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MP Board Class 6th Science Solutions Chapter 2 Components of Food

Components of Food Textbook Exercises

Question 1.
Name the major nutrients in our food?
Answer:
The major nutrients, in our food are proteins, fats, carbohydrates, vitamins and minerals. In addition to above, food also contains dietary fibres and water.

Question 2.
Name the following:

1. The nutrients which mainly give energy to our body.
2. The nutrients that are needed for the growth and maintenance of our body.
3. A vitamin required for maintaing good eyesight.
4. A mineral that is required for keeping our bones healthy.

Answer:

1. Fats and carbohydrates
2. Proteins
3. Vitamin A
4. Calcium

Question 3.
Name two foods each rich in:

1. Fats
2. Starch
3. Dietary fibre
4. Protein.

Answer:

1. Fats: Groundnuts, til, milk, ghee.
2. Starch: Peanuts, dal (cooked), rice (cooked), raw potato.
3. Dietary fibre: Fresh fruits, grains, pulses, potatoes.
4. Protein: Gram, soyabeans, eggs, paneer.

Question 4.
Tick (✓)  the statements that are correct?

1. By eating rice alone, we can fulfill nutritional requirement of our body.
2. Deficiency diseases can be prevented by eating a balanced diet.
3. Balanced diet for the body should contain a variety of food items.
4. Meat alone is sufficient to provide all nutrients to the body.

Answer:

1. (x)
2. (✓)
3. (✓)
4. (x)

Question 5.
Fill in the blanks:

1.  ……………………….. is caused by deficiency of Vitamin D.
2. Deficiency of ………………………. causes a disease known as beri – beri.
3. Deficiency of Vitamin C causes a disease known as ……………………….
4. Night blindness is caused due to deficiency of ………………….. in our food.

Answer:

1. Rickets
2. Vitamin B
3. Scurvy
4. Vitamin A.

Projects And Activities

Activity 1.
Prepare a table to show that means from different regions/states?
Answer:
Some common meals of different regions/states

Activity 2.
Prepare a table to show that various nutrients present in some food items?
Answer:
Nutrients present in some food items

Activity 3.
Make a list of uncooked items of food that are found around you. Indicate the importance of each one of these in your diet. How can these items be protected from spoiling or contamination?
Answer:
Uncooked food items, their importance in diet and method of protection.

Activity 4.
Define diagrammetically the daily requirement of carbohydrates, fats and proteins in an adult?
Answer:
Daily requirement of carbohydrates of an adult is about 400 g to 500 g per day and that of proteins is 65 g to 75 g per day. The daily requirement of fats for females is 50 g to 55 g per day and for males 60 g to 70 g per day. During pregnancy and lactation period, protein requirement in females is greater than that for the males.

Activity 5.
Define diagrametically the daily requirement of some minerals in an adult?
Answer:
The minerals present in our body are mainly in the form of compounds of sodium phosphorous, calcium, chlorine, iron, potassium, sulphur, copper and iodine. Only small amount of minerals are required in our daily diet. Each one of these minerals is necessary for a proper growth of the body and to maintain good health. Following figure shows the daily requirement of some minerals for adults.

Activity 6.
Define diagrametically the daily requirements of some vitamins in adults?
Answer:
Vitamins are an essential component of our diet as they perform specific functions in our body. Different types of vitamins have been given specific names, like vitamin A, vitaminC, vitamin D, vitamin E and vitamin K. Some of the vitamins are soluble in water while some others dissolve only in fats. Daily requirement (in mg) of various vitamins are shown in following diagram: vitamin A = 0.5 mg

Activity 7.
What is the basic functions of food and what it does to our body?
Answer:

Components of Food Intex Questions

Question 1.
Our body also prepares Vitamin D in the presence of Sunlight?
Answer:
Yes.

Question 2.
Paheli wonders whether animal food also consists of these different components and do they also need a balanced diet?
Answer:
Yes, animals also need a balanced diet.

Components of Food Additional Important Questions

Objective type Questions Components of Food

Question 1.
Choose the correct answer:

Question (a)
Which one of the following foods provides energy to the body:
(a) carbohydrates
(b) proteins
(c) minerals
(d) vitamins.
Answer:
(a) carbohydrates

Question (b)
Salad in our diet mainly contains:
(a) carbohydrates
(b) fats
(c) proteins
(d) roughage.
Answer:
(d) roughage.

Question (c)
Which one of the following is an example of fats?
(a) Banana
(b) Wheat
(c) Butter
(d) Lemon.
Answer:
(c) Butter

Question (d)
Kishmish is a dry form of:
(a) grapes
(b) watermelon
(c) mango
(d) none of the above.
Answer:
(a) grapes

Question (e)
Which one of the following represents a balanced diet?
(a) Leafy vegetables
(b) Mango
(c) Milk
(d) Wheat and rice.
Answer:
(c) Milk

Question (f)
The total requirement of fats for an adult is about –
(a) 60 g to 80 g per day
(b) 50 g to 70 g per day
(c) 80 g to 100 g per day
(d) None of the above.
Answer:
(a) 60 g to 80 g per day

Question (g)
Communicable diseases are caused by:
(a) virsues
(b) bacteria
(c) fungi
(d) all of these.
Answer:
(d) all of these.

Question (h)
Which is a water – soluble vitamin?
(a) Vitamin A
(b) Vitamin C
(c) Vitamin D
(d) all of these.
Answer:
(b) Vitamin C

Question (i)
Beri – Beri is caused due to the deficiency of vitamin:
(a) A
(b) B
(c) D
(d) K
Answer:
(b) B

Question (j)
One of the following is not a communicable disease –
(a) malaria
(b) scurvy
(c) typhoid
(d) dysentery.
Answer:
(b) scurvy

Question 2.
Fill in the blanks:

1. Organism feeding on flesh are called ………………………….
2. Less intake of proteins in diet causes ………………………….
3. Less intake of nutrients than required is …………………………..
4. Carbohydrates and fats are composed of ………………………..
5. Pulses are rich in …………………………..
6. Starch is a ………………….. sugar.
7. ………………….. is the essence of life.
8. A …………………….. includes both of these components in required proportion.
9. …………………….. is an example of saturated fat.
10. A bad taste and fuel smell indicates that the food has been infested with ………………………..
11. Vitamin K helps in ……………………….
12. Deficiency of B₁₂ causes ………………………..
13. Deficiency of iron causes ………………………….
14. ……………………….. and ……………………… are essential nutrients.
15. Lack of vitamins causes and leads to ………………………..

Answer:

1. Carnivorous
2. Kwashiorker
3. Malnutrition
4. Carbon, hydrogen and oxygen
5. Proteins
6. Polymers
7. Water
8. Balance diet
9. Vanaspati
10. Micro – organisms,
11. Clotting of blood
12. Anaemia
13. Anaemia
14. Vitamins, minerals
15. Specific diseases.

Question 3.
Which of the following statements are true (T) or false (F):

1. Proteins supply the maximum calories to our bodies.
2. We can live without proteins.
3. A diet that supplies enough calories is a balanced diet.
4. Protein is a staple food.
5. Potato is rich in carbohydrates.
6. Tomatoes contain Vitamin C.
7. Milk, meat, pulses and fish are sources of proteins.
8. Fats contain more energy than carbohydrates.
9. Expensive food is not always the best food.
10. Roughage contains all the food components.
11. High dose of vitamins to children may not harmful.
12. Rice is one of the major basic foods.
13. Over – eating does not causes any diseases.
14. Vitamin D deficiency may occur during pregnancy and lactation.
15. Vitamin C is not soluble vitamin.
16. Phosphorus is very important for the development of body.
17. Deficiency of Vitamin A makes our bones weak.
18. Deficiency of iron causes paleness.
19. Deficiency of Vitamin D cause swollen and bleeding gums.
20. Deficiency of vitamin B help to increase for palliate.

Answer:

1. False
2. False
3. False
4. False
5. True
6. True
7. True
8. True
9. True
10. False
11. False
12. True
13. False
14. True
15. False
16. True
17. False
18. True
19. False
20. False.

Question 4.
Match the items of Column A with the items of Column B:

Answer:

(i) – (c)
(ii) – (e)
(iii) – (a)
(iv) – (f)
(v) – (b)
(vi) – (d)

Components of Food Very Short Answer Type Questions

Question 1.
Which of the following produce energy: Fat or Carbohydrates?
Answer:
Carbohydrates.

Question 2.
What is calorie?
Answer:
Calorie is the unit of heat. The food which we take is oxidised, in the presence of oxygen with the liberation of energy.

Question 3.
Which one offers you more energy 100g of grapes or one banana?
Answer:
One banana.

Question 4.
Which has more vitamins 100 g of grapes or 100g of spinach?
Answer:
100g of spinach.

