass 100 gram is kept at a height ofS metre. What kind of energy does it possess and find out its value.
Solution:
The stone has potential energy because of height.
Here. m = 100g = \(\frac {100}{1000}\) = 0.1 kg
g = 9.8 ms^-2
h = 5m
We know that.
Potential Energy (P.E.) m x g x h = 0.1 x 9.8 x 5 = 4.9 Joule

Question 4.
To what height can an object of mass 1 kg be raised by giving it an energy of 980 joule?
Solution:
Here.
m = 1 kg
g = 9.8 ms^-2
h = ?
Potential Energy (RE.) = 980 J
We know that,
Potential Energy (RE.) = m.g.h.
or 980 = 1 x 9.8 x h
or h = \(\frac{980}{1 \times 9.8}\)
= 100 m

Question 5.
What is the increase in kinetic energy of a particle, if its velocity is increased to four times?
Solution:
Suppose the initial velocity of the particle = v
and its final velocity = 4 v
Suppose the mass of the particle = m
∴Initial kinetic energy (K₁) of the particle = \(\frac {1}{2}\) mv² ………….(i)
Final kinetic energy (K₂) of the particle = \(\frac {1}{2}\)m (4v)²
= \(\frac {1}{2}\)m x 16 v² …………..(ii)
From equations (i) and (ii),
\(\frac{\mathrm{K}_1}{\mathrm{~K}_2}=\frac{\frac{1}{2} \mathrm{mv}^2}{\frac{1}{2} \mathrm{~m} \times 16 \mathrm{v}^2}=\frac{1}{16}\)
or K₂ = 16K₁
Thus, the increase in kinetic energy is 16 times the initial kinetic energy.

Question 6.
The kinetic energy of a ball of mass loo g is 20 joule. Calculate its velocity.
Solution:
Here,
m = 100 g = \(\frac {100}{1000}\)kg = 0.1 kg
v = ?
Kinetic Energy (K.E.) = 20 J
We know that, Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
Or 20 = \(\frac {1}{2}\) x 0.1 x v²
v² = \(\frac {20 x 2}{0.1}\) = 400
or v = \(\sqrt{400}\) = 20 ms^-1

Question 7.
A coolie lifts an object of mass 20 kg from the earth to 1.5 m and keep It on his head. Calculate the amount of work done by him on the object.
Solution:
Here,
Mass of the object (m) = 20kg
Displacement (s) = 1 .5m
Work done (W) = F x s = mg x s
= 2okg x 10 ms^-2 x 15 m
= 300kgms^-2 m
= 300Nm= 300J
The amount of work done by the coolie on the object is 30 joules.

Question 8.
A man of mass 50 kg jumps to a height of 1.2 m. What is its potential energy at the highest point?
Solution:
Here,
m = 50kg
g = 9.8 ms^-2
h = 12m
We know that,
Potential Energy (P.E.) m x g x h
= 50 x 9.8 x 1.2
= 588 J
Thus, at the highest point the potential energy of the man is 588 J.

Question 9.
With what velocity should a man of mass 50 kg run so that his kinetic energy becomes equal to 625 J.
SolutIon:
Here,
m = 50kg
v = ?
Kinetic Energy (K.E.) = 625 J
We know that,
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
Or 625 = \(\frac {1}{2}\) x 50 x v²
v² = \(\frac{625 \times 2}{50}\) = 25
or v = \(\sqrt{25}\) 5ms

Question 10.
A man throws a stone of mass 10 kg from a ladder of height 5m. When ¡t reaches the ground, what will be its kinetic energy ? What will be its velocity at the nearest point to the ground?
Solution:
Here,
m = 10kg
h = 5m
v = ?
Kinetic Energy (K.E.) = ?

We know that,
The kinetic energy (K.E.) of the stone = The loss in potential energy due to falling of the stone
= mgh
= 10 kg x 9.8 ms^-2 x 5m = 490J ……………(i)
Because kinetic energy = \(\frac {1}{2}\) mv² .(ii)
From eqns. (i) and (ii),
\(\frac {1}{2}\) mv^-2 = 490J
or \(\frac {1}{2}\) x 10kg x v² = 490J
or v² =\(\frac{490 \times 2}{10}=\frac{980}{10}\) = 98
or v = \(\sqrt{98}\) = 9.9 ms^-1 (approx.)

Question 11.
A car of mass 1000 kg is moving with a velocity of 30 ms^-1. After applying brakes it stops with uniform acceleration at a distance of 50m. Calculate the force and the work done by the brakes?
Solution:
Here,
m = 1000kg
u = 30 ms^-1
v = 0 ms^-1
s = 50m
F = ?
W = ?
We know that, v² – u² = 2as
a = \(\frac{v^2-u^2}{2 s}=\frac{(0)^2-(30)^2}{2 \times 50}=\frac{-900}{100}\) ms^-2
= – 9 ms^-2 So retardation = 9ms^-2
F = m.a.l000kg x 9ms² = 9000 N
The force applied by the brakes on the car (F) = 9000 N
Work done(W) = F x s = 9000 N x 50 m = 450000 J = 450 kJ

Question 12.
A man of mass 50kg climbs a vertical height of 10m on a mountain in 20 seconds. What ¡s his power ? (Given that g 9.8 ms-2)
Solution:
Here,
F = m x g
= 50 x 9.8 = 490N
Now, W = F x S = 490N x 10m = 4,900 Nm
Time (t) = 20 seconds
Power (P) = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{4,900}{20}\) = 245 watt

Question 13.
A boy of mass 40 kg runs up a staircase of 45 steps in 9 sec. If the height of each step is 15 cm, Calculate his power ? (The value of g is 10 ms²)
Solution:
Weight of the boy (mg) 40kg x 10ms² = 400N
Total height of the steps (h) = 45 x 15/100m = 675m
Time taken to climb the steps (t) = 9s
We now that
Power (P) = Time taken = \(\frac{\mathrm{mgh}}{t}\) = \(\frac{400 \mathrm{~N} \times 6.75 \mathrm{~m}}{9 \mathrm{~s}}\) = 300 W
The power of the boy is 300 W.

Question 14.
A man of mass 50 kg climbs 30 steps in 30 sec. If height of each step is 20 cm. Then calculate the power to climb the total steps.
Solution:
Mass of the man (m) = 50 kg
Total number of steps = 30
height of each step = 20 cm = \(\frac {20}{100}\) m = 0.2 m
Total height covered by the man = 30 x 0.2 m = 6m
Acceleration due to gravity (g) = 10 ms^-2
Thus, work done by the man = m.g.h = 50 x 10 x 6J = 3000J
Time taken to climb the steps = 30 s

= \(\frac{3000 \mathrm{~J}}{30 \mathrm{~s}}\)
= 100 Js^-1 = 100 W(watt)

Question 15.
An electric bulb of 100 Wis used for 2 hours, how much electrical energy is consumed?
Solution:
Power of the bulb = 100W
Time = 2 hours 2 x 3600 = 7200 sec.
Energy consumed = Power x Time
= 100 x 7200Ws = 720000 joule

Question 16.
A rocket has a mass of 3 x 10⁶ kg. Calculate (a) its potential energy (b) its kinetic energy, at a height of 25 km moving with a velocity of 1 km/s. (Take g = 10 ms^-2)
Solution:
Here,
Mass of the rocket (m) 3 x 10⁶ kg
Velocity of the rocket (v) = 1 k ms^-1 = 1000 ms^-1
Height of the rocket (h) = 25 km = 25000 m
Acceleration due to gravity (g) = 10 ms^-2

(a) Potential Energy = m.g.h.
3 x 10⁶ x 10 x 25000 J
= 7.5 x 10¹¹

(b) Kinetic Energy = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 3 x 10⁶ x (1000)² J
= 7.5 x 10¹¹J

Question 17.
A horse is pulling a cart moving with a velocity of 18 kmh^-1 with a force of 300 N. Find out the power of the horse in watt and also in horse power?
Solution:
Here,
Velocity of the cart(v) = 18 kmh^-1
= \(\frac{18 \times 1000}{3600}\) ms^-1 = 5 ms^-1
The force exerted by the horse on cart (F) = 300 N
m.g. = 300N
Work done = m.g.h
= 300 x 5 = 1500J

= \(\frac{15000}{1s}\) = 1500 w
We know that, 746 W = 1 horsepower
1 waft = \(\frac {1}{746}\) horsepower
1500 watt = \(\frac {1}{746}\) x 1500 horsepower
= 2 horsepower

Do Yourself

(1) A car is moving with a uniform velocity of 54 kmh^-1 What will be the kinetic energy of a child of mass 40 kg sitting inside it?
Answer:
[4500J]

(2) Calculate the power of that pump which stores the water of loo kg in 25 seconds into a tank kept at a height of 19m. (Let g = 10 ms²)

(3) What will be the energy of an object of mass 10 kg at a height of 40 m from the ground level?
Answer:
[Take g 100 ms²] [40003]

(4) A force of 5 N is acting on a body. The body displaces to a distance of 2m in the direction of the force. Find out the work done on the object.
Answer:
[103)

(5) An object of mass 15 kg is moving uniformly with a velocity of 4 ms². Calculate the kinetic energy of the object.
Answer:
[120J]

(6) What will be the height from the ground level where an object of mass 1 kg will have a potential energy of 1 joule if g= 10 ms²?
Answer:
[0.1 m]

(7) If 4900 joule of energy is utilised in lifting a body of mass 50 kg to a certain height, then what will be the height of the object if g = 9.8 ms²?
Answer:
[10m]

(8) A body of mass 3 kg starts falling down from rest position. After 3 sec., what will be its kinetic energyifgl0m.c2?
Answer:
[1350m]

(9) An object of mass 10 kg is lifted to a height of 6m from the ground level. Calculate the energy present in the object. The value of g is
9.8 ms².
Answer:
[588 J]

(10) An object of mass 12 kg is kept at a certain height. 1f its potential energy is 480 J; calculate its height from the ground level.
Answer:
[Take g = 10 ms²] [4 m]

(11) A man of mass 50kg takes a packet of mass 20kg upto the terrace of a building of height 14m, what is the work done by the man ?
Answer:
[Take g = 10 ms²] [9604 J]

(12) A vehicle of mass 1800 kg is moving with a velocity of 40 ms. After applying brakes it stops with a uniform acceleration at a distance of 200 m. Calculate the force applied and work done by the brakes.
Answer:
[1440 kJ]

(13) What will be the increase in kinetic energy of a particle if its velocity is doubled?
Answer:
[4 times ]

(14) How much amount of work is done on lifting an object of mass 2 kg to a height of 2m if the value of g is 9.8 ms²?
Answer:
[39.2J1

(15) The height of’Kutubminar’ is 70m. What is the work done by a man of mass 50 kg on reaching the top?
Answer:
[Take g 9.8 ms^-2] [3430011

(16) A stone of mass 10g is kept at a height of 5m. What kind of energy is present in it and calculate its value?
Answer:
[Take g 9.8 ms^-2] [0.49J]

(17) Two girls A and B of mass 400 N each, climbs a rope upto a height of 8m. IfA takes 20 sec. to do this work while B takes 50 sec.,
then calculate the individual power of both the girls.
Answer:
[A 160 W, B = 64W]

(18) Find the energy in KWh consumed inS hours by four devices of power 500 W each.
Answer:
[10Kwh]

Essay Type Questions

Question 1.
Define potential energy. Derive an expression for the potential energy of a body.
Answer:
Potential Energy:
The energy possessed by a body due to its position or change in shape is known as potential energy.

Expression for potential energy:
Consider a body of mass m kept on the.ground level at point ‘A’. it is raised to a point ‘B’ by applying a force F on it. The distance between A and B is ‘h’.
As we know that, the force F acting on the body to lift it, is equal and opposite to the gravitational force
F = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) …………(i)

Where, M_e = Mass of the earth
R_e = Radius of the earth
But, g = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) (g = acceleration due to gravity) ………..(ii)
From equations (i) and (ii),
F = mg
The work done in raising the body to a height ‘h’
W = F x S (∵ s = h) = mgh
Because work done is same as the energy. Therefore, at a height ‘h’ the potential energy of the object is,
Potential Energy (P.E.) = mgh

Question 2.
Define kinetic energy. Obtain an expression for it.
Answer:
Kinetic Energy:
The energy possessed by a body due to its motion is called kinetic energy.

Expression:
Let us suppose a body of mass m is dropped from a height h and its velocity becomes V.
on reaching the ground level Then –
Initial velocity of the body (u) = 0
Acceleration (a) = g (acceleration due to gravity)
Final velocity = v
Distance (s) = Height (h)
We know that, v² – u² = 2as
v² – 0 = 2gh
v² = 2gh
h = \(\frac{v^2}{2 g}\) …………. (i)
As the object comes down, its potential energy decreases and kinetic energy increases. When the body strikes to the ground level, its hoIe potential energy is converted into kinetic energy. According to the law of conservation of energy, Energy can neither be increased nr be decreased.”
Thus, kinetic energy at ground level = Potential energy at height h.
Kineiic Energy = mgh
or K.E. = mg x \(\frac{v^2}{2 g}\) (∵ h = \(\frac{v^2}{2 g}\))
K.E. = \(\frac {1}{2}\) mv²

Question 3.
Define Law of conservation of energy. Give an example of conservation of energy.
Answer:
Law of conservation of energy-According to this law, energy can neither be created nor be destroyed. ¡t can only be transfèrred from one form to another. When the energy is released in any form, the same amount of energy appears in other forms. Thus, total energy of the universe remains constant. Suppose, an object of mass m starts falling from a height h, on the ground level.

(a)EnergyatpointA, PotentialEnergy = mgh
Kinetic Energy = 0
∴ Total energy = mgh + 0 = mgh ……………(i)

(b)Energy at point B:
Suppose the object falls from A to B at a distance s, then its height from ground level will be (h-S).
∴ Potential energy = mg (h-s)
and v²-u² = 2gs
v² = 2gs [∵ u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\)m x 2gs = mgs
Total Energy = mg (h-s) + mgh = mgs – mgs + mgs = mgh …………(2)

(c) Energy at point C; v²-u² = 2gh
v² = 2gh [∵u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\) m x 2gh = mgh
and Potential Energy = 0
Total Energy = mgh + 0 = mgh ……………(iii)
in this way we observe from eQuestion (i), (ii) and (iii) that the energy of the object remains constant at every point. There is potential energy in the object at highest point. When it falls litle downs same part of potential energy is converted into kinetic energy and on ground level whole potential energy is converted into kinetic energy.

Practical Work

Experiment 1.
Describe by an experiment that the wark done by a force can either be positive or negative.

Procedure:
A child is pulling a toy car parallel to the ground as shown in figure. The child has exerted a force in the direction of displacement of car. In this situation, the work done will be equal to the product of the force and displacement. In this situation the work done by the force is taken as positive. Consider a situation in which an object is being displaced by the action of forces and we identit one of the forces, F acting Opposite to the direction of the displacement s, thus, the angle between the two directions is 1800.

In such a situation, the work ‘ done by the force F is taken as negative and denoted by the minus sign. The work done by the force is F x (-s) or (-F x s). When the force acts opposite to the direction of the displacement, the work done is negative. When the force acts in the direction ofdisplacernen, the work done is positive.

Experimenf 2.
Describe by an activity in which both positive and negative forces are working.

Procedure:
When an object is lifted up, both positive and negative forces work, because the force applied by us, displaces the object upward and called positive force, while the force of gravity applied by earth pulls the object downward which is called negative force.

Experiment 3.
Describe by an experiment that the object at more height has more energy.

Procedure:
(1) Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from height of about 25 cm. The ball creates a depression.
(2) Repeat this activity from heights of 50 cm, I m and 1.5 m.
(3) Ensure that all the depressions are distinctly visible.
(4) Mark the depressions to indicate the height from which the ball was dropped.
(5) Compare their depths.

We will observe that when the ball was dropped from the height of 1.5 m, the depression was the deepest; which proves that the object at more height has more energy which makes the deepest depression.

Expereiient 4.
Describe by an experiment that the kinetic energy of an object increases by increasing its mass.

Procedure:
(1) Set up the apparatus as shown in figure
(2) Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
(3) Place a known mass on the pan so that the trolley starts moving.
(4) The trolley moves forward and hits the wooden block and displaces the block.
(5) Note down the displacement of the block. It is, clear that the work is done on the block by the trolley as the block has gained energy.
(6) Repeat this activity by increasing the mass on the pan.

You will observe that the displacement will increase with the increase in mass, which proves that the kinetic energy increases by increasing the mass.

Quick Review of the Chapter

1. What is required for doing a work?
(A) displacement in object
(B) force should be applied on the object
(C) both (A) and (B)
(D) no displacement in object
Answer:
(C) both (A) and (B)

2. What type of energy is there inside a body kept at height?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) no relation with energy
Answer:
(A) potential

3. The device that converts the electrical energy In heat energy:
(A) Electrical iron
(B) Electrical bulb
(C) Radio
(D) Electrical motor
Answer:
(A) Electrical iron

4. Which type of energy is there in a flying bird?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) neither of the two
Answer:
(C) both (A) and (B)

5. Rado transfer the electrical enegry in:
(A) light energy
(B) sound energy
(C) mechanical energy
(D) heat energy
Answer:
(B) sound energy

6. Which formula is correct for work?
(A) W = \(\frac {F}{s}\)
(B) W = \(\frac {s}{F}\)
(C) W= Fs
(D) W = \(\frac {1}{Fs}\)
Answer:
(C) W= Fs

7. The unit of force Is:
(A) Newton
(B) Newton metre
(C) Joule
(D) both (B) and (C)
Answer:
(A) Newton

8. If 5n force acts on an object, then displacement is 2m, what will be the work done?
(A) lONm
(B) IOJ
(C) lOJm
(D) both(A)and(B)
Answer:
(D) both (A) and (B)

9. A coolie lifts an object of mass iS kg from the earth to 1.5 m keep it on his head. Calculate the
amount of work done by the coolie.
(A) 22.5J
(B) 2.25 J .
(C) 225J
(D) 225N
Answer:
(C) 225 J

10. The potential energy of an object of mass 1kg kept at the height &f 2m will be:
(A) 9.8J
(B) 19.6J
(C) 29.4 J
(D) 39.2 J
Answer:
(B) 19.6 J

11. Whose unit is Newton metre (Nm)?
(A) work
(B) force
(C) acceleration
(D) power
Answer:
(A) work

12. How many watts are in 1 Kilowatt?
(A) 10³
(B) 10⁴
(C) 10⁵
(D) 10⁶
Answer:
(A) 10³

13. How many joules are in 1 Kilowatt?
(A) 36 lakhs
(B) 36 thousands
(C) 36 hundreds
(D) 36 crores
Answer:
(A) 36 lakhs

14. What is another name of newton metre?
(A) Erg
(B) Coulomb
(C) Joule
(D) Hertz
Answer:
(C) Joule

15. ……….. is vector.
(A) potential energy
(B) kinetic energy
(C) work
(D) none of these
Answer:
(D) none of these

16. A man stands for 10 minutes by carrying a weight of 20 kg the work done s ill be:
(A) 200joule
(B) onejoule
(C) zero
(D) 400joule
Answer:
(C) zero

17. The unit of energy is:
(A) Nm
(B) Joule
(C) both (A) and (B)
(D) no unit
Answer:
(C) both (A) and (B)

18. The example of kinetic energy is not:
(A) stone kept on mountain
(B) revolving wheel
(C) blowing air
(D) bullet shot out of a gun
Answer:
(A) stone kept on mountain

19. The example of potential energy is not :
(A) an arrow kept in the bow
(B) a stone on a mountain
(C) falling coconut
(D) a bullet in a gun
Answer:
(C) falling coconut

20. An object of 10 kg mass is carried to a height of 6 m. Calculate the energy of the object. (g = 9.8 mc²)
(A) 98 J
(B) 58.8 J
(C) 588 J
(D) 5880 J
Answer:
(C) 588 J

21. The law of conservation of mass is:
(A) all works require energy
(B) energy can be produced
(C) energy can be destroyed
(D) energy neither can be produced nor destroyed
Answer:
(D) energy neither can be produced nor destroyed

22. Which statement is not correct?
(A) Potential energy + kinetic energy constant
(B) Kinetic energy + potential energy = mechanical energy
(C) As much decrease is there in potential energy at a point so is in kinetic energy
(D) As much decrease is there in potential energy so much increase takes place in kinetic energy.
Answer:
(C) As much decrease is there in potential energy at a point so is in kinetic energy

23. Which formula is correct?
(A) P = W x t
(B) P = \(\frac {W}{t}\)
(C) P = \(\frac {t}{W}\)
(D) P = \(\frac{1}{W \times t}\)
Answer:
(B) P = \(\frac {W}{t}\)

24. What is the unit of power?
(A) Joule
(B) Nm
(C) erg
(D) Watt
Answer:
(D) Watt

25. How much energy will be consumed by a person of 600 newton weight for climbing a height of 10m?
(A) 60 joule
(B) 600 jouie
(C) ó000joule
(D) 588 joule
Answer:
(C) 6000 joule

26. When will be the minimum amount of potential energy in your body ? When you:
(A) are standing
(B) are sitting in a chair
(C) ire lying on the earth
(D) are sitting on the earth
Answer:
(C) are lying on the earth

27. When ¡s work not done?
(A) on fixing a nail in a wood
(B) on sliding a box on the floor
(C) when there is no constituent of parallel force in the direction of motion
(D) a weight hanging on a peg
Answer:
(C) when there is no constituent of parallel force in the direction of motion

28.
10 seconds are taken to fetch a bucket of 5 kg mass from a well of 10 m depth. The power used in the power will be:
(A) 5ONm
(B) 50)
(C) 50 W
(D) 500 W
Answer:
(C) 50 W

29. How much energy will be consumed by 5 fans of 100 watt each in 4 hours?
(A) 20 kwh
(B) 10 kwh
(C) 2 kwh
(D) 1 kwh
Answer:
(C)2 kwh

30. The rate of doing work is called:
(A) Power
(B) Force
(C) Energy
(D) None of these
Answer:
(A) Power

31. What is the momentum of an object of mass m, moving with a velocity u?
(A) (mυ)²
(B) mυ²
(C) \(\frac {1}{2}\) mυ²
(D) mυ²
Answer:
(C) \(\frac {1}{2}\) mυ²

Haryana State Board HBSE 9th Class Science Important Questions Chapter 10 Gravitation Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 10 Gravitation

Very Short-Answer Type Questions

Question 1.
What is meant by gravitation?
Answer:
Everybody on earth attracts each other. Such type force of attraction between bodies is called gravitation.

Question 2.
Who discovered the gravitational force firstly?
Answer:
Sir Isaac Newton.

Question 3.
In which direction the gravitational force between two objects works?
Answer:
The gravitational force between two objects works in the direction of a line joining the centre of their masses.

Question 4.
Which force is responsible for the existence of our solar system?
Answer:
Gravitational force is responsible for our solar system.

Question 5.
What ‘G’ indicates?
Answer:
‘G’ indicates gravitational constant.

Question 6.
What is the value of ‘G’ ? ’
Answer:
VMtte of‘G’ is 6.673 x 10¹¹ Nm²/kg²

Question 7.
How much is the mass of earth ?
Answer:
Mass of earth (M_e) = 6 x 10²⁴ kg (approx.).

Question 8.
How much is the radius of earth ?
Answer:
Radius of earth (R_e) = 6.4 x 1o⁶m.

Question 9.
At which part of earth, the value of g is maximum and minimum?
Answer:
The value of g is maximum on poles of earth and minimum on the equator. ‘

Question 10.
Write the formula for the force of attraction between two objects of masses m₁ and m₂ respectively kept at a distance of R.
Answer:
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{R}^2}\)

Question 11.
What is acceleration due to gravity (g) ?
Answer:
The acceleration by which earth attracts other object towards its centre is called acceleration due to gravity.

Question 12.
Does the acceleration of an object falling towards earth depend upon its mass ?
Answer:
No, the acceleration of an object falling towards earth does not depend upon its mass.

