Category: Class 8

Std 8 Science Chapter 2 Health and Diseases Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 8 Science Solutions Chapter 2 Health and Diseases Notes, Textbook Exercise Important Questions and Answers.

Class 8 Science Chapter 2 Health and Diseases Question Answer Maharashtra Board

Class 8 Science Chapter 2 Health and Diseases Textbook Questions and Answers

1. Distinguish between- Infectious and non – infectious diseases.

2. Identify the odd term. 

Question a.
Malaria, hepatitis, elephantiasis, dengue.
Answer:
Hepatitis (All other diseases are caused by the carrier mosquito.)

Question b.
Plague, AIDS, Cholera, T.B.
Answer:
AIDS [All others are bacterial (caused by bacteria) diseases.]

3. Answer in one to two sentences.

Question a.
Which are various media of spreading the infectious diseases?
Answer:
Infectious diseases spread through contaminated air, water, food, vectors such as insects, animals and man.

Question b.
Give the names of five non-infectious diseases other than given in the lesson.
Answer:
Asthma, cataract, diseases of kidney such as kidney stones and renal failure, arthritis, Alzheimer which is a condition during old age, hypertension, migraine, etc.

Question c.
Which are the main reasons of diabetes and heart diseases?
Answer:
Improper lifestyle, wrong type of diet, lack of exercise, excessive mental stress and strain, imbalance in secretion of hormones, etc. are the main reasons for diabetes and heart diseases.

4. What can be achieved/can be prevented?

Question a.
Drinking boiled and filtered water.
Answer:

1. When water is boiled, all the disease causing pathogens present in it are killed.
2. Different diseases like cholera, enteritis, diarrhoea, dysentery, hepatitis, typhoid are caused by such water-borne pathogens.
3. If we boil the water we are protected against all such diseases.
4. If the i water is filtered we can avoid infections from nematode Dracunculus which causes Naru. By boiling and filtering water, even the epidemics by such infectious diseases can be controlled.

Question b.
Avoiding smoking and alcoholism.
Answer:
1. Smoking and alcoholism are two dangerous habits that cause addictions. Alcoholism causes disorders of liver.
2. The addict suffers from malnourishment.

3. His I mental and psychological conditions become abnormal. There is social and familial) impacts too due to alcoholistn.

4. Smoking is an invitation for cancer. In cigarettes/ bidis, there is hazardous nicotine. Nicotine is not only toxic but is also carcinogenic. Cancers of oral cavity, tongue, respiratory tract, lungs is very common among smokers. Therefore, addictions like smoking and alcoholism should always be avoided.

Question c.
Regular balanced diet and exercise.
Answer:
1. Regular and balanced diet results into perfect health. The disease fighting power or immunity of a person is increased due to healthy diet.
2. One can avoid of frequent infections and also mental well¬being is established.
3. Heart disease, diabetes, obesity, etc. can be avoided by not consuming high caloric junk food.
4. Exercise improves blood circulation. Many disorders which are caused by faulty lifestyle can be prevented by having regular balanced diet and exercise, thus one must always follow these for well-being.

Question d.
Proper checking of blood before blood donation.
Answer:
1. There are definite blood groups.
2. During blood transfusion, the donor’s and recipient’s blood should be well-matched with each other. Otherwise, blood gets clumped inside the body of recipient.
3. Through infected blood, the viruses such as those causing hepatitis B or AIDS are transmitted to other healthy persons. Thus, for prevention of transmission of such diseases, blood should be checked before blood donation.

5. Read the passage and answer the questions.

Master X’ is a 3-year-old child. He is living with his family in a slum. The public toilet is present near his house. His father is drunkard. His mother does not know the importance of balanced diet.

Question a.
Master ‘X’ can suffer from which different possible diseases in above conditions?
Answer:
Master ‘X’ stays in an area which is devoid of cleanliness. The public toilet is near his house. This indicates that he will have to fight against many infectious diseases. Because his father is a drunkard, there will always be a dearth of money in r the house. He may not be able to afford, enough and balanced food. Thus his diet must be deficient in vitamins and minerals.

Moreover, his mother is ignorant about the importance of health and a balanced diet. This must be causing Master ‘X’ malnourishment and loss of immunity. The financial conditions and the addition of the father must be causing stress in the house.

This will further add stress on Master ‘X’ resulting into susceptibility to infections. As it is due to the location the germs are around and thus they may be attacking Master ‘X’. He will on and off suffer from digestive disorders! such as typhoid, hepatitis, cholera, enteritis, etc.

Question b.
How will you help him and his family in this situation?
Answer:
Initially, we shall help Master ‘X’ to procure healthy and balanced diet having fruits, vegetables, milk, etc. His mother will be taught importance of the balanced diet. The surroundings should be clean and hygienic. Food should be covered, houseflies should not be allowed to contaminate the food. Germicidal floor cleaners and disinfectants should be used to keep off the houseflies and cockroaches. His father will be taken to de-addiction centres such as ‘Alcoholics Anonymous’. He will be persuaded to leave alcoholism.

Question c.
Which disease can occur to the i father of master ‘X’?
Answer:
Father of Master ‘X’ can develop diseases of liver and kidney.

6. Give the preventive measures of following diseases:

Question a.
Dengue:
Answer:
Dengue is transmitted through bite of mosquito of Aedes aegypti. DEN-1, 2 virus belonging to the type – flavivirus is the causative pathogen for Dengue. Wherever there is stagnant or accumulated water, there is possibility of mosquito breeding. Therefore, care is to be taken to drain of such water. Thus this is a very important preventive measure. Especially, in the man¬made containers and in clean water, the Aedes mosquito prefers to breed.

Therefore, such water storages should be either covered or should be decanted. Another way to keep off from dengue is to increase our immunity to fight against the virus. There is vaccine called CYD-TDV or Dengvaxia which is synthesised in 2017. But it is still not considered to be completely safe.

Question b.
Cancer:
Answer:
The most important preventive measure is to remain away from the carcinogenic substances. Tobacco, gutkha, cigarette, bidi, etc. are addictive things which are very bad for our health. The nicotine present in these cause cancers of oral cavity and of respiratory system. Radiations can also cause cancer.

We should not expose ourselves to hazardous radiations. Balanced and healthy diet, proper exercise and living stress free with mental balance are some of the preventive measures of the cancer. Only for few types of cancer like cervical cancer in women, vaccine has been developed.

Question c.
AIDS:
Answer:
When blood transfusion is done, the blood should be checked for the presence of HIV. The used syringes, needles, etc. should not be used without sterilization. Through blood and blood products and body fluids HIV finds its way into the body. Therefore, these precautions should be taken. Unsafe sexual contact is the most important mode of transmission of HIV. Thus one should never deal with such dangerous acts. Awareness about AIDS and HIV is the real preventive measure.

7. Explain the importance.

Question a.
Balanced diet.
Answer:
The diet that contains all the nutrients in the balanced proportion is called balanced diet. One can avoid malnutrition by taking balanced diet. The immunity increases due to balanced diet containing good proportion of vitamins and minerals. Some diseases can be avoided due to raised immunity. Wrong lifestyle and wrong diet can cause hypertension, diabetes or heart disease. To maintain our health and keep the body in equilibrium, we have to take balanced diet.

Question b.
Physical exercise/Yogasanas.
Answer:
By exercise and yoga, the blood circulation of body is improved. Body and joints remain flexible. Mental strain and stress is reduced. Insomnia (inability to fall asleep), arthritis, indigestion, and some other disorders can be avoided. The person who performs exercise, always remains away from the addictions. Yoga makes levels of hormones, enzymes, etc. in equilibrium. By keeping control over breathing through pranayama many respiratory and circulatory disorders can be prevented.

8. Make a list. 

Question a.
Viral diseases.
Answer:
AIDS, Hepatitis, Influenza, Rabies, Polio.

Question b.
Bacterial diseases.
Answer:
Typhoid, Tuberculosis, Cholera, Leprosy.

Question c.
Diseases spread through insects.
Answer:
Malaria, Dengue, Elephantiasis.

Question d.
Hereditary diseases.
Answer:
Diabetes, Haemophilia, Muscular dystrophy, Colourblindness.

9. Write the information on modern diagnostics and treatments of cancer.

Question a.
Answer:
1. Following methods are used as diagnostic methods to detect the cancer. Techniques like CT scan, MRI scan, mammography, biopsy.
Treatment of the cancer is done by the following methods:

2. For treatment of cancer, some conventional methods are used. Along with these methods, chemotherapy, radiation therapy and surgery are commonly used to treat cancerous growth and tumours. New and modern techniques of robotic and laparoscopic surgery are also used for the treatment.

10. Enlist the names and composition of the medicines present at your home.

Question 1.
Prepare posters giving information about various diseases, public awareness and arrange exhibition in school.

Question 2.
Visit the public health center/clinic nearby and collect the information about vaccination.

Question 3.
Compose a street-play to increase public awareness about dengue, malaria, swine flu and present ¡tin the area nearby your school.

Project:

Question 1.
Prepare posters giving information about various diseases, public awareness and arrange exhibition in school.

Question 2.
Visit the public health center/clinic nearby and collect the information about vaccination.

Question 3.
Compose a street-play to increase public awareness about dengue, malaria, swine flu and present it in the area nearby your school.

Class 8 Science Chapter 2 Health and Diseases Additional Important Questions and Answers

Rewrite the sentences after filling the blanks:

Question 1.
For eradication of tuberculosis everybody should be vaccinated by …………. vaccine.
Answer:
For eradication of tuberculosis everybody j should be vaccinated by B.C.G vaccine.

Question 2.
Hepatitis B can be transmitted by ………….
Answer:
Hepatitis B can be transmitted by blood transfusion.

Question 3.
For treating diarrhoea/dysentery…………. is given to the patient.
Answer:
For treating diarrhoea/dysentery O.R.S. is given to the patient.

Question 4.
…………. virus is responsible for the infection of swine flu.
Answer:
Influenza A (H1N1) virus is responsible for the infection of swine flu.

Question 5.
…………. mosquito spreads dengue.
Answer:
Aedes aegypti mosquito spreads dengue.

Question 6.
Malaria is caused by female mosquito of …………. genus.
Answer:
Malaria is caused by female mosquito of Anopheles genus.

Question 7.
…………. test is performed on blood to diagnose AIDS.
Answer:
ELISA test is performed on blood to diagnose AIDS.

One word in the following sentences is wrong. Change it to make the sentences correct:

Question 1.
Culex mosquito is seen in clean water.
Answer:
Culex mosquito is seen in dirty water.

Question 2.
Transmission of swine flu is done by dogs and human beings.
Answer:
Transmission of swine flu is done by pigs and human beings.

Question 3.
Hydrophobia is the main symptom of typhoid.
Answer:
Hydrophobia is the main symptom of rabies.

Question 4.
The maximum number of patients of diabetes are found in America.
Answer:
The maximum number of patients of diabetes are found in India.

Question 5.
Generic medicines are also called branded medicines.
Answer:
Generic medicines are also called general medicines.

Question 6.
If the secretion of insulin is increased, then the diabetes is caused.
Answer:
If the secretion of insulin is decreased, then the diabetes is caused.

Question 7.
Virus causing rabies enter the brain through food.
Answer:
Virus causing rabies enter the brain through neurons.

Match the columns:

Question 1.

Answer:

Question 2.

Answer:

Identify the odd term:

Question 1.
Chemotherapy, Radiation therapy, Surgery, Pranayama
Answer:
Pranayama (All others are treatment for cancer.)

Question 2.
Chronic cough, Hoarse voice, Uncontrolled blood sugar level, Difficulty in swallowing.
Answer:
Uncontrolled blood sugar level (All others are symptoms of cancer of throat.)

Question 3.
Mantra, Vaccination, Black magic, Hatred.
Answer:
Vaccination (This is the only way for prevention of diseases. Rest of the others are blind faiths which are unscientific.)

Consider the relation between the items in the first pair and write the correlation for second pair:

1. Swine flu : Virus : :
Tuberculosis : : …………….
Answer:
Bacteria

2. Anopheles : Malaria :
Culex : …………………
Answer:
Elephantiasis

3. Housefly : Typhoid : : Dog : ………….
Answer:
Rabies

4. Decrease in blood platelets : :
Dengue : : Difficulty in breathing : ……………
Answer:
Swine flu

5. Cancer : Laparoscopic surgery : :
Heart disease :……………
Answer:
By-Pass surgery.

Distinguish between the following:

Question 1.
Typhoid and Tuberculosis :
Answer:

What can be achieved/can be prevented?/Give reasons.

Question 1.
People suffering from communicable diseases should avoid going to public places.
Answer:

1. Communicable diseases spread when people share space with a diseased person.
2. If people suffering from l communicable diseases go to a public place, they would spread the disease causing germs in the air.
3. These germs can be transmitted to other healthy persons.
4. If the resistance power of the persons is less, they will fall sick by getting infected.
5. This may break into an epidemic. Therefore, people suffering from communicable diseases should avoid going to public places.

Question 2.
Pet animals should be given anti-rabies vaccine.
Answer:

1. We keep animals like cats and dogs at home as pets.
2. They may get infected with the virus of rabies.
3. If such an infected pet bites us, we may also get rabies.
4. Rabies is a lethal disease.
5. In order to protect the animals and to prevent rabies being transmitted, they should be given anti-rabies vaccine.

Question 3.
Tuberculosis is considered as the most communicable disease.
Answer:

1. Tuberculosis is caused due to bacterial infection.
2. It is an air-borne infection and spreads through the spittle of T.B. patient.
3. When personal and public hygiene is not followed, this disease spreads very rapidly.
4. It is estimated that every two minutes, one patient dies of T.B. in India. Tuberculosis, therefore, is considered as the most communicable disease.

Question 4.
ORS gives temporary relief to the patient of diarrhoea.
Answer:

1. In diarrhoea, due to loose motions the patient becomes dehydrated.
2. The eyes get sunken and the mouth becomes dry.
3. If immediate help is not given, the condition can become critical. The absorption function of the intestine is disturbed.
4. But by giving ORS, enough water, sugar and salt enter the body saving the life of the patient.
5. The sugar and salt in the ORS gets absorbed in the intestine and reduce the dehydration.
6. However, for eradication of infection, proper medical treatment is to be given, therefore, ORS gives only a temporary relief to the patient of diarrhoea.

Answer the following questions in one sentence:

Question 1.
Which virus causes dengue and how does it spread?
Answer:
Dengue is caused by DEN-1, 2 virus belonging to the type – flavivirus. Aedes aegypti type of mosquito spreads the disease by transferring this virus through its bite.

Question 2.
What are the reasons for spread of swine flu?
Answer:
Swine flu is caused by the virus influenza A (H1N1). Its carrier is pig. The contact with such pigs or infected persons can cause infection of swine flu. The infection spreads through sweat and through secretions of nose, throat and saliva of the diseased person.

Question 3.
What are the symptoms of swine flu?
Answer:
Symptoms of swine flu : Palpitations, difficulty in breathing, sore throat, body pain along with high fever.

Question 4.
In whom was HIV first reported?
Answer:
HIV was first reported in an African species of monkeys

Question 5.
What is a malignant tumour?
Answer:
Due to uncontrolled cell division
there is a formation of lump of cancerous cells. If this lump is without the covering, then it is called malignant tumour.

Question 6.
How is efficiency of the heart reduced?
Answer:
When the blood circulation for the heart muscles is obstructed, it does not receive enough oxygen and nutrients. Due to this condition heart has to exert more and gets stressed resulting into loss of efficiency.

Question 7.
What are the different types of treatment for heart diseases?
Answer:
Angioplasty, By-Pass surgery, open heart surgery, installation of stents, installation of pacemaker, heart transplant are some of the treatment for heart diseases.

Question 8.
What is Pradhan Mantri Jan Aushadhi Yojana? When was it started?
Answer:
Government of India started the Pradhan Mantri Jan Aushadhi Yojana on 1st July 2015. Under this scheme people who are not able to afford best quality medicines are given generic medicines of the same quality at much reduced price.

Question 9.
When can eye donation be done? What is the advantage of eye donation?
Answer:
Anybody can donate eyes posthumously. Due to eye donation, some blind person can regain the sight.

Explain the importance:

Question 1.
Generic medicines.
Answer:
Generic medicines are also called general medicines. They are affordable for the common citizens of India. These medicines are manufactured and distributed without any patent. They are similar in quality and composition as the branded i medicines. The proportion of compounds in these medicines and its formula of preparation is readily available. Thus the money spent on the research is reduced. Therefore, generic medicines are much cheaper than the expensive branded medicines.

Question 2.
Blood donation.
Answer:
Blood donation is said to be the greatest donation that one can give. Blood donation can save someone’s life. By one unit blood of a single donor, three different patients can be saved. From such blood red blood cells, white blood cells and blood i platelets can be separated and given to different patients who are in need with corresponding component. Blood cannot be manufactured artificially and hence blood donation is the only way to collect blood. One healthy person can donate blood four r times in a year, thus saving 12 patients.

Question 3.
First Aid for Heart Disease.
Answer:
A person who is suffering from cardiac problems can get a heart attack suddenly. In such case, the patient should be given immediate and appropriate treatment to save his or her life. First and foremost one should call 108 for cardiac ambulance. Then the patient’s pulse or heartbeats should be checked to know his/her condition. Check the consciousness of the patient. Then make the patient to lie on a hard surface in horizontal position.

Perform C.O.L.S. or, ‘Compression Only Life Support’ on the patient. During this procedure, a pressure is given in the centre of the thorax of a patient for at least 30 times at the rate of 100 to 120 r strokes per minute. This is a first aid for the r heart attack. Before the arrival of doctor or proper medical aid, the life of a patient is saved.

Answer the following questions:

Question 1.
Write a brief account about tuberculosis.
Answer:
Mycobacterium tuberculae is the bacteria which causes tuberculosis (TB) infection. This infectious and communicable disease passes through spittle of the patient. The sneezing and coughing of patient cause f droplet infections. With prolonged contact with the patient and using the objects of the patient there is greater chance of acquiring TB infection. Different body organs such as bones, uterus and lungs can be affected by tuberculosis.

The pulmonary TB i.e. TB of the lungs is more common. The symptoms of TB are prolonged cough, blood through spittle, decrease in the body weight and difficulty in breathing. B.C.G. vaccine can be given in young age for prevention of TB. The patient needs to be isolated if TB is detected. Regular antibiotics have to be taken. DOTS process of medication needs to be followed.

Question 2.
Write the symptoms of dengue.
Answer:
Symptoms of dengue are as follows:

1. High and severe fever.
2. Severe headache accompanied with vomiting.
3. Pain in the eye sockets as if someone is pushing the eyes out.
4. Rapid reduction in the number of blood platelets.
5. Internal haemorrhages in the body.