Question 5.
What are nutrients?
Answer:
Nutrients are the components of food that the body needs in adequate amount for growth to reproduce and lead a normal healthy life.

Question 6.
Write the sources of car body drates?
Answer:
The carbohydrates are mainly found in sugar, wheat, maize and cereal etc.

Question 7.
What are the sources of fats?
Answer:
The sources of fats are ghee, butter, nuts and vegetable oils.

Question 8.
How much energy is produced from one gram of carbohydrates?
Answer:
16.8 kJ of energy is produced from one gram of carbohydrates.

Question 9.
Write the names of any two water soluble vitamins?
Answer:
Vitamin B and C.

Question 10.
What are the sources of Vitamin A?
Answer:
The sources of Vitamin A are milk, carrot, fish, oil etc.

Question 11.
Name the fat soluble Vitamins?
Answer:
Vitamin A and D.

Question 12.
What are the sources of Vitamin D?
Answer:
The sources of Vitamin D are eggs, fish, oil and milk products.

Question 13.
Which mineral is vital for bones and teeth?
Answer:
Calcium and phosphorus.

Question 14.
Name the main constituent of roughage?
Answer:
Cellulose is the main constituent of roughage.

Question 15.
What do you mean by staple food?
Answer:
The main food that we eat to provide us energy is called staple food. For example, chapati, rice, bread, etc.

Question 16.
Name the two biotic factors that damage the food – grains?
Answer:

1. Temperature, and
2. Moisture content.

Question 17.
Grapes get spoiled faster as compared to apples. Why?
Answer:
Because grapes contain more water content than apples, so they spoil faster.

Question 18.
Write the cause for food poisoning?
Answer:
Food poisoining is caused by micro – organisms like bacteria which can reproduce rapidly.

Question 19.
What is dehydration?
Answer:
Removal of water from fruits and vegetables is called dehydration.

Question 20.
Name two sources each of animal and vegetable proteins?
Answer:
Sources of animal proteins are Egg, Meat, Fish and Milk. Sources of vegetable proteins are Pulses, Peas, Bean and Soyabean.

Question 21.
What are the symptoms of Vitamin C deficiency?
Answer:
Deficiency of Vitamin C causes trouble in gums.

Question 22.
Name the disease caused by deficiency of Vitamin A?
Answer:
Deficiency of Vitamin A causes weakness in eyes and night blindness.

Question 23.
Same mass of which nutrient gives more energy fats or carbohydrates?
Answer:
The fats produce more energy than carbohydrates because they have less oxygen percentage. A gram of carbohydrate produce 4.2 kcal While a gram of fat produce 9.1 kcal of heat.

Question 24.
What is a balanced diet?
Answer:
A meal which contains various constituents of food which are necessary to keep the body healthy. A balanced meal has an appropriate food ratio of carbohydrate, protein, fat, vitamins and minerals.

Question 25.
Mention the factors which effect our health?
Answer:
The factors which effect our health are:

1. Unbalanced food
2. Diseases caused by infection.

Question 26.
What are the major factors affecting the human health?
Answer:
The major factors affecting the human health are:

1. Intrinsic (or internal) factors
2. Extrinsic (or external) factors.

Question 27.
Define intrinsic factors?
Answer:
The disease causing factors which exist within the humun body are called intrinsic factors?

Question 28.
Define extrinsic factors?
Answer:
The disease causing factors which come from outside the human body are called extrinsic factors.

Question 29.
Name the disease caused by deficiency of Vitamin C?
Answer:
Scurvy.

Question 30.
Name the disease caused by deficiency of Vitamin D?
Answer:
Rickets.

Question 31.
How much proteins do you need in your daily diet?
Answer:
We need proteins according to our body weight which is 2.5 gm per kilogram weight of the body.

Question 32.
When you fry your food in oil, which Vitamins are generally lost?
Answer:
Vitamin C.

Question 33.
Name some water – borne diseases?
Answer:
Water – borne disease are jaundice, cholera, polio, diarrhoea, typhoid.

Question 34.
Name some air – borne diseases?
Answer:
Air – borne diseases are whooping cough, common cold.

Question 35.
Name some of the diseases caused by extrinsic factors?
Answer:
Kwashiorkor, goitre, obesity, malaria, T.B., AIDS etc.

Question 36.
List some food – borne diseases?
Answer:
Some food – brone diseases are diarrhoea, dysentery and cholora during the rainy season.

Components of Food Short Answer Type Questions

Question 1.
How can we say that the fats are like an energy bank in living organism?
Answer:
We know that the fats have more calories of energy than carbohydrates in a unit mass. Fats have less oxygen and give more energy. Only fat can be stored for the future use as the polar bear does. It takes food before winter and then hibernates for several months. During this period the stored fat is consumed and thus fat acts as an energy bank.

Question 2.
How will you test for carbohydrate?
Answer:
To test the carbohydrate we take the given material and heat it with water. Then we put two drops of iodine solution in it and see the result. If the colour is changed to blue black, then carbohydrate is there otherwise not.

Question 3.
Name any three sources of carbohydrates?
Answer:
Carbohydrates are the compounds of carbon, hydrogen and oxygen. These are one of the compound of our food which provide us energy. Various sources from where we get carbohydrates are rice, wheat, cereals, sugar etc.

Question 4.
What are proteins?
Answer:
Proteins are the polymers of amino acids. There are only twenty amino acids known to us. They link together to form proteins. The amino acids are made up of carbon, hydrogen, oxygen and nitrogen. However some also contains phosphorous, sulphur etc. The important sources of proteins are meat, fish, egg, milk and all pulses.

Question 5.
What are minerals?
Answer:
Minerals are the chemical elements present in our food. They help to regulate various metabolic activities in our body. The importatant minerals required are calcium, phosphorous, iron, iodine, potassium sodium and magnesium. Calcium and phosphorous are required for bone and teeth formation, iron is required for the formation of haemoglobin and sodium and potassium are required for normal functioning of the nerve cells.

Question 6.
Why does living organisms required food?
Answer:
Every living being needs energy for its life processes. This energy can be obtained only from the food. So to meet out the energy requirement of the body, food is necessary. Food is also needed for growth and control of various life activities of the body.

Question 7.
What are the three important qualities of balanced diet?
Answer:
The three important qualities of balanced diet are as follows:

1. It should be rich in essential nutrients such as vitamins, minerals, amino acids, etc.
2. It should be able to provide enough raw material to meet the basic needs of growth, repair and replacement of cells in our body.
3. It should provide energy required by the body.

Question 8.
How can you vary your diet without making it costlier?
Answer:
We can vary our diet by adopting the following instructions without making it costlier:

1. We should take seasonal vegetables and fruits because they are cheap at that time.
2. Rice and wheat should be eaten alternately so that a balanced diet may be obtained at low cost.
3. We should use cheap and nutritious fruits e.g., banana, guava, are more nutritious than grapes.

Question 9.
Name the foods needed?

1. For strong bones and teeth.
2. To prevent scurvy.
3. To avoid constipation.
4. For warmth.
5. For growth.

Answer:

1. For strong bones and teeth. Milk, Fish, Oils, Eggs,
2. To prevent scurvy. All citrus fruits, Amla, Orange, Lemon, etc.
3. To avoid constipation. Water, Juicy fruits, Fresh vegetables, etc
4. For warmth. Meat, Fish,
5. For growth. Green leafy vegetables, Milk.

Question 10.
Explain why you should:

1. Eat less cakes and ice – cream.
2. Remove most of the fat from meat.
3. Eat fresh food instead of processed food.
4. Eat more fruits and vegetables.

Answer:

1. Cakes and ice – cream have too much carbohydrates and sugar.
2. Fat is not digested quickly. It is used as fuel in the deficiency of carbohydrates only otherwise in deposits in the inner side of blood vessels.
3. The perishable food items are processed to keep them edible for a long time but many food nutrients are destroyed during processing, so we should eat fresh food. Moreover the nutrients present in the food can easily be processed food.
4. Fruits and vegetables contain more food nutrients than the preserved and non-perishable food items, so we should eat more fruits and vegetables.

Question 11.
What is the main difference between vitamins and minerals?
Answer:
Vitamins:

1. Vitamins are chemical substances which help the proteins particularly enzymes in their proper functioning.
2. Their main source are fruits, vegetables, milk and other food items. They cannot be extracted from the earth.

Minerals:

1. Minerals are the main constituents of our body parts such as teeth, bone blood etc.
2. The minerals can be extracted from the earth. But animals obtain them from fruits, vegetables, milk, etc.

Question 12.
What are deficiency diseases?
Answer:
There are many vitamins which are used as nutrients. They are named by alphabet letters such as A, B, B₁, B₂, B₆, B₁₂, C, D, K. Each of them plays its own role. If any one of them is not present, an abnormality in the body can be seen. Such abnormalities are known as vitamin deficiency diseases.