Question 13.
What will be the force of attraction between two objects if the distance between them becomes twice ?
Answer:
Force of attraction will become one-fourth.

Question 14.
What will be the weight of an object on moon in comparison to earth ?
Answer:
\(\frac {1}{6}\) times.

Question 15.
What is centripetal acceleration ?
Answer:
When an object moves on a circular track’ ah acceleration is produced due to change in velocity, which is called centripetal acceleration.

Question 16.
What is centripetal force ?
Answer:
The force which produces centripetal acceleration is called centripetal force. It is always directed towards centre. That is, F = \(\frac{\mathrm{mv}^2}{\mathrm{r}}\)

Question 17.
What ‘g’ indicates ?
Answer:
‘g’ indicates acceleration due to grayity.

Question 18.
What is the value of ‘g’ near the syrfepe of earth ?
Answer:
The value of ‘g’ is 9.8 m/s².

Question 19.
On which place, the value of ‘g’ is zero ?
Answer:
The value of‘g’is zero at the centre of earth.

Question 20.
Write the formula for the relation between ‘g’ and ‘G’
Answer:
g = \(\frac{\mathrm{G} \times \mathrm{M}_{\mathrm{e}}}{\mathrm{R}^2}\)

Question 21.
What are the factors on which the gravitational force depends ?
Answer:
(i) On the mass of objects.
(ii) On the distance between objects.

Question 22.
What is the distance of moon from earth ?
Answer:
60 times of the radius of earth – 384000 km.

Question 23.
If an object is dropped from height A with zero velocity, then how much distance it will cover in one second ?
Answer:
Distance covered, h = \(\frac {1}{2}\) gt²

Question 24.
Whose value, out of ‘g’ and ‘G’ is constant and whose not ?
Answer:
The value of ‘G’ is constant at every place on earth, but value of ‘g’ is more on poles and less on equators.

Question 25.
How much time does the earth take to move around the sun ?
Answer:
The earth takes 365 \(\frac {1}{4}\) days to move around the sun.

Question 26.
How much time does the moon take to move around the earth ?
Answer:
The moon takes 27 days 8 hours to move around the earth.

Question 27.
What is meant by mass ? Write its unit.
Answer:
Mass is the quantity of matter contained in it. Its unit is kilogram.

Question 28.
What is meant by weight ? What is its measurement unit ?
Answer:
The force by which earth attracts an object towards itself is called the weight. Its measurement unit is Newton.

Question 29.
What is meant by weight lessness ?
Answer:
The state of body when gravity of planet (earth) stops acting, the person in spaceship is in the state of weightlessness.

Question 30.
What is meant by centre of mass ?
Answer:
The centre of mass is that point at which whole mass of object is directed.

Question 31.
What is meant by centre of gravity ?
Answer:
The centre of gravity is that point of the object at which the whole force of gravity works.

Question 32.
Where is the centre of gravity of objects having regular shape ?
Answer:
The centre of gravity of die objects having regular shape and equal density lies on geometrical centre.

Question 33.
What will happen if the mass and acceleration of two objects become equal ?
Answer:
If the mass and acceleration of two objects become equal, then they will appear to move towards ‘ each other.

Question 34.
What is meant by the force of gravity ?
Answer:
The force between earth and an object is called the force of gravity.

Question 35.
Which balance is used to measure weight ?
Answer:
Spring balance is used to measure weight.

Question 36.
Which balance is used to measure mass ?
Answer:
Physical balance is used to measure mass.

Question 37.
Why is the weight of the object less on the moon in comparison to the earth ?
Answer:
Because the acceleration due to gravity on the ftibon is less than that on the earth.

Question 38.
Two stones of masses 25 kg and 2 kg are dropped at the same time from the top of a building. Which stone will reach first on earth ?
Answer:
The acceleration due to gravity does not depend upon the mass of the object. So both stones will reach on the earth in same time.

Question 39.
There is gravitational force between two objects kept on table. Why they don’t move towards each-other even in the presence of this force ?
Answer:
The gravitational force between two objects is very weak so that they do not move.

Question 40.
The earth attracts an iron pin downward with gravitational force, but the magnet attracts the iron pin upwards against the direction of attraction of earth, why ?
Answer:
Because attractive force applied by magnet on iron pin is more than the gravitational force on pin.

Question 41.
Where the gravitational force will be more-between two objects on earth or between object and earth ?
Answer:
The gravitational force between object and earth is more.

Question 42.
Where is the weight of the object zero ?
Answer:
The weight of the object is zero at the centre of earth.

Question 43.
What is the relation among u, v, g and t ?
Answer:
v = u + gt

Question 44.
What is the sign of ‘g’ according to Cartesian Coordinate system ?
Answer:
‘g’ is always taken as negative.

Question 45.
What is meant by projectile range ?
Answer:
The maximum horizontal distance covered by projectile is called projectile range.

Question 46.
What is the angle between velocity and acceleration at the highest point of projectile ?
Answer:
Right angle (90°).

Question 47.
How far is the sun from the earth ?
Answer:
1.5 x 10¹¹ metres.

Question 48.
What is meant by thrust ?
Answer:
The force acting perpendicular to the surface of object is called thrust.

Question 49.
Give an example of thrust in daily life.
Answer:
When we press a pin on drawing board, the force applied by our thumb acts in perpendicular direction on board. Thus, it is thrust.

Question 50.
What is pressure ? What is its SI unit ?
Answer:
The thrust per unit area is called pressure. That is pressure  Its SI unit is Pascal (Pa).

Question 51.
What is the SI unit of thrust ? Africa
Answer:
Newton (N).

Question 52.
In regard of which scientist, the SI unit of pressure is Pascal (Pa) ?
Answer:
In honour of scientist Blaise Pascal.

Question 53.
What is the bigger unit of pressure ?
Answer:
Kilopascal (k Pa).

Question 54.
Why cutting, tearing and grooving apparatus are always sharp ?
Answer:
The cutting, tearing and grooving apparatus are sharp so that thrust affects minimum area to increase pressure and the object can be cut, tear or groove easily.

Question 55.
When is the pressure of man more -sitting on ground or standing at one place and why ?
Answer:
The pressure of man is more when standing at one place because the area in standing position is less than that in sitting position at one place.

Question 56.
What is the effect of depth on liquid pressure ?
Answer:
The liquid pressure increases with depth.,

Question 57.
Why do the egg floats in salty water ?
Answer:
The density of water increases by adding salt in it, due to that buoyant force of water increases and the egg floats.

Question 58.
In which stage, the object floats ?
Answer:
If the weight of the body is equal to the weight of fluid displaced, then the object floats on the surface of fluid.

Question 59.
Write the name of two devices that work on Archimedes’ principle.
Answer:
(i) Ship and
(ii) submarine.

Question 60.
Which device is used to measure the purity of the milk and what is its principle ?
Answer:
The lactometer is used to measure the purity of the milk which works on Archimedes’ principle.

Question 61.
Write the name and principle of that device which is used to measure the density of liquids.
Answer:
Hydrometer, which works on Archimedes’ principle.

Question 62.
What is the mass of unit volume called ?
Answer:
Density.

Question 63.
What is the SI unit of density ?
Answer:
Kilogram per metre cube (kg m^-3).

Short Answer Type Questions

Question 1.
If a piece of stone is allowed to fall freely from a certain height, then what will be the changes in velocity ?
Answer:
The following are the changes :
(1) The initial velocity of stone is zero and it increases at coming down.
(2) The stone nearest to earth has maximum velocity.
(3) The velocity of stone becomes zero after striking the ground. –

Question 2.
What is gravitation ? In which direction it works ?
Answer:
Everybody of universe attracts other body towards itself. This attraction between two bodies is gravitation. This attraction between two bodies works along the line joining the centres.

Question 3.
Prove that acceleration produced in the object dropped towards the earth does not depends upon its mass.
Answer:
If an object of mass, m is dropped from the centre of earth at a distance of R, then the force exerted by the earth on object,
F = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}\)
Where, M_e = mass of earth, G = gravitation constant
∴ Acceleration produced in the object = \(\frac {F}{m}\) (F = ma; a = \(\frac {F}{m}\))
= \(\frac{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}{\mathrm{~m}}\) = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2} \times \frac{1}{\mathrm{~m}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\)
∴ The acceleration produced in an object does not depend upon the mass of dropped object.

Question 4.
What is Newton’s third law of motion ? Is this law applicable on gravitational force ? Explain.
Answer:
According to Newton’s third law, “To every action, there is equal and opposite reaction.” For example, we act on ball when we throw it on wall and ball comes back by the reaction of wall. Thus, an object A exerts a force on object B, then object B is also exerts equal and opposite force on A. This fact is also true for gravitational force. When a stone falls on earth, it happens due to the force of gravity of the earth on stone. In fact, the piece of stone, attracts the earth towards itself with equal force of gravity. But the mass of stone is less than that of the earth so the force exerted by the stone on the earth is less. Therefore the stone seems to attracted towards earth.

Question 5.
Discuss the motion of the object thrown upward.
Answer:
When a body is thrown upward, the earth attracts that body continuously towards itself. Therefore, the gravitational force works in opposite direction of motion and the body moves slowly. The speed reduces at a rate of 9.8 ms-2 in every second. At last when the body reaches at highest point, the speed becomes zero. The body again starts falling towards earth with an acceleration of 9.8 ms-2.

Question 6.
Find the unit of ‘G’
Answer:
‘G’ is the ratio of force (F) and \(\frac{M_1 M_2}{R^2}\) Therefore, its unit depends upon the units offorce, mass and distance. If the mass is taken in kilogram, distance in metre and time in second, then the uqit offorce is newton (N).
Unit of G = \(\frac{M_1 M_2}{R^2}\) = Nm²/kg²

Question 7.
Why is G called universal constant ?
Answer:
We know that F = G \(\frac{m_1 m_2}{R^2}\)
‘G’ is constant among diem. The value of ‘G’ does not depend upon the Value of mj, m2 or R. Its value does not depend upon that how, when and where the force is measured. The value of‘G’ remains consant for any two bodies in universe. The ratio of F and \(\frac{m_1 m_2}{R^2}\) is equal to G which is equal for any two bodies. That is why, G is called universal constant.

Question 8.
What is the difference between ‘g’ and ‘G’ ?
Answer:
The difference between ‘g’ and ‘G’ is as follows:

Question 9.
Prove that the bodies of different masses when dropped together from the same height, strikes in same time (assume, air resistance is negligible).
S0lution:
Suppose two objects of different masses were dropped together from same height h,
In this state s = h, u = 0
According to formula s = ut + \(\frac {1}{2}\) gt²
h = 0 x t + \(\frac {1}{2}\) gt²
Or h = \(\frac {1}{2}\) gt²
Or 2h = gt²
Or gt² = 2h
t² = \(\frac {2h}{g}\)
Or t = \(\sqrt{\frac{2 h}{g}}\)
It is clear from the equation that the time taken by objects does not depend upon masses. Thus, both objects fall in same time on earth.

Question 10.
What is meant by vertical direction of earth?
Answer:
If we suppose the earth as a ball of uniform density, then its centre of mass will be at its centre. Thus, the force of attraction of the earth on a body works in the direction of its (earth) centre. This is the direction which we call vertical direction.

Question 11.
Which are equations related to the freely falling bodies?
Solution:
(i) v = u + gt
(ii) v² – u² = 2gs
(iii) s = ut + \(\frac {1}{2}\) gt²
Here,
u = initial velocity
y = final velocity
g = acceleration due to gravity
s = distance or height
t = time

Question 12.
What is meant by weight ? Write the formula for weight.
Answer:
The force of attraction applied by earth on an object is called its weight. if earth produces an acceleration ‘g’ on an object of mass, m then, the force of artraction applied by earth on object, F = mg
But, the force of attraction applied by the earth on object (F) = Weight of object (W) = W = mg

Question 13.
Find a formula to compare the weight of a body on earth and on any other heavenly body.
Solution:
Suppose, mass of the object = m
Mass of the earth = M_e
Radius of earth = R_e
The attractive force applied by earth on object, F_e = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}{ }^2}\) ………..(i)
Suppose, mass of hçavenly body = M_m
Radius of heavenly body = R_m
The force of attraction applied by other heavenly body on object
F_m = \(\mathrm{G} \frac{\mathrm{M}_{\mathrm{m}} \times \mathrm{m}}{\mathrm{R}_{\mathrm{m}}^2}\) …………..(ii)

Dividing eq (ii) by eq (1)
\(\frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{F}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}}{\mathrm{M}_{\mathrm{e}}} \times\left(\frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{m}}}\right)^2\)

Question 14.
Find the relation between acceleration due to gravity (g) and universal gravitational constant (G).
Answer:
Suppose an object of mass m is kept at a distance of R from the centre of the earth. If M is the mass of the earth, then the force applied by earth on that object:
F = \(\mathrm{G} \frac{\mathrm{Mm}}{\mathrm{R}^2}\) ………..(i)
The force by which the earth attracts the object towards its centre, is cal led weight. It is also denoted by’ F’.
F = mass (m) x acceleration due to gravity (g)
or F = mg ……….(ii)
From equations (i) and (ii) mg = \(\mathrm{G} \frac{\mathrm{Mm}}{\mathrm{R}^2}\)
Or g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
This is the required relation. The value of’g’ can be found with the help of this.

Question 15.
What is mass ? What is its measurement unit ? WrIte the law to measure mass.
Answer:
Mass: The quantity of matter contained in an object is called its mass. It is a scalar and constant quantity.
The unit of mass is kilogram. It is found by physical balance.

Law to measure mass:
Suppose we have two objects of masses m₁ and m₂, kept at the same distance from the centre of the earth, then the force of attraction between the earth and m₁:
F₁ = R₂
Here, M_e = mass of earth and R = distance of mass m₁ from the centre of the earth.
Like this, the force of attraction between the earth and m₂
F₁ = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_1}{\mathrm{R}^2}\)
Here, M_e = mass of earth and R = distance of mass m₁ from the centre of the earth.
Like this, the force of attraction between the earth and m₂
F₂ = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_2}{\mathrm{R}^2}\)
1f both the forces are same, then F₁ = F₂
\(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_1}{\mathrm{R}^2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_2}{\mathrm{R}^2}\)
∴ m₁ = m₁
Thus, if two forces are equal then the masses of two objects will be equal and parts of physical balance will be equal.

Question 16.
Define projectile and path of projection and give some examples of projection motion.
Answer:
Projectile:
A body thrown in horizontal direction is called projectile.

Path of projection:
The curved path travelled by projectile is called path of projection.

The examples of projection motion are following:
(1) The path of a body dropped from aeroplane, i.e., a ball falling from a building of 80m high is shown in figure.
(2) The path of a bullet fired from gun.
(3) The path of sphere thrown by player.

Question 17.
Why do astronauts feel weightlessness in spaceship?
Answer:
When the astronauts sitting in spaceship move around the earth, the gravitational force between the earth and spaceship is balanced by the force used to move around the earth, due to that the astronauts feel weightlessness. The value of acceleration due to gravity becomes zero in spaceship.

Question 18.
Generally, if the cotton ball and stone of same size are allow ed to fall from a certain height, then which one will strike the surface of earth first? What Galello discussed about this fact?
Answer:
Generally, the piece of stone reaches first on the surface of earth in comparison to cotton. But first of all Galelio told that the air resistance becomes zero, cotton ball and piece of stone will reach in same time on the surface of earth with equal acceleration. According to Galelio. “the objects the acceleration of falling on the acceleration of the earth does not depend upon their masses.”

Question 19.
How does acceleration due to gravity change at any place (i) due to height, (ii) due to the shape of the earth?
Answer:
In the following way the acceleration due to gravity changes at any place:
(i) Due to height: The value of ‘g’ is less on the mountains situated at some height on earth surface, whereas it ¡s more on plain surface.
∴ \(\frac{W_h}{W}=\frac{g_h}{g}=\frac{R^2}{(R+h)^2}\)
Like this, the value of ‘g’ goes on decreasing by increasing height.

(ii) Due to the shape of earth: The earth is not completely round like a ball, but it is convexed at the equator line and concaved at the polar line. There is a difference of 22 km between the radius of the equator and polar line of the earth. That’s why
the value of’g’ is more on poles and less on equators.
∴ g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) (Where, G and M are constant.)
∴ g ∝ \(\frac{1}{R^2}\)

Question 20.
“The moon falls towards the earth.” What does it mean ? Why does it not fall on the surface of the earth?
Answer:
The moon falls towards the earth means that the earth attracts the moon towards itself due to its gravitational force. But, moon moves around the earth in a circular path and does not fall, because the velocity of the object moving on a circular path changes at every point. The change in velocity or acceleration is due to the change in the direction of motion of the object. The centripetal force which produces this acceleration maintains the body to move in circle, which is always directed towards the centre.

Question 21.
The earth attracts the apple, does the apple also attract the earth ? If yes, then why the earth does not move towards the apple?
Answer:
Definitely the apple attracts the earth towards itself, but the mass ofapple is negligible in comparison to earth so the earth does not move towards apple.

Question 22.
What are Kepler’s three laws of motion?
Answer:
Kepler’s is three laws of motion are following:
(i) The orbit of a planet is an ellipse with the sun at one of the foci. As the position of sun is shown by in the figure.
(2) The line joining the planet and the sun sweep equal area in equal intervals of time. Thus, if the time of travel from A to B is same as that from C to D, then the areas OAB and OCD are equal.
(3) The cube of the mean distance of a planet from the sun is proportional to the square of its orbital period, T, or constant. It is important to know that Kepler could not give a theory to explain the motion ofplanets. It was Newton who showed that the cause of the planetory motion is the gravitational force that the sun exerts on them.

Question 23.
Why are the tyres of buses or trucks wider than that of cars?
Answer:
The thurst on buses or trucks is more than that on cars. If their tyres are made less wider than the tyres of cars, the pressure on them will increase and they will sink ¡n earth. Therefore, the tyres of buses or trucks are made wider, so that the area will increase and pressure on earth decrease.

Question 24.
The camel can walk in desert easily, but ¡t is difficult for horse to walk, why?
Answer:
The area of feet of camel is more than that of horse. So the affected area is more and pressure decreases. The camel can walk easily in desert due to less pressure. But the area of feet of horse is less, and the pressure increases. Due to large pressure, it is difficult to walk by horse in desert.

Question 25.
What is buoyant force? [low is it measured?
Answer:
When a body is immersed in a fluid, an upward force works on the body which is called buoyant force. Measurement of Buoyant Force: Take a metal’s sphere. Hang it on spring balance and note its weight. Let its weight is W1 gm. Now put some water in a beaker and take the sphere in the beaker slowly. You will see that the weight of sphere seems decreasing when put in water slowly, which is indicated by spring balance. Note the indicator of spring balance when the solid sphere immersed completely in water.
Let it is W₂ gm.
The difference between two weights will be the buoyant force of water.
Buoyant force = weight of sphere in air – weight of sphere in water.
= (W₁ – W₂)gm

Question 26.
What is the reason that, an iron needle sinks in water, but a ship made up of iron floats in water?
Answer:
The iron needle sinks in water, but a ship made of iron floats in water because weight of water displaced by needle is less than the weight of needle. Due to heavy weight, the needle sinks in water. But the weight of water displaced by ship is more than the weight of ship and it floats in water.

Question 27.
Why is it easy to swim in sea water than river?
Answer:
It is easy to swim in sea water than river because the density of sea water is more than the density of river water. Thus, the sea water due to being heavy, applies larger buoyant force on the body of person, duc to this reason it is easy to swim in sea water.

Question 28.
What is density ? What is its SI unit ? What is its importance?
Answer:
The mass per unit volume of a body is called density. Its SI unit is kilogram per metre cube. The density of the substance always remain same in certain circumstances. The density of gold ¡s 19300 kg m^-3 and the density of water is 1000 kg m^-3.
Importance: The density of the sample of any substance helps in the measurement of its purity.

Question 29.
What is relative density ? What is the condition of sink or float in water on the basis of it?
Answer:
The ratio of density of substance and density of water is called relative density of the body. Thus density of substance

∴Relative density is a ratio, so, it has no unit.
The relative density ofgold is 19.3. If the relative density of any solid or liquid is more than I, it will sink in water, if it is less than 1, it will float in water.

Question 30.
Why the ice floats on water?
Answer:
The ice floats on water because the density of ice is less than the density of water. when a piece of the ice is put in the water, then its weight is less than the weight of water displaced by it.

Question 31.
Define Archimedes’ principle and explain it. What arc uses of this principle? Or Define Archimedes’ Principle. Write Iwo applications ofArchimedes’ Principle.
Answer:
Statement of Archimedes’ principle: When a body is immersed fully or partially ¡n a fluid, it expirences an upward force that is equal to the weight of the fluid displaced by it.

Explanation:
When a solid is immersed in a fluid, the fluid pushes it upwards by its buoyant force, and its wekht decreases, because buoyant force is equal to the weight of water displaced by the solid. So the decrease in weight of solid when immersed ¡n fluid is equal to the weight of the water displaced.

Uses of Archimedes’ principie in daily life:
(I) It is used in designing of ships and submarines.
(2) The lactometer, which is used to determine the purity of milk, is based on this principle. .
(3) The hydrometer which is used for determining density of liquid, is also based on this principle.

Question 32.
What happens to the gravitational force between two objects If:
(i) the distance between the objects ¡s one-third (1/3)?
(ii) the masses of both the objects are halfed (1/2)?
Answer:
(i) As we know that gravitational force between two objects is inversely proportional the square of the distance between them. So, if the distance between them is then the gravitational force between them will be 9 times.
(ii) As the gravitational force between two objects is proportional to the product of masses of two objects. So, if the masses of two objects are halfed (1/2) then the gravitational force between them will be one tourth (1/4).