Question 3.
How is rabies caused? What are the important symptoms of the rabies?
Answer:
Rabies is a viral disease. The bite of a rabid animal (the animal who is already infected with rabies virus) transfers the virus in the human blood through the bite wound. Dog, rabbit, monkey and cat can be infected with rabies if they are not vaccinated. The virus of rabies enters the brain through neurons. If not treated in time, rabies can be fatal.

Symptoms of rabies: There is a fever which lasts for 2 to 12 weeks. The victim becomes hyper-excited. He /She shows exaggerations in behaviour. Hydrophobia is very unique symptom in which person is scared and phobic of water. The symptoms can be seen within 90 to 175 days after the animal-bite.
When the dog or any animal bites, before the symptoms are seen, the victim should immediately take anti-rabies vaccine. This vaccination can save a person.

Question 4.
What are the causes of cancer?
Answer:
The carcinogenic substances can cause cancer. Thus such substances should never be used. Tobacco, gutkha and other nicotine containing products can cause cancer. Smoking and alcoholism can also induce cancer development. Cancer is also hereditary disease which is said to be caused by oncogenes. The lack of high fibre food in the diet and junk food in great quantities can induce the cancer development in some part of the alimentary canal.

Question 5.
Write the symptoms of cancer.
Answer:

1. If part of the respiratory system is under the attack of cancer, it can cause, long term cough, hoarse voice, difficulty in swallowing or difficulty in breathing.
2. The wounds do not become cured and the scar too is incurable. Inflammations does not subside.
3. There is formation of tumour like lump in the breast of the females.
4. In all the types of cancer there is unexplained weight loss.

Question 6.
What are the preventive measures to avoid the cancer?
Answer:
Complete control over diet should be followed. It should be’ healthy and nutritive. Such diet protects one from the cancer. The modern treatment processes should be immediately carried out, if the cancer is detected. With time-lapse, the growth of the cancer keeps on increasing. Physical exercises should be regularly done. Tobacco consumption, smoking are some addictions that can induce cancer. These should never be tried.

Question 7.
What are the causes and symptoms of diabetes?
Answer:
Causes of diabetes: The pancreas in normal persons secrete hormone insulin which maintains the sugar level in the blood. When there is deficiency of insulin, the symptoms of diabetes develop due to uncontrolled blood sugar level. Hereditary causes also can develop diabetes.

Obesity, lack of physical exercise, mental stress can also lead to diabetes. Overconsumption of fried food and over-indulgence in eating sweetmeats increase the chances of diabetes.

Symptoms of diabetes: The person who develops diabetes has frequent urination and continuous thirst. The body weight also reduces rapidly. Sometimes it may rise suddenly. There is continuous feeling of fatigue.

Question 8.
What are the causes of heart diseases?
Answer:
Addictions such as smoking, alcoholism and junk food lead to heart disease. Diabetes, hypertension, obesity are responsible for developing heart disease. Lack of physical exercise, no physical work, continuous sedentary mode are some causes which reduce the efficiency of the working of heart. Mental strain and stress, anger, frustration, anxiety are some psychological reasons that can lead to heart problems

Question 9.
What happens when medicines are missed?
Answer:
When medicines are taken without the consultation of doctor, there are certain problems. When the medicines and drugs are taken without any cause or symptom, then it is a misuse. Overdose of pain killers produce side effects. Some of these cause damage to nervous system, excretory system and liver. Overdose of antibiotics develops nausea, stomachache, loose motions and rash, white patches on tongue, etc.

Questions based on charts/tables:

Question 1.
Complete the table: Prepare similar table of information as the table given on page 7 about various disease like enteritis, malaria, plague and leprosy.
Answer:

Question 2.
Draw a flow diagram to show treatment for heart diseases.
Answer:

Activity-based questions:

Question 1.
Enlist the names and composition of the medicines present at your home.
Answer:
Students are expected to write this answer based on the medicines in their own homes.

Question 2.
Observe and discuss :

Answer:
Observe the following pictures and write description in the boxes.

Open-Ended Questions:

Question 1.
Poor people do not afford the costly medicines. Is there any alternative for this? Which one?

Answer:
Some medicines are very costly. The poor people cannot afford such expensive medicines. In such cases, some social organizations may give some help to these patients. Some trusts like Tata trust can also offer help to the needy for the treatment.

Some hospitals have social workers who help the needy for raising the funds for poor patients. Another way is to use generic medicines.

These medicines are at par in quality with the branded medicines but they are quite less in the price. They are available in the generic medicine shops. One can also take benefit of Pradhan Mantri Jan Aushadhi Yojana.

Question 2.
What measures will you take to stop the breeding of mosquitoes?
Answer:
We have to take care that there is no water stagnated in and around the house. The garden plants and pots in the house should be checked for the mosquito larvae and eggs if any. If water is not receding and remaining accumulated, then the spraying of insecticides should be done with the help of elders. With the help of elders in the house, the complaint can be lodged in the Malaria eradication department in the nearby Municipality or Gram Panachyat office.

If the water is in ponds, we can release guppy fishes to eradicate the mosquito larvae. The fish feed on the larvae and automatically control the mosquito population.

Question 3.
Our behaviour with HIV infected person must be normal or should not be normal? What is your opinion? Write it with correct explanation.
Answer:
Our behaviour with HIV infected person must be normal. There are no chances of acquiring AIDS by having normal behaviour with the HIV positive person. HIV can pass only through body fluids and blood from HIV infected person to the other normal person. Similarly, the needles or syringes used by HIV infected person can cause the infection of HIV to the other person.

Such ways of contamination should always be avoided. The sexual contact or the blood transfusion with HIV positive person should always be avoided. But the normal behaviour such as hand-shake, eating together, studying in the same class will not matter at all in the transmission of AIDS.

Balbharati Maharashtra State Board 8th Std Science Textbook Solutions 

• Living World and Classification of Microbes Class 8 Science Textbook Solutions
• Health and Diseases Class 8 Science Textbook Solutions
• Force and Pressure Class 8 Science Textbook Solutions
• Current Electricity and Magnetism Class 8 Science Textbook Solutions
• Inside the Atom Class 8 Science Textbook Solutions
• Composition of Matter Class 8 Science Textbook Solutions
• Metals and Nonmetals Class 8 Science Textbook Solutions
• Pollution Class 8 Science Textbook Solutions
• Disaster Management Class 8 Science Textbook Solutions
• Cell and Cell Organelles Class 8 Science Textbook Solutions

Std 8 Science Chapter 3 Force and Pressure Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 8 Science Solutions Chapter 3 Force and Pressure Notes, Textbook Exercise Important Questions and Answers.

Class 8 Science Chapter 3 Force and Pressure Question Answer Maharashtra Board

Class 8 Science Chapter 3 Force and Pressure Textbook Questions and Answers

1. Write proper word In the blank space:

Question a.
The SI unit of force is the ………..
(dyne, newton, joule)
Answer:
The SI unit of force is the newton.

Question b.
The air pressure on our body is equal to the …………. pressure.
(atmospheric, sea bottom, space)
Answer:
The air pressure on our body is equal to the atmospheric pressure.

Question c.
For a given object, the buoyant force in liquids of different ………… is ………….. .
(the same, density, different, area)
Answer:
For a given object, the buoyant force in liquids of different density is the same.

Question d.
The SI unit of pressure is ………………
(N/m³, N/m², kg/m², Pa/m²)
Answer:
The SI unit of pressure is N/m².

2. Make a match.

Question a.

Answer:

3. Answer the following questions in brief. 

Question a.
A plastic cube is released in water. Will it sink or come to the surface of water?
Answer:
It will come to the surface of water.
[Note: This is because its density is less than that of water. When it floats, the unbalanced force acting on it is zero.]

Question b.
Why do the load carrying heavy vehicles have large number of wheels?
Answer:
The pressure produced by a given force depends on the area of the surface on which the force acts. The greater the surface area, the less is the pressure produced. Load carrying heavy vehicles have large number of wheels so that the load (weight, force) is distributed over large surface area of the wheels in contact with the road. Hence, the pressure decreases and the tyres do not burst.

Try this :
Pressure of a liquid:

Activity 1:
Take a plastic bottle. Take a 10 cm long piece of a glass tube on which a rubber balloon can be fitted. Warm up one end of the glass tube and gently push it into the bottle at about balloon inflates. What is observed? The pressure of water acts on the side of the bottle as well.

[Note: Here, the area of cross section of the tube remains the same. As the level of the water in the bottle rises, the mass of the water increases resulting in increase in the weight. As the applied force increases, the pressure increases. Therefore, the balloon increases in size.]

Activity 2:
Take a plastic bottle. Pierce it with a thick needle (or with a hot nail) at the points 1, 2, 3 as shown in the Fig. Fill water in the bottle up to full height. A shown in the figure, water jets will be seen emerging and projecting out. The water jet emerging from the hole at the top will fall closest to the bottle. The jet from the lowest hole falls farthest from the bottle.

Also, jets coming out from the two holes at the same level fall at the same distance from the bottle. What is understood from this? At any one level, the liquid pressure is the same. Also, the pressure increases as the depth of the liquid increases.

Question c.
How much pressure do we carry on our heads? Why don’t we feel it?
Answer:
The air pressure at the sea level is about 101 × 10³ Pa. This is the pressure that we carry on our heads. The cavities in our body are filled with air, and arteries and veins are filled with blood. Their pressure balances the pressure due to the atmosphere. Hence, we don’t feel the atmospheric pressure.

4. Why does it happen?

Question a.
Why does it happen? A ship dips to a larger depth in freshwater as compared to marine water.
Answer:
The buoyant force acting on a body is proportional to the density of the fluid in which the body is immersed. The density of freshwater is less than that of marine water. Hence, the buoyant force on a body in freshwater is less than that in marine water. Therefore, a ship dips to a larger depth in freshwater as compared to marine water.

Question b.
Why does it happen? Fruits can easily he cut with a sharp knife.
Answer:

1. It is easy to cut vegetables. fruits with a sharp knife. A blunt knife does not work here.
2. The force exerted perpendicularly on a unit area is called ‘pressure’
   
3. Presently we are considering only the force acting on an area in a direction perpendicular to it.

Question c.
Why does it happen? The wall of a dam is broad at its base.
Answer:
1. The pressure at a point in a liquid is proportional to the height of the liquid column above it. Hence, the pressure of water in a dam is much greater at the bottom of the dam than at the top.
2. To withstand this high pressure, the wall of a dam is made stronger and thicker (broad) at the base than at the top.

Question d.
Why does it happen? If a stationary bus suddenly speeds up, passengers are thrown in the backward direction.
Answer:
1. When passengers sit or stand in a stationary bus, they are in a state of rest. When the bus suddenly speeds up, the lower (parts of their body in contact with the bus acquire the speed of the bus.
2. The upper parts of their body, however, continue to be in the state of rest due to inertia. Hence, they are thrown in the backward direction.

5. Complete the following tables. 

Question a.

Answer:
Using the formula,
density = mass/volume:

Question b.

Answer:
Using the formula, relative density = density of a metal/density of water:

Question c.

Answer:
Using the formula, pressure = weight/area:

6. The density of a metal is 10.8 × 10³ kg/m³. Find the relative density of the metal.

Question a.
The density of a metal is 10.8 × 10³ kg/m³. Find the relative density of the metal.
Solution:
Data: Density of the metal = 10.8 × 10³ kg/m³
density of water = 10³ kg/m³
relative density of the metal = ?

The relative density of the metal = 10.8.

7. The volume of an object is 20 cm³ and the mass is 50 g. The density of water is 1 gm^-3. Will the object float on water or sink in water?

Question a.
The volume of an object is 20 cm³ and the mass is 50 g. The density of water is 1 gm^-3. Will the object float on water or sink in water?
Solution :

It is greater than the density of water.
Hence, the object will sink in water.

8. The volume of a plastic-covered sealed box is 350 cm³ and the box has a mass 500 g. Will the box float on water or sink in water? What will be the mass of water displaced by the box?

Question a.
The volume of a plastic-covered sealed box is 350 cm³ and the box has a mass 500 g. Will the box float on water or sink in water? What will be the mass of water displaced by the box?
Solution:

It is greater than that of water.
Hence, the box will sink in water.
The volume of water displaced by the box (V) = the volume of the box = 350 cm³

∴ The mass of water displaced by the box = 1 g.cm^-3 × 350 cm³ = 350 g.

Project:

Question a.
Video record all the experiments (Try it) in this chapter with the help of mobile phone and send to others.

Class 8 Science Chapter 3 Force and Pressure Additional Important Questions and Answers

1. Rewrite the sentences after filling the blanks

Question 1.
The tendency of an object to remain in its existing state is called its …………. .
Answer:
The tendency of an object to remain in its existing state is called its inertia.

Question 2.
Pressure = ……………
Answer:

Question 3.
1 bar = ……………….. N/m².
Answer:
1 bar = 10⁵ N/m².

Question 4.
1 atmosphere = ………….. Pa.
Answer:
1 atmosphere = 101 × 10³ Pa.

Question 5.
The SI unit of density is …………. .
Answer:
The SI unit of density is kg/m³

Write proper word In the blank space:

Question 1.
According to Archimedes’ principle, the magnitude of the force of buoyancy acting on a body is …………. .
(Vρg, Vρ/g, Vρm, mρg)
Answer:
According to Archimedes’ principle, the magnitude of the force of buoyancy acting on a body is Vρg.

Question 2.
The pascal is the unit of ……………. .
(velocity, pressure, mass, force)
Answer:
The pascal is the unit of pressure.

Question 3.
Keeping the surface area constant, if the applied force is doubled, the pressure …………….. .
(becomes double, remains the same, becomes four times, becomes half)
Answer:
Keeping the surface area constant, if the applied force is doubled, the pressure becomes double.

State whether the following statements are True or False:

Question 1.
The density of water is 1000 g/cm³.
Answer:
False. [The density of water is 1000 kg/m³ (or 1 g/cm³)]

Question 2.
Force and weight have the same units.
Answer:
True.

Question 3.
Atmospheric pressure at sea level is about 10⁶ dynes/cm².
Answer:
True.

Question 4.
The buoyant force due to a liquid is proportional to the acceleration due to gravity.
Answer:
True.

Question 5.
Atmospheric pressure increases with altitude.
Answer:
False. (Atmospheric pressure decreases with altitude.)

Question 6.
Pressure due to a given force is directly proportional to the area on which the force acts.
Answer:
False. (Pressure due to a given force is inversely proportional to the area on which the force acts.)

Question 7.
When a body is completely immersed in a liquid, the buoyant force acting on it due to the liquid is proportional to the volume of the liquid displaced by the body.
Answer:
True.

Question 8.
The density of a material is useful to determine its purity.
Answer:
True.

Question 9.
One tends to slip over a banana peel on the street and one can slip due to mud are events that occur due to reduced friction.
Answer:
True.

Question 10.
Frictional force is electromagnetic in origin.
Answer:
True.

Identify the odd term:

Question 1.
Density, Pressure exerted by a gas, Mass, Force.
Answer:
Force. (Force is a vector quantity; other quantities are scalar quantities.)

Question 2.
Lactometer, Hydrometer, Voltmeter, Submarine.
Answer:
Voltmeter. (Its working is not based on Archimedes’ principle. The working of a lactometer, hydrometer and submarine is based on Archimedes’ principle.)

Rewrite the following table in such a way that Column 2 and Column 3 will match with Column 1:

Question 1.

Answer:

Answer the following questions in one sentence:

Question 1.
Which of the following has more inertia? A ₹ 10 coin and a ₹ 1 coin.
Answer:
A ₹ 10 coin has more inertia than a ₹ 1 coin.

Question 2.
Name the physical quantity expressed in pascal.
Answer:
Pressure is expressed in pascal.

Question 3.
State the SI unit of pressure.
Answer:
The SI unit of pressure is N/m², also called the pascal.

Question 4.
Name the property of a liquid due to which it exerts an upward force on an object immersed in it.
Answer:
Buoyancy is the property of a liquid due to which it exerts an upward force on an object immersed in it.

Question 5.
Name the principle used in designing ships and submarines.
Answer:
Archimedes’ principle is used in designing ships and submarines.

Question 6.
What is specific gravity?
Answer:
The specific gravity of a substance is another name used for relative density, i. e., the ratio of the density of the substance to the density of water.

Question 7.
State any one factor on which the pressure exerted by a liquid at a point inside the liquid depends.
Answer:
The pressure exerted by a liquid at a point inside the liquid depends on the density of the liquid.

Question 8.
State any one factor on which the buoyant force due to a liquid depends.
Answer:
The buoyant force due to a liquid depends on the density of the liquid.

Question 9.
Name the device used to determine the purity of a sample of milk.
Answer:
The lactometer is used to determine the purity of a sample of milk.

Question 10.
Name the device used to determine the density of a liquid.
Answer:
The hydrometer is used to determine the density of a liquid.

Question 11.
Name two instruments whose working is based on Archimedes’ principle.
Answer:
Working of the lactometer and hydrometer is based on Archimedes’ principle.

Answer the following questions:

Question 1.
Give three examples to show that a force acts on two bodies through an interaction between them.
Answer:

1. Consider a car at rest on a level (plane) road. If it is pushed from behind, it moves in the forward direction.
2. Iron nails get attracted to the poles of a magnet and stick to the magnet.
3. The moon revolves around the earth.

Question 2.
What is a contact force? Give one example.
Answer:
A force that acts through a direct contact of two objects or via one more object, is called a contact force.
Example: If a ball at rest on the ground is kicked, it starts moving.

Question 3.
What is a non contact force? Give one example.
Answer:
A force that acts between two objects even if the two objects are not in contact, is called a non contact force.
Example: The earth revolves around the Sun.

Use your brainpower!
Make a list of some more examples in which contact and non contact forces are applied. Write the types of force.
Answer:
1. Some examples in which contact forces are applied:

• to cut an apple with a knife (muscular force, frictional force)
• to lift a ball lying on the ground (muscular force, frictional force)

2. Some examples in which non contact forces are applied:

• the motion of the earth around the Sun (gravitational force)
• the motion of an electron around the nucleus of an atom (mainly the electric force).

Question 4.
In the following examples, state whether the force is a contact force or non contact force:

1. a reluctant dog is being pulled by his master
2. a boy playing football is kicking the ball away
3. when iron nails are brought near a magnet, they are attracted to the poles of the magnet and stick to the magnet
4. a coconut is falling from the coconut tree
5. when a comb is rubbed against hair, small pieces of paper kept on a table get attracted to the comb
6. when brakes are applied to a moving bicycle, it stops after some time

Answer:

1. contact force
2. contact force
3. non contact force
4. non contact force
5. non contact force
6. contact force.