Question 13.
List some diseases that are caused by vitamin deficiency in the body.
Answer:
Vitamin deficiency diseases are:

Components of Food Long Answer Type Questions

Question 1.
What are the various functions of protein?
Answer:
The various functions of proteins are:

1. They form enzymes which are very important for living organisms.
2. They are able to repair cells of the body which have undergone wear and tear.
3. They are also used in making new cells.
4. Proteins help in building of the body.
5. Proteins help in digestion of body.
6. Haemoglobin is a kind of protein which helps in the transportation of oxygen and carbon dioxide.
7. Muscle, skin, hair and nails are all proteins.
8. Proteins also act as the body materials.

Question 2.
What are the roles of each of these components?
Answer:
Components of Food

1. Carbohydrates
2. Proteins
3. Fats
4. Vitamins
5. Minerals.

Question 3.
How can you balance your diet without adding to its cost? Suggest any one method to do so?
Answer:
We should eat the things which are easily available and cheap and having the food nutrients in equal quantity as that of costly food items. We should find out the nutrients and their percentage in the edible things and also their cost and then suggest the people to eat those things.
A list is given for average daily calorie needs of people of different ages.

Question 4.
What are carbohydrates?
Answer:
The carbohydrates are components of carbon, hydrogen and oxygen. The simple carbohydrate is glycose. The other carbohydrates include sucrose, lactose, sugar, starch etc. The staple foods like rice, wheat, maize are rich sources of carbohydrates together with potato, sugarcane, grapes etc.

Question 5.
Draw a neat diagram of some sources of fats?
Answer:

Question 6.
Write a short note on Vitamin B complex?
Answer:
Vitamin ‘B’ complex is not a single vitamin but it is a group of many vitamins. The main vitamins of this group are Vitamin B₁, Vitamin B₂, Vitamin B₁₂.

Vitamin B₁:
It is found in egg, meat, cereals, yeast, cabbage, soyabean. This is important in helping the digestive system and the nervous system. The deficiency disease of this vitamin is known as Beri – Beri.

Vitamin B₂:
The chief sources of this vitamin are green leafy vegetables, peas, beans, cheese. It is helpful in keeping our mouth and skin healthy and for normal growth.

Vitamin B₁₂:
This vitamin is available in milk, cheese, etc. and is responsible for proper growth of the body. The deficiency disease due to this vitamin is the anaemia.

Question 7.
Write down the sources, importance and deficiency diseases of Vitamin A, Vitamin C, Vitamin D and Vitamin K?
Answer:

Vitamin A:
The main sources of this vitamin are milk, butter, cheese, egg, liver oils, green and yellow vegetables. This is important for eyes, hair and skin and the deficiency disease due to Vitamin ‘A’ is night blindness.

Vitamin C:
Citrus fruits as the lemon, organges are the main sources and is also available in plenty amount in goose berries, guava and amla. This vitamin is helpful in keeping teeth, gums and joints healthy. The deficiency disease is known as the scurvy.

Vitamin D:
The main sources are fish, liver oil, milk. Our body can prepare it in sunlight. It is essential for the normal growth of the bones. The deficiency disease is known as the rickets.

Vitamin K:
It is available in green leafy vegetables, tomatoes and egg yolks in sufficient amount. Due to its deficiency there is excessive bleeding after injury. Vitamin D helps in clotting of the blood.

Question 8.
What are the functions of iron and iodine in the body?
Answer:
The function of iron in our body are:
It gives red colour to the blood and transmits oxygen to body. It is needed for the formation of blood cells (RBC). The functions of iodine in our body are: It is very important component of thyroxin, a hormone secreted by the thyroid gland situated in the neck. Iodine deficiency can cause disorders resulting in retarded growth and mental disability. It also causes abnormal enlargment of the thyroid gland commonly known as goitre.

MP Board Class 6th Science Solutions

MP Board Class 9th Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

Question 1.
A traffic signal borard, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’ Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig.) the advertisements yield an earning of ₹ 5000/m² per year. Acompany hired one of its walls for 3 months. How much rents did it pay?

Solution:

s = \(\frac{122+120+22}{2}\) = \(\frac{264}{2}\) = 132 m
s – a = 132 – 122 = 10 m
s – b = 132 – 22 = 12m
s – c = 132 – 22 = 110 m
area of triangular portion of wall = \(\sqrt{32x10x12x110}\)
= \(\sqrt{2x2x3x11x10x2x2x3x11x10}\)
= 10 x 2 x 2 x 3 x 11 = 1320 m²
Rate = ₹ 5000/m² per year
Rent for 3 months = 1320 x \(\frac{5000×3}{12}\)
= 330 x 5000
= ₹ 16,50,000

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “Keep The Park Green And Clean” (see Fig.). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Solution:
a = 15 m
b = 11 m
c = 6m
p = a + b + c
= 15 + 11 + 6
= 32 m P

s = \(\frac{P}{2}\) = \(\frac{32}{2}\) = 16m
s – a = 16 – 15 = 1 m
s – b = 16 – 11 = 5 m
s – c = 16 – 6 = 10m
Area of triangular park = \(\sqrt{16x1x5x10}\)
= \(\sqrt{2x2x2x2x1x5x2x5}\)
= 2 x 2 x 5√2 = 2o\(\sqrt{2m^{2}}\)

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:

a = 18 cm
b = 10 cm
Let the third side be c
p = 42 cm
18 + 10 + C = 42
C = 14
S = \(\frac{P}{2}\) = \(\frac{42}{2}\) = 21 cm
s – a = 21 – 18 = 3
s – b = 21 – 10 = 11
s – c = 21 – 14 = 7
Area of ∆ = \(\sqrt{21x3x11x7}\) = \(\sqrt{3x7x3x11x7}\)
= 3 x 7\(\sqrt{11}\) = 21\(\sqrt{11}\) cm²

Question 5.
Sides of triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area
Solution:
a = 12x
b = 17x
c = 25x
p = 12x + 11x + 25x = 540
54x = 540
x = \(\frac{540}{54}\) = 10
a = 12 x 10 = 120 cm
b = 17 x 10 = 170 cm
c = 25 x 10 = 250 cm
s = \(\frac{P}{2}\) = \(\frac{540}{2}\) = 270 cm
s – a = 270 – 120 = 150 cm
s – b = 270 – 170 = 100 cm
s – c = 270 – 250 = 20 cm
Area of ∆ = \(\sqrt{270x150x100x20}\)
= \(\sqrt{3x3x3x10x3x5x10x10x10x2x10}\)
= 3 x 3 x 10 x 10 x 10
= 9000 cm²

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:

a = 12 cm
b = 12cm
Let the third side be c.
p = a + b + c
30 = 12 + 12 + c
c = 30 – 24 = 6 cm
s = \(\frac{P}{2}\) = \(\frac{30}{2}\) = 15
s – a = 15 – 12 = 3
s – b = 15 – 12 = 3
s – c = 15 – 6 = 9
Area of ∆ = \(\sqrt{5x3x3x9}\)
= \(\sqrt{5x3x3x3x3x3}\)
= 3 x 3\(\sqrt{15}\) = 9\(\sqrt{15}\) cm²

Area of Quadrilaterals:
To find the area of quadrilaterals divide the quadrilateral into two triangles using a diagonal and then use heron’s formula.

Example 1:
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, And the height of the parallelogram.
Solution:

Here, a = 13 cm,
b = 14 cm,
c = 15 cm
Base of parallelogram =14 cm.

= 7 x 3 x 2 x 2
= 84 cm²
Let h be the height of the parallelogram ADEC.
Area of parallelogram ADEC = Area of ∆ABC (given)
Base x height = 84
14 x height = 84
h = \(\frac{84}{14}\)
= 6 cm.

Example 2:
The sides of a quadrilateral, taken in order are 5, 12, 14 and 15 meters respectively, and the angle contained by the first two sides is a right angle. Find its area.
Solution:

Here, AB = 5m
BC = 12m
CD = 14m
DA = 15 m
Join AC. The ABCD is divided into two triangles ABC and ACD. The area of the quadrilateral is equal to sum of areas of ∆ABC and ∆ADC.
In ∆ABC

area of ABCD
= ar (∆ABC) + ar (∆ACD) –
= (84 + 30)m² = 114m²

Example 3:
In a parallelogram measure of adjacent sides are 34 cm and 20 cm. One of the diagonals is 42 cm. Find the area of the parallelogram.
Solution: In Fig.