Numerical Questions

Question 1.
A bullet is fired with a velocity of 25 ms1 in horizontal direction. How much distant it will fall in first two seconds?
Solution:
Vertical distance is independent to horizontal velocity.
∴ Initial velocity (u) = 0
Acceleration due to gravity (g) = 9.8 ms^-2
Time (t) = 2s
Distance (s) = ?
We know that
s = ut + \(\frac {1}{2}\) gt²
= 0 x 2 +\(\frac {1}{2}\) x 9.8 x (2)²
= 19.6 m

Question 2.
Find the weight of an object of mass 5 kg.
Solution:
Mass of the object (m₁) = 35 kg
Acceleration due to gravity (g) = 9.8 ms^-2
Weight of the object (W) = ?
We know that, W = m x g = 5 x 9.8 = 49 N

Question 3.
If the mass of you and your friend is 35 kg, and the distance between you and your friend is 1 m, then what will be the force of attraction between two?
Solution:
Here,
Mass(m₁) = 35kg
Ma.ss(m₂) = 35kg
Distance(r) = 1 m
Gravitational constant (G) = 6.67 x 10^-11 Nm² kg^-2
Force of attraction (F) = ?
We know that, Force of attraction, (F) = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\)
= \(\frac{6.67 \times 10^{-11} \times 35 \times 35}{(1)^2}\)
= 8.17 x 10^-8 N Ans

Question 4.
FInd the weight of an object of mass 10 kg on the surface of that planet whose mass is two times and radius Is three times that of earth.
Solution:
For planet, M = 2M_e and R = 3 R_e
∴ g₁ = \(\frac{\mathrm{G}\left(2 \mathrm{M}_{\mathrm{e}}\right)}{\left(3 \mathrm{R}_{\mathrm{e}}\right)^2}=\frac{2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}^2}\)
= \(\frac {2}{9}\) g
= \(\frac {2}{9}\) x 9.8 = \(\frac {19.6}{9}\)m^-2 [∵ g = 9.8 m^-2]
∴ Weight of the object of 10 kg on the planet (W) = m x g₁
= 10 x \(\frac {19.6}{9}\) = \(\frac {19.6}{9}\) = 21.78 N

Question 5.
The weight of an object is 49 N, what will be its mass?
Solution:
Weight of object (W) = 49 N
Acceleration due to gravity (g) = 9.8 ms^-2
Weight = mass x acceleration due to gravity
W = mg
49 = m x 9.8
m = \(\frac {49}{9.8}\) = 5 kg

Question 6.
A car fall off a ledge and drops to the ground in 0.5g. Let g = 10 ms^-2
(i) What is the speed of car on striking the ground?
(ii) What ¡s its average speed during \(\frac {1}{2}\) s?
(iii) How high is the ledge from the ground?
Solution:
Time (t) = \(\frac {1}{2}\) s
Initial velocity (u) = 0 ms^-1
Acceleration due to gravity (g) = 10 ms^-2
Acceleration of car (a) = + 10 ms^-2 (downward)
(i) Speed (v) = at
v = 10 ms^-2 x \(\frac {1}{2}\) s = 5 ms^-1

(ii) Average speed = \(\frac {u + v}{2}\) = \(\frac{\left(0 \mathrm{~ms}^{-1}+5 \mathrm{~ms}^{-1}\right)}{2}\) = 2.5 ms^-1

(iii) Distance travelled, s = \(\frac {1}{2}\) at²
= \(\frac {1}{2}\) x 10 ms^-2 x (1/2s)²
= \(\frac {1}{2}\) x 10 ms^-2x (1/4s)² = 1.25 m
Thus
(i) its speed on striking the ground = 5 ms^-1
(ii) its average speed during the \(\frac {1}{2}\) s = 2.5 ms^-1
(iii Height of the ledge from the ground = 1.25 m

Question 7.
A stone ¡s dropped from the edge of the roof:
(i) how long time does It take to fall 4.9 m?
(ii) what is the speed of stone at that time?
(iii) how fast does It move at the end of 1.9 m?
(iv) what is the acceleration after Is and after 2s?
Solution:
(i) Initial velocity of stone (u) = 0
The distance covered by the stone (s) = 4.9 m
g = 9.8 ms^-2
Time, t = ?
Now, S = ut + gt²
4.9 = 0 x t +\(\frac {1}{2}\) x 9.8 x t²
Or 4.9 = 0 + 4.9 t²
Or 4.9 t² = 4.9
Or t² = \(\frac {4.9}{4.9}\) = 1s

(ii) Suppose the speed of stone at that time = v
Now, v = u + gt
v = 0 + 9.8 x 1
v = 0 + 9.8 = 9.8 ms^-1

(iii) initial velocity of stone (u) = 0
The distance covered by the stone (s) = 79 m
g = 9.8 mr^-2, v = ?

Now, v² – u² = 2gs
v² – (0)² = 2 x 9.8 x 7.9
v² – 0 = 154.84
v = 154.84
v = \(\sqrt{154.84}\) = 12.44 ms^-1

(iv) Because the body is faLling freely under the action of gravity.
∴ Acceleration of the body after 1 s = 9.8 ms^-2
Acceleration of the body after 2 s = 9.8 ms^-2

Question 8.
A ball is thrown up vertically upto a height of 100 m, find its initiai speed.
Solution:
Here,
Initial speed (u) = ?
Final speed (v) = 0
Distance (s) = 100m
Acceleration due to gravity (g) = -9.8 ms^-2
Weknowthat, v² – u² = 2gs
0² – u² = 2 x (-9.8) x 100
– u² = – 1960
u = \(\sqrt{1960}\) = 44.2 ms^-1 Ans

Question 9.
What will be the height of building Ifa piece of stone takes 4 seconds when dropped freely from the roof of a building?
Solution:
Here, Initial speed (u) = 0
Time(t) = 4s
Height or distance (s) = ?
Acceleration due to gravity (g) = 9.8 ms²
We know that, s = ut + \(\frac {1}{2}\) gt²
= 0 x 4 + \(\frac {1}{2}\) x 9.8 x (4)²
= \(\frac {1}{2}\) x 9.8 x 4 x 4 = 78.4 m

Question 10.
The volume of a 500 g sealed tin (or box) is 350 cm³. What is the density of sealed tin ? Will it float or sink in water? What will be the mass of the water displaced by this tin?
Solution:
Mass of tin (m) = 500 g = \(\frac {500}{1000}\) kg = 0.5 kg
Volume of tin = 350 cm³
= \(\frac{350}{100 \times 100 \times 100}\) m³
= 35 x 10^-5 m³
Density of tin = \(\frac{\text { mass }}{\text { volume }}=\frac{0.5}{35 \times 10^{-5}}\) kg m^-3
= \(\frac{5 \times 10^5}{35 \times 10}=\frac{10^4}{7}\) kg m³
= 1.429 x 10³ kg m^-3
We know that the density of water = 1000 kg m^-3
= 10³ kg m^-3

∴ Relative density of tin density of water = \(\frac{1.429 \times 10^3}{10^3}\) = 1.429
Because the relative density of tin (is more than 1). Therefore, tin will sink in the water. Thus, mass of the water displaced wilt be less than that of tin.

Question 11.
The volume of a piece of iron is 20cm3 and mass ¡s 156 g. What is its density?
Solution:
Mass of the iron = 156 g
Volume of the iron = 20 cm³

= \(\frac {156}{20}\) = 7.8g cm^-3.

Do Yourself

(1) Write down the expression for the acceleration experienced by a particle on the surface of moon due to the gravitational force on moon. Find the ratio of this acceleration to that experienced by the saine particle on the surface of the earth. if the acceleration due to gravity on earth is 98 ms² what is the acceleration of a particle on the surface of moon?
Answer:
[(i) a = \(\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}}\) (ii) 0.16 (iii) 1.57 ms^-2]

(2) A particle is thrown up in vertical direction with a velocity of 50 ms. What will be its velocity at highest point? What is the maximum height attained by this particle? How much time will it take to reach at highest point?
Answer:
[(i) v = 0 (ii) 125 m (iii) 5 s]

(3) What will be the force due to gravity on the surface of that planet whose mass and radius, both are
half than that of earth ?
Answer:
[Take g = 9.8 ms2 1 [19.6 ms^-2]

(4) A force of 20 N is applied on an object of weight 9.8 N. Calculate the mass of the object and the
acceleration produced.
Answer:
[Take g = 9.8 ms^-2] [1 kg; 20 ms^-2]

(5) The mass of the earth is 6 x 10²⁴ kg and that of the moon is 7.3 x 1o²² kg. 1f the distance between the
earth and the moon is 3.84 x 10⁵ km, calculate the force exerted by the earth on the moon.
Answer:
[19.8 x 10¹⁹ N]

(6) Calculate the force of attraction acting on person of mass 35 kg on the surface of earth. The mass of the earth is 6 x 10²⁴ kg and radius is 6.4 x 10⁶m.
Answer:
[342 N]

(7) Find the force of attraction between two objects having mass 50 kg and 10 kg and the distance between them is 1 metre.
Answer:
[3.34 x 10^-8 N]

(8) The mass of earth is 6 x kg and mass of moon is 7.4 x 10²² kg. 1f the distance between the earth and the moon is 3.84 x 10⁵ km, find the force exerted by the earth on the moon
Answer:
[G = 6.7 x 10^-11 Nm² kg^-2]
[22.01 x 10²⁰N]

(9) An object is thrown up ¡n vertical direction and attains a height of 10 m. Calculate
(i) the velocity with which the object is thrown up and
(ii) time taken by the object to reach the highest point.
[u = 14 ms^-1 t = 1 .43s]

(10) A boy dropped a stone from the edge of a rock of 49 m high. After is, he dropped second stone. What was the speed of second stone if both stones reached the eanh in same time ?
[ Take g = 9.8 ms^-2]
[12.08 ms^-1]

(11) By which initial velocity, the ball will be thrown up so that it can go upto a height of 40 m.
Answer:
[28 ms^-1] [10 kg]

(12) Find the mass of that object whose weight is 196 N.
Answer:
[Take -9.8 ms^-2] (10 kg]

(13) An aeroplane is flying at a height of 1000 m in horizontal direction with a velocity of 100 ms^-1. A bomb is dropped from it. Calculate:
(i) the time taken by the bomb to reach earth.
(ii) the velocity before striking the ground.
(iii) the distance of target.
Answer:
[(1) 14.28 s ((ii) 172.4 ms^-1 ((iii) 1428.5 m]

(14) The mass of an object ¡s 10 kg. What will be its weight on the earth?
Answer:
[98 N]

(15) The weight of an object is 10 N on weighing. What will be its weight on moon?
Answer:
[98 N]

(16) A block of wood is kept on a table top. The mass of wooden block is 5 kg and its dimcnsions are 40 cm x 20 cm x 10cm. Find the pressure exerted by the wooden block on the table top ¡fit is made to lie on the table top with its sides of dimensions (a) 20 cm x 10 cm and (b) 40 cm x 20 cm.
Answer:
[(a) 2450 Nm^-2 (b) 612.5 Nm^-2]

(17) The relative density of silver is 10.8 g. The density of water is 1 kgm3. What will be the density
of silver in SI unit?
Answer:
[10.8 x 10³ kg m³]

Essay Type Questions

Question 1.
Find the deflnition of universal gravitational constant ‘G’. Write the unit of ‘G’ In SI system.
Answer:
According to Newton’s Universal Law of Gravitation,

F = \(\mathrm{G} \frac{\mathrm{M}_1 \mathrm{M}_2}{\mathrm{R}^2}\)
Or G = \(\frac{\mathrm{FR}^2}{\mathrm{M}_1 \mathrm{M}_2}\)
If M₂ = M₂ = 1 unit and R = 1 unit, then
G = \(\frac{F \times 1^2}{1 \times 1}=\frac{F \times 1}{1 \times 1}\) = F
∴ Thus, universal gravitational constant is a gravitational force which is placed between the objects of unit mass which are at unit distance from each other.

SI unit of G:
G = \(\frac{\mathrm{FR}^2}{\mathrm{M}_1 \mathrm{M}_2}\)
Because G is the ratio ofF and M1 M2/R2 . Therefore, its unit depends upon the units of force, mass and distance. If mass is taken in kilogram, distance in metres and time in seconds, then unit of force is newton (N). Therefore, unit of G will be Nm^-2 kg^-2. The value of G found by the experiments on golden balls is:
G = 0.000,000,000,066,734 Nm² kg²
= 6.673 x 10^-11 Nm² kg²

Question 2.
State the universal law of gravitation. Derive the formula for gravitational force between two bodies. Aand B of masses M and m, placed at a distanced.\(1 \frac{1}{2}+2 \frac{1}{2}=4\)
Answer:
Universal Law of Gravitation:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the çentres of two objects.

Let two objects A and B of masses M and ‘m’ lies at a distance ‘d’ from each other as shown in figure. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses, that is,
F ∝ M x m …………(i)
And the force between two objects is inversely proportional to the square of the distance between them, that is,
F ∝ \(\frac{M \times m}{d^2}\) ………..(ii)
Combining eqs. (i) and (ii), we get
F ∝ \(\frac{M \times m}{d^2}\) or F = G \(\frac{M \times m}{d^2}\)
Where G is the constant of proportionality and is called universal gravitational constant.

Question 3.
Compare the weights of body if that are weigh on earth and moon.
Answer:
The value of acceleration due to gravity of each planet (g) depends upon its mass and radius. Due to this reason, weight of an object on moon is \(\frac {1}{6}\) th times than on the earth. Suppose the mass of object is ‘m’_e. If the mass of earth is Me and its radius is R_e, then weight of the object on earth (F_e).
F_e = \(\frac{\mathrm{GM}_{\mathrm{e} \mathrm{m}}}{\mathrm{R}_{\mathrm{e}}^2}\) ………..(i)
In same way, the mass of the moon is M_m and its radius is Rm then weight of the object on moon (F_m)
F_m = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^2}\) ………(ii)
Dividing eQuestion (i) by eQuestion (ii) \(\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{m}}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^2} \times \frac{\mathrm{R}_{\mathrm{m}}^2}{\mathrm{GM}_{\mathrm{m}} \times \mathrm{m}}=\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{M}_{\mathrm{m}}}\left(\frac{\mathrm{R}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}}}\right)^2\)
∴ The mass of the earth is 100 times than that of moon, and radius is 4 times more
M_e = 100 M_m and
R_e = 4 R_m

By putting the value of M_e and R_e
\(\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{m}}}=\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{M}_{\mathrm{m}}}\left(\frac{\mathrm{R}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}}}\right)^2=\frac{100 \mathrm{M}_{\mathrm{m}}}{\mathrm{M}_{\mathrm{m}}} \times\left(\frac{\mathrm{R}_{\mathrm{m}}}{4 \mathrm{R}_{\mathrm{m}}}\right)^2\)
\(\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{m}}}\) = 100 x \(\frac {1}{6}\)
= 6 approx.
F_e = 6 F_m
Thus, it is clear that the mass of the object on the moon will be m, but its weight on the earth is 6 times more in compare to moon.

Question 4.
Write the difference between:
(i) Gravitational force and force of gravity
(ii) Centre of mass and centre of gravity.
Answer:
(i) The difference between gravitational force and force of gravity is as follows:

(ii) Centre of mass and centre of gravity.
Suppose a giant body is made of many particles. Then it is possible for us to define that point where the whole of the mass of the body is concentrated. This point is called ‘centre of mass’. On the surface of earth or nearby it, where force of gravity is constant, centre of mass is only that point where the force of gravity on whole body can be experienced. In this state, the centre of mass is called centre of gravity.

Thus ‘centre of gravity’ of a body is that point, where the force of gravity is acting on the whole body as shown in fig. The centre of gravity of objects having regular size and uniform density lies on their geometrical centre. This can be balanced on a needle at the geometrical centre. For example, the centre of gravity of a spherical ball, or rectangular piece of wood or circular disc of metal is at their centre.

Practical Work

Experiment 1.
Describe an activity to understand the motion of moon from earth. Procedure:
(i) Take a piece of thread. Tie a small stone at one end.
(it) Hold the other end of the thread and whirl it round as shown in fig.
(iii) The direction of the motion of stone will be circular motion.
(iv) Now release the thread.
(v) Again, note the direction of motion of the stone.
Before the thread is released, the stone moves in a circular path with a certain speed and changes direction at every point. The change in direction involves change in velocity or acceleration. The force that cause this acceleration and keeps the body moving along the circular path is acting towards the centre. This force is called the centripetal force.

Experiment 2.
Explain an activity to understand the meaning of free fall of objects.

Procedure:
(i) Take a stone.
(ii) Throw it upwards.
(iii) It reaches a certain height and then it starts falling down.
We know that the earth attracts objects towards it. This is due to force of gravity. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

Experiment 3.
Prove with experiment that the gravitational force between two objects of general masses is weak.

Procedure:
Take a magnet. Take it near the iron pin kept near the piece of stone. You will see that the iron pin is not attracted towards piece of stone but towards the magnet. It proves that the gravitational force between stone and pin is very weak. So the iron pin is not attracted towards stone. In fact, the gravitational force between piece of stone and iron pin is very weak as compared to magnetic force. So the iron pin attracted towards magnet but not towards piece of stone. It is concluded that the gravitational force between two objects of general masses is weak.

Experiment 4.
Explain with an experiment that why iron sinks into the water while cork floats ?

Procedure:
(i) Take a beaker filled with water.
(ii) Take an iron nail and a cork of equal mass.
(iii) Place them on the surface of water.
(iv) You will observe that the cork floats and the nail sinks.
This is due to the difference in their densities. The density of a substance is defined as the mass per unit volume. The density of cork is less than the density of water. This means that the up thrust of water on the cork is greater than the weight of the cork. So it floats.

Quick Review of the Chapter

1. What is the force of attraction called between the two particles of the universe ?
(A) force of gravity
(B) gravitational force
(C) centripetal force
(D) magnetic force
Answer:
(B) gravitational force

2. The force exerted by the earth on an object is called :
(A) force of gravity
(B) gravitational force
(C) centripetal force
(D) magnetic force
Answer:
(A) force of gravity

3. The acceleration which maintains the circular motion always directed towards the centre of circle, that is called :
(A) force of gravity
(B) gravitational force
(C) centripetal force
(D) buoyant force
Answer:
(C) centripetal force

4. Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. This law is called :
(A) universal law of gravitation
(B) law of gravity
(C) universal law of centripetal
(D) Archimedes’ principle
Answer:
(A) universal law of gravitation

5. The magnitude of the gravitational force on an object on the surface of earth is :
(A) \(\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\)
(B) \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
(C) \(\mathrm{GM}_{\mathrm{e}} \mathrm{m}\mathrm{R}_{\mathrm{e}^2}\)
(D) \(\frac{\mathrm{M}_{\mathrm{e} m}}{\mathrm{GR}^2}\)
Answer:
(C) \(\mathrm{GM}_{\mathrm{e}} \mathrm{m}\mathrm{R}_{\mathrm{e}^2}\)

6. The value of gravitational constant is :
(A) different at each place on earth
(B) same at all planets and satellites
(C) vary from place to place
(D) all above statements are false
Answer:
(B) same at all planets and satellites

7. On which part of earth the value of acceleration due to gravity (g) is maximum?
(A) on poles
(B) on equator
(C) on centre
(D) on all of them
Answer:
(A) on poles

8. What is the weight of an object on moon in comparison of earth ?
(A) 6 times
(B) 3 times
(C) \(\frac {1}{6}\) times
(D) \(\frac {1}{3}\) times
Answer:
(C) \(\frac {1}{6}\) times

9. The value of acceleration due to gravity (g) is :
(A) same at every place on earth
(B) same at all planets and satellites
(C) vary from place to place
(D) all above statements are false
Answer:
(C) vary from place to place

10. If the weight of a person is 24 N on the earth, then what will be his weight on moon ?
(A) 6N
(B) 8N
(C) 4N
(D) 96 N
Answer:
(C) 4 N

11. The value of acceleration due to gravity (g) is least on which part of earth ?
(A) on poles
(B) on equator
(C) on mountains
(D) on all above
Answer:
(B) on equator

12. The formula of the attractive force (F) between two objects of masses ntj and m2 respectively kept at a distance of R is :
(A) \(\frac{\mathrm{Gm}_1}{\mathrm{~m}_2 \mathrm{R}^2}\)
(B) \(\frac{\mathrm{Gm}_2}{\mathrm{~m}_1 \mathrm{R}^2}\)
(C) \(\frac{\mathrm{R}^2}{\mathrm{Gm}_1 \mathrm{~m}_2}\)
(D) \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{R}^2}\)
Answer:
(D) \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{R}^2}\)

13. The value of acceleration due to gravity (g) on the surface of earth is :
(A) 9.8 ms^-2
(B) 4.9 ms^-2
(C) 19.6 ms^-2
(D) 29.4 ms^-2
Answer:
(A) 9.8 ms^-2

14. The value of acceleration due to gravity (g) is zero on :
(A) poles
(B) equator
(C) centre
(D) everywhere
Answer:
(C) centre

15. The value of acceleration due to gravity (g) at a distance of 12800 km from the centre of earth is:
(A) 1/4 th that of earth
(B) half of the acceleration due to
(C) 4 times that of earth
(D) gravity of earth double that of earth
Answer:
( A) 1/4 th that of earth

16. The distance of the moon from the earth is :
(A) 384000 m
(B) 384000 km
(C) 38400 km
(D) 38400 m
Answer:
(B) 384000 km

17. The quantity of matter present in the body is called :
(A) weight
(B) mass
(C) pressure
(D) thrust
Answer:
(B) mass

18. The weight of an object is zero at:
(A) poles
(B) equator
(C) mountains
(D) centre of earth
Answer:
(D) centre of earth

19. What is the relation among u, y, g and t?
(A) v = u + gt
(B) v + u = gt
(C) v + gt = u
(D) \(\frac {v}{g}\) = u + t
Answer:
(A) v = u + gt

20. The distance of sun from earth is approximately:
(A) 1.5 x 10¹¹ m
(B) 1.5 x 10¹¹ km
(C) 1.5 x 10¹¹ cm
(D) 1.5 x 10¹¹ mm
Answer:
(A) 1.5 x 10¹¹ m

21. By which symbol pascal is denoted?
(A) Pa
(B) Ps
(C) Pc
(D) P
Answer:
(A) Pa

22. On what angle a player throws ball with horizontal line so the ball falls at maximum distance?
(A) 900
(B) 0°
(C) 45°
(D) 600
Answer:
(C) 45°

23. The free fall means:
(A) the objects fàlling towards earth
(B) the objects going upward from the earth
(C) the objects moving parallel to earth
(D) none of these
Answer:
(A) the objects falling towards earth

24. SI unit of pressure is:
(A) Newton
(B) kilogram
(C) metre
(D) Pascal
Answer:
(D) Pascal

25. What will be the produced pressure 11300 n force is applied on the area of 20m2?
(A) 6000 Pa
(B) 3OPa
(C) 15 Pa
(D) 7.5 Pa
Answer:
(C) 15 Pa

26. It is difficult to hold a school bag having a strap made of a thin and strong string, because thrust affects:
(A) large area
(B) more area
(C) small area
(D) longer area
Answer:
(C) small area

27. When an object is immersed in a liquid, the liquid exerted an upward force, which is called:
(A) buoyant force
(B) gravitational force
(C) frictional force
(D) opposing force
Answer:
(A) buoyant force

Haryana State Board HBSE 9th Class Science Important Questions Chapter 9 Force and Laws of Motion Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 9 Force and Laws of Motion

Very Short-Answer Type Questions

Question 1.
Define force.
Answer:
Force is that external factor that changes or tends to change the position of rest or speed and direction of moving object.

Question 2.
What is balanced force?
Answer:
When two or more forces acting on a body have zero resultant, then such forces are called balanced force.

Question 3.
What is unbalanced force ?
Answer:
When the resultant of two or more forces is not zero, such forces are called unbalanced force.

Question 4.
Which force works when a rubber ball is pressed between palms ?
Answer:
It is an example of a balanced force.

Question 5.
Which scientist has established the laws to control the motion of objects firstly?
Answer:
Sir Isaac Newton.

Question 6.
What is the angle of inclination?
Answer:
The angle made by an inclined plane with horizontal is called the angle of inclination.

Question 7.
Define frictional force.
Answer:
It is the opposing force that comes between the two surfaces in contact when a body moves or tends to move on another body.

Question 8.
Write the effect of frictional force.
Answer:
Frictional force reduces the motion of body.

Question 9.
Why the bicycle stops when we stop pedaling?
Answer:
The frictional force between the tyre of the bicycle and road reduces its motion and stops the bicycle.

Question 10.
Why powder is spread on the carom board in carom game?
Answer:
The powder is spread on the carom board to reduce the frictional force between carom board and coin.

Question 11.
What is the law of conservation of momentum?
Answer:
According to law of conservation of momentum, in absence of any external force, the total momentum of a system remains conserved.

Question 12.
Write Newton’s second law of motion?
Answer:
The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. ‘

Question 13.
What is meant by momentum?
Answer:
The momentum of an object is the product of mass and velocity. It is denoted by p.
or p = mv

Question 14.
What is SI unit of momentum?
Answer:
Kilogram metre per second (kg ms^-1).

Question 15.
Momentum is a scalar quantity or vector.
Answer:
Momentum is a vector quantity, because it has both magnitude and direction.

Question 16.
In which direction, the momentum works ?
Answer:
Momentum works in the direction of velocity.

Question 17.
What is the relation among force, mass and acceleration ?
Answer:
Force = Mass x Acceleration (F = ma)

Question 18.
What is the SI unit of force ?
Answer:
The SI unit of force is Newton.

Question 19.
Define one Newton force.
Answer:
One Newton force is that force which produces an acceleration of 1 ms^-2 in a body of mass 1 kg.

Question 20.
What is negligible action ?
Answer:
When a body affects the other by applying force, then we say that there is negligible action between first and second body.

Question 21.
How many types of inertia is ?
Answer:
There is two types of inertia:
(i) Inertia of rest
(ii) Inertia of motion.