Question 5.
Give one example in which frictional force is useful.
Answer:
While walking, we push the ground behind with our feet. In the absence of friction between the ground and the lower surface of our feet, we will slip and will not be able to walk.
[Note: Frictional force is electromagnetic in origin.]

Try this: 
Take two plastic bottles with rectangular shape. Close their openings by fitting the lids tightly. Keep two small bar magnets on them and fix them neatly using a sticking tape.

Fill a big plastic tray with water and leave the two bottles floating with magnets at the top. Take one bottle near the other. If the north pole of the magnet is near the south pole of the other magnet, the bottles will head towards each other, because unlike poles attract each other.

Observe what will happen when the directions of the bottles are changed. We can observe change in the motion of the bottles without any direct contact. This means that there exists a non contact force between the two magnets.

Observation: When the directions of the bottles are changed, if the north pole of one magnet is near the north pole of the other magnet (or the south pole of one magnet is near the south pole of the other magnet), the bottles will move away from each other because like poles repel each other.

Use your brain power!

Question.
You have learnt about static electricity in the previous standard. Electrostatic force is a non contact force. To verify this, which experiment will you perform?
Answer:

Do not switch on the fan in the room. Keep very small pieces of paper on the table. Rub a plastic comb against hair and bring it near the bits of paper. You will find that the bits of paper are attracted by the comb. The comb, on rubbing acquires electrostatic charge. It induces opposite charges on the bits of paper. Hence, the bits of paper are attracted by the comb.

Try this: 
Balanced forces and unbalanced force:

Take a cardboard box, tie thick string to its two sides and keep it on a smooth table as shown in Fig. Take the strings on both sides of the table. Tie weighing pans to the two ends. Keep equal masses in both the pans. The box does not move on the table.

If more mass is kept in one of the pans than in the other, the box starts moving in the direction of that pan. Equal gravitational force acts on both the pans when equal masses are kept in them. This means balanced forces act on the box, with effective force equal to zero as these are acting in opposite directions.

On the contrary, if more mass is kept in one pan than in the other, the box starts moving in the direction of the pan with more mass. When unequal forces are applied to the box on the two sides, an unbalanced force acts on the box resulting in imparting motion to the box.

Children playing tug of war pull the rope in their respective directions. If the pull of the force is equal on the two sides, the rope does not move. If the force is more on one side, the rope moves in that direction. This means that initially, the two forces are balanced; the rope moves in the direction of higher force when the forces become unbalanced.

Let us see one more example. When a big grain storage container is required to slide on the ground, it becomes easier if two persons push it rather than one person. When the force is applied by both in the same direction, the movement is easy. You may have experienced this. What do we understand from this example?

1. If several forces are applied on an object in the same direction, a force equal to their addition acts on that object.
2. If two forces are applied on one object in directions opposite to each other, a force equal to their difference acts on the object.
3. A force is expressed in magnitude and direction.

Force is a vector quantity. If more than one force are acting on a body, then the effect on the body is due to the net force. When a force is applied on a stationary object it moves, its speed and direction change. Similarly, a force is required to stop an object in motion.

An object can change its shape due to force. While kneading a dough made from flour, the dough changes its shape when a force is applied. A potter applies a force in a specific direction while shaping the pot. Rubber, when stretched, expands. There are many such examples. ,

Question 6.
What are balanced forces?
Answer:
If a body is acted upon by two forces, equal in magnitude, opposite in direction and having the same line of action, the forces are called balanced forces. Here, the net force acting on the body is zero.

Question 7.
What is an unbalanced force?
Answer:
If two or more forces act on a body such that their resultant is not zero, the resultant is an unbalanced force.
[Note: Unbalanced force acting on a body = mass of the body × acceleration of the body.]

Question 8.
Explain: Force has magnitude as well as direction. OR Force is a vector quantity.
Answer:
The effect of force applied to a body depends upon how much force we supply, i.e., the magnitude of the force, and the direction in which the force is applied. Consider a ball at rest on the ground. When ; we push it, it starts rolling. The greater the applied force, the greater is the speed acquired by the ball.

Consider a body moving in a straight line, If we apply a force in the direction of motion of the body, the speed of the body increases. On the contrary, if we apply a force in the direction opposite to that of motion of the body, the speed of the body decreases. These ( examples show that force has magnitude as well as direction, i.e., force is a vector quantity.

Question 9.
Explain the term balanced forces.
Answer:
Consider a rigid body acted upon by two forces, equal in magnitude, opposite in . direction and having the same line of action. These forces are called balanced forces as their net effect on the body is zero.
Example: A glass slab kept on a table is acted upon by two balanced forces: (i) the weight of the slab acting downward and (ii) the upward force on the slab due to the table. Their net effect on the slab being zero, the slab remains at rest.

Question 10.
Explain the term unbalanced force.
Answer:
A single force acting on a body is an unbalanced force. It produces acceleration in the body. If two or more forces act on a body such that their resultant is not zero, the resultant is an unbalanced force responsible for accelerating the body.
Example: When a ball lying on the ground is hit with a bat, the ball is set in motion by the applied force.

Question 11.
What will happen if the force is removed completely when an object acquires a certain speed?
Answer:
If the force is removed completely when an object acquires a certain speed, the object will move with the velocity it has at the instant the force is removed.

For example, a body moving with constant speed along a circular path in a horizontal plane will fly tangentially in the sense of motion if the centripetal force (the force directed towards the centre of the circle) is removed completely.

Always remember:

1. The tendency of an object to remain in its existing state is called its inertia.
2. This is why an object in stationary state remains in the same state and an object in motion remain in the state of motion in the absence of an external force.

Types of inertia:

1. Inertia of the state of rest: An object in the state of rest cannot change its state of rest due to its inherent property. This property is called the inertia of the state of rest.

2. Inertia of motion: The inherent property of an object due to which its state of motion cannot change, is called its inertia of motion. For example, a revolving’ electric fan continues to revolve even after it is switched off, passengers sitting in the running bus get aerk in the forward direction if the bus suddenly stops.

3. Directional inertia: The inherent property of an object due to which the object cannot change the direction of its motion, is called directional inertia. For example, if a vehicle in motion along a straight line suddenly turns, the passengers sitting in it are thrown opposite to the direction of turning.

Try this:
Activity 1:
Take a postcard and keep it on a glass. Keep a 5 Rupee coin on it. Now skilfully push the card. The coin straight away falls in the glass. Have you ever done this?
Answer:
Yes. (Explanation: The postcard moves forward due to the applied force and then falls due to the earth’s gravitational force. In the absence of adequate frictional force between the coin and the postcard, the coin does not move forward with the postcard, but straightaway falls in the glass due to the earth’s gravitational force.)

Activity 2: Hang a half a kg mass from a stand, with a string 1. Tie another string 2 to the mass and keep it hanging. Now pull the string 2 with a jerk. The string 2 breaks but the mass does not fall. Heavy mass does not move. Now pull the string 2 slowly. The string 1 breaks and the mass fall down. This is because of the tension developed in the string 2 due to the mass.

(Explanation: (1) As no force acts on the mass, it remains at rest due to inertia. (2) The transmission of force results in the tension in the string 1. As the string 1 cannot withstand it, it breaks and the mass falls down.)

Question 12.
What are the three types of inertia?
Answer:
Types of inertia:

• inertia of rest
• inertia of motion
• inertia of direction.

Question 13.
What is inertia of rest ? Give two examples of inertia of rest.
Answer:
The inherent property of a body by virtue of which it cannot change its state of rest is called the inertia of rest.
Examples:

1. When we dust a carpet, the carpet moves but the dust particles in it remain at rest due to inertia and hence get separated from the carpet. Hence, the carpet becomes clean.
2. When a bus starts suddenly, the passengers experience a backwarderk due to inertia.

Question 14.
What is inertia of motion? Give two examples of inertia of motion.
Answer:
The inherent property of a body by virtue of which it cannot change its state of motion is called the inertia of motion.
Examples:

1. When a fan is switched off, its blades continue to rotate for some time. Due to internal friction and friction with air, the blades of the fan stop rotating after some time.
2. Passengers in a bus experience a forwarderk when the bus stops suddenly due to application of brakes.

Question 15.
What is inertia of direction? Give two examples of inertia of direction.
Answer:
The inherent property of a body by virtue of which it cannot change its direction of motion is called the inertia of direction.
Examples:

1. While sharpening a knife, sparks fly off tangentially in the sense of motion from the grinding stone.
2. When a vehicle moves, the mud particles sticking to its wheels fly off tangentially in the sense of motion. Hence, mudguards are fitted to vehicles.

Question 16.
Why do we fall sideways when we are sitting in a bus and it takes a sharp turn?
Answer:
When we sit in a bus and the bus is in motion, we are in a state of motion in the same direction. When the bus takes a sharp turn, our body tends to maintain the state of motion in the straight line due to inertia. The portion of our body in firm contact with the seat acquires the motion along the curved path, but the upper portion of our body, tends to move in the initial direction of motion. Hence, we fall sideways.

Question 17.
What happens when you shake a wet piece of cloth? Explain your observation.
Answer:

1. When we shake a wet piece of cloth, water droplets come out.
2. Initially, the wet piece of cloth is at rest. When the cloth is shaken, it is accelerated, but the water droplets in it, due to inertia, tend to maintain the state of rest. Hence, the droplets come out.

Question 18.
If brakes are suddenly applied to a moving car, the passengers in the car are pushed in the forward direction. Explain why.
Answer:
1. The passengers in a moving car have the same velocity as that of the car. When brakes are suddenly applied to the car, it stops suddenly and the lower parts of the passengers’ bodies in contact with the seats, come to rest.

2. The upper parts of their bodies, however, continue to be in a state of motion due to inertia. Hence, the passengers are pushed in the forward direction.

Question 19.
A person alighting from a moving train is likely to fall in the direction of motion of the train. Explain why.
Answer:

1. A person in a moving train has the same velocity as that of the train. After alighting from the moving train his feet come to rest on the platform.
2. However, due to inertia, the upper part of his body continues to be in a state of motion in the direction of motion of the train. Hence, he is likely to fall in the direction of motion of the train.

Try this: 

Activity 3: Take some sharp pointed nails and push them into a wooden plank by hammering on their heads. Now take? a nail and hold it with its head on the plank and hammer it down from the pointed end. When pressing the drawing pins into a drawing board, they get into the board easily. By applying a force using the thumb one can push the pins into the boards. On the contrary, while pressing ordinary pins into the board with a thumb, the thumb may get hurt.

What does this simple experiment tell?
The nail easily penetrates into wood from its pointed end. From this you will notice that when a force is applied on the head of the nail, it is easy to hammer it into the plank.
Explanation: The less the area of the surface on which the force is applied, the greater is the effect of the force.

Question 20.
Define pressure.
Answer:
The force exerted perpendicularly on a unit area is called pressure.

Use your brain power!

Question 1.
It is easy to cut vegetables, fruits with a sharp knife. A blunt knife does not work here. Why does this happen?
Answer:
The effect of a given force varies l inversely as the area of the surface on ( which the force is applied. The less the surface area, the greater is the effect of the force. The cutting edge of a sharp ’ knife has less cross sectional area relative to that of a blunt knife. Hence, it is easy to cut vegetables, fruits with a sharp knife, rather than with a blunt knife. For a given force, pressure is inversely proportional to the area of the surface on which the force acts.

Question 21.
State the formula for pressure, Hence, determine the unit of pressure.
Answer:

The SI unit of force is the newton (N) and that of area is m². Therefore, the SI unit of pressure is N/m². It is called the pascal (Pa). 1 Pa = 1 N/m².
[Note: The unit pascal is named in honour of Blaise Pascal (1623-62), French mathematician, physicist and philosopher.]

Question 22.
State the CGS unit of pressure. State the relation betweeen the SI and CGS units of pressure.
Answer:
The CGS unit of pressure is the dyne/cm².
IN = 105 dynes, 1m² = (102 cm)² = 104 cm² 1 Pa = 1 N/m² = 105 dynes/104 cm² = 10 dynes/cm².

Question 23.
State the factors on which pressure depends.
Answer:
Pressure depends on the applied force and the area of the surface on which the force is applied.

Question 24.
Explain with a suitable example that pressure varies inversely as the area of the surface on which the force is applied, if the force remains constant.
Answer:
The tip of a pointed nail has an extremely small area, while that of a blunt nail has a comparatively large area. A given force creates a large pressure on the pointed nail and it can be easily hammered into the wood, while a very less pressure is created on the blunt nail and it cannot be easily hammered into the wood.

This shows that pressure varies inversely as the area of the surface on which the force is applied, if the force remains constant. If the same force is applied to surfaces having different areas, the pressure is more on the surface having a smaller area.

[Note: The bottom surface of a camel’s feet is broad. Hence, the camel’s weight acts on a large surface area. This reduces the pressure on the sand. Hence, the camel’s feet do not penetrate deep into the sand. Therefore, it becomes easier for the camel to walk on sand.]

Question 25.
With neat diagrams, describe an experiment to show that pressure increases if the surface area is decreased, keeping the applied force the same.
Answer:
Take a brick measuring 20 cm x 10 cm x 5 cm. Take some clay in a glass trough. Add water to it and knead it into a soft dough. Place the brick on the dough with one of its faces measuring 20 cm x 10 cm in contact with the dough. Observe how deep the brick penetrates into the dough.

Clean the brick and place it on the dough with one of its faces measuring 10 cm x 5 cm in contact with the dough. Observe how deep the brick penetrates into the dough. You will find that the brick penetrates deeper in this case than that in the first case.

• In the first case, the weight of the brick acts on a surface area of 200 cm².
• In the second case, the weight of the brick acts on a surface area of 50 cm².
• This shows that pressure increases if the surface area is decreased, keeping the applied force the same.

Question 26.
State the unit for pressure used in atmospheric science. How is it related to the unit pascal?
Answer:
In atmospheric science, the unit used for pressure is the bar. 1 bar = 10 Pa (pascal).

Try this:

Question 1.
Do the activity as depicted in Figure What is seen?
Answer:
In Fig.(a), the plank is horizontal. In Fig.(b), four books are placed side by side on the plank. The plank bends slightly due to the pressure produced by the weight of the books.

In Fig.(c), the four books are placed one above the other in the middle of the plank. Here the area of the surface of the plank on which the force acts is reduced by a factor of four relative to the earlier case. Hence, the plank bends considerably.
[Note: This shows that for a given force, pressure varies inversely as the area of the surface on which the force acts.]

Use your brainpower! 

Question 1.
You must have seen a vegetable vendor carrying a basket on her head. She keeps a twisted piece of cloth on the head, below the basket. How does it help?
Answer:
Keeping a twisted piece of cloth on the head increases the area of the surface on which the weight of the basket containing vegetables acts. Hence, the pressure produced by the force (weight) is reduced and it becomes easier to carry the basket.

Question 2.
We cannot stand at one place for a long time. How then can we sleep on a place for 8 and odd hours?
Answer:
When we stand, our weight acts on relatively small surface area, resulting in increased tension on the muscles of our legs. Hence, we cannot stand at one place for a long time. When we sleep, our weight acts on relatively large surface area, resulting in comparatively reduced tension. Therefore, we can sleep on a place for 8 and odd hours.

Question 3.
For skiing on ice, why are long flat skis used?
Answer:
The pressure produced by a given force depends on the area of the surface on which the force acts. The greater the surface area, the less is the pressure produced. The skis used to slide over snow are long and flat so that the area is increased and hence the pressure is decreased. This makes it easier to slide over snow.

Question 27.
Why do the blades of a pair of scissors have sharp edges?
Answer:
The pressure produced by a given force depends on the area of the surface on which the force acts. The less the surface! area, the greater is the pressure produced, The blades of a pair of scissors have sharp edges so that the area is decreased and hence the pressure is increased. This makes cutting • an object such as cloth easier.
[Note: Answers to questions such as why is the blade of an axe sharp? or why is the blade of a saw sharp? can be written on the basis of the answer given above.]

Question 28.
Why does a needle have s sharp point?
Answer:
The pressure produced by a given force depends on the areas of the surface on which the force acts. The less the surface area, the greater is the pressure produced. A needle has a sharp point so that the area is decreased and hence the pressure is increased. This makes the action of piercing easier.

Question 29.
Why do school bags have broad shoulder straps?
Answer:
The pressure produced by a given force depends on the area of the surface on which the force acts. The greater the surface area, the less is the pressure produced. School bags have broad shoulder straps so that the weight of the bag is distributed over a large surface area thereby decreasing the pressure on the shoulders of the student carrying the bag.

Question 30.
How will the pressure change if the area is doubled keeping the force the same?
Answer:
If the area is doubled, keeping the force the same, the pressure will become half the initial pressure.

Question 31.
How will the pressure change if the force is doubled, keeping the area the same?
Answer:
If the force is doubled, keeping the area the same, the pressure will become double the initial pressure.

Question 32.
State the characteristics of the pressure due to a liquid (or a fluid in general).
OR
Write a short note on the pressure due to a liquid (a fluid in general).
Answer:
Characteristics of the pressure due to a liquid (or a fluid):

1. The pressure at a point in a liquid (or a fluid) is due to the weight of the liquid (fluid) column above that point.
2. It acts on all sides of the container.
3. At a given depth it is the same in all directions.
4. It is independent of the size and shape of the container.
5. It is proportional to the height of the liquid (fluid) column above the given point.
6. It is proportional to the density of the liquid (fluid).
7. It is proportional to the acceleration due to gravity at the given place.

[Note: The pressure exerted by a liquid (or gas or fluid) at a depth h below the free surface of the liquid = hpg, where p is the density of the liquid (or gas or fluid) and g is the acceleration due to gravity.]

Question 33.
Give two examples to show that air exerts equal pressure in all directions.
Answer:

1. When air is filled in a balloon, it acquires its characteristic shape such as round or oval.
2. When a bicycle tube is filled with air, it acquires its characteristic (tube-like) shape throughout. This shows that air exerts equal pressure in all directions.

Question 34.
Whatisafluid?Givetwoeamples.
Answer:
A fluid is a substance which can flow.
Examples:

1. Water (liquid)
2. Air (gas)

[Note: Liquids and gases together are called fluids. Gases have very low viscosity compared to liquids. A liquid with low viscosity flows easily. A liquid with high viscosity does not flow easily.]

Question 35.
Take two rubber balloons. Fill one with water and blow air into the other. Now prick both balloons with a pin. What do you observe?
Answer:
Water and air both come out of the balloons. Air escapes quickly compared to water and produces a loud sound.