AB = DC = 34 cm
AD = BC = 20 cm
AC =42 cm.
We know that the diagonal of a parallelogram divides it into two tri¬angles of equal area.
∴ Area of parallelogram ABCD = 2 x Area of (∆ABC)
Consider ∆ABC
a = 34cm
b = 20cm
c = 42cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{34+20+42}{2}\)
= \(\frac{96}{2}\) = 48 cm
s – a = 48 – 34 = 14 cm
s – b = 48 – 20 = 28 cm
s – c = 48 – 42 = 6 cm
By Heron’s formula,

= 2 x 2 x 6 x 14
= 336 cm²
∴ Area of (∥^gm ABCD) = 2 x 336
= 672 cm²

Example 4:
A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of? 5 per m2. Find the cost of painting.
Solution:

Given P = 32m
BD = 10m
Rate of painting = ₹ 5/m²
Let the sides of rhombus be x m.
P = Ax
⇒ 32 = 4x
∴ x = 8m
So, AB = BC = CD = DA = 8 m
We know that the diagonal of a rhombus divides it into two triangles of equal area. .
∴ Area of rhombus ABCD = 2 x area of ∆ABD
Consider ∆ABD
a = 8m
b = 8m
c = 10m
S = \(\frac{a+b+c}{2}\) = \(\frac{8+8+10}{2}\)
= 13 m.
s – a = 13 – 8 = 5 m
s – b = 13 – 8 = 5m
s – c = 13 -10 = 3 m
By Heron’s formula,
Area of ∆ABD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{3x5x5x3}\)
= 5 x \(\sqrt{39}\)
= 5 x 6.24 = 31.2 m²
Area of rhombus ABCD = 2 x 31.2 = 62.4 m²
Area of rhombus to be painted = 2 x area of rhombus (∴ Painting is to be done on both sides)
= 2 x 62.4 = 124.80
Cost of painting = Rate x Area
= 5 x 124.80
= ₹ 624.

Example 5:
Two parallel sides of a trapezium are 60 cm and 77 cm other sides are 25 cm and 26 cm. Find the area of trapezium.
Solution:

Given AB = 60 cm
DC = 77 cm
AD = 25 cm
BC =26 cm
Draw a line BE ∥ AD from point B.
In ABED
AB ∥ DE
AD ∥ BE
ABED is a parallelogram.
AB = DE = 60
EC = 77 – 60 = 17 cm.
AD = BE = 25 cm
In ∆BEC
a = EC = 17 cm
b = BE = 25 cm
c = BC = 26cm
s = \(\frac{17+25+36}{2}\)
s – a = 34 – 17 = 17 cm
s – b = 34 – 25 = 9 cm
s – c = 34 – 26 = 8cm
By Heron’s formula,

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 8 Winds, Storms and Cyclones

Winds, Storms and Cyclones Intext Questions

Question 1.
I wonder why the winds shown in the figure are not in the exact north – south direction?

Answer:
The winds would have flown in the north – south direction from north to south or from south to north. A change in direction from however, caused by the rotation of the earth.

Question 2.
I want to know what these winds do for us?
Answer:
The winds from the oceans carry water and bring rain. It is a part of the water cycle.

Activities

Activity – 1
Blow the balloons:
Take two balloons of approximately equal size. Put a little water into the balloons. Blow up both the balloons and tie each one to a string. Hang the balloons 8 – 10 cm apart on a cycle spoke or a stick. Blow in the space between the balloons.

Question 1.
What did you expect? What happens?
Answer:
We expected that balloons would move apart. But the balloons come closer.

Activity – 2
Can you blow and lift?
Hold a strip of paper, 20 cm long and 3 cm wide, between your thumb and forefinger as shown in the paper. Paheli Thinks that the strip will be lifted up. Boojho thinks that the strip will bend down.

Question 1.
What do you think Will happen to the paper?
Answer:
Paper strip will be lifted up.

Question 2.
Were the observations along the lines you thought?
Answer:
Yes.

Question 3.
Do you get the feeling that the increased wind speed is accompanied by a reduced air pressure?
Answer:
Yes.

Activity – 3
Take two paper bags or empty paper cups of the same size. Hang the two bags in the inverted position on the two ends of a metal or wooden stick. Tie a piece of thread in the middle of the stick. Hold the stick by the thread (See Fig.) as in a balance. Put a burning candle below one of the bags as shown in the figure. Observe what happens.

Question 1.
Why is the balance of the bags disturbed?
Answer:
The bag below which the candle is lighted, is pushed up by the rising hot air from above the candle flame.

Question 2.
Does this activity indicate that warm air rises up?
Answer:
Yes.

Question 3.
Does the disturbance of the balance suggest that the warm air is lighter than the cold air?
Answer:
Yes.

Winds, Storms and Cyclones Text Book Exercises

Question 1.
Fill the missing word in the blank spaces in the following statements:

1. Wind is ……………. air.
2. Winds are generated due to ……………. heating on the earth.
3. Near the earth’s surface ……………. air rises up whereas air comes down.
4. Air moves from a region of ……………. pressure to a region of pressure.

Answer:

1. Moving
2. Uneven
3. Warm, cooler
4. High, low.

Question 2.
Suggest two methods to find out wind direction at a given place?
Answer:

1. By wind direction indicator.
2. By watching the direction of movement of a paper released in air.

Question 3.
State two experiences that made you think that air exerts pressure (other than those given in the text).
Answer:

1. Compressed air is used in the brake system for stopping trains.
2. Blowing air in a balloon makes it expand.

Question 4.
You want to buy a house. Would you like to buy a house having windows but no ventilators ? Explain your answer.
Answer:
No, a house which has no ventilators is not a healthy house to live in. Basically ventilators provide a path for warm air to go out of the rooms.

Question 5.
Explain why holes are made in hanging banners and hoardings?
Answer:
We know that air exerts pressure, so that due to this pressure banners and hoardings flutter when the wind is blowing. The holes are made in the banners and hoardings as wind pass through that holes and they does not become loose and fall down.

Question 6.
How will you help your neighbours in case cyclone approaches your village/town?
Answer:
I will help by following ways:

1. By warning everyone about the coming danger.
2. Searching for shelter.
3. Moving people fast to safe places.
4. Managing first aid facility.

Question 7.
What planning is required in advance to deal with the situation created by a cyclone?
Answer:
The following planning is required in advance to deal with the situation created by a cyclone:

1. Listening carefully to warnings being transmitted on TV and radio.
2. Setting up cyclone warning system,
3. Moving to cyclone shelter.
4. Storing food in water – proof bags.
5. Keeping an emergency kit ready.

Question 8.
Which one of the following place is unlikely to be affected by a cyclone.

1. Chennai
2. Mangaluru (Mangalore)
3. Amritsar
4. Puri.

Answer:
3. Amritsar.

Question 9.
Which of the statements given below is correct?

1. In winter the winds flow from the land to the ocean.
2. In summer the winds flow from the land towards the ocean.
3. A cyclone is formed by a very high – pressure system with very high – speed winds revolving around it.
4. The coastline of India is not vulnerable to cyclones.

Answer:
1. In winter the winds flow from the land to the ocean.

Extended Learning – Activities and Projects

Question 1.
You can perform the Activity 8.5 (of textbook) in the chapter slight differently at home. Use two plastic bottles of the same size. Stretch one balloon on the neck of each bottle. Keep one bottle in the sun and the other in the shade. Record your observations. Compare these observations and the result with those of Activity 8.5 of text book
Answer:
Do yourself.

Question 2.
You can make your own anemometer?
Answer:
Collect the following items:
4 small paper cups (used ice cream cups), 2 strips of cardboard (20 cm long and 2 cm wide), gum, stapler, a sketch pen and a sharpened pencil with eraser at one end. Take a scale draw crosses on the cardboard strips as shown in the Fig. (a). This will give you the centres of the strips.

Fix the strips at the centre, putting one over the other so that they make a plus (+) sign. Now fix the cups at the ends of the strips. Colour the outer surface of one cup with a marker or a sketch pen. All the 4 cups should face in the same direction.

Push a pin through the centre of the strips and attach the strips and the cups to the eraser of. the pencil. Check that the strips rotate freely when you blow on the cups. Your anemometer is ready. Counting the number of rotations per minute will give you an estimate of the speed of the wind.

To observe the changes in the wind speed, use it at different places and different times of the day. If you do not have a pericil with attached eraser you can use the tip of a ball pen. The only condition is that the strips should rotate freely. Remember that this anemometer will indicate only speed changes. It will not give you the actual wind speed.

Question 3.
Collect articles and photographs from newspapers and magazines about storms and cyclones. Make a story on the basis of what you learnt in this chapter and the matter collected by you?
Answer:
Do with the help of your subject teacher.

Question 4.
Suppose you are a member of a committee, which is responsible for creating development plan of a coastal state. Prepare a short speech indicating the measures to be taken to reduce the suffering of the people caused by cyclones?
Answer:
Do with the help of your subject teacher.