Question 22.
Write the concept of force on the basis of momentum.
F = \(\frac{p_2-p_l}{t_2-t_l}\)

Question 23.
How much force will be required to produce an acceleration of 2mS“2 in 6 kg object ?
Answer:
Force (F) = Mass (m) x Acceleration (a) = 6 x 2 = 12 N

Question 24.
A person pushes a box of 40 kg with a force of 80 N. What will be the acceleration of box ?
Answer:
Acceleration (a) = = 2 ms^-2

Question 25.
Write Newton’s third law of motion.
Answer:
According to Newton’s third law of motion, “The action and reaction are always equal and in opposite direction.”

Question 26.
What is inertia ?
Answer:
That property of objects which opposes the change in its motion or state of rest is called inertia.

Question 27.
Is the existence of a single force is possible ?
Answer:
No, the existence of a single force is not possible. The force is always in pairs.

Question 28.
Which force offers us the motion in a desired direction ?
Answer:
The force which offers us the motion in a desired direction is the reaction of the earth opposite to the force applied by our body.

Question 29.
Are the action and reaction imposed on an object ?
Answer:
No, action and reaction (force) are imposed on two different objects.

Question 30.
Write the name of two forces which arise without the contact of two objects.
Answer:
(i) magnetic force
(ii) force of gravity.

Question 31.
How the fruit from tree falls towards the earth ?
Answer:
Due to force of gravity.

Question 32.
Why is it difficult to climb on a smooth pole ?
Answer:
Due to less reaction.

Question 33.
Why oil is used in machines ?
Answer:
To reduce the friction.

Question 34.
Why is it difficult to walk on sand ?
Answer:
Due to more friction between feet and sand.

Question 35.
Compare the frictional force between the surfaces of solids and liquids.
Answer:
The frictional force between the surfaces of liquids is much less than the frictional force between the surfaces of solids.

Question 36.
A small piece of stone and a truck are moving with same motion. On which, the larger force will be imposed to stop both of them ?
Answer:
The larger force will be required to stop the truck, because it will have large momentum.

Question 37.
By which force, we can walk on road without slipping.
Answer:
Due to frictional force, we can walk on road without slipping.

Question 38.
Why it becomes difficult to walk on a green mossy road in rainy season ?
Answer:
The frictional force becomes less on mossy road so it becomes difficult to walk on such a road.

Question 39.
Which Newton’s law of motion helps us in the measurement of force ?
Answer:
Newton’s second law of motion helps us in the measurement of force.

Question 40.
According to which Newton’s Law of motion, jet aeroplanes fly ?
Answer:
They fly according to Newton’s third law of motion.

Question 41.
Why the player runs before jump ?
Answer:
To increase the momentum so that the jump can be high.

Question 42.
Force is a scalar quantity or Vector.
Answer:
Force is a vector quantity.

Question 43.
What is the total momentum of rifle and bullet after firing from rifle ?
Answer:
Zero.

Question 44.
How much acceleration will be produced if 1 newton (N) force is applied on an object of 1 kg mass ?
Answer:
1 ms^-2

Short-Answer Type Questions

Question 1.
What is meant by force ? What can be the effect on object by applying force on that ?
Answer:
Force is that external factor which changes or tends to change the position of rest or speed and direction of moving object.
Effects:
Following are the effects on body, when force is applied :
(i) Speed of object can be changed due to which the object moves fast or slow.
(ii) The direction of motion of object can be changed and can move body in rest.
(iii) Shape of the object can be changed.

Question 2.
Give five examples of force in daily life.
Answer:
Examples of force in daily life are as follows :
(i) When a force is applied on a rubber ball, it elongated. It is due to force.
(ii)When a spring is stretched, its length increases due to muscular force.
(iii) The steam gives its energy to steam engine due to that the steam engine moves forward.
(iv) A rolling ball can be stopped by muscular force.
(v) Petrol produces mechanical force by its energy, due to that bus, car etc. moves.

Question 3.
What is inclinded plane ? What are its uses ?
Answer:
A smooth plane which makes an angle with horizontal plane is called inclinded plane, e.g. knife, axe, screw, hilly roads, stairs etc. A man lifting a heavy weight with the help of inclinded plane made of wooden piece is shown in below fi.

Uses of inclinded plane:
The uses of inclinded plane are as follows:
(i) A heavy weight can be lifted by small force by using inclinded plane.
(ii) Heavy weight is loaded or unloaded in trucks by using it.
(iii) Screw also works on the principle of inclinded plane.

Question 4.
Distinguish between balanced and unbalanced forces.
Answer:
The difference between balanced and unbalanced forces are as follows :
Balanced Force Unbalanced Force
1. When balanced forces act together on a body, their resultant is zero.
2. The balanced force can’t move a body which is in rest position.
3. This force cannot change the speed and direction of moving body.
4. This force always changes the size of body.

1. When unbalanced forces act together on a body, their resultant is not zero.
2. The unbalanced force can move a body which is in rest position.
3. This force can change the speed and direction of a moving body.
4. This force does not always change the size of body.

Question 5.
What is meant by inertia ? Clarify by example.
Answer:
The property of bodies which opposes the change in motion of body or state of rest is called inertia.

Experiment:
Set a coin on a stiff playing card covering an empty glass tumbler. Give the card a sharp horizontal flick with a finger. You will see the card moves ahead but the coin will fall in glass. When the card is shoot fast, then due to property of inertia, the coin tries to remain in rest position. In this way, when the card moves, the coin falls in glass.

Question 6.
What are the types of inertia ? Explain with example.
Answer:
Inertia is of two types:
1. Intertia of rest: The property of body by which the body resists its rest position until an external force is not applied on it.
Example: A book on table does not move due to inertia of rest.

2. Inertia of motion: The property of body by which it resists its state of motion, until an external force is not applied on it.
Example: A passenger sitting in a bus or cart falls forward when the moving bus or cart stops suddenly. In such situation, the lower portion of body of passenger comes in rest as the bus or cart stops, but the upper portion tries to remain in motion, due to this passenger falls forward. –

Question 7.
What happens when a wet cloth is jerked ? Clarify your explanation.
Answer:
When we jerk the wet cloth, the cloth displaced from its position to inertia and they detached from cloth and escape in environment. Thus, the cloth dries fast.

Question 8.
In javelin throw, if a player crosses the fixed line, he fouls, but the playes usually fail to stop on this line, why ?
Answer:
When in javelin throw competition, a player runs and reaches the fixed line, his whole body comes in motion. When he tries to stop his feet on fixed line, the upper portion remains in motion due to inertia of motion and the player crosses the fixed line and fouls.

Question 9.
Why the rider falls backward when the horse starts running suddenly?
Answer:
The rider falls backward when the horse starts running suddenly because due to suddenly running the horse, it comes in motion but the rider tries to remain in rest due to inertia of rest and falls backward.

Question 10.
Why the rider falls forward when a running horse stops suddenly?
Answer:
The rider falls forward when a running horse stops suddenly because the horse and rider are in motion when it is running and when horse
stops suddenly it comes in rest but the rider tries to remain in motion due to inertia of motion and falls forward.

Question 11.
Why are the tyres of vehicles made rough and treaded ?
Answer:
The movement of vehicles is due to the frictional force between road and tyres. If their tyres are smooth then the friction will be less betw een road and tyre, so that it will be difficult to control the vehicle. In order to control vehicles, the tyres are made rough and treaded so that the friction between tyre and road increases and grip of tyres with road will be firm.

Question 12.
Define and explain Newton’s first law of motion.
Answer:
Newton’s first law of motion:
An object continuous to be a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force. This law is also known as “Galileo’s Law of Inertia.” e.g. a ball rolling on floor can move far due to less friction. If the friction reduces to zero by any mean, the ball will move with initial velocity for a certain time.

Question 13.
Prove that Newton’s first law of motion can be called the law of inertia.
Answer:
According to Newton’s first law of motion, an object continuous to be a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force. In other word, all objects oppose the change in their state of motion. This tendency of the objects to oppose the change in their state of motion is called inertia. Thus, Newton’s first law of motion is also called the law of inertia.

Question 14.
Prove that the mass of a body is the measure of its inertia.
Answer:
The quantity of matter in a body is its mass. In this way, the mass is the measure of inertia of body. A body having large mass has large inertia.

Explanation:
If we kick a football it moves so far. but if we kick a stone, it will not move and our foot gets wound, because in comparison of football, stone has larger capacity to oppose the change in motion due to its larger mass. The mass of a body is the measure of its inertia.Thus, the inertia of a stone is more.

Question 15.
Define and explain Newton’s second law of motion with the help of this law, obtain equation of force.
Answer:
Newton’s second law of motion : The force applied on a body is proportional to the product of its mass and acceleration. If we kick a ball, it will move quickly because its acceleration becomes more. In this way, a cricketer stops a moving ball slowly, so that the change in motion of ball takes much time. He moves his hands in the direction of ball and the acceleration of ball becomes less in comparison of initial and less force is required to stop the ball. Let us consider a body has mass m and a force F produces an acceleration a in it. According to Newton’s second law of motion,
F ∝ m x a
or F = K ma ………..(i)
here, K , is a constant, whose value depends upon the units of mass, force and acceleration.
Let m = 1 kg, a = 1 m s^-2 and f = 1 N
then, putting the value of F, m and a in eQuestion (i)
1 = K x 1 x 1
or K = 1
F = 1 x ma
Force = Mass x Acceleration

Question 16.
Define and explain Newton’s third law of motion.
Answer:
Newton’s third law of motion:
To every action, there is an equal and opposite reaction and they act on two different bodies. Action and reaction are equal and in opposite direction.

For example:
(i) A ball presses the earth with a force, the earth pushes the ball upwards with same force.
(ii) When a person swims, he pushes the water backward, he moves forward, by this reaction.

Question 17.
Explain why does cricket player pulls his hands backward while catching a fast moving cricket ball ?
Answer:
The momentum of moving ball is more. When player catches the ball, he pulls his hands backward so that momentum of ball takes some time to become zero. In this way, momentum of ball, rate of change and the force applied on hand will become less and there is no hurt on hands.

Question 18.
Why larger force is required to stop a ship than car moving on road ?
Answer:
The frictional force between the tyres of car and road is more. Therefore the car stops quickly by applying lesser force. The frictional force between the surfaces of liquids is much less than the frictional force between the surfaces of solids. Due to this reason, the ship experiences less friction in water. Thus, larger force is required to stop a ship than car.

Question 19.
When there is negligible reaction between two bodies how the direction changes when two moving objects strike? Clarify with figure.
Answer:
When there is negligible reaction between two bodies, one body affects the other by force. The collision between two marbles is an example of negligible reaction. In figure two marbles are shown moving towards each-other. The direction of their motion is shown by arrow. It is clear from figure that the direction of marble changes after collision. Two forces act in this action. One is that which has accelerated one marble and second is that which has accelerated the other marble.

Question 20.
What is meant by action and reaction ? What is the relation between them ? In which law of motion, these are described ?
Answer:
If a body applies some force on other, then it is the active force on other body. In response, the force which second body applied on first is the reactive force on first body. The relation between action and reaction : The force of action and reaction are equal and in opposite directions. An important fact is that the force of action and reaction always act on different bodies. These are described in third law of motion.

Question 21.
Give some examples of Newton’s third law of motion in daily life.
Answer:
Following are the examples of Newton’s third law of motion :
(i) The bullet moves forward when trigger is pressed and due to reaction force, the gun recoils backward.
(ii) The person pushes the water backward while swimming and moves forward due to tne reaction or
water.
(iii) When a person jumps from boat at river-shore, the boat displaced far from shore due to reaction of water on that.
(iv) We press the earth backward while walking and
due to force of reaction the earth pushes our feet forward and we move. .
(v) The ores push the water backward and due to reaction, the boat move forward.
(vi) When a player jumps up, he pushes the surface of earth downward, so that he gets reaction.

Question 22.
Why the boat moves in opposite direction when a person jumps to shore from the boat ? Explain.
Answer:
A person applies force in upward to get the necessary motion at the time of jumping from boat. His reaction is on boat, which according to Newton’s third law of motion is in opposite direction. As a result of this, the boat move far on opposite direction.

Question 23.
Explain in brief, the relation between Newton’s first and second law of motion.
Answer:
The Newton’s first law of motion is the part of second law. It can be clarify on the basis of following fact:
According to second law of motion,
F = m a
if F = 0, then
m a = 0 m = 0,
a = 0
But mass, m cannot be zero, because it is a constant quantity. So, if acceleration a is zero, the change in velocity will also be zero. Therefore, there is no change in the position of body. If it is in rest, it will remain in rest, and if it is in motion, it will remain in motion unless external force is not applied. This is the statement of Newton’s first law of motion. Thus Netwon’s first law of motion is the part of second law.

Question 24.
What is the reason that a fast moving bullet makes a round hole in glass but on throwing the stone, the glass damages ?
Answer:
The size of bullet is small and that of stone is big so that the glass does not move when bullet is striked, only that part move where the bullet is striked and makes a hole, but there is no hole when stone is striked because whole glass moves and damaged.

Numerical Questions

Question 1.
Speed-time graph of a moving coin on table is shown in fig. . The weight of coin is 10 g.
How much force the table exerts to stop the coin?
Solution:
Here,
Initial velocity (u) = 24 ms^-1
Final velocity (v) = 0 ms^-1
Time (r) = 8 sec
Mass (m) = 10g = \(\frac {10}{1000}\) kg

We know that,
Acceleration (a) = \(\frac{v-u}{t}=\frac{0-24}{8}\) = \(\frac {-24}{8}\) = -3 ms^-2
∴ Force(F) = m.a
\(\left(\frac{10}{1000} \times-3\right)\) N
= – 0.03 N
(- ve sign shows that force decreases the motion.)

Question 2.
A cricket ball of mass 70g is moving with a velocity of 0.5 ms^-1. A cricket player stops it in 0.5 s. Calculate the force applied by the player to stop
the ball.
Solution:
Mass of the ball (m) 70g =\(\frac {70}{1000}\)= 0.07 kg
Initial velocity of ball (u) = 0.5 ms^-1
Final velocity of ball (v) = 0
The time taken to stop the ball, (t) = 0.5s
We know that, v = u + at
a = \(\frac{v-u}{t}=\frac{0-0.5}{0.5}=\frac{-0.5}{0.5}\) = -1 ms^-2
Thus, the force applied by the player to stop the ball
(F) = ma
= 0.07 x -1N = -0.07 N
(-ve sign shows that force is working against momentum.)

Question 3.
How much acceleration will be produced in the object ¡fa force of 200 N is applied on an object of 5 kg mass?
Solution: Here,
Mass(m) = 5kg
Force (F) = 200N
Acceleration (a) = ?
We know that, F = ma
a = \(\frac{F}{m}=\frac{200}{5}\) = 40 ms^-2

Question 4.
A constant force is working on an object of 5 kg mass for 2 seconds. It increases the velocity of object from 3ms1 to 7ms^-1. Calculate the value of applied force. Now, what will be the final velocity if the force works for 5 seconds on object?
Solution:
Here,
Mass of object (m) = 5 kg
Initial velocity of object (u) = 3 ms^-1
Final velocity of object (v) = 7 ms^-1
Time (t) = 2s
Acceleration (a) = ?
We know that.
a = \(\frac{v-u}{t}=\frac{7-3}{2}\)ms^-2 = \(\frac {4}{2}\) = 2 ms^-2
The force applied on object (F) = ma
= 5 x 2 = 10 N

In second stage,
Initial velocity of object (u) = 3 ms^-1
Final velocity of object (v) = ?
Time (t) = 5s
Acceleration (a) = 2 ms^-2
We know that,
v = u + a = 3 + 2 x 5 = 13 ms^-1

Question 5.
How much force is required to stop a car of mass 1000 kg and a loaded truck of mass 10,000 kg moving both with a velocity of 5ms in 2 seconds?
Solution:
Here.
Initial velocity of both (u) = 5ms
Final velocity of both (v) = O
time (t) = 2 s
Acceleration (a) = ?
We know that, v = u + at
a = \(\frac{v-u}{t}=\frac{0-5}{2}=\frac{-5}{2}\) = ms^-1

Mass of the car (m₁) = 1000 kg
Mass of the truck (m₂) = 10,000 kg
The necessary force to stop the car (F₁) = m₁ a
= 1000 x \(\frac {-5}{2}\) N
= – 2500 N
Thus, necessary force to stop the car = – 2500 N
(-ve sign shows that force is working against momentum.)
In same way, the necessary force to stop the truck (F₂) = m₂ a
= 10000 x \(\frac {-5}{2}\)N
= – 25000 N
Therefore, the necessary force required to stop the truck is more than that of car.

Question 6.
Which will require more force – to accelerate a body of mass 2 kg with a rate of 5 ms^-2 or to accelerate a both of mass 4kg with a rate of 2 m^-2?
Solution:
Here,
m₁ = 2kg
a₁ = 5 ms^-2
Force (F₁) = m₁a₁ = 2 x 5 = 10 N
m₂ = 4 kg
a₂ = 2 ms^-2
Force(F₂) = m₂a₂ = 4 x 2 = 8N
In this way F₁> F₂
∴ Large force will required to accelerate the object of 2 kg mass with a velocity of 5 ms^-2

Question 7.
A bullet of mass 20 g is horizontally fired with a velocity of 150 ms from a pistol of mass 2 kg. What is the recoil velocity of the pistol?
Solution:
Mass of bullet, m₁ = 20 g ( = 0.02 kg)
Mass of pistol. m₂ = 2 kg
The initial velocity of bullet (u₁) and pistol (u₂) is zero, respeciively.
That is u₁ = u₂ = 0
Final velocity of bullet, v₁ = + 150 ms^-1
The direction of bullet is taken + ve according to left to right system. Suppose, recoil velocity of pistol is v.
Total momentum of bullet and pistol before firing = (2 + 0.02) kg x o ms^-1
= 0 kg ms^-1
Total momentum after firing = 0.02 kg x (+ 150 ms^-1) + 2 kg x v ms^-1
= (3 + 2v)kg ms^-1
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing
3 + 2v = 0
2v = -3
v = \(\frac {-3}{2}\)
v = -1.5 ms^-1
(-ve sign shows that the pistol recoils opposite to the direction of bullet i.e. right lo left.)

Question 8.
A coin is moving with a velocity of 24 cm s rolls with an acceleration of 3 cm² and stops in 8 s. Find the distance covered by the coin.
Solution:
Here,
Initial velocity (u) = 24 cm s
Final velocity (v) = 0
Acceleration (a) = – 3 cm s²
Time (t) = 8s
Distance (s) = ?
We know that, Distance (s) = ut + \(\frac {1}{2}\) at
= 24 x 8 + \(\frac {1}{2}\) x (-3) x (8)²
= 192 – 96
= 96 cm

Question 9.
In which the more force will require-to produce an acceleration of 5 ms in mass 10 g or to produce an acceleration of 2 ms2 in mass 20 g?
Solution:
In first stage, Mass (m) = 10 g = \(\frac {10}{1000}\) kg
Acceleration (a) = 5 ms^-2
The necessary force (F) = m x a
= \(\frac {10}{1000}\) x 5
= 0.05 N
In second stage, Mass (m) = 20 g = \(\frac {20}{1000}\) kg
Acceleration (a) = 2 ms^-2
The necessary force (F) = m x a = \(\frac {20}{1000}\) x 2 = 0.04N
Thus, larger force will be required to produce an acceleration of 5 ms2 in an object of mass 10 g. Answer:

Question 10.
A bullet of mass 20 g moving with a velocity of 16 ms^-1 enters ¡n a sand bag and comes in rest after 0.05 s. Find,
(a) depth of penetration ¡n sand
(b) the average resistive force applied by sand.
Solution:
Here,
(a) Initial velocity (u) = 16 ms^-1
Final velocity (v) = 0
Time (t) = 0.05 s
Distance (s) = ?
Average resistive force (F) = ?

We know that,
Acceleration (a) = \(\frac{v-u}{t}=\frac{0-16}{0.05}\) = 0.05
= \(\frac{v-u}{t}=\frac{0-16}{0.05}\)
= \(\frac{-16 \times 100}{5}\) = – 320 ms^-2
s = ut + \(\frac {1}{2}\) at²
= 16 x 0.05 + \(\frac {1}{2}\) x (-320) x (0.05)²
or s = 0.8 – 0.4 = 0.4m
(b) Mass (m) = 20 g = \(\frac {20}{1000}\) kg
Average resistive force (F) = m x a = \(\frac {20}{1000}\) x -320 = -6.4 N

Question 11.
How much will be the change in momentum of an object of mass 5kg, If Its speed:
(i) reduces from 20 ms^-1 to 0.20 ms^-1
(ii) Increases from 30 ms¹ to 40 ms^-1.
Solution:
Here,
Mass of the body (m) = 5kg
Initial velocity (u) = 20 ms¹
Final velocity (v) = 0.20 ms¹
Initial momentum (p₁) = m x u = 5 x 20 = 100 kg ms^-1
Final momentum (P₂) = m x v = 5 x 0.20 = 1.00 kg ms^-1
Thus, decrease in momentum = p₁ – p₂ = 100 – 1 = 99 kg ms^-1
Initial velocity (u) = 30 ms¹
Final velocity (v) = 40 ms¹
initial momentum (p₁) = m x v = 5 x 30 = 150 kg ms^-1
Final momentum (p₂) = m x v = 5 x 40 = 200 kg ms^-1
Thus, increase in momentum = 200 – 150 = 50 kg ms^-1

Question 12.
A driver accelerates a car In first stage with 1.8 ms^-2 and in second stage with 1.2 ms^-2 acceleration. Find the ratio of forces applied by engine in both stages.
Solution:
Let,
The mass of car = m kg
Acceleration in first stage = (a₁) = 1.8 ms^-2
∴Force in first stage (F₁) = m x a₁ = m x 1.8m = 1.8 m N
Acceleration in second stage (a₂) = 1.2 ms^-2
∴Force in first stage (F₂) = m x a₂ = m x 1.2 = 1.2m N
\(\frac{F_1}{F_2}=\frac{1.8 \mathrm{~m}}{1.2 \mathrm{~m}}=\frac{18}{12}=\frac{3}{2}\)
F₁:F₂ = 3:2

Question 13.
On applying a force for 1.2 s on an object, its speed increases from 1.8 ms^-1 to 4.2 ms^-1. The same force is applied for 2 s again. How much change in speed will be in 2s?
Solution:
Initial velocity in first stage (u) = 1.8 ms^-1
Final velocity (v) = 4.2 ms^-1
Time (t) = 1.2 s
According to formula, v = u + at
or u + at = v
1.8 + a x 1.2 = 4.2
or 1.2a = 4.2 – 1.8
or 1.2a = 2.4
or a = \(\frac {2.4}{1.2}\) = 2 ms^-1
∴ Acceleration = 2 ms^-2
initial velocity in second stage (u) = 4.2 ms
Time (t) = 2s
Acceleration (a) = 2 ms2
Final velocity (v) = u + at = 4.2 + 2 x 2 = v = 4.2 + 4 = v = 8.2 ms^-1
Thus, change in speed in 2s = 8.2 – 4.2 = 4 ms^-1

Do Yourself

(1) How much force will require to produce an acceleration of 2ms2 in an object of mass 12 kg? What
will be the acceleration, if force is doubled?
Answer:
24N, 4 ms^-2

(2)Aperson pushes a box of mass 50kg by a force of 80 N. What will be the acceleration of box ? What will be the acceleration of box if its mass is reduced to half?
Answer:
1.6 ms^-2, 3.2 ms^-2

(3) A motorcycle is moving with a velocity of 90 kmh1 and on applying brakes, stops in 5s. Calculate the force applied by brake on motorcycle, if its mass including driver is 200 kg.
Answer:
1000 N

(4) A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 ms. ¡f the bullet takes 0.003 second to move through the barrel, calculate the force experienced by the rifle due to its recoil.
Answer:
1000 N

(5) Calculate that force which offers a velocity of 30 ms in lOs to a car, if the mass of car is 1500 kg.
Answer:
4500 N

(6) A car starts from rest and moves for 2 minutes with an uniform acceleration of 0.1 ms2. Calculate the distance covered by the car.
Answer:
720 m

(7) Calculate that force which will produce an acceleration of 0.40 ms^-2 in a toy car of mass 0.25 kg.
Answer:
0.1N

(8) An object is dropped froma height of 20 metres. 1f g = 10 ms^-2 then, calculate the velocity when it strikes to earth. Find the time taken to fall from that height.
Answer:
20 ms^-1, 2s

(9) The velocity-time graph of a ball of mass 20 g moving in straight line on a long table is giyen in figure. How much force will be table offer to ball to take it in rest?