Question 36.
State the characteristics of pressure exerted by a fluid.
OR
Write a short note on pressure exerted by a fluid.
Answer:
Characteristics of pressure exerted by a fluid:

1. A fluid due to its weight, exerts pressure on the base as well as the walls of the container that holds it.
2. A fluid exerts pressure on a body immersed in it.
3. The pressure exerted on any confined mass of fluid is transmitted undiminished in all directions.

[Note: The pressure exerted by a fluid is a scalar quantity.]

Question 37.
Take an empty can. Pour small quantity of water in it. Boil this water for a few minutes until the steam has driven out most of the air. Now close the can with the stopper tightly. Allow it to cool by pouring cold water over it. What do you observe?
Answer:
The can gets gradually crushed.
[Note: The steam inside the can condenses to form water as the can cools. Therefore, the pressure inside the can becomes much less than the external pressure of the air. Hence, the can gets crushed.]

Question 38.
Put a folded newspaper on a plastic bag. Blow air into the bag. What do you observe?
Answer:
The plastic bag inflates as air is blown into it. This raises the folded news¬paper put on the bag.

Question 39.
What is meant by atmospheric pressure?
Answer:
The earth is surrounded by air from all sides. This layer of air is called the atmosphere. Its density is high up to about 16 km from the earth’s surface. Beyond that, up to about 400 km, its density is very low. Air, due to its weight, exerts pressure on the surface of the earth.

The pressure exerted by air or the atmosphere surrounding the earth is known as the atmospheric pressure. It is the ratio of the weight of the air to the area of the surface of the earth. It decreases with altitude as the density of air decreases with altitude and also the weight of the air column above a given place.
[Note: At sea level the atmospheric pressure is about 105 Pa. We do not feel it because the pressure of blood and other fluids in our body balances it.]

Question 40.
State the relation between 1 atmosphere and the pascal.
Answer:
1 atmosphere = (about) 101 × 10³ Pa (pascal)
[Note: The air pressure at the sea level is (about) 1 atmosphere.
1 atmosphere = 101325 Pa.
1 bar = 10³ mbar (millibar)
1 mbar w 10³ Pa (hectopascal)
Atmospheric pressure is expressed in mbar or hectopascal (hPa).]

Use your brain power!

Question 1.
At the sea level the atmospheric pressure 101 × 10³ Pa is acting on a table top of size 1 m². Under such a heavy pressure, why doesn’t the table top crumble down?
Answer:
The air below the table top exerts pressure 101 × 10³ Pa on it in the upward direction. Hence, the table top doesn’t crumble down.

Question 41.
Think – Why?
Question i.
Some people feel their ears popping at the top of a mountain.
Answer:
Atmospheric pressure decreases with altitude. At the top of a mountain, it becomes less than the internal pressure in the ear. Hence, some people feel their ears popping at the top of a mountain.

Question ii.
Some people feel breathless as they climb higher and higher on a mountain.
Answer:
Atmospheric pressure decreases with altitude. Hence, some people feel breathless as they climb higher and higher on a mountain.

Question iii.
We can enjoy a cold-drink or fruit juice with the help of a straw but can we imagine drinking a cold-drink or fruit juice on the moon using a straw?
Answer:
When we suck air in the straw, the pressure of the air in the straw becomes less than that of the outside air on the cold drink or fruit juice in the bottle (or the glass). Hence, the cold drink or fruit juice rises in the straw and enters our mouth. We can then drink it. There is no atmosphere on the moon. Hence, it is not possible to enjoy a cold drink or fruituice on the moon by using a straw.

Question iv.
People are often advised not to carry fountain pens while travelling by air.
Answer:
The ink in a fountain pen (filled at sea level at atmosphere pressure) may come out through its mouth while travelling by air if the outside pressure becomes less than the pressure in the ink holder of the pen. This can spoil the clothes/purse/bag. Hence, people are often advised not to carry fountain pens while travelling by air.

Question 42.
Why do some people have earache when they travel by an aeroplane?
Answer:
When an aeroplane descends at a high speed, there is an increase in air pressure. This increases the pressure on the ear drum. Hence, some people have earache when they travel by an aeroplane.

Question 43.
Explain why a person may bleed from the nose when at a great height above the sea level.
Answer:
The pressure exerted by the blood in blood capillaries is slightly more than the atmospheric pressure and acts in a direction opposite to that of the atmospheric pressure. Atmospheric pressure decreases with height and at a great height above the sea level, it is very low.

As a result, there arises a difference in the internal and external pressures on the walls of the cells and blood capillaries. If the difference is large, it may cause the cell wall and the blood capillaries to burst. Thus, the capillaries in the nose (and ear) may burst causing bleeding.

Question 44.
When a rubber sucker is pressed onto a flat glass surface, it sticks tightly on the surface. Why? You need a large force to separate it from the surface. Why?
Answer:
When a rubber sucker is pressed onto a flat glass surface, practically all the air between the surfaces of the sucker and the glass is pushed out. The air pressure there becomes much less than the atmospheric pressure. Hence, the sucker sticks to the glass due to the external atmospheric pressure. The atmospheric pressure is about 105 Pa. It is very large. Hence, to work against it to separate the sucker from the glass, a large force is needed.

Question 45.
Describe a simple experiment to demonstrate atmospheric pressure.
Answer:
Fill a glass completely with water (to its brim) and cover it with a flat and stiff card paper (or a piece of glass). Holding your palm on the card, turn the glass upside down and take the palm away from the card.

You will find that the water does not spill. The atmospheric pressure on the card (acting upward) is greater than the pressure of the water in the glass (acting downward). Hence, the water in the glass does not spill.

Question 46.
Explain the working of an ink dropper.
[Application]
Answer:
An ink dropper consists of a tube of glass or plastic, with one end tapering to a narrow opening and the other end fitted with a small rubber bulb. When the narrow open end is dipped into the ink and the rubber bulb is pressed, some air in the tube escapes through the open end. This reduces the air pressure inside the dropper.

On releasing the bulb, the atmospheric pressure on the ink pushes the ink into the dropper. The dropper is then taken out and its open end is held over the open barrel of the pen. The bulb is then pressed so that the ink in the dropper enters the pen.
[Note: The medicine dropper works in the same manner.]

Question 47.
Why is the opening of a dropper very narrow?
Answer:
The pressure produced by a given force is inversely proportional to the area of the surface on which the force acts. The opening of a dropper is very narrow. Hence, its area of cross-section is very small. As a result, even if the dropper has a small amount of ink it, its pressure can equal the atmospheric pressure. As the opening is narrow, it is easier to transfer the ink to the pen. The possibility of ink spilling is very low.

Question 48.
What is the characteristic of the cap of eye drop bottles?
Answer:
The cap of an eye drop bottle is fitted with a dropper.

Question 49.
Explain the working of a syringe
used by children when they play with coloured water. [Application]
Answer:
As shown in the figure, a syringe used by children when they play with coloured water consists of a cylinder made of plastic or metal fitted with a piston. One end of the cylinder is in the form of a narrow

tube. The snugly fitting piston can slide in , and out smoothly. The rod connected to the i piston passes through a hole in the centre of the lid and has a handle at the other end. When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure.

When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure. Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out.

To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.

Question 50.
How does the doctor’s syringe work?
Answer:
As shown in the figure, a syringe used by children when they play with coloured water consists of a cylinder made of plastic or metal fitted with a piston. One end of the cylinder is in the form of a narrow

tube. The snugly fitting piston can slide in , and out smoothly. The rod connected to the i piston passes through a hole in the centre of the lid and has a handle at the other end.

When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure. When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure.

Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out. To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.

The tip of a syringe is fitted with a very fine and hollow needle. The required quantity of medicine can be taken in the syringe with the help of the piston. The medicine can then be injected into the body of a patient using the needle and the piston.

Question 51.
Explain the origin of pressure due to a gas enclosed in a container.
Answer:
Molecules of a gas are in a state of continuous random motion. These molecules possess energy due to motion. When the molecules collide with a wall of the container, they rebound. A large number of molecules collide with the wall every second. Hence, a significant force is exerted on the wall. Pressure is the force per unit surface area. Thus, a pressure is exerted by the gas on a wall of the container.

Try this: 

Buoyant force:
Take a plastic bottle and fix the lid tightly. Now place it in wafer and see. It will float on water. Try and push it into the water. Even when pushed, it continues to float. This experiment can also be done with a plastic hollow ball. Now fill the bottle with water to the fullest capacity and close the lid, and release in water. The bottle will float inside the water. Why does this happen?

The empty plastic bottle floats on the surface of water. On the contrary, the bottle full of water floats inside water but does not go to the bottom. The weight of the empty bottle is negligible as compared with the weight of the water inside.

Such a bottle with water neither floats on the surface, nor does it go to the bottom. This means the force due to gravity acting downwards (fg), must have been balanced by an opposing force in the upward direction (fb) on the bottle filled with water. This force must have originated from the water surrounding the bottle. The upward force acting on the object in water or other fluid or gas is called buoyant force (fg).

Question 52.
Define buoyant force.
Answer:
The upward force acting on the object in water or other fluid or gas is called the buoyant force.

Question 53.
State the factors on which the buoyant force depends.
Answer:
The buoyant force depends upon the volume of the object immersed in the fluid (V), the density of the fluid (ρ1) and the acceleration due to gravity (g) at that place.
[Note: Magnitude of the buoyant force = Vρ1 g.]

Use your brain power!

Question.
While pulling a bucket from a well, the bucket full of water immersed fully in water appears to weigh less than when it has been pulled out of water. Why?
Answer:
1. When a bucket full of water is immersed in water, the net force acting on the bucket = weight of the bucket full of water-the buoyant force exerted by the water on the bucket. The buoyant force due to a fluid is proportional to the density of the fluid.

2. The density of water is much greater than that of air. Therefore, the buoyant force acting on a bucket full of water while it is in water is much greater than that when it is out of water, i.e., in air. Hence, it appears to weigh less, while it is in water than when it has been pulled out of water.

Try this:

Question 1.
Take a piece of thin aluminum sheet and dip it in water in a bucket. What do you observe?
Answer:
It sinks in the water.

Question 2.
Now shape the same piece of aluminium into a small boat and place it on the surface of water. It floats, isn’t it?
Answer:
Yes, the boat floats on the surface of water.

Question 54.
A lemon sinks in a glass filled with water but it floats when we stir in two spoons of salt in the water. Explain why.
Answer:
When salt is dissolved in water, the density of the water increases. The buoyant force is proportional to the density of the liquid. Hence, when the lemon is immersed in the water containing salt, the magnitude
of the buoyant force acting on the lemon becomes greater than the magnitude of the weight of the lemon. Therefore, it floats in water.

Do you know?

How is it decided that an object left in a liquid will get sink in the liquid, will float on the surface, or will float inside the liquid?
1. The object floats if the buoyant force is larger than its weight.
2. The object sinks if the buoyant force is smaller than its weight.
3. The object floats inside the liquid if the buoyant force is equal to its weight. Which forces are unbalanced in the above cases?
Answer:
Unbalanced force:
1. Magnitude of the unbalanced force acting on the object = magnitude of the buoyant force on the object-magnitude of the weight of the object. The direction of the unbalanced force is the direction of the buoyant force.

2. Magnitude of the unbalanced force on the object = magnitude of the weight of the object – magnitude of the buoyant force on the object. The direction of the unbalanced force is the direction of the weight of the object.

3. Here, the unbalanced force is zero.

Try this:

Take a long rubberband and cut it at one point. At one of its ends tie a clean washed stone or a 50 g weight as shown in Figure
Now hold the other end of the rubberband and make a mark there. Keep the stone hanging in air and measure the length of the rubberband from the stone to the mark made earlier.

Now take water in a pot and hold the rubberband at such a height that the stone sinks in it. Again measure the length of the rubberband up to the mark. What is observed? This length is shorter than the earlier length. While dipping the stone in water, length of the stretched rubber gets slowly reduced and is minimum when it sinks completely. What could be the reason for a shorter length of the rubberband in water?

When the stone is sunk in water, a buoyant force acts on it in the upward direction. The weight of the stone acts downwards. Therefore, the force which acts on it in the downward direction is effectively reduced.
(This decreases the downard force on the rubberband. Hence, its length decreases.)

Question 55.
State Archimedes’ principle.
Answer:
Archimedes’ principle: When an object is partially or fully immersed in a fluid, a force of buoyancy acts on it in the upward direction. This force is equal to the weight of the fluid displaced by the object.
[Note: The two forces mentioned here are equal in magnitude and opposite in direction.]

Question 56.
Using Archimedes’ principle, determine the magnitude of the buoyant force.
Answer:
Let m = mass of the body (object) immersed in the fluid, V = volume of the body, ρ = density of the body, ρ1 = density of the fluid, g = magnitude of the acceleration due to gravity. Suppose that the body is completely immersed in the fluid. Then the volume of the fluid displaced by the body = V. According to Archimedes’ principle, magnitude of the buoyant force = magnitude of the weight of the fluid displaced by the body = mass of the displaced fluid × g = volume of the displaced fluid × density of the fluid × g
(as density = mass/volume)

If the body is partially immersed in the fluid, the volume of the fluid displaced by the immersed part of the body = xV ; here 0 < x < 1.
In this case, the magnitude of the buoyant force

Question 57.
Using the formula for the magnitude of the buoyant force, state under what conditions the body will
1. sink in the fluid
2. float in the fluid
3. remain in equilibrium anywhere within the fluid.
Answer:
Magnitude of the buoyant force on the body = mg \(\left(\frac{\rho_{1}}{\rho}\right)\)
= magnitude of the weight of the body × \(\left(\frac{\rho_{1}}{\rho}\right)\)
1. If the density of the fluid (ρ1) is less than the density of the body (ρ), the magnitude of the buoyant force on the body will be less than the magnitude of the weight of the body. Therefore, the body will sink in the fluid.|

2. If the density of the fluid is greater than that of the body, the magnitude of the buoyant force on the body will be greater than that of the weight of the body. Therefore, the body will float in the fluid.

3. If the density of the fluid is equal to that of the body, the magnitude of the buoyant force on the body will be equal to that of the weight of the body. Therefore, the body will remain in equilibrium anywhere within the fluid.

Use your brain power!

Question.
Explain the observations in the earlier experiments according to the Archimedes’ principle.
Answer:
The increase in the length of the rubberband (y) is proportional to the downward force (f) acting on it.
1. When the stone tied to the rubberband is hanging in air, the magnitude of the buoyant force exerted by air on the stone is negligible compared to that of the weight of the stone. Hence, ignoring it, we have f = fg = mg, where m is the mass of the stone and g is the acceleration due to gravity.

2. When the stone is immersed partially in water, f = fg – fb, where fb is the magnitude of the buoyant force exerted by water on the stone.
Now, fg = Vρg and fb = xVρwg, where V = volume of the stone, ρ = density of the stone, xV= volume of the water displaced by the part of the stone immersed in water = volume of the part of the stone immersed in water (x < 1) and ρ = density of water.

This shows that as the stone is gradually lowered in water, x goes on increasing and hence/goes on decreasing. Therefore, elongation (y) of the rubberband goes on decreasing, i.e., the length of the rubberband goes on decreasing.

3. When the stone is completely immersed in water, x becomes maximum, equal to 1. Here, f = fg \(\left(1-\frac{\rho_{\mathrm{W}}}{\rho}\right)\). This is the minimum value of. Here, the elongation of the rubberband is minimum, i.e., the length of the rubberband is minimum.

Question 58.
State the applications of Archimedes’ principle.
Answer:
Archimedes’ principle is used in the construction of ships and submarines. The working of the lactometer and hydrometer is based on Archimedes’ principle.
[Note: The hydrometer is used to determine the density or relative density of a liquid.]

Question 59.
If a spring balance is used to weigh a body, will the weight of the body be the same in vacuum and air? Explain why.
Answer:
When a body is suspended in air, the buoyant force acts on the body. Hence, the magnitude of the net downward force on 1 the body = the magnitude of the weight of the body – the magnitude of the buoyant force on the body. Hence, when a spring balance is used to weigh a body, the weight of the body in air is less than that in vacuum.

Question 60.
What is density of a substance? I Obtain its SI unit.
Answer:
The density of a substance is the ratio of its mass to volume.
The SI unit of density = \(\frac{\text { the SI unit of mass }}{\text { the SI unit of volume }}\) = kg/m³
[Note: Density is useful in determining the purity of a substance. The CGS unit of density is g/cm³.
1 kg/m³ = 10³g/(100 cm)³ = 10^-3g/cm³
∴ 1g/cm³ = 1000 kg/m³]

Question 61.
What is relative density of a substance?
Answer:
The relative density of a substance is the ratio of its density to the density of water.
[Note: Relative density is also called specific gravity. It is a ratio of two equal (same) physical quantities. Hence, it has no unit.]

Question 62.
A piece of wood floats both in water and kerosine. In which liquid does it sink more during floating? Why?
Answer:
The piece of wood sinks more in kerosine than in water during floating. The density of kerosine is less than that of water. The buoyant force on a body is proportional to the density of the liquid in which the body is immersed. When a body floats, the magnitude of the buoyant force acting on the body is equal to that of the weight of the body.

Hence, the volume of the liquid displaced by a floating body is inversely proportional to the density of the liquid. As a result, when a piece of wood floats in kerosine, it displaces greater volume of kerosine compared to the volume of water displaced when the piece of wood floats in water. Hence, it sinks more in kerosine than in water.
[Note: When a body floats in a liquid fb = fg
∴ Vρ1g = Vbρbg
∴ The volume of the liquid displaced by the body is V = Vb \(\frac{\rho_{\mathrm{W}}}{\rho_{1}}\), where Vb is the volume of the body, ρb is the density of the material of the body and ρ1 is the density of the liquid. Thus, V ∝ 1 /ρ1]

Question 63.
State whether the body will float or sink in a liquid if the density of the body is 1. greater than that of the liquid 2. less than that of the liquid 3. equal to that of the liquid.
Answer:

1. The body will sink in the liquid if the density of the body is greater than that of the liquid.
2. The body will float in the liquid if the density of the body is less than that of the liquid.
3. The body will float inside the liquid if the density of the body is equal to that of the liquid.

Question 64.
If the relative density of a body is greater than 1, will it float in water?
Answer:
If the relative density of a body is greater than 1, it will not float in water.
[Note: The relative density of a pin is much greater than 1. But when kept gently on the surface of water, it floats. This is due to the surface tension of water.]

Question 65.
A glass of water has an ice cube floating in water. The water level must touches the rim of the glass. Will the water overflow when the ice melts? Give the reason.
Answer:
The water will not overflow when the ice melts. The water level will remain the same. Ice floats on water because its density is less than that of water. When ice melts, the volume of the water formed is less than the volume of the ice which has melted.