Question 5.
Interview eyewitness to collect the actual experience of people affected by a cyclone?
Answer:
Do with the help of your subject teacher.

Question 6.
Take an aluminium tube about 15 cm long and 1 to 1.5 cm in diameter. Cut slice of a medium – sized potato about 2 cm thick. Insert the tube in the slice, press it, and rotate it 2 – 3 times. Remove the tube. You will find a piece of potato fixed in the tube like a piston head. Repeat the same process with the other end of the tube. Now you have the tube with both ends closed by potato pieces with an air column in between.

Take a pencil with one end unsharpened. Place this end at one of the pieces of potato. Press it suddenly to push the potato piece in the tube. Observe what happens. The activity shows rather dramatically how increased air pressure can push things.
Answer:
Do yourself.

Winds, Storms and Cyclones Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (i)
A storm is marked by –
(a) Strong winds
(b) Rain
(c) Thunder the lightning
(d) All the above.
Answer:
(d) All the above.

Question (ii)
The moving air is called –
(a) Wind
(b) Strong winds
(c) Storm
(d) None of these.
Answer:
(a) Wind

Question (iii)
The amount of water on the earth remains more or less the same because of –
(a) Thunder
(b) Storm
(c) Flood
(d) Water cycle.
Answer:
(d) Water cycle.

Question (iv)
The word monsoon is derived from –
(a) Arabic word
(b) English word
(c) Hindi word
(d) Urdu word.
Answer:
(a) Arabic word

Question (v)
The diameter of the eye of the cyclone varies from –
(a) 10 km to 15 km
(b) 10 km to 20 km
(c) 10 km to 30 km
(d) 10 km to 40 km.
Answer:
(c) 10 km to 30 km

Question (vi)
A violent tornado can travel at speeds of about –
(a) 200 km/h
(b) 300 km/h
(c) 350 km/h
(d) None of these.
Answer:
(b) 300 km/h

Question 2.
Fill in the blanks :

1. Orissa was hit by a cyclone with wind speed of 200 km/h on …………….
2. On 29 October, 1999, a second cyclone with wind speed of ……………. hit Orissa again.
3. The greater the difference in pressure, the ……………. the air moves.
4. The worm air is lighter than the ……………. air.
5. At the poles, the air is colder than that at latitudes about ……………. degrees.
6. The word monsoon is derived from the Arabic word …………….
7. Clouds bring ……………..
8. Farmers in our country depend mainly on rains for their …………….
9. A large cyclone is a violently rotating mass of ………… in the atmosphere.
10. A tornado is a ……………. funnel shaped cloud that reaches from the sky to the ground.

Answer:

1. 18 October 1999
2. 260 km/h
3. Faster
4. Cold,
5. 60
6. Mausam
7. Rain
8. Harvests
9. Air
10. Dark.

Question 3.
Which of the following statements are true (T) or false (F):

1. The cyclone affected agriculture, communication, transport and electricity supply.
2. On heating the air expands and occupies more space
3. In winter, the direction of the wind flow gets reversed.
4. The winds from the oceans carry water arid bring rain.
5. Water cycle is not a continuous phenomenon.
6. Thunderstorms are caused by violent air current inside the cumulus clouds.
7. The cyclones are called hurricane in America.
8. The storms are called typhoons in China.
9. Uneven heating on the earth is the main cause of wind movement.
10. We must stand under a high-rise building or a tree when caught in a thunderstorm.
11. Tropical cyclones occur throughout the year.
12. Lightning, rains and storms are always harmful for the earth.

Answer:

1. True (T)
2. True (T)
3. True (T)
4. True (T)
5. False (F)
6. True (T)
7. True (T)
8. True (T)
9. True (T)
10. False (F)
11. False (F)
12. False (F)

Question 4.
Match the items in Column A with Column B:

Answer:

(i) (b)
(ii) (c)
(iii) (d)
(iv) (a).

Winds, Storms and Cyclones Very Short Answer Type Questions

Question 1.
What is a wind?
Answer:
The moving air is called wind.

Question 2.
Define the term cycle?
Answer:
A cycle is an event or phenomenon which repeats it selfs after sometime.

Question 3.
Define the term evaporation?
Answer:
The process of changing water from its liquid form to its vapour is known as evaporation.

Question 4.
When you fly a kite, does the wind coming from your back help?
Answer:
Yes.

Question 5.
If you are in a boat, is it easier to row it if there is wind coming from behind you?
Answer:
Yes.

Question 6.
Define tornadoes?
Answer:
A tornado is a dark funnel shaped cloud that reaches from the sky to the ground. In our country tornadoes are not very frequent.

Question 7.
Which region gets maximum sunlight?
Answer:
Regions close to the equator get maximum sunlight.

Question 8.
What do you mean by the “eye” of a storm?
Answer:
The centre of a cyclone is calm area. It is called the eye of the storm.

Question 9.
Can you imagine what would happen if high speed winds blow over the roofs of buildings?
Answer:
If the roofs were weak, they would be lifted and blown away.

Question 10.
What do you mean by “hurricane”?
Answer:
“Hurricane” is the term used for storm in West Indies and America.

Question 11.
When is cyclone alert issued?
Answer:
A cyclone alert or cyclone watch is issued 48 hours in advance of any expected storm.

Question 12.
Which factors contribute to the development of cyclone?
Answer:
Facters like wind speed, wind direction, humidity and temperature contribute to the development of cyclones.

Question 13.
When is cyclone warning issued?
Answer:
A cyclone warning is issued 24 hours in advance.

Question 14.
Why smoke always rises up?
Answer:
Smoke is hotter than air, so it is also lighter than air. That is why smoke always moves up.

Question 5.
How do “high – speed winds” harm us ?
Answer:
High – speed winds accompanying a cyclone can damage houses, telephones and other communication systems, trees, etc. causing tremendous loss of life and property.

Question 16.
What is “beaufort scale”?
Answer:
The number and name of a wind is determined by the speed at which it flows on an internationally accepted scale, called beaufort scale.

Question 17.
Is our body a conductor?
Answer:
Yes.

Question 18.
How are high building protected from lightning?
Answer:
High buildings are protected from lightning by fixing lightning conductor on the building.

Winds, Storms and Cyclones Short Answer Type Questions

Question 1.
How is storm caused?
Answer:
When the wind blows gently, it is called a breeze. But, when it blows very fast it cause storm. Storm may be defined as something taking place in the weather of a violent nature. At sea, a storm may be a strong wind or gale. On land, a storm usually means a weather situation marked by heavy rain and often with strong winds, lightning and thunder.

Question 2.
Explain the structure of a tornado?
Answer:
The diameter of a tornado can be as small as a metre and as large as a kilometer, or even wider. The funnel of a tornado sucks dust, debris and everything near it at the base (due to low pressure) and throws out near the top.

Question 3.
How is lightning useful in nature?
Answer:
Lightning is useful in nature because during lightning in tense heat and high temperature are produced. As a result, nitrogen combines with oxygen to form its oxides. These oxides of nitrogen further get dissolved in water to form a dilute solution of nitric acid that comes to the ground with rain. This is how nature provides nitrogenous compounds to plants that are important for their growth.

Question 4.
How are lightning and thunder caused?
Answer:
When two oppositily charged clouds are near each other, the air between them becomes good conductor because charges begin, to move in air very speedily. The presence of electric charges in very large quantities in the air causes to appear as sleaks of lightning and thunder.

Question 5.
Explain the terms thunderstorms and cyclones.
Answer:
Thunderstorms develop in hot, humid tropical areas like India very frequently. The rising temperatures produce strong upward rising winds. These winds carry water droplets upwards, where they freeze, and fall down again. The swift movement of the falling waterd roplets along with the rising air create lightning and sound. It is this event that we call a thunderstorm.

Question 6.
Suggest precautios if a storm is accompanied by lightning?
Answer:
If a storm is accompanied by lightning, we must take the following precautions:

1. Do not take shelter under an isolated tree. If you are in a forest take shelter under a small tree. Do not lie on the ground,
2. Do not take shelter under an umbrella with a metallic end.
3. Do not sit near a window. Open garages, storage sheds, metal sheds are not safe places to take shelter.
4. A car or a bus is a safe place to take shelter.
5. If you are in water, get out and go inside a building.

Question 7.
Suggest some effective safety measures for cyclone.
Answer:
Some Effective Safety Measures:

1. A cyclone forecast and warning service.
2. Rapid communication of. warnings to the Government agencies, the ports, fishermen, ships and to the general public.
3. Construction of cyclone shelters in the cyclone prone areas, and Administrative arrangements for moving people fast to safer places.