Answer:
0.0004 N

(10) A bullet of mass 10 g leaves the barrel with a velocity of 300 ms1 in 0.0035 s. How much force was exerted by the rifle on bullet?
Answer:
857.14 NJ

(11) A car is moving with a speed of 108 kmh1 and takes 4s to stop after applying brakes. Calculate the force on car after applying brakes. Total mass of car including passengers, is 1000 kg.
Answer:
750 N

(12) A force of SN produces an acceleration of 10 ms^-2 in mass m1 and 20 ms^-2 in mass ms^-2. What will be the acceleration produced by this force, if both the masses are combined?
Answer:
6.67 ms^-2

(13)A resultant force of 15 N is applied on an object kept on a frictionless horizontal plane produces an acceleration of 5 ms^-2. Find the mass of object.
Answer:
3kg

(14) On applying a force of 5 N on mass in1, an acceleration of 8 ms2 is produced and by applying same force on mass ms^-2, produces an acceleration of 24 ms^-2 . How much acceleration will be produced by in combined mass by this force. 1f both masses becomes one by tieing?
Answer:
6 ms^-2

(15) Two hockey players of opposite teams, while trying to hit a hockey ball in the ground collide and immediately get entangled. One has a mass of 60 kg and is moving with a velocity of 5 ms^-2 while the other has a mass of 55 kg and is moving faster with a velocity of 6 ms towards the first player. In which direction and with what velocity will they move after they get entangled ? Assume the frictional force between the feet of the two players and the ground is negligible.
Answer:
Right to left and with velocity of 0.26 ms

Essay Type Questions

Question 1.
How Galileo proved that the object will move with a ceaseless speed if a balanced force is not applied on that object?
Answer:
On the basis of experiments written ahead, Galileo first of all told that the object will move with a ceaseless speed if a balanced force is not applied on that object. Galileo studied the motion of objects on a double inclined plane. He observed that velocity of a body increases when it rolls down the inclined plane and decreases when it climbs upon as shown in fig.

Thus, he concluded that if the body moves on smooth plane then its speed neither increases nor decreases but its speed will remain ceaseless.
Now, he imagine that if a ball is allowed to go down by touching a hollow semi-spherical surface, it will reach the same height to which it was dropped. He concluded on the basis of motion of bob pendulum because the bob of pendulum attains that height from where it started moving. Semi-spherical surface works the same which thread does in pendulum.

After that he told that the shape of pot is taken as showm in fig., then the ball or object reaches the same height in this situation also. Yet, it has to cover large distance to do so.

Now, if the slope on the other side of pot is decreased, the ball covers the more distance to reach that height. In this way he concluded that if the slope is decreasing to one side of pot, the ball will cover the large distance. If there is no slope the surface becomes plane the ball will cover infinite distance and move with constant velocity.

Question 2.
What is the momentum of body ? Establish a relation between the momentum and force of a moving body.
Answer:
Momentum: The momentum of body is the product of its mass (m) and velocity (v). It is denoted by ‘p’
Momentum = Mass x Velocity
or p = mv
We know that a body accelerates if some force is applied to it. According to Newton’s second law of motion, the acceleration produced in a body depends upon the force applied on that. If by applying a force, ‘F’ on a body of mass m, produces an acceleration of ‘a’ then according to second law of motion
F = ma
a = \(\frac {F}{m}\) ………….(1)
Suppose, Initial velocity of body = v₁ms^-1
After time (t) the final velocity = v₂ ms^-1
Then a = \(\frac{v_2-v_1}{t}\)
From eqs. (i) and (ii) is
\(\frac{F}{m}=\frac{v_2-v_1}{t}\)
F = \(\frac{m\left(v_2-v_1\right)}{t}\)
F = \(\frac{m v_2-m v_1}{t}\)
But mv₁ = p₁ (Initial momentum)
and mv₂ = P₂ (Final momentum)
F = \(\frac{p_2-p_1}{t}\)
F = rate of chance of momentum.
Thus, the force applied on a body is equal to the change in momentum in unit time.

Question 3.
What do you understand by conservation of momentum ? Prove the law of conservation of momentum in mathematical way.
Answer:
Conservation Law of Momentum : According to this law, the total momentum of a group of objects remains constant until it is not act upon by external force.
Activity: To prove this law, suppose two objects A and B of mass m1 and m2 are moving with velocities u₁ and u₂ respectively. Let both react with each other till time t. After action, their velocities become v₁ arid v₂as
shown in fig.
if any external force does not act, then the rate of change in momentum of object A is m₁ \(\frac{\left(v_1-u_1\right)}{t}\) In the same way the rate of change in momentum of object B is m₂ \(\frac{\left(v_2-u_2\right)}{t}\) If a force F₁ is exerted on A by B and F₂ on B by A, then according to Newton’s second law of motion,

F₁ = m₁\(\frac{\left(v_1-u_1\right)}{t}\) ………….(1)
and F₂ = m₂\(\frac{\left(v_2-u_2\right)}{t}\) ………….(2)
According to Newton’s third law of motion, F₁ and F₂ act equally and ¡n opposite direction.
F₁ = -F₂ ………….(3)
From eqs. (1), (2) and (3), is
m₁ = \(\frac{\left(v_1-u_1\right)}{t}\) = m₂
or m₁(v₁ – u₁) = m₂ (v₂ – u₂)
or m₁v₁ – m₂u₁ = m₂v₂ + m₂u₂
or m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
or m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂ ………(4)

Thus, total momentum before collision m₁u₁ + m₂ u₂ is equal to total momentum after collision m₂v₂ . This is the law of conservation of momentum.

Question 4.
What is meant by friction ? Prove that it is a boon as well as curse.
Answer:
Friction: The opposing force that comes into play when a body moves on another surface or through a fluid (liquid or gas) is called friction.
Friction as a boon: The friction is a necessary evil because nothing is possible without it, that is:
(i) We cannot move if there is no friction between road and feet.
(ii) The vehicles cannot move if there is no friction between road and tyres.
(iii) The nail can be insisted in wall only due to friction.
(iv) Only due to friction, the match stick burns.
(v) It is possible to write on blackboard due to friction between chalk and blackboard.
(vi) Due to friction, we can write on copy.
(vii) The vehicles can be stopped only by friction.
(viii) The belts related to electric motors can move on wheels due to friction.

Friction as a curse: Besides a necessary evil, friction is harmful to us as:
(i) Due to friction the parts of machine get bear and tear.
(ii) Due to friction, the efficiency of machines is not 100%.
(iii) More energy is spent due to friction.
(iv) Due to friction, our shoes tear.
(v) Due to friction, the tyres of vehicles reduces.

Question 5.
Write the methods of reducing friction. Explain any one out of them.
Answer:
Methods of Reducing Friction : Friction can he reduced by following methods:
(i) The frictional force can be reduced by planing the depresions on the surface of body
(ii) Oil or lubricant educes the friction between two surfaces. Due to this the oil is put in the part machines and the friction can be reduced.
(iii) The mixture of water and soap can minimise the friction between two surfaces.
(iv) The friction can be minimize by using ball-bearings.
(v) When we sprinkle powder on a carom board, the powder tills up the depressions and the coin
(vi) In order to reduce friction, special shape is given to vehicles.
(vii) By polishing two surfaces and by putting coaltar on roads, the friction can be reduced.

To reduce the friction by using ball-bearIng : Ball-bearings are used in machines to reduce friction. Small balls are there in ball-bearings which are kept between the sliding surfaces of machines. You will have seen them in the wheels of cycle. These reduces the friction which save energy and work.

Practical Work ‘ .

Activity I.
Two activities to show the effect of force which proves that the force can change the size of body.

Procedure:
(i) Take a spring and pull it towards outside by holding its two ends. You, will see the size of spring increase, which shows the force applied on spring by us. It proves that the force can change the size of body.
(ii) Take a spherical rubber ball and press it between your palms. You will see that on pressing, the size of ball becomes some long. It proves that the force can change the size of body.

Activity 2.
Perform on activity to show that a body starts moving only when the force applied by us is more than frictional force.

Procedure:
Take a box and tell to children to push it. You will see that the box does not move if they apply less force, because the frictional force between the lower surface of box and rough surface of floor balances the force applied to push and due to this the box does not move. The children push the box with large force, but box does not move. This is because of that the friction force is still balances the pushing force. If the children push the box with much more force, then pushing force will become more than frictional force. Here, the unbalanced force is working and the box is moving. It proves that if the force applied by us is more than frictional force, the body moves.

Activity 3.
Perform an activity to show the effect of action and reaction. .

Procedure:
(i) Take a big rubber balloon and inflate it fully.
(ii) Tie its neck using a thread. Also using adhesive tape, fix a straw on the surface of this balloon.
(iii) Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.
(iv) Now remove the thread tied on the neck of balloon.
(v) Now the air escape from the mouth of the balloon and the straw and balloon will move in opposite direction which shows reaction.

Procedure 2:
(i) Take a test tube of good quality glass material and put a small amount of water in it.
(ii) Place a stop cork at the mouth of it. Now suspend the test tube horizontally by two strings.
(iii) Heat the test tube with a burner until water vaporises and the cork blows out.
(iv) We observe that the test tube recoils in the direction

Quick Review of the Chapter

1. When a force is applied on a body, its :
(A) motion can change
(B) size can change
(C) direction of motion can change
(D) all of the above
Answer:
(D) all of the above

2. If the resultant of all forces acting on a body is zero, then all forces acting on body is called :
(A) balanced force
(B) unbalanced force
(C) frictional force
(D) none of these
Answer:
(A) balanced force

3. Unit of force is:
(A) joule
(B) metre per second square
(C) Newton
(D) metre per second
Answer:
(C) Newton

4. If the resultant of all forces acting on a particle is not zero, then forces are called:
(A) balanced force
(B) unbalanced force
(C) frictional force
(D) none of these
Answer:
(B) unbalanced force

5 Which scientist has established the laws of controlling the motion of objects firstly ?
(A) Sir Isaac Newton
(B) C.V. Raman
(C) Rutherford
(D) Neils Bohr
Answer:
(A) Sir Isaac Newton

6. Newton’s laws of motion are:
(A) 2 ‘
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

7. Newton first law of motions is called:
(A) Law of friction
(B) Law of conservation of momentum
(C) Law of inertia
(D) Law of gravity
Answer:
(C) Law of inertia

8. The rate of change of momentum of a body is proportional to the force applied and in the direction in which force is applied. This is Newton’s:
(A) first law
(B) second law
(C) third law
(D) fourth law
Answer:
(B) second law

9. The product of mass and velocity of an object is called:
(A) momentum
(B) inertia
(C) friction
(D) force
Answer:
(A) momentum

10. SI unit of momentum is:
(A) kg^-2
(B) kg ms^-1
(C) kg m
(D) kg mh^-1
Answer:
(B) kg ms

11. The relation among F, m and a is:
(A) F = \(\frac {m}{a}\)
(B) Fa = m
(C) F = ma
(D) none of these
Answer:
(C) F = ma

12. The force required to produce an acceleration of 1 mc2 in a body of mass 1 kg is:
(A) 1 Newton
(B) 2 Newton
(C) 3 Newton
(D) 4 Newton
Answer:
(A) I Newton

13. The force which acts without the contact of two objects is:
(A) magnetic force
(B) inertia
(C) friction
(D) all of the above
Answer:
(A) magnetic force

14. According to which force, the apple faits on earth from tree?
(A) force of gravity
(B) inertia
(C) force of magnet
(D) frictional force
Answer:
(A) force of gravity

15. It is difficult to walk on sand because the friction between feet and sand is:
(A) less
(B) more
(C) normal
(D) not there
Answer:
(B) more

16. A piece of stone and a truck is moving with same speed. In order to stop both, which one will require more force to stop?
(A) truck
(B) stone
(C) none on them
(D) both require equal.
Answer:
(A) truck ‘

17. Which of the following force move the object?
(A) Gravitational force
(B) Unbalanced force
(C) Balanced force
(D) Frictional force
Answer:
(B) Unbalanced force

18. Why the lubricant is used in machines?
(A) To increase friction
(B) To decrease friction
(C) To zero friction
(D) To clarity
Answer:
(B) To decrease friction

19. The dust detached from carpet when beaten with stick because of:
(A) inertia of rest
(B) motion of carpet
(C) motion of stick
(D) force
Answer:
(A) inertia of rest

20. If an external force ¡s working on a body, It accelerates in the dxrection of force. The acceleration produced is proportional to:
(A) applied force
(B) velocity
(C) mass
(D) inertia
Answer:
(A) applied force

21. If the mass of two objects A and B is 6 kg and 34 kg respectively, then:
(A) the inertia of object A is more than B
(B) the inertia of B is more than A
(C) inertia of A and B are same
(D) inertia of A and B are zero
Answer:
(B) the inertia of B is more than A

22. According to Newton’s third law of motion, the force related to action and reaction should apply on:
(A) one body
(B) same bodies
(C) always different bodies
(D) is not necessary to have equal result but their direction should be same
Answer:
(C) always different bodies

23. The explanation of falling forward when get down from bus can be done by:
(A) Newton ‘s first law of motion
(B) Newton ‘s second law of motion
(C) Newton’s third law of motion
(D) all of the above
Answer:
(A) Newton’s first law of motion

24. Launching of rocket ¡s based on the principle of:
(A) Newton’s first law of motion
(B) Newton’s second law of motion
(C) Newton’s third law of motion
(D) all of the above
Answer:
(C) Newton’s third law of motion

25. If the mass of an object m, velocity y, acceleration a, then the momentum (p) will be:
(A) p = mv
(B) p = \(\frac {m}{a}\)
(C) p = ma
(D) p = \(\frac {m}{a}\)
Answer:
(A) p = mv

26. Which one of the following has mo inertia?
(A) a rubber ball
(B) a piece of stone
(C) a cycle
(D) a train
Answer:
(D) a train

27. The branch of a tree sheds its leaves when it is shedding fastly:
(A) due to mass
(B) due to inertia
(C) due to weight
(D) due to acceleration
Answer:
(B) due to inertia

28. If an object of mass 30 kg has the momentum of 120 kg ms’, then its velocity will be:
(A) 4 ms^-1
(B) 8 ms^-1
(C) 2 ms^-1
(D) 1 ms^-1
Answer:
(A) 4 ms^-1

29. When a running bus stops suddenly, you falls forward and when starts running suddenly from rest you falls backward, It Is
(A) due to mass
(B) due to weight
(C) due to motion
(D) due to inertia
Answer:
(D) due to inertia

30. The mass of a cart is 1500 kg. If the cart is to bring in rest with negative acceleration (retardation) of 1.7 ms^-2, then the force between the cart and road will be
(A) 882.35 N
(B) 2550 N
(C) 1500N
(D) 2550 kgm
Answer:
(B) 2550 N

31. A bullet of mass 10 g moving in a straight line with a velocity of 150 ms^-1 strikes with a wooden piece and stops after 0.03 s. How far the bullet will penetrate the wood?
(A) 2.25 km
(B) 2.25 m
(C) 22.5m
(D) 15m
Answer:
(B) 2.25 m

32. A body of mass 1 kg moving in a straight line with a velocity of 10 ms^-1 strikes with a wooden piece of mass 5 kg which is in rest. After that both move in straight line together. The total momentum before collision will be
(A) 10 kg ms^-2
(B) 2 kg ms^-1
(C) 50kgms^-1
(D) 10kg ms^-1
Answer:
(D) 10kg ms^-1

Haryana State Board HBSE 9th Class Science Important Questions Chapter 8 Motion Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 8 Motion

Very Short-Answer Type Questions

Question 1.
What do you mean by rest ?
Answer:
When a body does not change its position with respect to its surroundings, then the body is said to be at rest.

Question 2.
What is meant by motion ?
Answer:
When the position of the body continuously changes with respect to its surroundings, then the body is said to be in motion.

Question 3.
What is meant by relative motion ?
Answer:
The increase or decrease in distance between two moving bodies is known as relative motion.

Question 4.
How the motion of air can be measured ?
Answer:
The motion of air can be measured by its effect.

Question 5.
Passengers are sitting in a moving train. Tell, about which objects the passengers are in rest position and about which objects the passengers are in motion ?
Answer:
A passenger sitting in a moving train is in the state of rest in comparison of other passengers and the luggage kept in train. But the same person is in motion as compared to outsider trees and buildings.

Question 6.
Give one difference betw een the motion of living and non-living things.
Answer:
Living organisms move by themselves but external factor is required for the motion of non-living things.

Question 7.
What is distance ?
Answer:
The total path travelled by a body between initial and final points is called distance.

Question 8.
What is displacement?
Answer:
The shortest distance between the initial and final position of the body is called displacement.

Question 9.
What is the SI unit of distance and displacement ?
Answer:
The SI unit of distance and displacement is metre.

Question 10.
What is meant by reference point or origin point ?
Answer:
A fixed point with respect of which a body is at rest or in motion is known as reference point or origin point.

Question 11.
What those quantities are called, which require magnitude and direction both for their expression ?
Answer:
Those quantities which require magnitude and direction both for their expression are called vector quantities.

Question 12.
Which quantities are expressed by magnitude only ?
Answer:
Scalar quantities are expressed by magnitude only.

Question 13.
Write two examples each of vector and scalar quantities.
Answer:
Vector quantities – Velocity and displacement.
Scalar quantities – Distance and speed.

Question 14.
When an object covers equal distance in equal intervals of time, then what its speed is called ?
Answer:
When an object covers equal distance in equal intervals of time, its speed is known as uniform speed.

Question 15.
Give two examples of uniform speed.
Answer:
(i) Motion of earth
(ii) Motion of pendulum of a watch.

Question 16.
When an object covers unequal distance in equal intervals of time, then what its speed is called ?
Answer:
When an object covers unequal distance in equal intervals of time, its speed is known as non¬uniform speed.

Question 17.
Write two examples of non-uniform speed.
Answer:
(1) Train starting from a station.
(2) Birds flying in the sky.

Question 18.
Write down the formula to find out the average speed of a moving object.
Answer:
The distance travelled by an object per unit time is known as average speed.

Question 19.
What is meant by velocity ?
Answer:
The distance covered by a moving body in one second in a certain direction is called velocity.

Question 20.
What is the unit of velocity ?
Answer:
The unit of velocity is metre per second (ms’1).

Question 21.
What is the relation among distance, speed and time ?
Answer:
Distance = Speed x Time

Question 22.
What will be the velocity of an athlete, if he takes 10 seconds to cover a distance of 100 metres ?
Answer:

Question 23.
What is meant by acceleration ?
Answer:
The rate of change of velocity of a moving body is called acceleration.
a = \(\frac {v – u}{t}\)

Question 24.
What is meant by retardation ?
Answer:
The rate of decrease in velocity in a certain direction is called retardation i.e. negative acceleration s called retardation.

Question 25.
What is SI unit of acceleration ?
Answer:
SI unit of acceleration is metre Der second sauare

Question 26.
Write example ot accelerated motion.
Answer:
A body moving in a circle with uniform speed is an example of accelerated motion.

Question 27.
Choose scalar and vector from following :
(i) Delhi is at a distance of 1450 km from Mumbai.
(ii) Delhi is at a distance of 1450 km in north from Mumbai.
Answer:
(i) It is an example of scalar quantity.
(ii) It is an example of vector quantity.

Question 28.
When, a body is said to be moving with uniform acceleration ?
Answer:
If there is equal change in velocity in equal intervals of time, then the body is said to be moving with uniform acceleration.

Question 29.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is known as uniform circular motion.

Question 30.
Which type of motion is shown, if the speed-time graph is a straight line ?
Answer:
It shows uniformly accelerated motion.

Question 31.
What will be the acceleration of the body, if the speed-time graph is parallel to time axis ?
Answer:
The speed of the body will uniform, therefore acceleration will be zero.

Question 32.
What shows the slope of speed-time graph of an object ?
Answer:
Acceleration of the object.

Question 33.
What the area under and between interval of speed-time graph of moving body shows ?
Answer:
The distance covered by the body in that time interval.

Question 34.
If the distance-time graph, parallel to time axis is a straight line, then average speed in different intervals will be equal or different.
Answer:
Equal.

Question 35.
If the distance-time graph of a body is a straight line, what you will say about the motion of the body ?
Answer:
The body will be in constant position.

Question 36.
If the distance-time graph of a body is a curve, then its average speed in different intervals will be different or uniform.
Answer:
The average speed will be different in different intervals.

Question 37.
If the speed of a moving body becomes 30 km/h from 0 in 2 hours, then what will be its acceleration?
Answer:
Acceleration =img= 15 kmh^-2

Question 38.
If a body is moving on circular track with uniform-speed, then which type of its motion will be ?
Answer:
Non-uniform and accelerate motion.

Question 39.
Speed of light is 3 x 10⁸ ms^-1. What will be its speed in kmh^-1 ?
Answer:
speed of light = 3 x 10⁸ms^-1 = \(\frac{3 \times 10^8 \times 3600}{1000}\)kmh^-1
= 1.08 x 10⁹kmh^-1

Question 40.
Rajdhani Express covers a distance of 1384 km in 17 hours. What will be its average speed?
Answer:

Question 41.
What will be the average speed of a car, if it covers a distance of 80 km in two hours ?
Answer:
Average speed = \(\frac {80}{2}\) = 40 kmh^-1

Question 42.
What is meant by circular motion ?
Answer:
When a body moves on circular track, its motion is known as circular motion.

Question 43.
Give two examples of circular motion.
Answer:
(1) The motion of the wheel of a vehicle.
(2) The motion of a stone tied on the end of a rope in plane circle.

Question 44.
What is meant by graph ?
Answer:
The figure representation of any information is called graph.

Question 45.
Which type of motion is shown by a freely falling object ?
Answer:
It shows a uniformly accelerated motion.

Question 46.
A stone is thrown in vertical direction, then its velocity is continuously decreasing. Why ?
Answer:
When a stone is thrown in vertical direction, its velocity continuously decreases because the acceleration due to gravity is always directed towards the centre of earth.

Question 47.
You have gone to school on 8 o’clock in morning and come back 2 o’clock in afternoon. What will be your displacement?
Answer:
Zero.

Question 48.
What will be the velocity of that object which is thrown vertically upward from the surface of earth and attains maximum height ?
Answer:
Zero.

Question 49.
What is the relation among s, u, a and t ?
Answer:
s = ut + \(\frac {1}{2}\) at².

Question 50.
What is the relation among v, u, a and t ?
Answer:
v = u + at

Question 51.
What is the relation among v, u, a and s ?
Answer:
v² – u² = 2as

Question 52.
What will be the speed of an athlete, when he completes a round in t seconds on a circular track of radius r ?
Answer:
Speed (v) = \(\frac {2πr}{t}\)

Question 53.
When there are clouds in the sky, the lightning and thundering happen in same time. First, you see the eightning After some time, the sound of thundering reaches to you. Can you tell, why it is so ?
Answer:
Because the velocity of light (3 x 10⁸ ms^-1) is much more than the velocity of sound (346 m/s).

Short Answer Type Questions

Question 1.
What is the difference between rest and motion ?
Answer:
When a body does not change its position with respect to its surroundings, it is called rest, e.g. a book lying on the table, a stone lying on the floor, etc. When the position of the body continuously changes with respect to its surroundings, it is called motion, e.g. a moving car, a running horse, etc.

Question 2.
What is meant by displacement ? When it is positive and negative ?
Answer:
The change in the position of body in a certain direction is called displacement.
The position of a body changes from x₁ to x₂, then the displacement (d) can be written as:
d = (x₂ – x₁)
The +ve and -ve signs of displacement shows the direction of change in position of body. If displacement is +ve, then body moves from left to right. If the displacement is -ve, then body moves from right to left. It is a vector quantity.