When ice in water melts, this difference equals the volume of the water formed when the part of ice above the surface of water melts. Therefore, the water level remains the same. Hence, there is no overflow of water when the ice melts.

Question 66.
A plastic ball is released underwater. State whether it will sink or come up to the surface of water. Give the reason.
Answer:
A plastic ball released under water will come up to the surface of water. The density of water is greater than that of plastic. Hence, when a plastic ball is under water, the magnitude of the buoyant force exerted by water on the ball is greater than the magnitude of the weight of the ball.

Therefore, the ball will start moving upward. As it comes up with part of the ball above the water surface, the volume of the water displaced by the ball becomes less and hence at a certain stage, the buoyant force and the weight balance each other. Then the ball continues to remain in that state, as the net force on the ball becomes zero.
[Note: Initially, the ball moves slightly up and down near the water surface. The force due to friction with water, opposing the motion of the ball, finally makes the ball steady.]

Write short notes:

Question 1.
Buoyant force.
Answer:
1. When a body is immersed partially or completely in a liquid, the liquid exerts forces on all sides of the body. This force is perpendicular to the surface of the body and equals the product of pressure and area at that point.

2. The resultant of all these forces acts upward. It is called the upthrust or buoyant force.

3. The buoyant force is proportional to (i) the volume of the liquid displaced by the body (ii) the density of the liquid (iii) the acceleration due to gravity. Its magnitude equals the magnitude of the weight of the liquid displaced by the body.

Question 2.
Applications of Archimedes’ principle.
Answer:
1. The working of a lactometer, a device used to determine the purity of a sample of milk, and a hydrometer, a device used to determine the density of a liquid, is based on Archimedes’ principle. The extent to which a lactometer floats (or sinks) depends on the density (and hence purity) of the milk. The same thing is true for a hydrometer. The greater the density of a liquid, the less is the extent to which a body sinks in it.

2. Archimedes’ principle is used in design of ships and submarines. A submarine is provided with large tanks at the front and the back. Its weight can be increased by filling the tanks with sea water or air from compressed air reservoirs. The weight can be decreased by pumping out water from the tanks by forcing compressed air in them. By controlling the weight, it can be made to sink or rise to the surface as desired.

3. The density of a body that floats or sinks in water or kerosine can be determined by. Archimedes’ principle.
4. The density of kerosine can be determined by Archimedes’ principle, using a body of material that is not affected by water and kerosine.

Give scientific reasons:

Question 1.
The tiles are placed over a slushy patch of ground to help cross It.
Answer:

1. Tiles have greater area than the area of our feet.
2. The weight of the person crossing the slushy patch is exerted over a large area of the tiles.
3. Therefore, there is a decrease in the pressure and hence the tiles do not sink much in the slushy patch of ground. This helps to cross the slushy patch of ground.

[Note: If there were no tiles, the feet will come in direct contact with the slushy ground. The area of the feet being less, the weight of the person will act over a smaller area. Therefore. there will be more pressure and hence the feet will sink into the slushy ground.]

Question 2.
Drawing pins have flattened heads.
Answer:

1. The head of a drawing pin is flattened and the other end is pointed.
2. When enough force is applied to the head of the pin, the pressure due to the force on the pointed end increases tremendously and the pin can be easily inserted in the drawing board.
3. When we press the flattened end, the force applied spreads over a larger area. This reduces the pressure of the reaction force acting on the thumb. Hence, the thumb is not injured.
4. If the head of the pin is sharp, then the pressure due to the force would be more and hence the pressure of the reaction force would also be more and the sharp end would prick the thumb causing injury.

Question 3.
An iron nail sinks in water but a ship made from iron floats on water.
Answer:
1. An iron nail sinks in water because its density is more than that of water.
2. A ship made from iron, due to the particular shape given to it, displaces a large amount of water so that the buoyant force acting on the ship due to water balances the weight of the ship. Hence, the ship floats on water.

Question 4.
A piece of iron sinks in water but floats on mercury.
Answer:
The density of iron is more than that of water but less than that of mercury. Hence, a piece of iron sinks in water but floats on mercury.

Question 5.
A sheet of metal that sinks in water can float if shaped like a pan.
Answer:

1. A sheet of metal sinks in water because its density is more than that of water.
2. If the sheet is shaped like a pan, it can displace a large amount of water such that the buoyant force on the pan due to water balances the weight of the pan. Hence, it can float on water.

Solve the following examples:

Problem 1.
(i) Calculate the pressure exerted by the wooden block when it is kept in the vertical position.
Given: The length of the wooden block is 80 cm, the breadth is 50 cm, the thickness is 20 cm and the weight is 500 N
(ii) Also calculate the pressure when the wooden block is kept in the horizontal position with its surface 80 cm × 50 cm touching the floor.

Solution:
Data: F = W = 500 N, l = 80 cm = 0.8 m, b = 50 cm = 0.5 m. h = 20 cm = 0.2 m
(i) A = bh = 0.5 m × 0.2 m = 0.1 m²

The pressure exerted in the vertical position of the block = 5000 N/m² or 5000 Pa.

(ii) A = lb = 0.8 m × 0.5 m = 0.4 m²

The pressure exerted in the horizontal position of the block = 1250 N/m² or 1250 Pa.

Problem 2.
Measure the length, breadth, height and mass of a rectangular tiffin box. Find the weight of the box and calculate the pressure in two different positions as in Ex. (1) above.
Solution:
Let l = 0.25 m, 6 = 0.1 m, h = 0.05 m, F = W= 0.5 N
(i) A = bh = 0.1 m × 0.05 m = 0.005 m²

(ii) A = lb = 0.25 m × 0.1 m = 0.025 m²

Problem 3.
A force of 1000 N is applied over an area 50 cm × 20 cm. Find the corresponding pressure.
Solution:
Data: F = 1000 N,
A = 50 cm × 20 cm = 0.5 m × 0.2 m = 0.1 m², pressure = ?

The pressure = 10 N/m².

Problem 4.
A metal block has thmensions 10 cm × 5 cm × 2 cm and the density of the metal is 8 × 10³ kg/m³. It is kept on a table with the face 10 cm × 5 cm in contact with the table. Find the force and pressure exerted by the block on the table. (g = 9.8 m/s²)
Solution :
Data : 1 = 10 cm, b = 5 cm,
h = 2 cm, p= 8 × 10 kg/m3, g = 9.8 m/s²,
A = lb = 10cm × 5cm = 50cm² = 50 × 104m²
= 5 × 10^-5m², force =?, pressure = ?
Volume of the block = lbh =
10 cm × 5 cm × 2 cm= 100 cm³
= 100 × 10^-6 m³ = 1 × 10^-4m³
Mass of the block = volume × density
(∵ density = mass/volume)
∴Mass of the block,
m = 1 × 10^-4 m³× 8 × 10³ kg/m³ = 0.8 kg
Weight of the block = mg = 0.8 kg × 9.8 m/s² = 7.84 N
∴ The force exerted by the block on the table = 7.84 N.
force 7.84 N

= 1.568 × 10³ N/m² or 1.568 × 10³ Pa
The pressure exerted by the block on the table = 1.568 × 10 N/m2 or 1.568 × 10³ Pa.

Problem 5.
A body of volume loo cm³ is immersed completely in water. Find the weight of the water displaced by the body. 1g = 9.8 m/s². p (water) = kg/m³]
Solution :
Data: V = 100 cm³ = 100 × 10^-6 m³
= 1 × 10 m³, p(water) = 10 kg/m³
g = 9.8 m/s², weight of the displaced water ?
Density = \(\frac{\text { mass }}{\text { volume }}\)
∴ Mass = volume × density
Volume of the water displaced by the body = l × 10^-4 m³
∴Mass of the water displaced,
m = l × 10^-4 m³ × 10 kg/m³ = 0.1 kg
∴Weight of the water displaced
= mg = 0.1 kg × 9.8 m/s² = 0.98 N.
The weight of the water displaced by the body = 0.98 N.

Problem 6.
A body of mass 200 g and volume 50 cm³ is put in a bucket containing water. Will it float or sink ?
[ρ(water) = 1 g/cm³]
Solution:
Data: m = 200 g, V= 50 cm³,
ρ (water) = 1 g/cm³
Density(ρ) = \(\frac{\text { mass }}{\text { volume }}\)
∴ ρ (body) = \(\frac{200}{50 \mathrm{~cm}^{3}}\)
It is greater than the density of water.
Hence, the body will sink in water.

Problem 7.
A body of mass 200 g and volume 400 cm³ is put in a bucket containing water. Will it float or sink?
[ρ (water)=1 g/cm³]
Solution:
Proceed as above.
ρ (body) = \(\frac{200 \mathrm{~g}}{400 \mathrm{~cm}^{3}}\) = 0.5 g/cm³
It is less than the density of water.
∴ The body will float in water.

Problem 8.
The mass of a tile is 500 g. If the density of the tile is 2.5 g/cm³, what will be the weight of the tile when it is completely immersed in water?
(g = 9.8 m/s², ρ(water) = 1000 kg/m³)
Solut10n:
Data: m = 500, g = 0.5 kg, ρ
(tile) = 2.5 g/cm³ = 2500 kg/m³, g = 9.8 m/s²,
ρ (water) = 1000 kg/m³, weight of the tile when completely immersed in water (also called the apparent weight) = ?
ρ = \(\frac{m}{V}\)

∴ Volume of water displaced by the tile
= 2 × 10^-4 m³
∴ Mass of water displaced by the tile
m’ = ρ (water) V = 1000 kg/m³ × 2 × 10^-4 m³
= 0.2kg
∴ Magnitude of the weight of this water
= mg = 0.2 kg × 9.8 m/s² = 1.96 N
∴ Buoyant force exerted on the tile = 1.96 N
Magnitude of the weight of the tile =
mg = 0.5 kg × 9.8 m/s² = 4.9 N
∴ Weight of the tile when completely immersed in water (apparent weight) = weight of the tile in air-buoyant force on the tile
= 4.9 N – 1.96 N = 2.94 N(downward)

Examples For Practice:

[g = 9.8 m/s², ρ (water) = 10³ kg/m³ = 1 g/cm³]

Question 1.
Calculate the relative density of a metal having density 7.5 g/cm³.
Answer:
7.5

Question 2.
Find the density of steel if its relative density is 8 and the density of water is 10 kg/m³.
Answer:
8 × 10³ g/cm³

Question 3.
A body has mass 200 g and volume 100 cm³. Find its density and relative density.
Answer:
2 g/cm³, 2

Question 4.
If the relative density of a material is 2.5, find its density.
Answer:
2.5 × kg/m³ or 2.5 g/cm³

Question 5.
A force of 100 N is applied on an area 40 cm × 25cm. Find the corresponding pressure.
Answer:
10³ N/m²

Question 6.
If the pressure exerted on an area 10 cm × 10 cm is 1000 dynes/cm², find the applied force.
Answer:
10⁵ dynes

Question 7.
A metal block of mass 10 kg is kept on a table. If the contact surface area Is 100 cm², find the pressure on the table.
Answer:
9.8 × 10³ N/m² or 9.8 × 10³ Pa

Question 8.
A body of volu.me 50 cm³ is immersed completely in water. Find the weight of the water displaced by the body.
Answer:
0.49 N

Question 9.
A block of mass 100 g and volume 20 cm³ is put in a bucket filled with water. Will it float or sink?
Answer:
The body will sink in water.

Question 10.
Will a block of mass 100g and volume 400 cm³ float or sink in water?
Answer:
The block will float in water.

Question 11.
The volume of a cube is 125 cm³ and its mass is 250 g. It is put in a tub containing water. Will it float or sink?
Answer:
It will sink in water.

Balbharati Maharashtra State Board 8th Std Science Textbook Solutions 

• Living World and Classification of Microbes Class 8 Science Textbook Solutions
• Health and Diseases Class 8 Science Textbook Solutions
• Force and Pressure Class 8 Science Textbook Solutions
• Current Electricity and Magnetism Class 8 Science Textbook Solutions
• Inside the Atom Class 8 Science Textbook Solutions
• Composition of Matter Class 8 Science Textbook Solutions
• Metals and Nonmetals Class 8 Science Textbook Solutions
• Pollution Class 8 Science Textbook Solutions
• Disaster Management Class 8 Science Textbook Solutions
• Cell and Cell Organelles Class 8 Science Textbook Solutions

Std 8 Science Chapter 10 Cell and Cell Organelles Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 8 Science Solutions Chapter 10 Cell and Cell Organelles Notes, Textbook Exercise Important Questions and Answers.

Class 8 Science Chapter 10 Cell and Cell Organelles Question Answer Maharashtra Board

Class 8 Science Chapter 10 Cell and Cell Organelles Textbook Questions and Answers

1. Who am I?

Question a.
I am ATP producing factory.
Answer:
Mitochondria

Question b.
I am single-layered but maintain cellular osmotic pressure.
Answer:
Vacuole

Question c.
I support the cell, but I am not cell wall. I have a body resembling net.
Answer:
Endoplasmic reticulum

Question d.
I am chemical factory of the cell.
Answer:
Chloroplasts in case of plant cells can be called chemical factory as they synthesise carbohydrates. Ribosomes also synthesise proteins, so those can also be called chemical factory. Golgi complex is secretory in function, hence it can be also called factory. Mitochondria though mainly referred to as powerhouse of the cell, it is also mentioned as chemical factory by some authors.

Question e.
Leaves are green because of me.
Answer:
Chloroplast.

2. What would have happened? If……….

Question a.
RBCs had mitochondria.
Answer:
Mitochondria continuously carry out oxidation and form energy inside the cell. They produce energy-rich compound, ATP. In this process, they utilize carbohydrates, fats and proteins present in the cell. If RBCs has mitochondria, they would have used oxygen for this purpose than carrying it to all the cells of the body. The cells would not have obtained oxygen.

Question b.
There had been no difference between mitochondria and plastids.
Answer:
Mitochondria carry out oxidation of carbohydrates, fats, etc. with the help of enzymes. Plastids are synthesising carbohydrates with the help of solar energy and chlorophyll. Both the cell organelles have their own sets of different enzymes as per their role. If there would have been no difference between mitochondria and plastids, the specific functions would not have been taken place.

Question c.
Genes had been absent on the chromosomes.
Answer:
Genes are functional segments on the chromosomes which are responsible for transmitting the hereditary information.

Question d.
Plasma membrane had not been selectively permeable.
Answer:
Selectively permeable membrane allows some substances to enter the cell, while prevents other unwanted or harmful substances. If plasma membrane would not have been selectively permeable, there would be no control over entry and exit of any substances. The process of osmosis would also be erroneous in such case.

Question e.
Plants lacked anthocyanin.
Answer:
If plants lacked anthocyanin, no part of the plant would display purple or blue colour. Anthocyanin attracts the insects for pollination and seed dispersal. These processes will be affected due to lack of anthocyanin. These pigments are also said to be protective in nature for the plant. This protection will not be given to the plant in absence of anthocyanin.

3. Who is odd man among us? Give reason.

Question a.
Nucleolus, mitochondria, plastids, endoplasmic reticulum
Answer:
Nucleolus. (All the others are cell organelles but nucleolus is not a cell organelle present in cytoplasm.)

Question b.
DNA, Ribosomes, Chlorophyll
Answer:
Chlorophyll. (DNA and Ribosomes are present in plant as well as in animal cells. Chlorophyll is present only in plant cells.)

4. Give functions.

Question a.
Plasma membrane.

Answer:

1. Plasma membrane acts as a selectively permeable membrane. It allows entry of those useful substances which are needed for the cell. It does not allow entry of the harmful and unwanted substances.
2. Plasma membrane keeps the homeostasis in the cell. The cell is kept in steady state even if the external environment changes.
3. Plasma membrane is responsible for processes of endocytosis and exocytosis.
4. The processes of diffusion and osmosis are possible only due to plasma membrane.
5. In animal cells, plasma membrane is the outermost protective covering of the cell.

Question b.
Cytoplasm.
Answer:

1. All the cell organelles are spread in the cytoplasm of a cell.
2. The cytoplasm is the medium for many cellular chemical reactions.
3. The cytosol which is the part of cytoplasm other than cell organelles stores many vital substances like amino acids, glucose, vitamins, etc.
4. Cytosol also helps in the cellular movements.

Question c.
Lysosome.

Answer:

1. Lysosome helps in the destruction of attacking viruses and bacteria and thereby help in the immune response.
2. Lysosomes act as demolition squads. They destroy worn-out cellular organelles and organic debris. This process is called autolysis which All Pagesmeans self-destruction.
3. They are also called suicide bags as in a worn out, damaged or old cell, lysosomes automatically burst. The lytic enzymes present in the lysosome digest their own cells.
4. Lysosomes can digest stored proteins, fats during starvation.

Question d.
Vacuole.

Answer:

1. Vacuoles maintain the osmotic pressure of the cell.
2. Various metabolic byproducts and end products such as glycogen, proteins, water, etc. are stored in the lysosome.
3. In food vacuole of amoeba, the food is temporarily stored till digestion. In other animal cells, vacuoles can store waste products and food.
4. Vacuoles of plant cells can provide turgidity and rigidity as it contains good amount of cell sap.

Question e.
Nucleus.

Answer:

1. Nucleus is the controlling centre for the entire cell.
2. It controls all metabolic activities of the cell.
3. The cell division is possible due to the nucleus.
4. The chromosomes present in the nucleus carry the genes. These genes are responsible for the transmission of hereditary characters from parental generation to the next generations.

5. Who gives me the colour? (Select the correct option).

Question a.

Answer:

Project:

Question 1.
Prepare model of a cell using different ecofriendly materials.

Question 2.
Study osmosis using parchment paper or a similar membrane.

Question 3.
Form a friends’ group in your class. Give each one role of a cell organelle. Present a skit accordingly.

Class 8 Science Chapter 10 Cell and Cell Organelles Important Questions and Answers

Rewrite the sentences after filling the blanks:

Question 1.
Cell wall is mainly composed of carbohydrates like ……….. and …………. .
Answer:
Cell wall is mainly composed of carbohydrates like cellulose and pectin.

Question 2.
Plasma membrane is said to be a ……………. …………….. membrane as it allows some substances to enter the cell, while prevents other substances.
Answer:
Plasma membrane is said to be a selectively permeable membrane as it allows some substances to enter the cell, while prevents other substances.

Question 3.
Homeostasis is maintained in the cell by ………….. .
Answer:
Homeostasis is maintained in the cell by plasma membrane.

Question 4.
An …………… is a specialized subunit having specific function within the cell.
Answer:
An organelle is a specialized subunit having specific function within the cell.

Question 5.
……………. has ribosome granules on its outer surface.
Answer:
Rough ER has ribosome granules on its outer surface.