Winds, Storms and Cyclones Long Answer Type Questions

Question 1.
How does a thunderstorm becomes a cyclone?
Answer:
Before cloud formation, water takes up heat from the atmosphere to change into vapour. When water vapour changes back to liquid form as raindrops, this heat is released to the atmosphere. The heat released to the atmosphere warms the air around. The air tends to rise and causes a drop in pressure. More air rushes to the centre of the storm. This cycle is repeated. The chain of events ends with the formation of a very low – pressure system with very high – speed winds revolving around it. It is this weather condition that we call a cyclone. Factors like wind speed, wind direction, temperature and humidity contribute to the development of cyclones.

Question 2.
Define the structure of a cyclone?
Answer:
Structure of a cyclone:
The centre of a cyclone is a calm area. It is called the eye of the storm. A large cyclone is a violently rotating mass of air in the atmosphere, 10 to 15 km high. The diameter of the eye varies from 10 to 30 km. It is a region free of clouds and has light winds. Around this calm and clear eye (See Fig.), there is a cloud region of about 150 km in size.

In this region there are high – speed winds (150-250 km/h) and thick clouds with heavy rain. Away from this region the wind speed gradually decreases. The formation of a cyclone is a very complex process.

Question 3.
With a neat diagram show the formation of a cyclone.
Answer:

Question 4.
Describe the action taken by the people and some precautions if you are staying in a cyclone hit area?
Answer:
Action on the part of the people:

1. We should not ignore the warnings issued by the meteorological department throught TV, radio, or newspapers.
2. We should make necessary arrangements to shift the essential household goods, domestic animals and vehicles, etc. to safer places.
3. We should avoid driving on roads through standing water, as floods may have damaged the roads.
4. We should keep ready the phone numbers of all emergency sendees like police, fire brigade, and medical centres.

Some precautions, if you are staying in a cyclone hit area:

1. Do not drink water that could be contaminted. Always store drinking water for emergencies.
2. Do not touch wet switches and fallen power lines.
3. Do not go out just for the sake of fun.
4. Do not pressurise the rescue force by making undue demands.
5. Cooperate and help your neighbours and friends

Question 5.
Show the phenomena that lead to the formation of clouds and falling of rain and creation of storms and cyclones with the help of a flow diagram or a flow chart.
Answer:

Question 6.
On a map, show the regions near the equator where cyclones form.
Answer:

MP Board Class 7th Science Solutions

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:

1. Draw a ray OA and with O as centre and any radius, draw an arc, cutting OA at B.
2. With B as centre and the same radius draw an arc cutting the arc drawn in step (i) at C.
3. With C as centre draw another arc with same radius cutting the arc drawn in step (i) at D.
4. with C as centre and the same radius draw an arc.
5. With D as centre and the same radius draw an arc, cutting the arc drawn in step (iv) at E.
6. Draw OE ∴ ∠AOF = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray justify the construction.
Solution:

1. Draw ∠AOF = 90° by following the same steps for constructing a 90° angle.
2. Draw OG, the bisector of ∠AOF.
3. ∠AOF= 45°

Question 3.
Construct the angle of the following measurements:
Solution:

1. 30°
2. 22 \(\frac{1}{2}\)°
3. 15°

1. 30°

1. Draw a ray OA.
2. With O as centre and any radius draw an arc which intersect OA at B.
3. With B as centre and same radius draw an arc cutting the arc drawn is step (ii) at C.
4. Join OC and produce upward
5. ∠BOC = 60°
6. Draw the bisector OD of ABOC.
7. ∠BOD = ∠COD = 30°

2. 22 \(\frac{1}{2}\)°

1. Draw a ray OA.
2. With O as centre and any radius draw an arc which intersect intersect CM at B.
3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii) at C.
4. With C as centre and same radius draw an arc cutting the arc drawn in step (ii) at D.
   
5. With C and D as centres and same radius draw arcs to intersect at E.
6. Join OE.
7. ∠AOG – 90°
8. Draw the bisector OE of ∠AOE, to get ∠AOF = 45°
9. Draw the bisector OG of ∠AOF.
10. ∠AOG = 22\(\frac{1}{2}\)°

3. 15°

1. Draw a ray CM.
2. With O as centre and any radius draw an arc which intersect CM at B.
3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii), at C.
4. Draw OD as the bisector of ZAOC.
5. ∠BOD = 30°
6. Draw the bisector OE of ∠AOD.
7. ∠AOE = 15°

Question 4.
Construct the following angles and verily by measuring them by a protractor.

(i) 75°
(ii) 105°
(iii) 135°

Solution:

Question 5.
Construct an equilateral triangle given its side and justify the construction.
Solution:

1. Draw a ray AY with intial point A.
2. With centred and radius equal to length of a side of the A draw an arc BY, cutting the ray AX at B.
3. With centre B and the same radius draw an arc cutting are BY at C.
4. Join AC and BC to obtain the required A.

Construction of Triangles:

Example 1:
Construction of a Triangle when its base, a base angle and sum of other two sides are given
Solution:
Step of Construction:

1. Draw the base PQ.
2. At point P draw an angle, MPQ equal to the given angle.
3. Cut a line segment PM equal to sum of sides i.e., (PR + RQ) from point P.
4. Join MQ.
5. Draw the perpendicular bisector of MQ which intersect PM at R.
6. Join QR. PQR is the required triangle.

Justification:
Mark points S as shown in Fig.
QS = MS (∴ RS is the perpendicular bisector of MQ)
RS = RS (Common)
∠QSR = ∠MSR (Each \({ 90 }^{ \underline { 0 } }\))
∆RSQ ≅ ∆RSQ (By SAS)
and so PQ = RM (By CPCT)
Now PR = PM – RM = PM – RQ
PM = PR + RQ.

Example 2:
Construct a AABC in which BC = 3.6 cm, AB + AC = 4.8 and A RSM QS – MS RS = RS cm and B = ∠60°.
Solution:
Steps of Construction:

1. Draw BC = 3.6 cm.
2. Draw ∠CBX= \({ 60 }^{ \underline { 0 } }\) at B.
3. From BX, cut off line segment BD = 4.8 cm.
4. Join DC.
5. Draw the perpcndicub: bisector ofDC meeting BD at A.
6. joinAC.
7. ABC is the required Mangle.

Justification:
‘A‘ lies on the perpendicular bisector of DC
∴ AD = AC
Now BD = 4.8 cm
⇒ BA + AD = 4.8
BA + AC = 4.8 (∴ AD = AC)

Example 3:
Construction of a Triangle when its Base Angle and the Difference of the other two sides are given.
Solution:
Case I:
Given the base BC, a base angle, say ∠B and AB – AC. Steps of Construction:

1. Draw the base BC.
2. At point B draw an angle ∠CBX equal to the given angle.
3. Cut a line segment BD = AB – AC from point B.
4. Join DC.
5. Draw the perpendicular bisector of . DC which intersect BX at A.
6. Join AC.
7. ΔABC is the required triangle.

Justification:
As point A lies on the perpendicular bisector of DC
∴ AD = AC
BD = AB – AD = AB – AC (∴ AD = AC)

Case II:
Given the base BC, a base angle say ∠B and (AC – AB).
Steps of Construction:

1. Draw the base SC.
2. At point S, draw an angle, ∠CBX equal to the given angle and extend the arm XB backward.
3. Cut a line segment BD equal to (AC – AB) from the extended arm.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersect BX at A.
6. Join AC.
7. ΔABC is the required triangle.

Justification:
As the perpendicular bisector of DC passes through A.
∴ AD – AC
BD = AD – AB
∴ BD = AC – AB

Example 4:
Construct a ∆ABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw base BC = 3.4 cm.
2. Draw ∠CBX = \({ 45 }^{ \underline { 0 } }\).
3. From BX, cut line segment BD =1.5 cm.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersect BX at A.
6. Join AC.
7. ABC is the required triangle.

Justification:
As A lies on perpendicular bisector of DC.
AD = AC
BD = 1.5 cm
⇒ AB – AD =1.5 cm
AB – AC =1.5 cm

Example 5:
Construct a ∆PQR in which QR = 5.8 cm, PR – PQ = 1.8 cm and ∠Q = \({ 45 }^{ \underline { 0 } }\). Justify your construction.
Solution:
Steps of Construction:

1. Draw base QR = 5.8 cm.
2. Draw ∠RQP = \({ 45 }^{ \underline { 0 } }\).
3. Produce arm XQ backward and cut a line segment QS = 1.8 cm.
4. Join SR.
5. bisector of SR which intersect QX at P.
6. Join PR.
7. PQR is the required triangle.