Question 3.
What is meant by speed ? Write its unit.
Answer:
Speed is the distance travelled by the body in unit time interval in any direction.

Suppose, a body covers a distance ‘s’ in time ‘ f, then its speed v = –
Unit of speed is metre per second (ms^-1). 1

Question 4.
What is meant by non-uniform motion ? Draw a graph for non-uniform motion.
Answer:
Motion of a body is said to be non-uniform, if it covers unequal distance in equal intervals of time, e.g. motion of a car on road, because its speed keeps on changing. Similarly, the motion of a train is an example of non-uniform motion, because train covers different distance in every second. Distance-time graph is shown to clear the motion of train. It is clear from the figure that speed changes with time and distance-time graph is not a straight line. It is an example of

Question 5.
How vector is shown ?
Answer:
Vector is represented by a straight line with an arrow head over it. The length of this line is proportional to the magnitude of vector and arrow head shows the direction, which is known as the ‘head’ of vector and other end is known as the ‘tail’ of vector. For example, a velocity of 30 ms^-1, 30° in south-west direction can be shown in the form of a line carrying arrow head, as shown in figure.

Question 6.
Two cars moving in opposite directions, cover an equal distance d in an hour. What will be displacement of each after an hour, if they are moving in north and south directions respectively ?
Answer:
Let O is the reference point of both cars.
The distance covered by car A from reference point O in hour in north direction = d
∴Displacement OA = d
The distance covered by car B from reference point O in hour in south direction = d
∴Displacement OB = d
Total displacement of both cars, AB = d + d = 2d

Question 7.
What is meant by graph ? What are its uses ?
Answer:
To draw the change in one physical quantity7 with respect to another physical quantity on two different axes is called graph.

Uses: The uses of graph are as follows:
(l) The relation among different quantities can be studied easily.
(2) The table of two or more than two quantities can be compared easily.
(3) It is more simple to explain quantity in graph than table.
(4) Graph represents data in less space in comparison to table which takes larger space to represent data.
(5) Quantity related to slope of graph can be found.

Question 8.
What is distance-time graph ? The distance covered by a bus in different time intervals is given in the table. Draw a distance-time graph with the help of it.

Answer:
A simple and successive method, which shows the different distances at different time of a moving body, is called distance-time graph. The distance covered by the bus in different time-intervals is shown in distance-time graph.

Question 9.
Draw distance-time graph for tw o bodies moving with uniform speed of 4 ms^-1 and 7 ms^-1 and compare it.
Answer:
We know that,
Distance = Speed x Time
(i) Distance-time graph for a body moving with uniform speed of 4 ms^-1.


(ii) Distance-time graph for a body moving with uniform speed of 7 ms^-1.

On drawing these values on graph paper, we see that distance-time graph of given body will be as the adjacent figure.

Comparison of Graphs:
The slope of the body moving with uniform speed of 7 ms^-1 will be more than that of the body moving with uniform speed of 4 ms^-1.

Question 10.
Distance-time graph for a body is shown in figure. Is it the graph of real situation ? Give reason.
Answer:
No, this is not the graph of real situation. Reasons are as follows:
(1) According to distance-time graph, the time taken to reach that point from where it started its journey is less than the time taken by it to complete that journey.
(2) Time (t) is such a quantity which is continuously increasing. This will be larger after t = 14.

Question 11.
Distance-time table of a moving car is as follows:

Draw distance-time graph for car on the graph paper. Answer: Distance-time graph for car:

Question 12.
The distances covered by Amitabh and Archna by their bicycles in different times:

Show both by distance-time graph.
Answer:
The distance-time graph for both is given below:

Question 13.
Draw velocity-time graph for a body moving with a constant velocity of 40 kmh^-1.
Answer:
To draw a velocity-time graph of a car moving with constant velocity’ of 40 kmh^-1. It means, car covers a distance of 40 km in an hour, 80 km in two hours, 120 km in 3 hours and so on, it can be seen from figure, that velocity-time graph is a straight line and parallel to time axis. It is true for all velocity-time graphs when motion is uniform.

Question 14.
How we can calculate the distance covered by a body in a given time by using velocity¬time graph ?
Answer:
Suppose the velocity-time graph of a car moving with speed of 40 kmh^-1 is as the given figure. Let us consider, we have to calculate the distance covered by the car between time and t₂. Draw perpendiculars from points t₁ and t₂ on time axis as shown in figure.

By this, we obtain a rectangle ABCD between graph and X-axis. In this rectangle side AD is equal to BC, which is equal to (t₂ -t₁), and side AB is equal to CD, which is similar to 40 kmh^-1. We know that, if a body moving with velocity (v) covers a distance (.s’) in time (?) then,
s = vt
∴ Distance covered by the car in time (t₂ – t₁) = [(40 kmh^-1) (t₂ – t₁)h]
= 40(t₂ – t₁) km
= Area of rectangle ABCD or the area under velocity-time graph shows the distance covered by the car. This is true for any velocity-time graph, whether velocity is uniform or non-uniform,

Question 15.
How velocity is calculated by straight-lined distance-time graph ? Explain.
Answer:
We can calculate the velocity of a car from its distance-time graph. In fig., think about the smaller part AB of distance-time graph of motion of car. We draw a straight-line from point A parallel to X-axis and from point B parallel to Y-axis to find the velocity of car. These two straight lines meet at a point C and make a AABC. Now, on graph, AC represents time-interval (t₂ – t₁), and BC is equal to distance (s₂ – s₁). It can be seen from the graph that when car move from point A to B, it covers a distance (s₂ – s₁) in time (t₂ – t₁)
Velocity of car = \(\frac{s_2-s_1}{t_2-t_1}\)

Question 16.
Show uniformly accelerated motion, non-uniformly accelerated motion and uniform retarded motion by graph.
Answer:
1. Uniformly Accelerated Motion: In uniformly-accelerated motion, speed-time graph is a straight a line.

2. Non-uniformly Accelerated Motion: In non-uniformly accelerated motion, speed-time graph is a

3. Uniformly Retarded Motion: In uniformly retardation, the acceleration is negative (-ve) and velocity time graph is a straight line.

Question 17.
Draw speed-time graph for a body, moving in a straight line. What the slope of this straight line show ?
Answer:
If there is equal change in the velocity of a body in equal intervals of time, then we can say that the body is moving with uniform acceleration. The speed-time graph for a body moving in a straight-line is shown in figure. If the speed of the body is v₁ at time t₁ and v₂ at time t₂, then
acceleration of the body, a = \(\frac{v_2-v_1}{t_2-t_1}\)

The time intervals (t₂ – t₁) and (t₄ -t₃) are equal as shown in figure, because the change in heights of speed are also equal. Therefore, this graph shows the motion of such a body, which is moving with uniform acceleration.
The speed-time graph of a body moving with uniform acceleration is a straight line. The acceleration of such body is the slope of this straight line.

Question 18.
How you will prove that the motion is relative ?
Answer:
In order to prove that the motion is relative, we can compare the passengers sitting in a moving bus or train with road or trees standing near the track, vehicle or persons, e.g. the passengers sitting inside the bus . or train are in rest as compared to the luggage lying inside, but are in motion as compared to outsided objects.

Question 19.
Write the difference between-
(i) vector and scalar.
(ii) uniform and non-uniform motion.
(iii) distance and displacement.
Answer:
The difference between vector and scalar:
Vector:
1. Those physical quantities which contains both direction and magnitude are known as vector quantities, e.g. velocity, displacement, force, etc.
2. Vector quantities are combined according to rule of triangle or parallelogram.
3. Vector quantities are represented by a straight line with an arrow head over it.

Scalar:
1. Those physical quantities which contain magnitude only is known as scalar quantities, e.g. distance, volume, mass, etc.
2. Scalar quantities can be combined algebraically.
3. There is no special method to represent scalar quantities.

(ii) The difference between uniform and non-uniform motion :
Uniform Motion:
When a body covers equal distance in equal intervals of time, its motion is called uniform motion, e.g. motion of earth, motion of pendulum of watch, the motion of moon around the earth.

Non-uniform Motion:
When a body covers unequal distance in equal intervals of time, its motion is called non¬uniform motion, e.g. motion of train starting from station, the motion of bus starting from bus stop, motion of cycle.

(iii) The difference between distance and displacement:

Distance:
1. The length of the path actually covered by the body is called distance.
2. It is a scalar quantity.
3. The distance between two points through different paths can be different.

Displacement:
1. The distance between the initial and final positions of the body is called displacement.
2. It is a vector quantity.
3. The value of the displacement between any two points is always same.

Question 20.
What is meant by uniform circular motion ? Justify your answer with example.
Answer:
When a body moves on circular track, it changes its direction at every point. Actually, circle is a polygon figure which has infinite sides. In figure any circular shaped track can be called a polygon of inner-sides. Therefore, a body moving in a circle with uniform speed is an example of accelerated motion. Although, there is no change in the speed of the body, but the direction of the motion of body is continuously changes. Example – Take a piece of thread and tie one stone on its one end and rotate it, it will be in a circular motion.

Numerical Questions

Question 1.
The distance between the home of a boy and school is 1.8 km, if he takes 5 minutes to reach school by cycle, then find his speed In metre/second.
Solution:
Distance (s) = 1.8 km = 1.8 x 1000 metres = 1800 metres
Time taken (t) = 5 minutes = 5 x 60 seconds = 300 seconds

= 6 ms^-1

Question 2.
A person takes 10 minutes to reach office from his home. If the office is at a distance of 3.6 km, then find the average speed of person.
Solution:
Total distance 3.6 km = 3.6 x 1000 m = 3600 metres
Total time = 10 minutes = l0 x 60seconds = 600 seconds
Total distance 3600 m

= 6 ms^-1

Question 3.
Trivendrum express moves from Guwahati at 23 : 30 hrs on Monday and after covering a distance of 3574 km, It reaches Trivendrum at 22 : 30 hrs on Thursday. Find out the average speed of
this train.
Solution:
The total distance covered by the train 3574 km
Total time between 23:30 hrs (Monday) to 22:30 hrs (Thursday) = 24 + 24 + 23 – 71h

= 50.34 kmh^-1

Question 4.
A car, starting from rest, acquire a velocity of 36 kmh’ in 10 seconds. Find out acceleration of the car.
Solution:
The car started from rest
∴ Initial velocity (u) = O
Final velocity (v) = 36 kmh^-1 = 36 x \(\frac {5}{18}\) = 10 ms^-1
Time (t) = 10 s
We know that. v = u + at
10 = 0 + a x 10
10 = 10 a
a = \(\frac {10}{10}\) = 1 ms^-1

Question 5.
A cycle moving with a velocity of 3 metre per second increase its motion with acceleration of 0.5 ms² . What will be Its velocity after 5 seconds and how much distance it will cover ¡n this time?
Solution:
Here, Initial velocity (u) = 3 ms
Acceleration (a) = 0.5 ms²
Time (t) = 5 s
Final velocity (v) = ?
Distance (s) = ?
We know that,
v = u + at = 3 + 0.5 x 5 = 3 + 2.5 = 5.5 ms^-1

(2) s = ut+ \(\frac {1}{2}\) at²
= 3 x 5 + \(\frac {1}{2}\) x 0.5 x 5 x 5
15 + 6.25 = 21.25 m

Question 6.
The uniform acceleration of a scooter is 4 ms2. How much distance will scooter cover after 10 seconds from its movement ? What will be its velocity after 10 s?
Solution:
Here,
Initial velocity of scooter (u) = 0
Acceleration (a) = 4 ms
Time (t) = 10 s
Distance (s) = ?
Final velocity (y) = ?
We know that
(1) s = ut + \(\frac {1}{2}\) at²
s = 0 x 10 + \(\frac {1}{2}\) x 4 x (10)²
s = 0 + \(\frac {1}{2}\) x 4 x 100 = 200m
(2) v = u + at = 0 + 4 x 10 = 40 ms^-1

Question 7.
A bicycle is travelling at a speed of 10 ms. The brakes are applied so as to produce a retardation of 0.5 ms2. Find, how far the bicycle travel before it is brought to rest?
Solution:
Initial velocity of bicycle (u) = 10 ms^-1
Final velocity (v) = 0
Retardation (a) = 0.5 ms^-2
We know that, v² – u² = 2as
(0)² – (10)² = 2(- 0.5) x s
– 100 = – 1s
s = 100 m

Question 8.
A car travels with uniform velocity 20 ms’ for 5 s. On applying brakes, the car comes to rest in 8s with uniform retardation. Draw, velocity-time graph. How far, the car will go after applying brakes?
Solution:
The velocity-time graph is shown in figure.
The distance covered by the car, after applying brakes
= area of ∆BDC = \(\frac {1}{2}\) x DC x DB
[ ∴ area of triangle = \(\frac {1}{2}\) x base x perpendicular]
= \(\frac {1}{2}\) x (13 – 5) x 20
= \(\frac {1}{2}\) x 8 x 20 = 80 m.

Question 9.
In a long running, the athletes have to move 4 rounds of running track The starting and end of race is at the same place. If the length of one round is 200 m then –
(a) how much distance is to be covered by athletes?
(b) what will be the total displacement when the athletes has completed the race?
(c) is the motion of athletes uniform or non-uniform?
(d) are the distance covered by the athletes and displacement equal, at the end of race?
Answer:
(a) The total distance to be covered by the athletes = 200 x 4 = 800 metres.
(b) Because the initial and final places of running track are at same place, therefore the total displacement of athletes = zero.
(c) The motion of athletes is non-uniform.
(a) The distance covered by the athletes at the end of race is not equal to displacement. Because –
Distance = 800 metres
Displacement = 0

Question 10.
Write the speed of following bodies in descending order –
(i) Bicycle running with a speed of 18 kmh^-1.
(ii) An athlete running with a speed of 7 ms^-1.
(iii)A car running with a speed of 2000 m min^-1
Solution:
We know that,
(i) 18 kmh^-1 = \(\frac {18000}{3600}\) ms^-1 = 5 ms^-1
(ii) 7 ms^-1
(iii) 2000 m min^-1 = \(\frac {2000}{60}\) ms^-1 = 33.33 ms^-1
∴ The descending order of speed of bodies is as follows:
(i) A car running with a speed of 2000 m min^-1.
(ii) An athlete running with a speed of 7 ms^-1.
(iii) A bicycle running with a speed of 18 kmh^-1.

Question 11.
A body moves with velocity of 2 ms^-1 for 5 seconds, in next 5 seconds, its velocity increases to 10 ms^-1 due to uniform acceleration. After that, its velocity decreases uniformly and comes in rest in 10 seconds, then
(a) draw velocity-time and distance-time graphs for the motion of the body.
(b) Show that portion in graph, where the motion is uniform and non-uniform.
(c) From graph, find the distance covered by the body after 2s and 12s and in the last lOs.
Answer:
(a) Velocity-time graph: The velocity of the body is 2 ms for first 5 seconds, which is marked
by AB in the graph. After that, it becomes 10 mst in another 5 seconds (between 5 and 10 s), which is marked
by BC in the graph. After that the body comes in rest in next 10 seconds, which is marked by CD in the graph.

(b) Distance-time graph: Because the speed of the body is not uniform. Therefore, distance-time graph is not a straight-line

From velocity-time graph, we find that the motion of the body is uniform in AB part, whereas the motion is non-uniform in BC and CD parts.
(c) (1)The distance covered by the body in 2s = 2 x 2 = 4 m.
(2) The distance covered by the body in 12 s = area of [OB’BA + BBC’C + CC’FF’]
Now, the area of rectangle OBBA = 2 x 5 = 10
Area of trapezium BB’C’C = \(\frac {1}{2}\) x (BB’ + CC’) x B’C
= \(\frac {1}{2}\) (2 + 10) x 5
= \(\frac {1}{2}\) x 12 x 5 = 30
Area of trapezium CC’FF’ = 4- (CC’ + FF’) x CF’
= \(\frac {1}{2}\) x (10 + 8) x 2 = \(\frac {1}{2}\) x 18 x 2 = 18
∴ The distance covered by the body in 12 sec 10 + 30 + 18 = 58 metres

(3) Distance covered in last lO seconds area of (trap. CC’ F’F + right-angle ∆FF’D
Now, area of trapezium CC’F’F = x (CC + FF’) x C’F’
= \(\frac {1}{2}\) x (10 + 8) x 2
= \(\frac {1}{2}\) x 18 x 2 = 18

Area of right-angle ∆FF’D = \(\frac {1}{2}\) x F’D x F’F = \(\frac {1}{2}\) x 8 x 8 = 32
∴ Distance covered in last 10 seconds = 18 + 32 = 50 metres.

Do Yourself

1. Rita takes 20 minutes to cover a distance of 3.2 km on a bicycle. Calculate her velocity in units of a kilometre minute and kilometre/hour.
Answer:
0.16 km/minute, 9.6 km/hour

2. Ahmed is driving his car with a velocity of 45 kmh^-1. How much distance will be covered in
(a) one minute and
(b) one second?
Answer:
(a) 750 metres
(b) 12.5 metres

3. Starting from rest, Saurabh paddles his bicycle to attain a velocity of 6 ms^-1 in 30 seconds. Then, he applies brakes after that the velocity of the bicycle comes down to 4 ms^-1 in next 5 seconds. Calculate the acceleration of the bicycle in both the cases.
Answer:
(a) 0.2 ms^-2
(b) -0.4 ms^-2

4. A train starting from rest attains a velocity of 108 kmh^-1 in 10 minutes. Calculate (I) acceleration (ii) distance covered by the train.
Answer:
(1) 0.5 ms²
(ii) 900 m

5. A train starting from rest attains a velocity of 72 kmh^-1 in 5 minutes. Assuming that the acceleration is uniform, find (j) acceleration of train and (ii) the distance travelled by the train while it attained this velocity.
Answer:
(i) ms^-2
(ii) 3 km

6. A car accelerates uniformly from 18 kmh^-1 to 36 kmh^-1 in 5 seconds. Calculate : (1) the acceleration and (ii) the distance covered by the car in that time.
Answer:
(i) 1 ms^-2
(ii) 37.5 m

7. The brakes applied to a car produces a negative acceleration of 6 ms2. if the car takes 2 seconds to stop after applying the brakes, calculate the distance it travels, during this time.
Answer:
12 m

8. A train covers a distance of first 30 km out of 60 km lengthy path with a uniform speed of 45 kmh^-1. By how much speed, the train will cover the remaining 30 km path so that its average speed will be 60 kmh^-1 ?
Answer:
90 kmh^-1

9. Choose right answer out of the following questions:
(a) How much distance will a car cover in 12 minutes moving with a speed of 35 kmh^-1’?
(i) 7 km
(ii) 35 km
(iii) 14 km
(iv) 28 km
Answer:
(i) 7 km

(b) If, the acceleration of the body moving with a speed of 20 ms^-1 in a straight line ¡s 4 ms^-2, then
how much speed of it will be after 2 seconds?
(1) 8 ms^-1
(ii) 12 ms^-1
(iii) 16 ms^-1
(iv) 28 ms^-1
Answer:
(iv) 28 mst

(C) The speed of a car becomes 50 kmh^-1 from 20 kmh^-1 in 10 seconds. The acceleration of this car is –
(i) 30 ms²
(ii) 3 ms²
(iii) 1 8 ms²
(iv) 0.83 ms²
Answer:
(iv) 0.83 mc²

10. A piece of wood slides down from rest on plane of 10m tong with an acceleration of 5 ms² . What will be the speed of piece of wood on reaching at the base of inclined plane?
Answer:
10 m /s

11.. A train is travelling at a speed of 90 kmh^-1 On applying brakes, a retardation of 0.5 mc² s produced. Find how far the train goes before it is brought to rest?
Answer:
625 m

12. A body covers a distance of 16 metres in 4 seconds and again covers a distance of 16 metres in 2 seconds. What will be the average speed of body?
Answer:
5.33 ms^-1

13. The odometer of a car represents 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 hours then, calculate the average speed of the car in kmh’ and ms.
Answer:
50kmh^-1, 13.9 ms^-1

14. Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Answer:
Speed 3 ms^-1, Velocity = O ms^-1

15. Starting from a rest position, Rahul paddles his bicycle to attain a velocity of 6 ms1 in 30s. Then he applies brakes such that the velocity of bicycle comes down to 4 ms^-1 in the next 5s. Calculate the acceleration of the bicycle ¡n both the cases.
Answer:
0.2 m² 0.4 ms

Essay Type Question

Question 1
Prove the following equations:
(i) v = u + at²
(ii) s = ut + ut²
(iii) v² – u² = 2as
Solution:
(i) Let the initial velocity of the body = u
Time = t
Final velocity = v
Change in velocity = v – u
Rate of change in velocity (acceleration), a = \(\frac {v – u}{t}\)
at = v – u
at+u = v
v = u + at
Let the initial velocity of the body = u
Acceleration = a
and after time t, final velocity = v
1f after t sec, the distance is s, then distance (s) = average velocity x time
But, average velocity = \(\frac {v – u}{t}\)
∴Distance (s) = \(\frac {v – u}{t}\)
s = \(\frac {v – u}{2}\) x t
s = \(\frac {u + u + at}{2}\) x t [∴v = u + at]
s = \(\frac {2u + 2t}{2}\) x t
s = \(\left(\frac{2 u t+a t^2}{2}\right)=\frac{2 u t}{2}+\frac{a t^2}{2}\)
s = ut + \(\frac {1}{2}\) at²

(iii) We know that, v = u + at
Squaring both sides
(v)² = (u + at)²
v² = u² + a²t² + 2uat
v² – u² = a²t² + 2uaI
v²– u² = 2uat + a²t²

v²– u² = 2a(ut + \(\frac {1}{2}\) at²)
v²– u² = 2as

Practical Work

Experiment 1:
Describe one activity for the difference between distance and displacement.

Procedure:
Take a metre scale and a long rope. Walk from one corner of a basket-ball court to its opposite corner way and measure with the help of scale. This will be the distance covered by you, whereas you will get displacement by putting a straight-rope from one corner to another. Suppose ABCD is basket-ball court. Distance to go from A to C AB + BC, whereas displacement = AC.

Experiment 2:
Clarify with an activity that direction of motion changes at every point on circular track

Procedure:
Take a piece of thread and tie a small piece of stone at one of its ends, move the stone to describe a circular path with constant speed by holding the thread at the other end as shown in figure. Now, let the stone go by releasing with thread and measure the direction of stone.

By repeating the activity for a few times and releasing the stone at different positions of circular path, check whether the direction in which the stone moves remains the same or not If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is’ released, it continues to move along the direction it has been moving at that instant. This shows that the direction of mot ion changes at every point when the stone was moving along the circular path.