Question 6.
During starvation, ………………… digest stored proteins, fats.
Answer:
During starvation, lysosomes digest stored proteins, fats.

Question 7.
……………….. is the secretory organ of the cell.
Answer:
Golgi complex is the secretory organ of the cell.

Question 8.
……………. compound ATP is produced in the mitochondria.
Answer:
Energy-rich compound ATP is produced in the mitochondria.

Given below are incorrect statements. Rewrite them after correcting them:

Question 1.
In mitochondria, the inner membrane is porous and the outer membrane is deeply folded.
Answer:
In mitochondria, the outer membrane is porous and the inner membrane is deeply folded.

Question 2.
Vacuole is bound by double membrane.
Answer:
Vacuole is bound by single membrane.

Question 3.
If fruit pieces are kept in thick saturated sugar solution, the water from fruit pieces enter the sugar solution resulting into their swelling.
Answer:
If fruit pieces are kept in a thick saturated sugar solution, the water from fruit pieces enter the sugar solution resulting into their shrinking.

Question 4.
Raisins kept in water shrink after an hour.
Answer:
Raisins kept in water swell after an hour.

Question 5.
Lysosome produces vacuoles and secretory vesicles.
Answer:
Golgi complex produces vacuoles and secretory vesicles.

Who gives me the colour? (Select the correct option)

Question 1.

Answer:

Find the odd one out by giving suitable reasons:

Question 1.
Demolition squads, Suicide Bags, Immune system, Powerhouse of the cell.
Answer:
A powerhouse of the cell. (All the others are descriptions of the lysosomes.)

Question 2.
Lignin, Suberin, Cutin, Iodine
Answer:
Iodine. (All the others are polymers present in the cell wall.)

Question 3.
Nucleolus, Genes, Chromosomes, Ribosomes
Answer:
Ribosomes. (All the others are inclusions in the nucleus.)

Write definitions/Give meanings:

1. Homeostasis: The tendency of the cell to keep the cellular environment constant in spite of changes in the outer: environment is called homeostasis.

2. Endocytosis: To take in the food or any other substance from outer environment into the cell is called endocytosis.

3. Exocytosis: To give out the unwanted substances from the cell to the outer « environment is called exocytosis.

4. Diffusion: The movement of the molecules from region of higher concentration to the region of lower concentration is called diffusion.

5. Osmosis: The movement of solute from low concentration to high concentration and the movement of solvent from high concentration to the region of low concentration across semipermeable membrane is called osmosis.

6. Plasmolysis: When the cell is kept ?! in hypertonic medium, the water exits through the process of exosmosis causing shrinkage of the cytoplasm, this is known as plasmolysis.

7. Isotonic solution: When the concentration of the cell and that of the medium in which the cell is kept is same, then such solution is called isotonic solution.

8. Hypotonic solution: When the concentration of the water in the cell is less than that of the concentration of the water in the surrounding medium in which the cell is kept, then such solution is called hypotonic solution.

9. Hypertonic solution: When the concentration of the water in the cell is more than that of the concentration of water in the surrounding medium then such solution is called hypertonic solution.

Distinguish between the following:

Question 1.
Prokaryotic and eukaryotic cell:
Answer:

Question 2.
Plant cell and animal cell
Answer:

Answer the following questions in one sentence:

Question 1.
What are the components of plasma membrane?
Answer:
In plasma membrane, protein molecules are embedded in two layers of phospholipids.

Question 2.
Which part of the cell maintains the homeostasis?
Answer:
Plasma membrane of the cell maintains the homeostasis.

Question 3.
What are genes?
Answer:
Genes are the functional segments on the chromosomes that carry hereditary i information from the parental generation to the offspring.

Question 4.
What is meant by rough ER?
Answer:
The endoplasmic reticulum that has ribosomes on its outer membrane is called rough ER.

Question 5.
Write the examples of plant pigments.
Answer:
Chlorophyll, Carotene, Xanthophyll, Anthocyanin, Betalains and Lycopene are some of the plant pigments.

Question 6.
What are the inclusions in the stroma of chloroplasts?
Answer:
Enzymes, DNA, ribosomes and carbohydrates that are necessary for photosynthesis are present in the stroma of the chloroplasts.

Question 7.
Which staining technique was developed by Camilio Golgi? Where was this technique used?
Answer:
Camilio Golgi developed the staining technique called ‘Black reaction’ which was used in the study of nervous system.

Question 8.
What type of work is done by National Centre for Cell Science?
Answer:
National Centre for Cell Science – NCCS is involved in research in cytology and research about cancer treatment and it also provides services for National Animal cell repository.

Question 9.
Ripe tomatoes appear red.
Answer:
When green tomatoes become ripe they lose chlorophyll and develop red pigment in them called lycopene. Therefore, ripe tomatoes appear red.

Give scientific reasons:

Question 1.
Raisins swell after keeping in plain water.
Answer:
When raisins are placed in plain water, there is action of endosmosis. The outer skin of raisins acts like selectively permeable membrane. Since the concentration of water inside the raisin is lesser than the concentration of water in the outer medium, water enters in the raisin. This causes raisins to swell after keeping them in plain water.

Question 2.
The fruit pieces kept in sugar syrup show shrinking.
Answer:
There is more concentration of water in the fruit pieces as compared to the concentration of water in the sugar syrup. Therefore, water is lost out by exosmosis. The membranes of the fruit pieces act as selectively permeable membranes. Thus the process of plasmolysis occurs resulting into shrinking of the fruit pieces.

Question 3.
The nucleus of the sieve tubes of the plant phloem is lost.
Answer:
The sieve tubes of the plant phloem conduct the food in plants. To make this transport easier, the nucleus of the sieve tubes of the plant phloem is lost,

Question 4.
Plant cells have less mitochondria than those of animal cells.
Answer:
Mitochondria are the cell organelles which are called powerhouse of the cell. They produce energy in the form of ATP. Animals are motile and need more energy for walking, running and moving. Plants are stationary. They do not need energy to greater extent. Therefore, they have lesser number of mitochondria.

Question 5.
Vacuoles do not have any typical size or shape.
Answer:
Vacuoles change their shape and size as per the need of the cell. Thus they do not have any fixed shape or size.

Question 6.
Ripe tomatoes appear red.
Answer:
When green tomatoes become ripe they lose chlorophyll and develop red pigment in them called lycopene. Therefore, ripe tomatoes appear red.

Give functions:

Question 1.
Endoplasmic reticulum.

Answer:

1. Endoplasmic reticulum or ER is the supporting framework of the cell.
2. The ribosomes attached to the membrane of the ER synthesize proteins. These proteins are conducted by ER.
3. The detoxification process is done by ER. The toxins that enter the cell through food, air and water are removed out by making them water soluble.

Question 2.
Golgi complex.

Answer:

1. Different secretions are prepared in the Golgi complex. Hence it is called the secretory organ of the cell.
2. The secretions are modified and sorted out as per their functions. They are further packed.
3. The enzymes, mucus, proteins, pigments, etc. are sorted and then dispatched to various target regions like plasma membrane, lysosome, etc.
4. Golgi complex also produces vacuoles and secretory vesicles.
5. Formation of cell wall, plasma membrane and lysosomes is aided by Golgi complex.

Question 3.
Plastids.
Answer:

1. Chloroplasts contain chlorophyll. They carry out the process of photosynthesis. They convert solar energy to chemical energy in the form of food.
2. Chromoplasts with different pigments can impart different colours to flowers and fruits.
3. Leucoplasts are responsible for the synthesis and storage of food like starch, oils and proteins.

Answer the following questions:

Question 1.
How does endosmosis and exosmosis occur in the cell?
Answer:

1. When the water concentration inside the cell is less as compared to the medium in which it is present, then the endosmosis takes place. This makes the water to enter inside the cell.
2. When water concentration inside the cell is more than the water concentration in the medium in which it is present, then the water comes out of the cell. This is called exosmosis.
3. Since the cell membrane acts as a semipermeable membrane, the processes of endosmosis and exosmosis takes place in the cell.

Question 2.
What is cytoplasm? What are the constituents of cytoplasm?
Answer:

1. The jelly like material present between the cell membrane and nucleus is called cytoplasm.
2. Cytoplasm without cell organelles is called cytosol.
3. All the cell organelles are spread in the cytoplasm.
4. Cytosol stores many vital constituents such as amino acids, glucose, vitamins, etc.
5. The cytoplasm of animal cells is dense and granular while that of plant cells is thin and peripheral. It is pushed to sides due to large central vacuole.

Question 3.
Describe the structure of the nucleus in the cell.

Answer:

1. Nucleus is the most important part of the eukaryotic cell.
2. Inside the nucleus there is round darkly stained nucleolus.
3. The nucleus is covered over by double membrane which is porous.
4. The nuclear pores allow the transport of different substances in and out of the nucleus to cytoplasm.
5. Inside the nucleus is the chromatin network which contains chromosomes. Chromatin fibres are thin which condense to form chromosomes. The chromosomes become clear and distinct at the time of cell division.
6. In every cell there are specific number of chromosomes. Chromosomes contain genes which are bearers of hereditary characters.

Question 4.
Why is endoplasmic reticulum compared with the pipelines?
Answer:

1. The endoplasmic reticulum works as pipelines to carry different substances in the cell.
2. It is a net like structure consisting of interconnected small tubes and sheets filled with fluid.
3. On the inner side the E.R. is connected to nucleus while at the outer side it is in contact with plasma membrane. Therefore, it works like a pipeline.

Question 5.
Write an account of the different structures seen in Golgi complex.
Answer:

1. Golgi complex is made up of 5-8 hollow and flat sacs called cisternae.
2. These are placed parallel to each other and are filled with different enzymes.
3. Golgi complex has two faces called forming face and maturation face.
4. The proteins packed in vesicles and coming from ER reach Golgi complex through cytoplasm.
5. They fuse with the formation face of the Golgi membranes for emptying their contents in the cisternae.
6. When these contents pass through the cisternae, they are chemically modified with the help of enzymes and are again packed in the vesicles.
7. These vesicles come out of Golgi ‘ complex at the maturation face.

Question 6.
How is energy produced in the mitochondria? How the structures of mitochondria help in this process?
Answer:

1. Around every mitochondrion there is a double membrane.
2. The outer membrane of these is porous while the inner membrane is deeply folded.
3. These folds or ‘cristae’ enclose the matrix filled with proteinaceous gel containing ribosomes, phosphate granules and DNA. Protein synthesis takes place in this matrix.
4. Mitochondria carry out oxidation of carbohydrates and fats in the cell. This produces energy in the form of ATP, s i.e. Adenosine Tri Phosphate

Question 7.
What is the benefit of foldings of inner membrane in mitochondria?

Answer:
The structure of the inner mitochondrial membrane is extensively folded and compartmentalized. The numerous imaginations of the membrane are called cristae. This folded inner membrane increases the area which is about 5 times more than that the outer membrane due to cristae. Cristae membranes have small round protein complexes known as Fx particles. In these particles the process of energy production goes on.

Research:

Question 1.
Keep 4 – 5 raisins in water and observe after an hour. Afterward, keep the same raisins in sugar solution and observe after an hour. Note down the observations and discuss in the classroom.
Answer:
When raisins are kept in water its outer skin acts as a semi-permeable membrane. The water content inside the raisin is lesser as compared to the water content in the outside medium. Therefore the water enters in the raisins due to process of endosmosis. Thus if raisins kept in plain water are observed after one hour, they are seen to be swollen.

On the other hand, if raisins are kept in sugar solution, they show plasmolysis and they shrink. The sugar solution acts as hypertonic medium. Water content in the raisin is higher than that present in the sugary solution. Thus water exists from raisins and its content thus shrinks.

Question 2.
Wooden doors fit very tightly in rainy season. Why does it happen?
Answer:
During rainy season there is more humidity in air. The doors get soaked in rain water. Though wood is non-living, it has the ability to absorb water. As the moisture is more in the surrounding area, it enters the wood. This is a type of endosmosis. It causes the doors to swell. The swollen doors then fit very tightly.

Diagram based questions:

Question 1.
Structure of the cell:

Answer:

(iii) Complete the chart:

Answer:

Question 2.
Sketch the diagrams to show how osmosis occurs in plant cell if kept separately in isotonic, hypotonic and hypertonic medium.
Answer:

Question 3.
Sketch the diagrams to show how osmosis occurs in animal cell if kept separately in isotonic, hypotonic and hypertonic medium
Answer:

Activity-based questions:

Question 1.
Experiment Activity – Take a drop of water on a clean glass slide. Using an ice-cream spoon, gently scrap the inner surface of your cheek. With a needle, transfer a little material from spoon to the water drop on the slide and spread it evenly. Put a drop of methylene blue stain on the smear. Put a cover slip and observe under microscope. Did you observe the cells with blue nucleus?
Answer:
Students should do this activity at school laboratory. There are squamous epithelial cells in the inner side of the cheek. When stained with methylene blue the nucleus takes up dark stain and can be seen clearly.

Question 2.
Experiment Activity – Take out a thin peel of Rheo or Croton leaf and observe the chromoplasts under the compound microscope.
Answer:
Students are expected to do the observations in the school laboratory.

Question 3.
Can you recall? Observe the cells of onion peel under the microscope. Have you seen the fully turgid, rectangular cells of onion peelings?

Answer:
Students are expected to do the observations in the school laboratory.

Balbharati Maharashtra State Board 8th Std Science Textbook Solutions 

• Living World and Classification of Microbes Class 8 Science Textbook Solutions
• Health and Diseases Class 8 Science Textbook Solutions
• Force and Pressure Class 8 Science Textbook Solutions
• Current Electricity and Magnetism Class 8 Science Textbook Solutions
• Inside the Atom Class 8 Science Textbook Solutions
• Composition of Matter Class 8 Science Textbook Solutions
• Metals and Nonmetals Class 8 Science Textbook Solutions
• Pollution Class 8 Science Textbook Solutions
• Disaster Management Class 8 Science Textbook Solutions
• Cell and Cell Organelles Class 8 Science Textbook Solutions

Std 8 Science Chapter 19 Life Cycle of Stars Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 8 Science Solutions Chapter 19 Life Cycle of Stars Notes, Textbook Exercise Important Questions and Answers.

Class 8 Science Chapter 19 Life Cycle of Stars Question Answer Maharashtra Board

Class 8 Science Chapter 19 Life Cycle of Stars Textbook Questions and Answers

1. Search and you will find:

Question a.
Our galaxy is called ………… .
Answer:
Our galaxy is called the Milky Way and Mandakini.

Question b.
For measuring large distances, ………… is used as a unit.
Answer:
For measuring large distances, the light year is used as a unit.

Question c.
The speed of light is …………. km/s.
Answer:
The speed of light is 300000 km/s.

Question d.
There are about ………… stars in our galaxy.
Answer:
There are about 100 billion stars in our galaxy.

Question e.
The end stage of the Sun will be …………. .
Answer:
The end stage of the Sun will be a white dwarf.

Question f.
Stars are born out of ………… clouds.
Answer:
Stars are born out of interstellar clouds.

Question g.
Milky way is a ………. galaxy.
Answer:
Milky way is a spiral galaxy.

Question h.
Stars are gigantic spheres of ………… gas.
Answer:
Stars are gigantic spheres of hot gas.

Question i.
The masses of other stars are measured relative to the mass of the …………. .
Answer:
The masses of other stars are measured relative to the mass of the Sun.

Question j.
Light takes ……………. to reach us from the Sun while it takes …….. to reach us from the moon.
Answer:
Light takes about 8 minutes to reach us from the Sun while it takes about 1 second to reach us from the moon.

Question k.
The larger the mass of a star, the faster is its ……….. .
Answer:
The larger the mass of a star, the faster is its evolution.

Question l.
The number of fuels used in the life of a star depends on its …………. .
Answer:
The number of fuels used in the life of a star depends on its mass.

2. Who is telling lies?

Question a.
Light year is used to measure time.
Answer:
False. (Light year is used to measure distance.)

Question b.
End stage of a star depends on its initial mass.
Answer:
True.

Question c.
A star ends its life as a neutron star when the pressure of its electrons balances its gravity.
Answer:
False. (A star ends its life as a neutron star when the pressure of its neutrons balances its gravity.)

Question d.
Only light can emit from the black hole.
Answer:
False. (Not even light can be emitted by a black hole.)

Question e.
The Sun will pass through the supergiant stage during its evolution.
Answer:
False. (The Sun will pass through the red giant stage during its evolution.)

Question f.
The Sun will end its life as a white dwarf.
Answer:
True.

3. Answer the following question:

Question a.
How do stars form?
Answer:
There are huge clouds of gas and dust in the empty spaces between the stars in a galaxy. These clouds are called interstellar clouds. The size of an interstellar cloud is about a few light years.

When an interstellar cloud starts contracting due to some disturbance, its density and temperature increase. This results in formation of a dense sphere of hot gas and nuclear energy generation starts at the centre of the star. Therefore, the gas sphere becomes self-luminous. Thus a star is formed, i.e., a star is born.
A huge interstellar cloud can produce thousands of stars at a time.

Question b.
Why do stars evolve?
Answer:
Although stars appear stable for quite a long period of time, their properties do change, though very slowly. A change in the properties of a star, leading to its passing through different stages, is called evolution of the star. Burning of the fuel at the centre of the star and a gradual decrease in its amount is the main reason of evolution of a star.

Question c.
What are the three end stages of stars?
Answer:

1. Stars having initial mass less than 8 times the mass of the Sun ultimately become white dwarfs.
2. Stars having initial mass between i 8 and 25 times the mass of the Sun ultimately become neutron stars.
3. Stars having initial mass larger than 25 times the mass of the Sun ultimately turn into black holes.

Question d.
Why was the name black hole given?
Answer:
When a star having initial mass larger than 25 times the mass of the Sun reaches its end stage, its gravitational force and density increase exponentially. All nearby objects get attracted towards the star and nothing can come out of it, not even light. All incident light is absorbed by the star. We can probably see a very minute black hole at its place. Hence, the name is given as black hole.

Question e.
Which types of stars end their life as a neutron star?
Answer:
Stars having initial mass between 8 and 25 times the mass of the Sun end up as neutron stars. When these stars pass through ( the supergiant stage, their size increases to 1000 times. Huge explosion that occurs in the last stage of these stars is very powerful and very high energy is given off. After the huge explosion, called the supernova explosion, their central portion contracts in size to about 10 km. In this stage, such stars are completely made up of neutrons and hence are called neutron stars.