Justification:
As P lies on perpendicular bisector of SR
PS = PR
QS = 1.8 cm
⇒ PS – PQ = 1.8 cm
∴ PR – PQ = 1.8 cm

Example 6:
Construct a ∆ABC in which BC = 5.8 cm, ∠C = \({ 60 }^{ \underline { 0 } }\) and AB – AC = 2.5 cm.
Solution:
Steps of Construction:

1. Draw base BC = 5.8 cm
2. Draw ∠ACB = \({ 60 }^{ \underline { 0 } }\)
3. Produce arm CX backward and cut a line segment CD = 2.5 cm.
4. Join BD.
5. Draw perpendicular bisector of BD which intersect CX at A.
6. Join AB.
7. ABC is the required triangle.

Justification:
A lies on the perpendicular bisector of BD.
∴ AB =AD
CD = 2.5 cm
⇒ AD – AC = 2.5 cm
⇒ AB – AC = 2.5 cm

Example 7:
Construct a ∆ABC in which AC – AB = 3.5 cm, BC = 6.2 cm and ∠C = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw BC = 6.2 cm
2. Draw ∠ACB = \({ 45 }^{ \underline { 0 } }\)
3. Cut CD = 3.5 cm from CX.
4. Join BD and draw the perpendicular bisector of BD which intersect CX at A.
5. Join AB.
6. ABC is the required triangle.

Justification:
As A lies on the perpendicular bisector of BD.
∴ AB = AD
CD = 3.5 cm
⇒ AC – AD = 3.5 cm
∴ AC – AB = 3.5 cm

Example 8:
Construction of a Triangle when its Perimeter and Two Base Angles are given.
Solution:
Given the base angles, ∠B and ∠C and perimeter, i.e., AB + BC + AC.
Steps of Construction:

1. Draw a line segment, DE equal to AB + BC + CA.
2. Draw ∠EDM and ∠DEN equal to the base angles ∠B and ∠C respectively.
3. Draw bisectors of ∠MDE and ∠NED which intersect at A.
4. Draw the perpendicular bisectors of AD and AE which intersect DE at B and C respectively.
5. Join AE and AC.
6. ABC is the required triangle.

Justification:
As B lies on the perpendicular bisector of AD
∴ AB = DB
In ∆ADB
AB = DB
∠ADB = ∠DAB
(∴ Angles opposite to equal sides of a A are equal)
Similarly, AC = CE
and ∠CAE = ∠CEA
Now DE = DB + BC + CE
= AB + BC + AC
In ∆ABD, ∠ABC = ∠DAB + ∠ADB = ∠ADB + ∠ADB = 2∠ADB
In ∆ACE, ∠ACB = ∠CAE + ∠CEA = ∠CEA + ∠CEA = 2∠CEA

Example 9:
Construct a triangle whose perimeter is 6.4 cm and angles at the base are \({ 60 }^{ \underline { 0 } }\) and \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw line segment DE equal to (AB + BC + CA) = 6.4 cm
2. Draw ∠EDM = \({ 60 }^{ \underline { 0 } }\) and ∠DEN = \({ 45 }^{ \underline { 0 } }\).
3. Draw AD and AE as bisectors of ∠MDE and ∠NED which inter sect at A.
4. Draw the perpendicular bisector of AD and AE which intersect DE at B and C respectively.
5. Join AB and AC.
6. ABC is the required triangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 11 Work and Energy

Work and Energy Intext Questions

Work and Energy Intext Questions Page No. 148

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

Answer:
The work done W on the body by the force is given by:
Work done = Force × Displacement
W = F × s
Given:
F = 7 N
s = 8 m
Hence, work done, W = 7 × 8
= 56 Nm
= 56 J.

Work and Energy Intext Questions Page No. 149

Question 1.
When do we say that work is done?
Answer:
We can say a work is done whenever the conditions given below are satisfied:

1. A force is applied over the body.
2. A displacement of the body is caused by the applied force, along the direction of the applied force.

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
When a force ‘F’ displaces a body by a distance ‘d’ in the direction of the applied force, then the work done ‘W’ on the body is given by:
Work done = Force × Displacement
W = F × d.

Question 3.
Define 1 J of work.
Answer:
1 J is the amount of work done when an object is provided with a force of 1 N that displaces it through a distance of 1 m in the direction of the applied force.

Question 4.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Answer:
Here,
Applied force, F = 140 N
Displacement, d = 15 m
We know,
Work done is given by the expression:
Work done = Force × Displacement
W = F × d
So,
W= 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the field.

Work and Energy Intext Questions Page No. 152

Question 1.
What is the kinetic energy of an object?
Answer:
The energy attained by or generated in a body due to its action or motion is called kinetic energy. Every object which possesses motion contain a kinetic energy. A body uses kinetic energy to do work. Kinetic energy can be used for any work to be performed. Kinetic energy is useful to generate other forms of energy too. It is expressed by KE and can be calculated by the following formula:
E_k = \(\frac { 1 }{ 2 }\) mv²
Here, m represents mass of object.
And ‘V’ gives the velocity by which object is shifting or working.

Question 2.
Write an expression for the kinetic energy of an object.
Answer:
Energy E_k is proportional to:

• Object’s mass and
• Square of its velocity.

Energy E_k due to a moving object with a body mass ‘m’ which is moving with a velocity v, can be given by the expression,
E_k = \(\frac { 1 }{ 2 }\) mv²
Its S.I. unit is joule (J).

Question 3.
The kinetic energy of an object of mass, m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
Given:
K.E. of the object = 25 J
Velocity of the object, v = 5 m/s
Putting value to formula:
∵ K.E = \(\frac { 1 }{ 2 }\) mv²

• m = 2 × K.E / v²
• m = 2 × \(\frac { 25 }{ 25 }\) = 2 kg

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

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

Work and Energy Intext Questions Page No. 156

Question 1.
What is Power?
Answer:
Work done is calculated by the amount of power consumption. Power can be understood by the term efficiency of an object to consume or generate energy. So, power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,
Power = \(\frac { Work }{ Time }\) = \(\frac { Energy }{ Time }\)
or
P = \(\frac { W }{ T }\)
It is calculated in watt (W).

Question 2.
Define 1 watt of power.
Answer:
As we know that:
Power = \(\frac { Work }{ Time }\)
Hence,
A body is said to have power of 1 watt if its work is equal to 1 joule in 1 s, i.e.,
1 W = \(\frac { 1J }{ 1s }\)

Question 3.
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
Power = \(\frac { Work }{ Time }\)
Given,
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
Putting values,
Power = \(\frac { 1000 }{ 10 }\) = 100 Js^-1
= 100 W.

Question 4.
Define average power.
Answer:
When efficiency of an operator is changed with time, average power is calculated.
The average power of an object is defined as the total work done by it in the total time taken.
Total time taken

Work and Energy NCERT Textbook Exercises

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind – mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.

Answer:
Work is done whenever the two given conditions are satisfied:

• A force is applied over the body.
• A displacement of the body is caused by the applied force, along the direction of the applied force.

Hence, work is done in case:

• Suma is swimming in a pond.
• A wind – mill is lifting water from a well.
• An engine is pulling a train.
• A sailboat is moving due to wind energy.

Work is not being done in case:

• A donkey is carrying a load on its back.
• A green plant is carrying out photosynthesis.
• Food grains are getting dried in the sun.

Explanation:

1. Suma applies a force to push the water backwards which causes a displacement. Hence, work is done by Suma while swimming.
2. While carrying a load, the donkey has to apply a force in the upward direction. But, displacement is exchanged with shifting, so the work done is zero.
3. A wind – mill works against the gravitational force to lift water. Hence, work is done.
4. In this case, chemical change occurs not a physical. Therefore, the work done is zero.
5. An engine applies force to pull the train. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.
6. This is an example of evaporation. Hence, the work done is zero.
7. Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

Question 2.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Gravitational forces are proportional to ‘h’ which is vertical displacement and work done by the force of gravity is considered only if vertical displacement occurs. Vertical displacement is given by the difference in the initial and final positions / heights of the object which is zero. In this case work done by gravity is given by thy expression,
W = mgh
Where,
h = Vertical displacement = 0
W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joule.

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
When a battery lights a bulb, then the chemical energy of the battery is converted into electrical energy. When the bulb receives this electrical energy, then it converts it into light and heat energy. Hence, the transformation of energy is as follows:
Chemical Energy → Electrical Energy → Light Energy + Heat Energy

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work done by the force.
Answer:
Kinetic energy is given by the expression, (E_k) = \(\frac { 1 }{ 2 }\) mv².
Where,
E_k = Kinetic energy of the object moving with a velocity,
v Kinetic energy when the object was moving with a velocity 5 ms^-1.
(E_k)₅ = \(\frac { 1 }{ 2 }\) × 20 × (5)² = 250 J
Kinetic energy when the object was moving with a velocity 2 ms^-1.
(E_k)₂ = \(\frac { 1 }{ 2 }\) × 20 × (2)² = 40 J

Question 5.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body.
Therefore, by the expression,
W = mgh
Here,
Vertical displacement, h = 0
W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
No. In freely falling object only potential energy decreases progressively, but at same time kinetic energy increases and total of both remains equal to initial energy.
Total energy = Potential energy + Kinetic energy
So, this process does not violate the law of conservation of energy. During the process, total mechanical energy of the body remains equal.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
When we ride a bicycle, the chemical energy of muscles of rider’s body gets transferred into heat energy and kinetic energy of the bicycle. Heat energy is changed to physical energy. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular energy → Kinetic energy + Heat energy
During the transformation, the total energy remains conserved.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Here, the energy is completely spent doing work (pushing) against friction between the ground and the rock.