Quick Review of the Chapter

1. A train is moving. The passengers sitting inside it, are ¡n rest compared to :
(A) other passengers
(B) the luggage in the train
(C) outsider objects
(D) other passengers and luggage in train
Answer:
(D) other passengers and luggage ¡n train

2. A train is moving. The passengers sitting inside it, are in motion as compared to :
(A) other passengers
(B) the luggage in the train
(C) outsider buildings and trees
(D) other passengers and luggage in train
Answer:
(C) outsider buildings and trees

3. If any object changes its position continuously with respect to other object, then objett is in state of
(A) motion
(B) rest
(C) constant
(D) none of these
Answer:
(A) motion

4. If any object does not change its position with respect to its surroundings, then the object is in state of
(A) motion
(B) rest
(C) circular motion
(D) movement
Answer:
(B) rest

5. Which of the following moves by themselves?
(A) living things
(B) non-living things.
(C) both living and non-living things
(D) neither living nor non-living
Answer:
(A) living things

6. SI unit of distance is:
(A) km
(B) m
(C) cm
(D) mm
Answer:
(B) m

7. WhIch of the following is a scalar quantity?
(A) momentum
(B) velocity
(C) displacement
(D) distance
Answer:
(D) distance

8. Rate of change of velocity is known as:
(A) Speed
(B) Acceleration
(C) Momentum
(D) Force
Answer:
(B) Acceleration

9. When a body covers equal distance in equal intervals of time, it is said to be
(A) uniform speed
(B) non-uniform speed
(C) average speed
(D) accelerated motion
Answer:
(A) uniform speed

10. If acceleration is in the direction of velocity, then it is called:
(A) Negative acceleration
(B) zero acceleration
(C) Positive acceleration
(D) none of these
Answer:
(C) Positive acceleration

11. If acceleration is ¡n the opposite direction of velocity then it is called:
(A) Negative acceleration
(B) zero acceleration
(C) positive acceleration
(D) none of these
Answer:
(A) Negative acceleration

12. If an object moves ¡n a circular path with uniform speed, Its motion is called: [H.B.S.E. 2020]
(A) Uniform circular motion
(B) Non-uniform circular motion
(C) Uniform curved motion
(D) Non-uniform curved motion
Answer:
(A) Uniform circular motion

13. SI unit of speed Is: [H.B.S.E. March, 2019]
(A) ms^-2
(B) ms^-1
(C) kmh^-1
(D) kms^-1
Answer:
(B) ms^-1

14. The distance covered by a moving body in one second in a certain direction is called:
(A) velocity
(B) acceleration
(C) non-uniform speed
(D) uniform speed
Answer:
(A) velocity

15. What is the relation among distance, speed and time?
(A) distance x speed = time
(B) time x distance = speed
(C) distance speed x time
(D) none of these
Answer:
(C) distance = speed x time

16. What will be the velocity of athlete if he takes 10 s to cover a distance of 100m?
(A) 100 ms^-1
(B) 10ms^-1
(C) 10 ms^-2
(D) 10 ms^-2
Answer:
(B) 10 ms^-1

17. The rate of decrease In velocity in a certain direction is:
(A) +ve acceleration
(B) – ve acceleration
(C) +ve and – ve acceleration
(D) none of these
Answer:
(B) – ve acceleration

18. The downward slope in velocity-time graph shows:
(A) accelerated motion
(B) equal motion
(C) retarded motion
(D) none of these
Answer:
(C) retarded motion

19. Rajdhanl Express covers a distance of 1384 km in 17 hours. Its average speed will be:
(A) 81.4 ms^-1
(B) 81.4 kms^-1
(C) 81.4 mh^-1
(D) 81.4 kmh^-1
Answer:
(D) 81.4 kmlf1

20. If a car covers a distance of 80 km in 2 hours, then its average speed will be:
(A) 40 kmh^-1
(B) 40 kms^-1
(C) 40 ms^-1
(D) 40 mh^-1
Answer:
(A) 40 kmh^-1

21. Cheetah is the fastest land animal and can achieve a peak velocity of 100 kmh upto a distance less than 500 m. H the cheetah spots his prey at a distance of 100 m, what ¡s the minimum time it will take to get the prey, if the average velocity attained by ¡t ¡s 90 kmh?
(A) 3s
(B) 4s
(C) 5s
(D) 6s
Answer:
(B) 4 s

22. The distance-time graph of a body is parallel to X-axis. It shows that:
(A) body is moving with non-uniform motion
(B) body is moving with accelerated motion
(C) the body is at rest
(D) body is moving with uniform motion
Answer:
(C) the body is at rest

23. By converting the speed of 2 km/h in m/s, we will get:
(A) \(\frac {5}{36}\) m/s
(B) \(\frac {5}{18}\) m/s
(C) \(\frac {5}{9}\)m/s
(D) \(\frac {50}{9}\) m/s
Answer:
\(\frac {5}{9}\)m/s

24. What is the speed of light in:
(A) 3 x 10⁸ m/s
(B) 3 x 10⁷ m/s
(C) 3 x 10¹⁰
(D) 3 x 10⁵ m/S
Answer:
(A) 3 x 10⁸ m/s

25. Which of the following is right for displacement?
(A) it cannot be zero
(B) its magnitude is more than the distance covered by the body
(C) it can be zero
(D) none of these
Answer:
(C) it can be zero

26. The odometer of any vehicle shows
(A) speed
(B) velocity
(C) distance
(D) time
Answer:
(A) speed

27. When an object Is In uniform motion, its path seems to be
(A) straight line
(B) curve
(C) irregular
(D) circular
Answer:
(A) straight line

28. During an experiment, a wireless signal from a spaceship reached ground station in 5 seconds. What Is the distance of spaceship from the ground station?
(A) 9 x 10¹⁰kjn
(B) 9 x 10¹⁰m
(C) 9 x 10¹⁰cm
(D) 9 x 10¹⁰ mm
Answer:
(B) 9 x 10¹⁰ m

29. A train is moving with a speed of 90 kmh^-1, on applying brakes, it produces a retardation of 0.5 ms². How much distance will it cover before coming to rest?
(A) 62.5 m
(B) 6.25 m
(C) 625 m
(D) 625 km
Answer:
(C) 625 m

30. A trolley while going down an inclined plane has an acceleration of 0.2 ms^-2. What will be its velocity In 3s after the starts?
(A) 0.6 ms^-1
(B) 0.6 cms^-1
(C) 0.6 mms^-1
(D) 0.6 ms^-2
Answer:
(A) 0.6 ms^-1

31. A racing car has a uniform acceleration of 4 ms^-2. What distance will it cover in lOs after the
start from rest?
(A) 20m
(B) 200m
(C) 2000m
(D) 2m
Answer:
(B) 200 m

32. A stone is thrown in vertically upward direction with a velocity of S ms⁴. If the acceleration of the stone during its motion is 10 ms², in downward direction, what will be the height aftalned by the stone?
(A) 12.5m
(B) 125m
(C) 1.25m
(D) 0.125m
Answer:
(C) 1.25 m

33. A stone Is thrown in vertically upward direction with a velocity of 5 ms. If the acceleration of the stone during its motion is 10 mr2, in downward direction, how much time will ¡t take to reach at maximum height?
(A) 0.5s
(B) 5s
(C) 5 min
(D) 0.5 mm
Answer:
(A) 0.5 s

34. An artificial satellite Is moving in a circular orbit of radius of 42,250 km. What will be Its speed, if it takes 24 hours to revolve round the earth?
(A) 3.07 ms^-1
(B) 3.07 kmh^-1
(C) 3.07 kms^-1
(D) 3.07 cms^-1
Answer:
(C) 307 kms^-1

35. An athlete completes one round of a circular track of radius r in time t, then its speed will be:
(A) 2πrt
(B) \(\frac{2 \pi r}{t}\)
(C) \(\frac{t}{2 \pi r}\)
(D) none of these
Answer:
(B) \(\frac{2 \pi r}{t}\)

36. The proper relation among s, u, t land a is:
(A) s = \(\frac{2 \pi r}{t}\)
(B) s = u + at
(C) s = ut + \(\frac {1}{2}\) at²
(D) s = uv + at
Answer:
(C) s = ut + \(\frac {1}{2}\) at²

37. The relation among y, u, a and t ¡s:
(A) v = u – at
(B) v = u + at
(C) v² = u² – at
(D) v² = u² + at
Answer:
(B) v = u + at

38. The proper relation between y, u, a and s is:
(A) v²= u² – 2as
(B) v² = 2as – u²
(C) v² – u² = 2as
(D) v² + u² = as
Answer:
(C) v² – u² = 2as

Haryana State Board HBSE 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Very Short-Answer Type Questions

Question 1.
Define Taxonomy.
Answer:
Science of classification of organism is known as Taxonomy.

Question 2.
Who classified the animals on the basis of their habitat ?
Answer:
Greek thinker Aristotle classified the animals into different groups based upon their habitat and habits.

Question 3.
Which type of cells have nucleus, cell organelles and cell wall ?
Answer:
Eukaryotic cells.

Question 4.
In which cell have the capacity to produce multicellular organisms ?
Answer:
Cell with nucleus.

Question 5.
Name the principle due to which some cell group together to form a single organism.
Answer:
Division of labour.

Question 6.
What are plants ?
Answer:
The organisms which produce their own food (autotrophs) are called plants.

Question 7.
What are animals ?
Answer:
The organisms which are heterotrophic known as animals.

Question 8.
Who proposed the concept of evolution ?
Answer:
Charles Darwin in his book “The Origin of Species” in 1859.

Question 9.
Which are primitive or lower organisms ?
Answer:
The group of organisms still possess their ancient body on earth is known as primitive or lower organisms.

Question 10.
Which are advanced or higher organisms ?
Answer:
The group of organisms who have undergone lot of changes in their body are called advanced or higher organisms.

Question 11.
Name the biologists who classified kingdom of organisms into broad categories.
Answer:
Ernst Haeckel (1894), Robert Whittaker (1959) and Carl Woese (1977).

Question 12.
What is the name of classification proposed by Whittaker ?
Answer:
Five kingdoms.

Question 13.
Which biologist classified the Monera Kingdom into Archaebacteria and Eubacteria ?
Answer:
Carl Woese.

Question 14.
Define species.
Answer:
Species is a group of organisms that are capable of breeding amongst themselves so as to produce fertile young ones. It is the smallest unit of classification.

Question 15.
What is the basic unit of classification ?
Answer:
Species.

Question 16.
What is nomenclature ?
Answer:
The system of naming animals and plants is termed as nomenclature.

Question 17.
Who is the father of Taxonomy ?
Answer:
Carolus Linnaeus.

Question 18.
Why does system of writing scientific names called bionomial system ?
Answer:
In this system every organism is given two names. One name is generic and the other is specific name.

Question 19.
What is the scientific name of human ?
Answer:
Homo sapiens.

Question 20.
What is the International System of nomenclature?
Answer:
Binomial system.

Question 21.
What are different categories of classification ?
Answer:
Species, Genus, Family, Order, Class, Phylum and Kingdom.

Question 22.
What is Genus ?
Answer:
Similar species together form genus.

Question 23.
How many kinds of kingdoms classified the living world in ancient time ?
Answer:
Two kingdoms:
(i) Plantae
(ii) Animalia.

Question 24.
What is classification ?
Answer:
Arranging of organism into groups on the basis of similarities and dissimilarities, which also show their relationship is called classification.

Question 25.
Write an importance of classification.
Answer:
It makes the study of organisms easy.

Question 26.
Which scientist classified the all organisms of the world into plantae and animalia kingdoms?
Answer:
Carolus Linnaeus in 1758.

Question 27.
Who raised the third kingdom Protista?
Answer:
E.H. Haeckel in 1866.

Question 28.
Who proposed the fourth kingdom Monera and fifty kingdom Animalia ?
Answer:
Robert H. Whittaker in 1969.

Question 29.
Which is the living organisms grow in whole life ?
Answer:
Plants.

Question 30.
How many sub-kingdoms of plant kingdom divided by Eichler ?
Answer:
Two: Cryptogamae and Phanerogamae.

Question 31.
Fungi cannot prepare their own food. Why ?
Answer:
Because chlorophyll is not present in them.

Question 32.
How many types of Angiosperms ?
Answer:
They are of two types:
(i) Dicot plants
(ii) Monocot plants.

Question 33.
Give two examples of Gymnosperms.
Answer:
(i) Cycas
(ii) Pinus.

Question 34.
Give one difference between algae and fungi.
Answer:
Chlorophyll present in algae but absent in fungi.

Question 35.
What is lichen?
Answer:
The symbiotic relationship of algae with fungi is called lichen.

Question 36.
Which sub-group of Cryptogams related to fungi ?
Answer:
Thallophyta.

Question 37.
Give one example of bryophyta.
Answer:
Mosses.

Question 38.
Give one example of pteridophyta.
Answer:
Ferns.

Question 39.
What are cryptogams ?
Answer:
The plants which do not bear seeds and spores located in their sex orgAnswer:

Question 40.
What is phanerogams ?
Answer:
Those plants in which seeds are produced after reproduction are called phanerogams.

Question 41.
What is meant by gymnosperms ?
Answer:
The plants bear naked seeds are called gymnosperms.

Question 42.
What is meant by angiosperms ?
Answer:
The plants which bear seeds into the fruit is called angiosperms. .

Question 43.
What is the main feature of plantae ?
Answer:
They are autotrophic.

Question 44.
What is the main feature of Animalia ?
Answer:
Heterotrophic.

Question 45.
What is porifera ?
Answer:
Those animals which bear pores all over their body are known as porifera.

Question 46.
Give two examples of coelenterata.
Answer:
Hydra and sea anemone.

Question 47.
What is the name of flatworm ?
Answer:
Platyhelminthes.

Question 48.
What is the scientific name of tapeworm ?
Answer:
Taenia solium.

Question 49.
Which group of animals is heterotrophic ?
Answer:
Nematoda.

Question 50.
Give one example of annelida.
Answer:
Earthworm.

Short-Answer Type Questions

Question 1.
What is the science of Taxonomy and Classification ?
Answer:
Taxonomy:
The branch of biology in which organisms are classified is called Taxonomy. First of all D. Candolli used this word Taxonomy involves identification, nomenclature and classification of organisms.

Classification:
It arranges the organisms in groups or set on the basis of their relations. Animals and plants are ranked in different categories as follow: Phylum, Class, Order, Family, Genus and Species. First of all, Carolus Linnaeus (1707-1778) gave idea about it in his book ‘Systema Naturae Therefore, Carolus Linnaeus is known as father of Taxonomy.

Question 2.
Why is it necessary to classify the living organisms ? Explain in your own words.
Answer:
The number of organisms in the world is very large. These organisms possess some similarities and some dissimilarties. The structure of some organisms is simple and that of others is complex. The organisms differ in their structure, nature, life-cycle, nutrition, respiration and reproduction, etc. Therefore, so as to make their study convenient, their classification is essential. The organisms are classified on the basis of similarities and dissimilarities. The science of classifiation provides a base for other branches of biology.

Question 3.
What is classification ? Explain its importance.
Answer:
Grouping organisms on the basis of similarities, dissimilarities and their inter-relationships is called classification.

Importance:
(1) Classification gives a base to the other branches of biology.
(2) Classification makes the study of organisms easy and convenient.
(3) Classification provides a clear picture of organisms.
(4) It provides information about inter-relationship of organisms.
(5) The study of geography is completely based on classification of plants and animals.
(6) The classification of organisms also contributes to the development of knowledge in other subjects.
(7) The branches of biology like ecology, cell biology, etc. have developed due to science of classification.

Question 4.
What is binomial nomenclature ? Explain with an example.
Answer:
Binomial Nomenclature: This system of nomenclature was given by Carolus Linnaeus. According to this system 4200 species of animals were named and were published in a book ‘SystemaNaturae’in 1758. In this system each plant or animal is given two names : The first name indicates the genera and is known as generic name whereas the second name indicates the specific name or species. Linnaeus is known as the father of Taxonomy.

For Example:
the scientific name of mango is Mangifere indica, that of mustard is Brassica compestris and that of man is Homo sapiens.
In all these the first name is generic name and the second name is that of species. All these names are universal and are used throughout the world.

Question 5.
What are the salient features of two kingdom system of classification ? Why are bacteria and fungi classified along with plants ?
Answer:
The living organisms are generally classified into two kingdoms : kingdom Animalia and kingdom Plantae. In both these kingdoms the characteristic features of organisms are same but they still differ in some characters. The organisms belonging to kingdom animalia are motile and cannot prepare their own food as they lack chlorophyll.

On the other hand the organisms belonging to kingdom plantae are (generally) autotrophs and prepare their own food. They are non-motile and remain confined to the same place. Their cell wall is made up of cellulose. They contain chlorophyll and are green. Bacteria and fungi do not contain green coloured pigment called chlorophyll, so they are not autotrophs. These obtain food/nourishment from dead or living organisms.

Still they are classified along with plants (kingdom Plantae) because:
(1) Their cell wall is made up of cellulose.
(2) They obtain their food/nourishment in soluble form.

Question 6.
Describe the different categories of classification ?
Answer:
Categories of the classification are the following:

Question 7.
What is Cryptogamae and Phanerogamae ?
Answer:
1. Cryptogamae: They are lowest group of plants. They lack flowers and seeds plants. They have bidden reproductive sex orgAnswer: Asexual reproduction takes place by the formation of spores.

2. Phanerogamae: They are highest group of plants. They bear roots, stem, leaves, flower, fruit and ’ seeds. They reproduce sexually by means of seeds. The seed contains embryo and nourishing material.

Question 8.
How is cryptogamae classified ?
Answer:
They are flowerless plants. Reproductive organs are hidden in them. There is no external flower and seed consists in them. According to classification, by Lindale and Eichler, cryptogamae is divided into three main groups:
(i) Thallophyta; e.g. Ulothrix, Penicillium, Lichens, etc.
(ii) Bryophyta; e.g. Funaria, Marchantia, etc.
(iii) Pteridophyta; e.g. Ferns, Selaginella, etc.

Question 9.
What is the difference between monocot and dicot plants ?
Answer:
These are the following differences between monocot and dicot plants :
Monocots:
1. Seeds with one cotyledon.
2. The leaves of these plants have parallel veination.
3. The root system is fibrous in them.
4. There is no secondary growth.
5. Stem divided into nodes.
Example: Wheat, maize, bajra, etc.

Dicots:
1. Seeds with two cotyledons.
2. Veination in their leaves is reticulate with a network of veins.
3. Tap root system is present in them.
4. Secondary growth is present.
5. Stem is not divided into nodes.
Example: Mustard (sarson), pea, gram, etc.

Question 10.
What is animalia ? Write names of main phylum and important features of this group. Answer: Animalia : The multicellular organisms with lack cell wall are called animalia. Their important features are following :
(i) They are heterotrophic and autotropic.
(ii) Their growth is definite.
(iii) These are free moving organisms.

They can be classified in the following :
(i) Porifera
(ii) Coelenterata
(iii) Platyhelminthes
(iv) Nematoda
(v) Annelida
(vi) Arthropoda
(vii) Mollusca
(viii) Echinodermata
(ix) Protochordata
(x) Vertebrata

Question 11.
Describe main features of porifera phylum.
Answer:
The main features of porifera are following:
(i) They mostly occur in saline water although some are found in fresh water.
(ii) They are multicellular.
(iii) Cells in them are loosely held together. Thus, they have cellular level of organisation.
(iv) Skeleton is not found in them.
(v) The shape of their body is the vase or sac like.
(vi) They bear pores all over body, with simple large opening known as osculum on the top.
(vii) They possess a unique canal system.
(viii) They reproduce by asexual and sexual methods.
Example:
Sycon, Spongia, Euplectella, Spongilla, etc.

Question 12.
Describe main features of Coelenterata.
Answer:
The main features of Coelenterata is following:
(i) They are solitary or colonial.
(ii) Anus is absent in them.
(iii) They possess radial symmetry.
(iv) They possess coelenterata.
(v) They bear tentacles.
(vi) Their body have stinging cells called cnidoblasts.
(vii) Polyp and medusa come into their life cycle one by one.
(viii) Reproduction is usually asexual in polyp form and sexual in the medusa form Examples : Hydra, Obelia, Sea anemone, Coral, etc.

Question 13.
Describe the main features of platyhelminthes.
Answer:
Platyhelminthes are characterised by the following main features:
(i) They are often dorso ventrally flattened.
(ii) They are usually endoparasites.
(iii) They are triploblastic, the body develops from three layers.
(iv) Their body having bilateral symmetry.
(v) Sexual reproduction takes place in them.
(vi) Only one hole is present in the alimentary canal.
(vii’) They have dorsoventrally flat and leaf like or ribbon like body.
Examples:
Tapeworm, planaria, liver- fluke, etc.

Question 14.
Write the main features of the nematoda phylum.
Answer:
Following are the main features of nematoda phylum:
(i) They are endoparasites.
(ii) They possess bilateral symmetry.
(iii) Alimentary canal is complete.
(iv) They are pseudocoelomates. Body cavity is not a true coelom in them.
(v) They are unisexual.
(vi) They are triploblastic.
(vii) Body size is microscopic and other are several centimetres in length.
Examples: Ascaris, Roundworm, Pinworm, etc.

Question 15.
Write main features of annelida phylum.
Answer:
Annelida phylum are characterised by the following main features:
(i) They live in moist soil, in freshwater or saline water.
(ii) They have long and palpented body.
(iii) They are first animals with true body- cavity or coelom.
(iv) Nephridia is present in them for the purpose of excretion.
(v) Animals have bilateral symmetry.
(vi) They are unisexual or bisexual.
(vii) Some forms have unjointed locomotary appendages, the parapodia. Others have setae for locomotion.
Examples: Earthworm, leech, nereis, sea mouse, etc.

Question 16.
What are the main features of arthropoda ?
Answer:
Arthropoda is characterised by the following main features :
(i) They occur everywhere, on land, in soil, in fresh and saline water.
(ii) They are heterotrophic.
(iii) They possess jointed legs.
(iv) Their body is segmented and divided into three regions: head, thorax and abdomen.
(v) Interior part of the body forms the distinct head, which bears well-developed sense orgAnswer:
(vi) Their body is covered externally by a chitinous exoskeleton.
(vii) Circulatory system is of open type.
(viii) Haemocoel present in them.
(ix) Male and female reproductive organs are separate.
Examples: Cockroach, crab, scorpion, bee, butterfly, mosquito, etc.

Question 17.
Write features of hemichordata phylum.
Answer:
Features of hemichordata are the following :
(i) They are worm-like unsegmented organisms.
(ii) They are all marine animals.
(iii) They may be solitary or colonial.
(iv) Their body is divisible into proboscis, collar, and trunk.
(v) They are bilaterally symmetrical.
(vi) Respiration takes place by gill slits.
(vii) Sexes are mostly separate.
Examples: Balanoglossus, Cephalodiscus, etc.

Question 18.
Write major differences of non-chordata and chordata.
Answer:
Non-chordata and chordata have following differences :
Non-Chordates:
1. There is no tail behind the anus.
2. Haemogloblin plasma dissolved in their blood.
3. Heart, if present, is dorsal.
4. Nervous system is ventral and solid.
5. Pharynx gill slits are absent

Chordates:
1. Tail is present behind their anus.
2. Haemoglobin is present in their red blood cells.
3. Heart is ventral and in a pericardial cavity.
4. Nervous system is dorsal and hollow.
5. Pharynx gill slits are present.

Question 19.
Write important features of chordata phylum and classified this phylum. Answer: Important features of chordata phylum are the following :
(i) Presence of nerve chord and hollow nervous system.
(ii) They all have paired gill slits.
(iii) Solid, elastic notochord occurs in all chordata and in any stage of life.
(iv) Presence of tail behind the anus stage.

Classification:
Chordata is divided into three subphyla:
(i) Urochordata
(ii) Cephalochordata
(iii) Vertebrata.

Essay Type Questions

Question 1.
Explain the classification of cryptogamae.
Answer:
Cryptogamae:
These are flowerless and seedless plants. They have hidden reproductive orgAnswer: They do not bear flowers and seeds externally.

They are divided into three divisions:

1. Thailophyta: They do not possess a stem, root or leaf and so called as thallus. Embryo formation after fertilization is absent. Their sex organs are single-celled. There is three clear group of plants in this division :
(A) Algae: Ulothrix, Cladophora, Ulva.
(B) Fungi: Aspergillus, Penicillium, Agaricus.
(C) Lichens: like Lichen perennail, Fruticose lichen.

2. Bryophyta:
The body of these plants are divided into structures like roots, stem or leaf. They grow on moist shady placed. Vascular tissue are absent in them. Their reproductive organs are multicellular. Embryo formation takes place after fertilisation.
Examples: Riccia, Funaria, Marchantia etc.

3. Pteridophyta:
The plant body is differentiated into root, stem and leaves. Vascular tissues are present in them. Their sex organs are multicellular. Embryo formation takes place after fertilization in them.
Examples: Fern (Pteris, selaginella).

Question 2.
Give the names of major phylurii with one or two features with examples.
Answer:
Major features with examples of the animal’s phylum are following:

Practical Work

Experiment 1.
To differentiate between monocots and dicots.