4. A. If you are the Sun, write about your properties in your own words.
B. Describe white dwarfs.

Question A.
If you are the Sun, write about your properties in your own words.
Answer:

If you are the Sun, write about your properties in your own words.
Answer:
There are billions of stars in the galaxy called the Milky Way and Mandakini. I am one of the small stars called the Sun. I have my own family called the solar system. There are planets, satellites, asteroids, comets and meteors in my family. My mass is 2 × 1030 kg which is about 3.3 lakh times that of the earth. With a radius of 695700 km, my size is about 100 times that of the earth. My surface temperature is about 5800 K while it is 1.5 × 107 K at the centre.

72 % of my mass consists of hydrogen while 26 % consists of helium. Rest 2 % is made up of elements heavier than helium. I am about 4.5 billion years old and the scientists on the earth have concluded that not much change has taken place in my properties during this period. According to the scientists, in the end stage of my life, I will turn into a red giant star. Thereafter, I will first explode and then contract to become as small as the earth. I will appear small as well as white and hence, I will be called a white dwarf. This will be my last stage for ever.

Question B.
Describe white dwarfs.
Answer:
Depending on the initial mass, stars can reach one of the three end stages. White dwarf is one such stage of stars having initial mass less than 8 times the mass of the Sun. These stars undergo huge expansion and their radius increases by a factor of 100 to 200. These stars appear reddish because of their large size and lower temperature. Hence, these stars are also called red giant stars.

At the end of their evolution, these stars explode, their outer gas envelope is thrown outward and the inner part contracts to the size of the earth. However, the density in the star becomes very high. In this stage, the pressure due to electrons becomes independent of temperature and sufficient to balance the gravitational force for ever.

In this stage, such stars look white and due to their small size they are called white dwarfs.

Can you recall?

Question 1.
What is a galaxy?
Answer:
A system of billions of stars, their planetary systems and interstellar clouds of gas and dust held together by gravitational attraction is called a galaxy.

Question 2.
What are the different constituents of our solar system?
Answer:
The different constituents of our solar system are as follows :

1. Sun as a star.
2. Eight planets, namely, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune; of which Mercury, Venus, Earth and Mars are made up of rocks and minerals, while Jupiter, Saturn, Uranus and Neptune are made up of gases.
3. Natural satellites of some planets revolving around the respective planets.
4. Asteroids located between Mars and Jupiter.
5. Comets made up of ice, dust and gases.
6. Meteors and meteoroids.

Question 3.
What are the major differences between a star and a planet?
Answer:

1. Nuclear/atomic explosions regularly take place at the centre of stars. This causes the discharge of heat and light. Hence, stars shine. Planets do not discharge any light. They are visible due to the light reflected by them.
2. Stars are very big in size and massive relative to planets.
3. Stars twinkle, planets do not.
4. Stars are made up of hydrogen, helium and other light elements.
5. Planets are made up of solid, liquid or gaseous substances or a combination thereof.

Question 4.
What is a satellite?
Answer:
An astronomical object orbiting a planet is called a satellite.

Question 5.
Which is the star nearest to us?
Answer:
The star nearest to us is the Sun.

Project:

Question 1.
Use your imagination and make models of the Milky Way and the solar system.
Answer:
N.B. Students can make these models and display the same in science exhibition. You can use sand, scrap material, stones, pebbles, marbles, pieces of glass, beads, etc.

Question 2.
Write the effects: If the Sun disappears ………..
Answer:
N.B. You can write the effects with the help of the following points:

1. No sunrise, no sunset, no beauty of morning and evening wee hours.
2. In the absence of the gravitational force due to the Sun, the earth will drift away in the galaxy.
3. Darkness, no source of energy, end of life on the earth.
4. Moon will be invisible.
5. The temperature of the earth will fall below the freezing point of water. All oceans will also freeze.
6. Even the atmosphere will freeze and fall on the earth. This will cause more cosmic rays to reach the earth.

Class 8 Science Chapter 19 Life Cycle of Stars Important Questions and Answers

Match the columns:

Question 1.

Answer:

Question 2.
Mass of the star:

Answer:
Mass of the star:

Answer the following questions in one sentence each:

Question 1.
Name the star nearest to the earth.
Answer:
The Sun.

Question 2.
Other than the Sun, which is the star nearest to the earth?
Answer:
Alpha Centauri.

Question 3.
What would be the last stage of the Sun?
Answer:
The last stage of the Sun would be white dwarf.

Question 4.
Name two forces that act on any star.
Answer:
Gravitational force and the force due to the pressure of hot gas are the two forces that act on any star.

Question 5.
What is a black hole?
Answer:
A black hole is the end stage of a quite big star, where due to the extremely high gravitational force, nothing, not even light, comes out.

Answer the following questions:

Question 1.
How is stability of stars maintained?
Answer:
Properties of a star remain unchanged for quite a long time. The gravitational force and the force due to the pressure of the hot gas act together on a star. The gravitational force acts towards the centre of the star and tries to bring the gas particles close together. Hot gas shows the tendency to spread and its force acts away from the centre of the star. This force tries to disperse the gas particles. A balance between the gravitational force and the force due to the hot gas keeps the star stable.

However, if the magnitude of any one force is more than that of the other force, the star either contracts or expands depending upon which force dominates.

Question 2.
Why is the Sun called an ordinary star?
Answer:
The star nearest to the earth is the Sun. Hence, it appears quite big and bright. There are billions of stars in our galaxy called the Milky Way and Mandakini which are greater or lower in mass, size and temperature than the Sun. Hence, the Sun is called an ordinary star.

Do you know?

When the Sun will become a red giant, its diameter will increase so much that it will swallow Mercury and Venus. It is possible that the earth will also be absorbed by the Sun. It will take about 4-5 billion years for the Sun to reach this state.

Question 3.
Describe various stages of evolution of star.
Answer:
Though the properties of a star remain unchanged for quite a long time, this situation is never static. A star passes through different stages. This process is called the evolution of a star.
Important stages of the evolution of a star are as follows:

1. Initial stage of stability: The gravitational force and the force due to the pressure of the hot gas act together on a star. The gravitational force acts towards the centre of the star and tries to bring the gas particles close together. Hot gas shows the tendency to spread and its force acts away from the centre of the star.

This force tries to disperse the gas particles. A balance between the gravitational force and the force due to the hot gases keeps the star stable, as long as the energy generation continues at the centre of the star.

2. Burning of the fuel: As a star continuously emits energy, its energy constantly decreases. When the fuel at the centre of the star is exhausted, the energy generation stops and the temperature of the star starts decreasing. Decreasing temperature causes the gas pressure to decrease and the balance between the gravitational force and the force due to the gas pressure is no more maintained.

As the magnitude of the gravitational force is now more than that of the force due to the gas pressure, the star starts contracting. This causes another fuel to start burning, e.g. on exhausting hydrogen, helium starts undergoing fusion. Availability of multiple fuels depends on the mass of the star.

The higher the mass of the star, the more is the number of fuels used. The star either contracts or expands during the course of using these fuels. This may cause the imbalance between the gravitational force and the force due to the hot gas.

3. Total exhaustion of the fuel: When all fuels are exhausted, the energy generation in the star finally stops completely and the temperature of the star starts decreasing. The balance between the gravitational force and the force due to the gas pressure can no more be maintained. The evolution of the star ends and the star proceeds to its end stage.

4. End stage of a star: Once the fuel in the star is totally exhausted, the energy generation in the star stops and subsequently the gas pressure decreases, the star starts contracting and its density starts increasing. When the density becomes very high, some new types of pressures are generated which are independent of the temperature of the gas.

In such a case, the pressure remains stable despite low temperature and absence of any energy generation and thus the star remains stable for ever. This stage is the end stage of the star. Depending on the initial mass, stars can reach one of the three end stages.

• Stars having initial mass less than 8 times the mass of the Sun ultimately become white dwarfs.
• Stars having initial mass between 8 and 25 times the mass of the Sun ultimately become neutron stars.
• Stars having initial mass larger than 25 times the mass of the Sun ultimately turn into black holes.

Write short notes on the following:

Question 1.
End stages of stars having initial mass less than 8 times the mass of the Sun.
Answer:
These stars undergo huge expansion and their radius increases by a factor of 100 to 200 during their various stages of evolution. These stars appear reddish due to their lower temperature. Hence, they are called red giant stars.

At the end of evolution, these stars explode, their outer gas envelope is thrown out and the inner part contracts roughly to the size of the earth. Hence, the density of the star becomes very high. In this stage, the pressure due to electrons becomes independent of temperature and sufficient to balance the gravitational force for ever. Such stars look white and due to their small size they are called white dwarfs.

Question 2.
End stage of the stars having mass between 8 and 25 times the mass of the Sun.
Answer:
These stars pass through the red giant stage and later super giant stage, during which their size may increase to 1000 times. The huge explosion, called the supernova explosion, occurs in the last stage of the evolution. It is very powerful and very high energy is given off in this case.

As a result, the stars are visible even during the day. Later their central portion contracts to about 10 km. In this stage, the stars are completely made up of neutrons and are called neutron stars. The pressure of these neutrons is independent of temperature and sufficient enough to balance the gravitational force for ever.

Question 3.
End stages of stars having mass larger than 25 times the mass of the Sun.
Answer:
After the supernova explosion, no pressure can balance the gravitational force. Hence these stars contract continuously and their gravitational force and density increase exponentially. All nearby objects get attracted towards these stars and not even light can come out of them. Light falling on these stars is completely absorbed by the star. We cannot see these stars. A very minute black hole is formed at the place of such a star. This is the end stage of these stars.

Balbharati Maharashtra State Board 8th Std Science Textbook Solutions

• Human Body and Organ System Class 8 Science Textbook Solutions
• Introduction to Acid and Base Class 8 Science Textbook Solutions
• Chemical Change and Chemical Bond Class 8 Science Textbook Solutions
• Measurement and Effects of Heat Class 8 Science Textbook Solutions
• Sound Class 8 Science Textbook Solutions
• Reflection of Light Class 8 Science Textbook Solutions
• Man-made Materials Class 8 Science Textbook Solutions
• Ecosystems Class 8 Science Textbook Solutions
• Life Cycle of Stars Class 8 Science Textbook Solutions

Std 8 Science Chapter 16 Reflection of Light Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 8 Science Solutions Chapter 16 Reflection of Light Notes, Textbook Exercise Important Questions and Answers.

Class 8 Science Chapter 16 Reflection of Light Question Answer Maharashtra Board

Class 8 Science Chapter 16 Reflection of Light Textbook Questions and Answers

1. Fill in the blanks:

Question i.
The perpendicular to the mirror at the point of incidence is called …………. .
Answer:
The perpendicular to the mirror at the point of incidence is called the normal.

Question ii.
The reflection of light from a wooden surface is ……….. reflection.
Answer:
The reflection of light from a wooden surface is irregular reflection.

Question iii.
The working of a kaleidoscope is based on the properties of …………… .
Answer:
The working of a kaleidoscope is based on the properties of reflection of light.

2. Draw a figure describing the following: The reflecting surfaces of two mirrors make an angle of 90° with each other. If a ray incident on one mirror has an angle of incidence of 30°, i draw the ray reflected from the second mirror. What will be its angle of reflection?

Question a.
Draw a figure describing the following: The reflecting surfaces of two mirrors make an angle of 90° with each other. If a ray incident on one mirror has an angle of incidence of 30°, draw the ray reflected from the second mirror. What will be its angle of reflection?
Answer:
For the ray C, the angle of reflection = 60°.

3. How will you explain the statement ‘We cannot see the objects in a dark room’?

Question a.
How will you explain the statement ‘We cannot see the objects in a dark room’?
Answer:
In a room that is completely dark, no light falls on objects. Hence, no light enters our eyes. Hence, there is no sensation of vision, i.e., we cannot see the objects.

4. Explain the difference between regular and irregular reflection of light.

Question a.
Explain the difference between regular and irregular reflection of light.
Answer:
For regular reflection of light, the angles of incidence as well as the angles of reflection are the same for all parallel rays of light incident on the plane and smooth surface. Hence, the reflected rays are also parallel to one another.

For irregular reflection of light, the angles of incidence for parallel rays of light incident on the rough surface are not equal, and hence the angles of reflection are also not equal. Here, the reflected rays are not parallel to one another and spread over a large surface.

5. Draw a figure showing the following:
(a) Incident ray, (b) Normal, (c) Angle of incidence, (d) Angle of reflection, (e) Point of incidence, (f) Reflected ray.

Question a.
Draw a figure showing the following:
(a) Incident ray
(b) Normal
(c) Angle of incidence
(d) Angle of reflection
(e) Point of incidence
(f) Reflected ray.
Answer:

6. Study the following incident.

Swara and Yash were looking in a water-filled vessel. They could see their images clearly in the still water. At that instant, Yash threw a stone in the water. Now their images were blurred. Swara could not understand the reason for the blurring of the images.
Explain the reason for blurring of the images to Swara by answering the following questions:

Question i.
Is there a relation between the reflection of light and the blurring of the images?
Answer:
Yes.

Question ii.
Which types of reflection of light can you notice from this?
Answer:
Regular reflection of light when light is incident on the still water and irregular reflection of light when light is incident on the water as ripples are produced on its surface when a stone is thrown in the water.
Still water behaves as a plane and smooth surface while oscillating water behaves as a rough surface.

Question iii.
Are the laws of reflection followed in these types of reflection?
Answer:
Yes.

7. Solve the following examples.

Question a.
If the angle between the plane mirror and the incident ray is 40°, what are the angles of incidence and reflection?
Solution:
The angle between the plane mirror and the incident ray is 40°. Therefore, the angle of incidence (i) = the angle made by the incident ray with the normal to the plane mirror = 90° – 40° = 50°. The angle of reflection, r – i – 50°.

Question b.
If the angle between the mirror and reflected ray is 23°, what is the angle of incidence of the incident ray?
Solution:
The angle between the mirror and the reflected ray is 23°. Therefore, the angle of reflection (r) = the angle made by the reflected ray with the normal to the plane mirror = 90° – 23° = 67°.
∴ The angle of incidence, i = r = 67°.

Project:

Question a.
Apollo astronauts who stepped on the moon have kept some large mirrors there. Collect information about how the distance to the moon is measured using these.

Class 8 Science Chapter 16 Reflection of Light Important Questions and Answers

Rewrite the following statements by selecting the correct option:

Question 1.
If the angle made by the incident ray with the surface of a plane mirror is 30°, the angle of reflection must be …….. .
(a) 30°
(b) 90°
(c) 60°
(d) 15°
Answer:
If the angle made by the incident ray with the surface of a plane mirror is 30°, the angle of reflection must be 60°.

Question 2.
If the angle of incidence is 40°, the angle made by the reflected ray with the surface of the plane mirror must be ……. .
(a) 40°
(b) 50°
(c) 20°
(d) 80°
Answer:
If the angle of incidence is 40°, the angle made by the reflected ray with the surface of the plane mirror must be 50°.

Question 3.
If the angle of incidence is 20°, the angle made by the reflected ray with the normal to the surface must be ……… .
(a) 20°
(b) 70°
(c) 10°
(d) 40°
Answer:
If the angle of incidence is 20°, the angle made by the reflected ray with the normal to the surface must be 20°.

Question 4.
In a kaleidoscope, the mirrors are inclined to each other at ……. .
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Answer:
In a kaleidoscope, the mirrors are inclined to each other at 60°.

Question 5.
In a periscope, the mirrors are ………….. .
(a) parallel to each other
(b) at right angles to each other
(c) inclined at 45° to each other
(d) inclined at 60° to each other
Answer:
In a periscope, the mirrors are parallel to each other.

Find the odd one out and give the reason:

Question 1.
Plane mirror, Plywood, Wood, Rough tile.
Answer:
Plane mirror. In this case, regular reflection of light takes place. In other cases, reflection of light is irregular.

State whether the following statements are True or False. (If a statement is false, correct it and rewrite it.)

Question 1.
The sense of vision is the most important among our five senses.
Answer:
True.

Question 2.
In a periscope, the angle between the incident ray and the normal to the mirror is 30°.
Answer:
False. (In a periscope, the angle between the incident ray and the normal to the mirror is 45°.)

Answer the following questions in one sentence each:

Question 1.
What is an incident ray?
OR
Define incident ray.
Answer:
A ray of light falling on a surface is called an incident ray.

Question 2.
What is the point of incidence?
OR
Define point of incidence.
Answer:
The point at which the incident ray strikes the surface is called the point of incidence.
[Note: It is also the point of reflection.]

Question 3.
What is the normal?
OR
Define normal.
Answer:
The perpendicular to a surface at the point of incidence is called the normal.

Question 4.
What is the reflected ray?
OR
Define reflected ray.
Answer:
The ray of light that leaves the surface at the point of reflection (the same as the point of incidence) is called the reflected ray.

Question 5.
What is the angle of incidence?
OR
Define angle of incidence.
Answer:
The angle between the incident ray and the normal is called the angle of incidence.

Question 6.
What is the angle of reflection?
OR
Define angle of reflection.
Answer:
The angle between the reflected ray and the normal is called the angle of reflection.

Try this:

Switch off the light in your room at night for some time and then turn it on again.

Question 1.
Could you see the objects in the room clearly when the light was switched off?
Ans.
No.

Question 2.
What did you feel when it was turned on again?
Answer:
We could see the objects clearly. From the above activity you can notice that there is some connection between the sense of vision and light. When we switch off the light at night, the objects in the room cannot be seen and they can be seen as before when the light is switched on again. Thus, we can see objects when the light coming from these objects enters our eyes.

Answer the following questions:

Question 1.
What is reflection of light?
Answer:
When light rays fall on an object, their direction changes and they turn back. This is called the reflection of light.

Try this:

Material:
Torch light, mirror, a stand for hanging the mirror, black paper, comb, white paper, drawing board.
Activity :
1. Fit a white paper tightly over a table or drawing board.
2. Leaving out some portion in the middle of the comb, cover the rest with black paper so that light can only pass through the open central portion.
3. Hold the comb perpendicular to the white paper and throw torch light on its central portion.
4. Adjust the comb and torch so as to get light rays on the white paper. Now keep a mirror in the path of this ray of light as shown in the figure.
5. What do you observe?

Answer:
Light rays which fall on the mirror get reflected and travel in a different direction.

Question 2.
State the laws of reflection of light.
Answer:

1. The angle of reflection is equal to the angle of incidence.
2. The incident ray, the reflected ray and the normal lie in the same plane.
3. The incident ray and the reflected ray are on the opposite sides of the normal.