Procedure:
Take some seeds of wheat, gram, pea, maize, rice, tamarind and put into water. When they swollen completely, then classified and note in table. Those have one cotyledon (wheat, maize, rice) are monocots and those have two cotyledons (gram, pea, tamarind) are dicots. Make table to see other difference in them.

Quick Review of the Chapter

1. Father of Taxonomy is:
(A) Aristotle
(B) Robert Hook
(C) De-Candoli
(D) Carolus Linnaeus
Answer:
(D) Carolus Linnaeus

2. Which is not the base for the classification of organisms like Monera?
(A) cell structure
(B) nutrition
(C) body organization
(D) habitat
Answer:
(D) habitat

3. Which animal is not Monera?
(A) diatom
(B) bacteria
(C) microplasma
(D) blue-green algae
Answer:
(A) diatom

4. Respiration in reptiles takes place:
(A) by gills
(B) by skin
(C) by lungs
(D) by skin and slits
Answer:
(C) by lungs

5. Mammalia animal is:
(A) seahorse
(B) chameleon
(C) ostrich
(D) whale
Answer:
(D) whale

6. Skeleton is hollow in:
(A) pisces
(B) amphibians
(C) reptiles
(D) aves
Answer:
(D) aves

7. Who does not contribute in the classification of animals in a large kingdom ?
(A) Ernst Haeckel
(B) Robert Whittaker
(C) Carl Woese
(D) Darwin
Answer:
(D) Darwin

8. Which scientist divided the Monera Kingdom into Archaebacteria and Eubacteria?
(A) Eichler
(B) Carl Woese
(C) Haeckel
(D) Carolus
Answer:
(B) Car! Woese

9. Which of the following is a gymnosperm plant?
(A) mango
(B) pea
(C) cycas
(D) guava
Answer:
(C) cycas

10. Which of the follow ing has outer covering?
(A) unio
(B) pila
(C) chiton
(D) all of the above
Answer:
(D) all of the above

11. Which of the following group have mammary glands?
(A) aves
(B) mammals
(C) reptiles
(D) none of the above
Answer:
(B) mammals

Haryana State Board HBSE 9th Class Science Important Questions Chapter 6 Tissues Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 6 Tissues

Very Short-Answer Type Questions

Question 1.
Are there same types of tissues in plants and animals?
Answer:
No, plants and animals have different types of tissues.

Question 2.
How are newly cells formed?
Answer:
Through cell division process.

Question 3.
What is called as unicellular organisms?
Answer:
All those organisms in which all the biological activities are fulfilled through single cell are called as unicellular organisms.

Question 4.
Which organisms undergo labour division?
Answer:
Multicellular organisms.

Question 5.
How many types of tissues are there in plants?
Answer:
Two types:
(i) Meristematic tissues
(ii) Permanent tissues.

Question 6.
Where are Meristematic cells found?
Answer:
Dividing cells (meristematic tissues) found at the tip of the shoot and at the tip of the root.

Question 7.
What is the function of meristematic tissue?
Answer:
Building up of new cells.

Question 8.
Do the permanent tissues have the ability of divide?
Answer:
No.

Question 9.
How many types of permanent plant tissues are there?
Answer:
Two types:
(i) Simple tissues
(ii) Complex tissues.

Question 10.
What are Parenchymatous tissues ?
Answer:
Parenchymatous tissues can be living and oval, rounded elongated or polygonal in shape. They have thin cell wall and the cells are isodiametric.

Question 11.
What are collenchymatous tissues ?
Answer:
Due to deposition of cellulose and Pectin at the angles of these tissues they get thinken in size. These tissues do not have intercellular spaces.

Question 12.
What are sclerenchymatous tissues ?
Answer:
These are dead cells. These cells do not have any intercellular space in them. They provide strength to the plant parts.

Question 13.
What are simple permanent tissues ?
Answer:
These tissues are formed of parenchymatous cells. Their origin, structural formation and function are same.

Question 14.
Which part of the plant consists of meristematic tissues ?
Answer:
Green part of the plant.

Question 15.
What are protective tissues made up of?
Answer:
By sclerenchymatous tissues.

Question 16.
Which parts of the plant consists of protective tissues.
Answer:
Leaves, flowers, stem and roots.

Question 17.
What is the other name of sclereids ?
Answer:
Brachysclereids or stone cells.

Question 18.
What is called as complex tissue ?
Answer:
Cells of different shape and size together form a group which works out a special function, is called as complex tissue.

Question 19.
How many types of complex tissues are there ?
Answer:
Two type:
(i) Phloem
(ii) Xylem.

Question 20.
Name the tissue that transports food in plants. [H.B.S.E. March, 2019]
Answer:
Sieve elements (phloem).

Question 21.
How many types of animal tissues are there ? Write their names.
Answer:
Four types:
(i) Epithelial tissues
(ii) Connective tissues
(iii) Muscular tissues
(iv) Nervous tissues.

Question 22.
Where are epithelial tissues found ?
Answer:
In the external and internal layers of animals.

Question 23.
Which tissue is called a protective covering in animals ?
Answer:
Epithelial tissue.

Question 24.
Which tissue is responsible for producing sperm and egg in animals ?
Answer:
Epithelial tissue.

Question 25.
What are the types of muscular tissues ?
Answer:
Three types:
(i) striated
(ii) unstriated
(iii) cardiac muscles.

Question 26.
Which muscles function according to owr will ?
Answer:
Voluntary muscles (Striated muscles).

Question 27.
Which type of muscles present in heart ?
Answer:
Cardiac muscles.

Question 28.
How many types of connective tissues are there ?
Answer:
Three.

Question 29.
Write two main functions of connective tissues ?
Answer:
Joining different organs and provide protection.

Question 30.
What are the types of skeletal connective tissues ?
Answer:
Two types:
(i) Bone
(ii) Cartilage.

Question 31.
Which tissue functions to connect bone to another bone?
Answer:
Ligaments.

Question 32.
What do tendons interconnect with?
Answer:
Muscle with bone.

Question 33.
Write down the name of the fluid tissue found in the body.
Answer:
Blood.

Question 34.
How many types of blood cells or corpuscles are found in blood?
Answer:
Three types of blood cells or corpuscles.
They are:
(i) Erythrocytes-Red Blood Cells (RBC)
(ii) Leucocytes: White Blood Cells (WBC) and
(iii) Blood Platelets.

Question 35.
What is the fluid part in blood called as ?
Answer:
Plasma.

Question 36.
What percentage of water is there in blood ?
Answer:
90%

Question 37.
Which tissues are brain, spinal cord and nerves formed of ?
Answer:
Nervous tissues.

Question 38.
What is nervous tissue called as ?
Answer:
Neuron.

Question 39.
What is the main function of neuron ?
Answer:
Sending impulses of messages.

Short-Answer Type Questions

Question 1.
Multicellular organisms are more superior in comparison of unicellular organisms. Why ?
Answer:
The unicellular organism can accomplish but a few functions and in doing that they are comparatively less expert to that of multicellular organisms. Among multicellular organisms by forming groups of lacs of cells not only those functions, rather they can perform more functions with complete skill, since they have the quality of labour-division in them.

Question 2.
Is there some difference between plants and animals tissues ?
Answer:
The plant tissues perform the function of supporting. Most of the tissues of plants are dead. The dead tissues too provide mechanical strength like the living tissues. They need little care of maintenance, where as the animals roam about here and there in search of companion, food and shelter. They consume more energy than plants. They possess maximum number of living tissues. Growth in some parts of the plants remains constrained, whereas in animals the case is different.

Question 3.
What are the different types of tissues found in plants ?
Answer:
Two types of tissues are found in plants:
(1) Meristematic tissues
(2) Permanent tissues

Question 4.
Write down the features of meristematic tissues.
Answer:
The features of meristematic tissues are as follows :
(1) Meristematic cells have same structure and their cell wall is thin.
(2) The shape of the cells is round, oval or polygonal.
(3) They are compactly arranged with each other and they don’t have any intercellular space in them.
(4) These cells have dense or sufficient cytoplasm (cell liquid) and a big nucleus.
(5) They have comparatively few vacuoles or no vacuole.

Question 5.
Where are meristematic tissues found in plants ?
Answer:
Meristematic tissues are found in the growing up parts of the plants only; Like-apical meristem and cambium (the region that increases thickness)

Question 6.
Write down the characteristics of permanent tissues.
Answer:
The characteristics of permanent tissues are as follows:
(1) Permanent tissues are formed of meristematic tissues.
(2) Their cells have no ability of division.
(3) They have a definite shapes.
(4) Their wall can be dead, thin or thick.
(5) They have enlarged cells and vacuolic cytoplasm.

Question 7.
Differentiate between meristematic tissue and permanent tissue.
Answer:
The difference between meristematic tissue and permanent tissue is as follows:
Meristematic Tissue:
1. They do not have intercellular space.
2. They have the ability of divide.
3. The are responsible for growth.
4. They are thin-walled and isodiametric.
5. They have an enlarged nucleus.

Permanent Tissue:
1. It can be or cannot be there in them.
2. They lack the ability of division.
3. They are no responsible for growth.
4. They are thin or thick-walled but not isodiametric.
5. They have small nucleus.

Question 8.
What are permanent tissues?
Answer:
Permanent tissues:
Permanent tissues are formed of meristematic tissues. Their tissues have no ability of division. Their size and volume is definite.

Permanent tissues are of two types:
(1) Simple tissues
(2) Complex or compound tissues.

Question 9.
Differentiate between parenchyma and collenchyma tissues.
Answer:
Following are the differences between parenchyma and collenchyma:
Parenchyma Tissue:
1. They are found in roots, stem and leaves.
2. Their walls are not thick.
3. They are formed of round and fine walled cells.
4. Their wall is made up of pectin and cellulose.
5. They helps in storing food and provide mechanical support.
6. They have intercellular space in them.

Collenchyma Tissue:
1. They are found in the middle vein of the stem and leaves.
2. Their walls are thick.
3. They have thick and elongated wall.
4. Their cell wall possesses collens of pectin.
5. They just provide mechanical support.
6. It is absent in them.

Question 10.
Enunciate different types of simple permanent tissues and write down the function of each. Answer: Simple permanent tissues are of three types :
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma

(a) Functions of Parenchyma:
(i) Storing of food and assimilation.
(ii) To provide mechanical strength to the parts of plant.
(iii) Storing of resin, tanin, gum particles, and resin of morganie waste material.
(iv) Due to presence of chloroplast or chlorenchyma, they prepare food themselves.

(b) Functions of Collenchyma:
(i) They provide flexibility and strength to the plants.
(ii) Due to presence of chloroplast or chlorenchyma, they prepare glucose and starch.

(c) Functions of Sclerenchyma:
(i) They provide more strength to cartex and hard seeds.
(ii) They get transformed into protective tissues and protect the interior parts.

Question 11.
Differentiate among parenchyma, collenchyma and sclerenehyma.
Answer:
Parenchyma:
1. Their cell wall is constructed ot pectin and cellulose.
2. These are living cells.
3. Their function is food-storage and providing support.
4. They are found in root, stem and leaves.
5. They are formed of rounded thin walled cells.

Collenchyma :
1. Their cell wall is also formed of pectin and cellulose.
2. They too have living cells.
3. Their function is to provide mechanical support.
4. They are found in the middle vein of leaf and stem.
5. They are formed of multiarmed cells.

Sclerenchyma:
1. Their cell wall is formed of lignin.
2. Their cells are dead.
3. Their function is only to provide support
4. They are found in the stem and pulp of the fruits.
5. They are formed of thick-walled cells.

Question 12.
Describe Sclerenchvmatous tissue.
Answer:
Sclerenchyma:
These ceils are elongated in size and thin but are equipped with lignin. These are dead cells and are attached together. These cells are pointed at both ends. There is a clear mid lamella between adjacent cells. Cells consist of skewed regions that are called as pit canals. The cells provide strength to the parts of the plant.

Question 13.
Name different components found in Xylem (xylem tissues) components.
Answer:
The four components found in Xylem (xylem tissues) are as follows:
(1) Tracheids,
(2) Vessels,
(3) Xylem or wood Parenchyma
(4) Wood or Xylem fibers.

Question 14.
Give the importance of phloem in plants.
Answer:
Importance of Phloem: It is planted conductive tissue. It transports the starch (food) prepared by leaves to different parts of the plant.

Question 15.
Write functions of the Xylem (xylem tissues).
Answer:
Functions of Xylem:
It is too plant conductive tissue. It’s function is to transport the absorbed water and dissolved mineral salts by the roots to the leaves of the
plants and the apex. It also provide mechanical power to the plant.

Question 16.
Write short note on protective tissue.
Answer:
Protective Tissue:
Sclerenchyma found in the outermost parts of plants gets transformed into protective tissues. These tissues remain in the external layer of leaves, stems and roots. These tissues protect the interior tissues of plants. These tissues are covered with cutin. The cells of cork are dead. Cork is a lighter and protective tissue, it is more compressible and does not catch fire quickly.

Question 17.
What is stomata? Write its functions.
Answer:
Stomata:
They are found in the epidermal cells of the lower surface of leaves of plants which are called as stomatal pore. Stomata get protected by two guard cells.

Functions:
(1) Stomata are the nostrils of plants.
(2) Stomata plays a vital role in the exchange of gases and through them water loss takes place in plants.

Question 18.
Write names of different tissues found in animals.
Answer:
Animal tissues:
There are four types of tissues found in animals.
(1) Epithelial tissue
(2) Connective tissue
(3) Muscular tissue
(4) Nervous tissue.

Question 19.
Give the names and important functions of animal tissues.
Answer:
Animal tissue:
(a) Epithelial tissues:
Functions: They participate in absorption, secretion, protection, excretion, etc.

(b) Connective tissue:
Functions: This tissue participates in interlink the various parts of the body, provide support, turgidity, storage, protection and transportation.

(c) Muscular tissue :
Functions: Muscular tissue participates in motion, conduction, contraction, etc.

(d) Nervous tissue:
Functions: These do the main work of controlling and management.

Question 20.
Write a typical note on epithelial tissue.
Answer:
This is a covering or outer protective tissue that cover the body of an animal. Also, it protects a number of organs and cavities inside the body. Skin, alimentary can be capillary tube’s coat, food pipe, hollow organs of lungs all are formed of epithelial tissues. There is very’ narrow space. In these tissues, the cells are compactly arranged.

Question 21.
Classify Epithelial tissues.
Answer:
On the basis of shape and functions of the cells, epithelial tissues are classified into following types:
1. Squamous epithelium: These flat and disc like cells help in covering the organs of the body providing protection to them.
2. Cubical epithelium: These cubical or isodiametric cells are found in ducts of salivary glands and kidney tubules.
3. Columnar epithelium: The pillar or column like wide cells found in the layer of stomach and intestine.
4. Ciliated epithelium: Columnar and cubical shaped ceils germinate hair like structures at their ends that are called as cilia.
5. Glandular epithelium: These cells are found in secretory glands.

Question 22.
Write down the functions of Epithelial tissue.
Answer:
The functions of Epithelial tissue are the following :
(1) It performs the function to protect and cover the body organs
(2) It performs the functions of secretion, (in the Stomach and salivary glands)
(3) It produces sensitivity. (Eye and retinal cells)
(4) It produces sperm and egg.
(5) These absorb water and other salts.
(6) They also play role in excretion.

Question 23.
Write down the functions of connective tissue.
Answer:
The functions of connective tissues are as follows :
(1) These tissues join different organs of body with one another.
(2) They perform the function to tieup up organs of the body.
(3) These tissues give support to the organs of the body.
(4) Bone and cartilage build up the skeleton of the body.
(5) Adipose tissues store fat.
(6) These tissue perform the function of blood circulation.
(7) White blood corpuscles (WBC) fight against harmful bacteria.

Question 24.
What is Blood?
Answer:
Blood:
Blood is a liquid connective tissue. It tissue drift in liquid plasma (matrix). Blood cells are callei
(i) Red blood corpuscles (RBC)-Erythrocytes
(ii) White blood corpuscles (WBC)-Leucocytes
(iii) Blood platelets
Blood runs in each part of the body and thus, it connects all the parts of the body.

Question 25.
What are the functions of blood ? Explain.
Answer:
Functions of Blood:
(1) Red blood corpuscles receive oxygen from lungs and send it to different parts of the body.
(2) Blood collects hormones from the hormone glands and sends it to different parts.
(3) It helps in excreting excretory products from the body.
(4) White blood corpuscles or leucocytes protects from harmful bacteria.
(5) Blood checks the body temperature.
(6) Blood platelets clots the blood and avoid the more discharge of blood out of the body.

Question 26.
What is Bone? What are its types ?
Answer:
Bone: Skeletal tissue is connective tissue. In this tissue, ’/3 part organic and 2/3 part inorganic material is found. Its matrix consists of a peculiar type of protein ‘ossein’ which is in the form of collagen fibres. Bone possesses matrix osteoblasts, cytus, fibers and matrix have calcium phosphate, calcium carbonate and calcium fluoride. It give strength to the bone.

Question 27.
What is compact bone ?
Answer:
Compact bone:
This bone used to be strong and hard. Inside the matrix of bone, calcium salts provide hardness to the bone. Compact bone is formed of round and irregular dense layers of bones. Compact bone is found in the long shaft of bones like-femur. It involves Haversian canals.

Question 28.
Write down two differences each between bone and cartilage.
Answer:
Following are the differences between bone and cartilage :
Bone Cartilage:
1. It’s matrix is hard and flexible.
2. It’s matrix is formed of condin.
3. It is covered with pericondrium.
4. Haversian system is absent in it.

bone:
1. It’s matrix is too rigid and strong.
2. It’s matrix is formed of ossein.
3. It is covered with peraosteium.
4. Haversian system is present in it.

Question 29.
Write down the difference between tendon and ligament. Answer: Following are the difference between tendon and ligament:
Tendon:
1. It is flexible and moulding.
2. It connects muscle with bone.
3. It is made up of yellow fibre.

Ligament:
1. it is inflexible and hard.
2. It connects bone with another bone.
3. It is made up of white fibred connective tissues.

Question 30.
Write down the functions of bone, cartilage tendon and ligament.
Answer:
(i) Bone : It constructs the structure of body and protects the soft orgAnswer:
(ii) Cartilage: It constructs the skeletons of cartilaginous fishes. In our body the tip of nose and external ear (pinna), is formed of it.
(iii) Tendon : It connects muscle with bone.
(iv) Ligament: It joins bone with another bone.

Question 31.
Write down the functions of skeletal tissue.
Answer:
Skeletal tissue: Skeletal tissue is rigid, there is sufficient quantity of calcium salts in bone-matrix. Following are the significant functions of skeletal tissue:
(1) Skeletal tissue enshapes the body.
(2) Skeletal tissue protect the sensitive organs of the body.
(3) Skeletal tissue helps in joining muscles together.
(4) This tissue provides movement and locomotion to the body.
(5) In the middle of the bones erythrocytes (RBC) are produced.
(6) It provides surface for attachment of muscles (tendons and ligaments).

Question 32.
What is areolar tissue?
Answer:
Areolar tissue:
This tissue is found below the skin, between the muscles, blood vessels and around the nervous tissue. These cells possess jelly types transparent and sticky matrix. This is connective tissue. It interlinks body cavity and other parts of the body.

Question 33.
What is muscular tissue ? What are its types ?
Answer:
Muscular tissue:
The cells of these tissues are elongated, spindle shaped and unbranched. Cardiac muscles are branched. Muscular tissues are found in different sizes. Muscular tissues provide locomotion and movement to different parts of body.

Muscular tissues are of three types :
(a) Striated muscles. (Skeletal or voluntary muscles)
(b) Unstriated muscles. (Smooth muscles or involuntary muscles)
(c) Cardiac muscles.

Question 34.
Make difference between striated muscles arid cardiac muscies.
Answer:
The differences between striated muscies and cardiac muscles are as follows:
Striated muscle:
1. These muscles work accordingly to our will i.e., they are voluntary muscles.
2. These muscles are jointed with bones.
3. They are unbranched.
4. They have a number of nuclei.
5. They have light or dark bands on them.

Cardiac muscle:
1. These muscles are involuntary, they do not work accordingly to our command.
2. These muscles are found in the heart.
3. They are branched.
4. They have one or two nuclei.
5. They have transverse indistinct light bands which show faint cross striations.

Question 35.
Describe the structure of skeletal and cardiac muscles.
Answer:
Skeletal Muscles:
We can move some muscles by the conscious will. Muscles present in our limbs move when we want them to, and stop when we so decide. Such muscles are called voluntary muscles. These muscles are also called skeletal muscles as they are mostly attached to bones and help in body movement. Under the microscope, these muscles show alternative light and dark bands or striations when stained appropriately. As a result, they are also called striated muscles. The cells of this tissue are long, cylindrical, unbranched and multinucleate (having many nuclei).

Cardiac Muscles: The muscles of the heart show rhythmic contraction and relaxation throughout life. These involuntary muscles are called cardiac muscles. Heart muscle cells are cylindrical, branched and uninucleate.

Essay Type Questions

Question 1.
What is called a tissue ? How many types of plant cells are there ? Describe meristematic tissues.
Answer:
Tissue:
A group of several cells that have common origin, structure and function is called a tissue. In plants there are two types of tissues.
(i) Meristematic tissues
(ii) Permanent tissues.

Meristematic Tissue:
In these tissues, division of cells takes places continuously and these cells that have the ability to divide, they keep on reproducing newly cells. Meristematic tissues are found only in those parts that participate in the growth process like tip of the shoot-root, and cambium. Due to these tissues, the length and thickness of the plants get increased.

On the basis of origination, meristematic tissues are of two types:
(i) Primary meristematic tissues.
(ii) Secondary meristematic tissues.

On the basis of position meristematic tissues are of three types:
(a) Apical meristem: They are found at the tip of the root, stem and branches; They are also of three types:
(1) Epidermis: These tissues buiid up the outer skin of the plant.
(2) Cortex: These tissues buiid up the filling up system.
(3) Hypodermis: These tissues make transport tissues, phioem and xylem,
(b) Intercalary meristem: This tissue is found in the stem of monocotyledones. Due to activity of this tissue plant grow up in height.
(c) Lateral meristem: This tissue is found in the stems and root of dicotyledones.

Question 2.
How many types of simple tissue of plant are there ? Explain them.
Answer:
Simple tissue When the meristematic tissues totally lose the ability of division, then they transform into permanent tissues. Permanent tissues are of two types: Simple tissues and complex tissues. Simple tissues are formed of only one type of same structural cells. Their function is also same. On the basis of structure these tissues are of three types:

(i) Parenchyma:
These tissues are found in the shoot system (stem, leaves and flower) and root in plants. In the green part of the plant their number used to be more. These cells can be round, elongated or polygonal shape. These are living cells. Their cell wall is thin and is made up of cellulose. These cells contain dense protoplasm. The main functions of these tissues are storage of food, assimilation, provide strength, storing of resin tanin and gum and many times preparing food too.

(ii) Collenchyma:
These cells too are living. These are multiarmed, oval and rounded shaped. Their cell wall is thin, but at the corners of the cell due to deposition of cellulose and pectin they get hard or stout. In these cells generally there are some chloroplasts. This tissue brings flexibility and strength in plants and prepare glucose and starch due to presence of chloroplast in it.

(iii) Sclerenchyma:
These cells are elongated in size and thin but are equipped with lignin. These are dead cells and are attached together. These cells are pointed at both ends. There is a clear mid lamella between adjacent ceils. Cells consist of askewed regions that are called as pit canals. The cells provide strength to the parts of plant.

Question 3.
What is called as complex tissues ? What are their types in plants ? Explain each of them.
Answer:
Complex Tissue:
Complex tissue is a group of more than one type of cells which perform a special function. Complex tissue is of two types :
(1) Xylem: Xylem is a transport tissue. It transports water and dissolved mineral salts from root to the shoot-system. All the cells of this tissue are dead, thick walled and lignified. This tissue is formed of with the combination of four cells.
(i) Trachieds
(ii) vessels
(iii) xylem or wood-parenchyma
(iv) xylem fibres.