Try this:

Verification of the laws of reflection of light.
Equipment: Mirror, drawing board, pins, white paper, protractor, scale, pencil.
Activity:

1. Fit a white paper on the drawing board tightly as possible.
2. On the paper draw a line PQ indicating the position of the mirror.
3. Draw a perpendicular ON to PQ at point O.
4. Draw a ray AO making an angle of 30° with ON.
5. Fix two pins S and R along AO.
6. Fix the mirror to a stand and place it along PQ perpendicular to the drawing board.
7. Fix pins at T and U along the line joining the bottom of the reflected images of the pins at S and R.
8. Remove the mirror and join the points T and U and extend it up to O.
9. Measure ZTON.
10. Repeat steps 4 to 9 for angle of incidence equal to 45° and 60° and write down the angles in the following table.

What relation do you find between the angle of incidence and the angle of reflection? If you have done the experiment carefully, you will find that the angle of incidence is equal to the angle of reflection in all three cases. This verifies the laws of reflection.

Question a.
What will happen when a light ray is incident perpendicular to the mirror?
Answer:

Here,
r = i = 90°.
Hence the light ray, on reflection, will retrace the path.

Question 3.
Figures (a) and (b) show three parallel rays, shown in grey, incident on smooth and rough surfaces. The reflected rays drawn using laws of reflection are shown in red.
1. Rays reflected from which surface are parallel to one another?
2. What conclusion can you draw from the figure?
Answer:
1. Rays reflected from the smooth surface are parallel to one another.

2.When the reflecting surface is plane and smooth, the angles of incidence (i) as well as of reflection (r) are the same for all parallel rays incident on the surface. If i₁, i₂, i₃, … are the angles of incidence for incident parallel rays, and r₁, r₂, r₃, …, are the corresponding angles of reflection, then, i₁ = i₂ = i₃ = ……. = r₁ = r₂ = r₃ = ….. This is called regular reflection. Here, the reflected rays are parallel to one another. If the reflecting surface is rough and parallel rays are incident on it, then the angles of incidence are not equal and hence the angles of reflection are also not equal. Here, i₁ ≠ i₂ ≠ i₃ … and r₁ ≠ r₂ ≠ r₃ …, but r₁ = i₁, r₂ = i₂, r₃ = i₃ … as laws of reflection are obeyed. This is called irregular reflection. [Fig.(b)]. Here, the reflected rays are not parallel to one another and spread over a large surface.

Question 4.
What is regular reflection of light?
Answer:
The reflection of light from a plane and smooth surface is called regular reflection of light.

Question 5.
What is irregular reflection of light?
Answer:
The reflection of light from a rough surface is called irregular reflection of light.

Always remember:

1. Laws of reflection are followed in both and regular and irregular reflection.
2. The reflection of light in irregular reflection has not been obtained because the laws of reflection are not followed. They are obtained because the surface is rough (irregular).
3. In irregular reflection, the angles of incidence at different points are different. But at any one point, the angles of incidence and reflection are equal, i.e. i₁ = r₁, i₂ = r₂ …..

Can you recall?

Reflection of reflected light :

Question 1.
How do you see if the barber m a saloon has cut the hair on your neck properly or not?
Answer:
In a saloon, there are mirrors in your front and at back. The image of the back of your head is formed in the mirror at the back. The image of this image is formed in the mirror in front of you. Thus you can see how the hair at the backside of your head is cut.

Question 2.
What type of image do we see in a mirror? What happens to the left and right sides?
Answer:
The image in a plane mirror is upright (erect) and of the same size as the object, but the left and right sides are interchanged. Our right hand appears to be the left hand in the image and the left hand appears to be the right hand in the image. (This is called lateral inversion.)

Question 3.
How do we see the image of the moon in water?
Answer:
The moon is not self luminous. The sunlight falling on the surface of the moon is reflected. This reflected light is again reflected by water to give us the image of the moon.

Try this:

Kaleidoscope:

Activity:

1. Take three rectangular mirrors of the same size.
2. Using sticking tape, stick the mirrors together making a triangle with the reflecting surface facing inwards (see Fig.).
3. Take a white paper of triangular shape and fix it with a tape at one end of the mirrors closing that end.
4. Insert 4 – 5 coloured glass pieces in the hollow of the mirrors.
5. Close the other end also with a paper and make a hole in it.
6. Look through the hole towards light. You will see innumerable images of the glass pieces. These are formed due to reflections by the three mirrors.

You can see different designs in the kaleidoscope. The speciality of a kaleidoscope is that the designs do not easily repeat themselves. Every time the design is different. People making wall papers which are used to decorate walls and cloth designers use a kaleidoscope for making new designs.

Periscope:
Activity:

1. Take a cardboard box. Make slits in the top and bottom sides of the box and place two mirrors so that they make an angle of 45° with the sides of the box and are parallel to each other. Fix them with a sticking tape.
2. Make two windows of 1 inch each near the two mirrors. Now see through the bottom window.
3. Make note of what you see.

From the bottom window, one can see what is in front of the top window. This device is called a periscope. This is used in submarines to see objects above the surface of water. It is also used to observe and keep a watch on the objects or persons on the ground from an underground bunker. The kaleidoscope and periscope both use the properties of reflection of light.

[Note: In a periscope, the angle of incidence is 45° and the two plane mirrors are parallel to each other. Hence, the emergent ray is parallel to the incident ray.]

Example questions for practice:

Question 1.
If the angle between the plane mirror and the incident ray is 20°, what is the angle between the reflected ray
and the plane mirror?
Answer:
20°.

Question 2.
See Fig. In terms of O, what are the angles (i) AON (ii) BON (iii) AOB (iv) BOQ?

Answer:
(i) 90° – θ (ii) 90° – θ (iii) 180° – 2θ

Balbharati Maharashtra State Board 8th Std Science Textbook Solutions

• Human Body and Organ System Class 8 Science Textbook Solutions
• Introduction to Acid and Base Class 8 Science Textbook Solutions
• Chemical Change and Chemical Bond Class 8 Science Textbook Solutions
• Measurement and Effects of Heat Class 8 Science Textbook Solutions
• Sound Class 8 Science Textbook Solutions
• Reflection of Light Class 8 Science Textbook Solutions
• Man-made Materials Class 8 Science Textbook Solutions
• Ecosystems Class 8 Science Textbook Solutions
• Life Cycle of Stars Class 8 Science Textbook Solutions

Rational and Irrational Numbers Class 8 Maths Chapter 1 Practice Set 1.1 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.1 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Std 8 Maths Practice Set 1.1 Chapter 1 Solutions Answers

Question 1.
Show the following numbers on a number line. Draw a separate number line for each example.
i. \(\frac{3}{2}, \frac{5}{2},-\frac{3}{2}\)
ii. \(\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}\)
iii. \(\frac{-5}{8}, \frac{11}{8}\)
iv. \(\frac{13}{10}, \frac{-17}{10}\)
Solution:
i. \(\frac{3}{2}, \frac{5}{2},-\frac{3}{2}\)

Here, the denominator of each fraction is 2.
∴ Each unit will be divided into 2 equal parts.

ii. \(\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}\)

Here, the denominator of each fraction is 5.
∴ Each unit will be divided into 5 equal parts.

iii. \(\frac{-5}{8}, \frac{11}{8}\)

Here, the denominator of each fraction is 8.
∴ Each unit will be divided into 8 equal parts.

iv. \(\frac{13}{10}, \frac{-17}{10}\)

Here, the denominator of each fraction is 10.
∴ Each unit will be divided into 10 equal parts.

Question 2.
Observe the number line and answer the questions.

i. Which number is indicated by point B?
ii. Which point indicates the number \(1\frac { 3 }{ 4 }\) ?
iii. State whether the statement, ‘the point D denotes the number \(\frac { 5 }{ 2 }\) is true or false.
Solution:
Here, each emit is divided into 4 equal parts.
i. Point B is marked on the 10th equal part on the left side of O.
∴ The number indicated by point B is \(\frac { -10 }{ 4 }\).

ii.

Point C is marked on the 7th equal part on the right side of O.
∴ The number \(1\frac { 3 }{ 4 }\) is indicated by point C.

iii. True
Point D is marked on the 10th equal part on the right side of O.
∴ D denotes the number \(\frac{10}{4}=\frac{5 \times 2}{2 \times 2}=\frac{5}{2}\)

Maharashtra Board Class 8 Maths Solutions

• Rational and Irrational Numbers Practice Set 1.1 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.2 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.3 Class 8 Maths Solutions
• Rational and Irrational Numbers Practice Set 1.4 Class 8 Maths Solutions

Statistics Class 8 Maths Chapter 11 Practice Set 11.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.2 8th Std Maths Answers Solutions Chapter 11 Statistics.

Std 8 Maths Practice Set 11.2 Chapter 11 Solutions Answers

practice set 11.2 8th class Question 1.
Observe the following graph and answer the questions.

i. State the type of the graph.
ii. How much is the savings of Vaishali in the month of April?
iii. How much is the total of savings of Saroj in the months March and April?
iv. How much more is the total savings of Savita than the total savings of Megha?
v. Whose savings in the month of April is the least?
Solution:
i. The given graph is a subdivided bar graph.
ii. Vaishali’s savings in the month of April is Rs 600.
iii. Total savings of Saroj in the months of March and April is Rs 800.
iv. Savita’s total saving = Rs 1000, Megha’s total saving = Rs 500
∴ difference in their savings = 1000 – 500 = Rs 500.
Savita’s saving is Rs 500 more than Megha.
v. Megha’s savings in the month of April is the least.

practice set 11.2 Question 2.
The number of boys and girls, in std 5 to std 8 in a Z.P. School is given in the table. Draw a subdivided bar graph to show the data. (Scale : On Y axis, 1cm = 10 students)

Solution:

Statistics class 8 practice set 11.1 Question 3.
In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

Solution:

Statistics class 8 Question 4.
In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale: On Y axis: 1 cm = 500 students)

Solution:

Maharashtra Board Class 8 Maths Solutions

• Discount and Commission Practice Set 9.1 Class 8 Maths Solutions
• Discount and Commission Practice Set 9.2 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.1 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.2 Class 8 Maths Solutions
• Statistics Practice Set 11.1 Class 8 Maths Solutions
• Statistics Practice Set 11.2 Class 8 Maths Solutions
• Statistics Practice Set 11.3 Class 8 Maths Solutions

Division of Polynomials Class 8 Maths Chapter 10 Practice Set 10.1 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Std 8 Maths Practice Set 10.1 Chapter 10 Solutions Answers

Question 1.
Divide and write the quotient and the remainder.
i. 21m² ÷ 7m
ii. 40a³ ÷ (-10a)
iii. (- 48p⁴) ÷ (- 9p²)
iv. 40m⁵ ÷ 30m³
v. (5x³ – 3x²) ÷ x²
vi. (8p³ – 4p²) ÷ 2p²
vii. (2y³ + 4y² + 3 ) ÷ 2y²
viii. (21x⁴ – 14x² + 7x) ÷ 7x³
ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²
x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³
Solution:
i. 21m² ÷ 7m

∴ Quotient = 3m
Remainder = 0

ii. 40a³ ÷ (-10a)

∴ Quotient = -4a²
Remainder = 0

iii. (- 48p⁴) ÷ (- 9p²)

∴ Quotient = \(\frac { 16 }{ 3 }\) p²
Remainder = 0

iv. 40m⁵ ÷ 30m³

∴ Quotient = \(\frac { 4 }{ 3 }\) m²
Remainder = 0

v. (5x³ – 3x²) ÷ x²

∴ Quotient = 5x – 3
Remainder = 0

vi. (8p³ – 4p²) ÷ 2p²

∴ Quotient = 4p – 2
Remainder = 0

vii. (2y³ + 4y² + 3 ) ÷ 2y²

∴ Quotient = y + 2
Remainder = 3

viii. (21x⁴ – 14x² + 7x) ÷ 7x³

∴ Quotient = 3x
Remainder = -14x² + 7x

ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²

∴ Quotient = 3x³ – 2x² + 4x + 1
Remainder = 0

x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³

∴ Quotient = 5m – 3
Remainder = 10m + 8

Maharashtra Board Class 8 Maths Chapter 10 Division of Polynomials Practice Set 10.1 Intext Questions and Activities

Question 1.
Fill in the blanks in the following examples. (Textbook pg. no. 61)

1. 2a + 3a = __
2. 7b – 4b = __
3. 3p × p² = __
4. 5m² × 3m² = __
5. (2x + 5y) × \(\frac { 3 }{ x }\) = __
6. (3x² + 4y) × (2x + 3y) = __

Solution:

1. 2a + 3a = 5a
2. 7b – 4b = 3b
3. 3p × p² = 3p³
4. 5m² × 3m² = 15m⁴
5. (2x + 5y) × \(\frac { 3 }{ x }\) = \(6+\frac { 15y }{ x }\)
6. (3x² + 4y) × (2x + 3y) = 6x³ + 9x²y + 8xy + 12y²

Maharashtra Board Class 8 Maths Solutions

• Discount and Commission Practice Set 9.1 Class 8 Maths Solutions
• Discount and Commission Practice Set 9.2 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.1 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.2 Class 8 Maths Solutions
• Statistics Practice Set 11.1 Class 8 Maths Solutions
• Statistics Practice Set 11.2 Class 8 Maths Solutions
• Statistics Practice Set 11.3 Class 8 Maths Solutions

Statistics Class 8 Maths Chapter 11 Practice Set 11.1 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.1 8th Std Maths Answers Solutions Chapter 11 Statistics.

Std 8 Maths Practice Set 11.1 Chapter 11 Solutions Answers

Question 1.
The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

Solution:

Question 2.
The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

i. How many families use 45 units electricity?
ii. State the score, the frequency of which is 5.
iii. Find N, and ∑ f_i × x_i .
iv. Find the mean of electricity used by each family in the month of May.
Solution:

i. 2 families used 45 units of electricity.
ii. The score for which the frequency is 5 is 75
iii. N = 25 and ∑ f_i × x_i = 1425
iv.

The mean of electricity used by each.

Question 3.
The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2,3, 5, 6, 4, 2. Prepare a frequency table and And the mean of members of 40 families.
Solution:


∴ The mean of the members of 40 families is 3.9.

Question 4.
The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exhibition is:
2, 3 ,4, 1, 2, 3, 1, 5, 4, 2, 3, 1, 3, 5, 4, 3, 2, 2, 3, 2. Prepare a frequency table and find the mean of the data.
Solution:


∴ The mean of the given data is 2.75.

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.1 Intext Questions and Activities

Question 1.
The number of pages of a book Ninad read for five consecutive days were 60, 50, 54, 46, 50. Find the average number of pages he read everyday. (Textbook pg. no. 67)
Solution:
\(\frac{60+[50]+[54]+[46]+50}{[5]}=\frac{260}{[5]}=[52]\)
∴ Average number of pages read daily is 52

Maharashtra Board Class 8 Maths Solutions

• Discount and Commission Practice Set 9.1 Class 8 Maths Solutions
• Discount and Commission Practice Set 9.2 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.1 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.2 Class 8 Maths Solutions
• Statistics Practice Set 11.1 Class 8 Maths Solutions
• Statistics Practice Set 11.2 Class 8 Maths Solutions
• Statistics Practice Set 11.3 Class 8 Maths Solutions

Discount and Commission Class 8 Maths Chapter 9 Practice Set 9.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.2 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Std 8 Maths Practice Set 9.2 Chapter 9 Solutions Answers

Question 1.
John sold books worth Rs 4500 for a publisher. For this he received 15% commission. Complete the following activity to find the total commission John obtained.
Solution:
Selling price of the books = Rs 4500
Rate of commission = 15%
Commission obtained = 15% of selling price
\(=\frac{[15]}{[100]} \times[4500]\)
= 15 × 45
∴ Commission obtained = 675 Rupees.
∴ The total commission obtained by John is Rs 675.

Question 2.
Rafique sold flowers worth Rs 15,000 by giving 4% commission to the agent. Find the commission he paid. Find the amount received by Rafique.
Solution:
Here, selling price of flowers = Rs 15,000,
Rate of commission = 4%
i. Commission = 4% of selling price
= \(\frac { 4 }{ 100 }\) × 15,000
= 4 x 150
∴ Commission = Rs 600

ii. Amount received by Rafique = selling price – commission
= 15,000 – 600
= Rs 14,400
∴ Rafique paid Rs 600 as commission and the amount received by him was Rs 14,400.

Question 3.
A farmer sold food grains for Rs 9200 through an agent. The rate of commission was 2%. How much amount did the agent get ?
Solution:
Here, selling price of food grains = Rs 9200,
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 9200
= 2 × 92
= Rs 184
∴ The agent got a commission of Rs 184.

Question 4.
Umatai purchased following items from a Khadi – Bhandar.
i. 3 sarees for Rs 560 each.
ii. 6 bottles of honey for Rs 90 each.
On the purchase, she received a rebate of 12%. How much total amount did Umatai pay?
Solution:
Here, number of sarees = 3,
Price of each saree = Rs 560
∴ Cost of 3 sarees = 560 × 3
= Rs 1680 …(i)
Also, number of honey bottles = 6,
Price of each bottle = Rs 90
∴ Cost of 6 honey bottles = 90 × 6
= Rs 540
Total amount of purchase
= cost of 3 sarees + cost of 6 honey bottles
= 1680 + 540 … [From (i) and (ii)]
= Rs 2220 …(iii)
Rate of rebate = 12%
Rebate = 12% of total amount of purchase
= \(\frac { 12 }{ 100 }\) × 2220
= 12 × 22.20
= Rs 266.40 ..(iv)
Amount paid by Umatai
= Total amount of purchase – Rebate
= 2,220 – 266.40 … [From (iii) and (iv)]
= Rs 1953.60
∴ The total amount paid by Umatai is Rs 1953.60.

Question 5.
Use the given information and fill in the boxes with suitable numbers.
Smt. Deepanjali purchased a house for Rs 7,50,000 from Smt. Leelaben through an agent. Agent has charged 2 % brokerage from both of them.
Solution:
i. Smt. Deepanjali paid 7,50,000 × \(\frac { 2 }{ 100 }\)
= 7,500 × 2 = Rs 15,000 brokerage for purchasing the house.

ii. Smt. Leelaben paid brokerage of Rs 15,000

iii. Total brokerage received by the agent is = 15,000 + 15,000 = Rs 30,000

iv. The cost of house Smt. Deepanjali paid is = 7,50,000 + 15,000 = Rs 7,65,000

v. The selling price of house for Smt.Leelaben is = 7,50,000 – 15,000
= Rs 7,35,000

Maharashtra Board Class 8 Maths Solutions

• Discount and Commission Practice Set 9.1 Class 8 Maths Solutions
• Discount and Commission Practice Set 9.2 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.1 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.2 Class 8 Maths Solutions
• Statistics Practice Set 11.1 Class 8 Maths Solutions
• Statistics Practice Set 11.2 Class 8 Maths Solutions
• Statistics Practice Set 11.3 Class 8 Maths Solutions