Question 9.
A certain, household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1 kWh
1 k Wh = 3.6 × 10⁶ J
Therefore, 250 units of energy
= 250 × 3.6 × 10⁶ = 9 × 10⁸ J.

Question 10.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fell, find its kinetic energy when it is half – way down.
Answer:
Gravitational potential energy is given by the expression,
W = mgh
Where,
h = Vertical displacement = 5 m
m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 ms^-2
∴ W = 40 × 5 × 9.8 = 1960 J.
At half – way down, the potential energy of the object will be \(\frac { 1960 }{ 2 }\) = 980 J.
At this point, the object has an equal amount of potential and kinetic energy.
This is due to the law of conservation of energy.
Hence, half – way down, the kinetic energy of the object will be 980 J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
Work is done whenever the two given conditions are satisfied:

1. A force acts on the body.
2. There is a displacement of the body by the application of force in or opposite to the direction of force.

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

Question 12.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer:
Yes. For a uniformly moving object, suppose an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. And since
displacement of the body by the application of force is required to prove that work is done, no work is done here.
Here, force of gravity is acting on the bundle, but the person 1 is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
Energy consumed by an electric heater can be obtained with the help of the expression,
P = \(\frac { W }{ t }\)
Where,
Power rating of the heater, P = 1500, W = 1.5 kW
Time for which the heater has operated, t = 10 h
Work done = Energy consumed by the heater
Therefore,
Energy consumed = Power × Time = 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
According to law of conservation of energy, energy can be neither created nor destroyed. It can only be converted from one form to another. Considering the case of an oscillating pendulum.

When a pendulum moves from its actual position P to either of its extreme point A or B, it rises through a height h above the mean level P. Here at this point, the kinetic energy of the bob changes into potential energy and the kinetic energy becomes zero, and the bob possesses only potential energy.

When it moves towards point P, its potential energy decreases progressively. And the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

The pendulum loses its kinetic energy to overcome atmospheric friction and stops after some time. Hence law of conservation of energy is not violated and the total energy of the pendulum and the surrounding system remain conserved.

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

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Answer:
Kinetic energy, E_k = \(\frac { 1 }{ 2 }\) mv²
Where,
Mass of car, m = 1500 kg
Velocity of car, v = 60 km/h
= 60 × \(\frac { 5 }{ 18 }\) ms^-1
E_k = \(\frac { 1 }{ 2 }\) × 1500 × [60 × \(\frac { 5 }{ 18 }\)]²
= 20.8 × 10⁴ J.
Hence, 20.8 × 10⁴ J of work is required to stop the car.

Question 18.
In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:
Case I: Here, the direction of force acting on the block is perpendicular to the displacement. Therefore, work done will be zero.

Case II: Here, the direction of force acting on the block is in the direction of displacement. Therefore, work done will be positive.

Case III: Here, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done will be negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
When all the forces cancel out each other, acceleration in an object will be zero even when several forces are acting on it. And as for a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

Question 20.
Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Answer:
We know,
P = \(\frac { W }{ t }\)
Given,
Power of the device (P) = 500 W = 0.50 kW
Total Time (t) = 10 h
Since,
Work done = Energy consumed by the device
Therefore, energy consumed
= Power × Time = 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 h will be:
4 × 5 kWh = 20 kWh
= 20 Units.

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
As the object hits the ground, its kinetic energy gets converted into heat energy and sound energy. Sometimes, it also deform the ground and itself depending upon the nature of the ground and the amount of kinetic energy of the object. Freely falling object towards the ground feels following changes:

• Its potential energy decreases and kinetic energy increases.
• When the object touches the ground, all its potential energy gets converted into kinetic energy.

Work and Energy Additional Questions

Work and Energy Multiple Choice Questions

Question 1.
S.I. unit of work is ____________ .
(a) Newton
(b) Joule
(c) Watt
(d) All.
Answer:
(b) Joule

Question 2.
Work done is applied and displacement occurred is acute ____________ .
(a) Negative
(b) Zero
(c) Positive
(d) None.
Answer:
(c) Positive

Question 3.
Work done is negative, if angle formed between force applied and displacement occurred is ____________ .
(a) Obtuse
(b) Right angle
(c) Acute
(d) Zero.
Answer:
(a) Obtuse

Question 4.
Work is done if displacement of object occurs in ____________ .
(a) Opposite the direction of force
(b) The direction of force
(c) Zero
(d) Infinite.
Answer:
(b) The direction of force

Question 5.
Force is to work done ____________ .
(a) Inverse
(b) Zero
(c) Proportional
(d) Infinite.
Answer:
(c) Proportional

Question 6.
If direction of applied force and displacement are perpendicular then resultant work will be ____________ .
(a) Zero
(b) Positive
(c) Negative
(d) Infinite.
Answer:
(a) Zero

Question 7.
Writing for two hours is equal to work ____________ .
(a) 1 joule
(b) 2 joule
(c) Zero
(d) None.
Answer:
(c) Zero

Question 8.
Falling ball will have ____________ .
(a) Kinetic energy
(b) Potential energy
(c) Both
(d) None.
Answer:
(c) Both

Question 9.
If mass of an object is doubled over a pully, force required to displacement upto previous destination will be ____________ .
(a) Double
(b) Half
(c) Four times
(d) Equal.
Answer:
(a) Double

Question 10.
Change in kinetic energy, if velocity of an object is doubled will be ____________ .
(a) Double
(b) Half
(c) Equal
(d) 4 times.
Answer:
(d) 4 times.

Question 11.
Rate at which work is done is called ____________ .
(a) Work
(b) Power
(c) K.E.
(d) P.E.
Answer:
(b) Power

Question 12.
Running wings of a fan shows an example of ____________ .
(a) K.E.
(b) P.E.
(c) Work
(d) Power.
Answer:
(a) K.E.

Work and Energy Very Short Answer Type Questions

Question 1.
Give a formula / expression to give total energy.
Answer:
Total energy = Potential energy + Kinetic energy.

Question 2.
How displacement is related to work?
Answer:
Displacement is proportionally related to work.

Question 3.
What is a positive work?
Answer:
When displacement occurs in direction of force applied and object shifted forms an acute angle with force, work is termed as positive.

Question 4.
What kind of force is applied in a lift?
Answer:
Upward and gravitational force is applied in a lift.

Question 5.
What kind of quantity is power?
Answer:
It is scalar.

Question 6.
What will be the potential energy of an object at ground?
Answer:
PE = mgh
Here, h = 0
So, PE = mg.

Question 7.
What will be the work done if no force is being applied?
Answer:
Since, W = F × S
If F = 0;
W = 0 × S
Hence, W = 0
∴ No work will be done.

Question 8.
Name three forms of energy.
Answer:
Potential energy, Kinetic energy and Gravitational energy.

Question 9.
Write expression to calculate kinetic energy if mass of an object is m and its velocity is v.
Answer:
K.E. = \(\frac { 1 }{ 2 }\) mv².

Question 10.
Write expression for potential energy for an object being shifted to ‘d’ height and ‘a’ mass.
Answer:
PE = agd.

Question 11.
If a machine consumes 1000 J of energy in a second and runs for 5 hours, then calculate total energy consumed.
Answer:
P = \(\frac { W }{ t }\)
= 1000 × (5 × 3600)
= 18000000 joule
or
= 1.8 × 10⁷ joule.

Question 12.
What happen to potential and kinetic energy of a falling object?
Answer:
P.E. = decrease to zero
K.E. = increase from zero
Total P.E. converts to K.E.

Work and Energy Short Answer Type Questions

Question 1.
If a pulley is pulling a box towards itself with a force equal to 10 N and have drawn the box upto 2 m. Calculate the work done by it.
Answer:
Given, F = 10 N and s = 2 m
We know W = F × s
Putting values = 10 × 2
W = 20 Nm
or
20 J.

Question 2.
If a hammer operates over a pully machine with 21 J and pulls the balls to 7 m, Calculate the force applied by the hammer.
Answer:
Given:
W = 21 J and s = 7 m
Using formula W = \(\frac { F }{ s }\)
or
F = \(\frac { W }{ s }\)
Putting values F = \(\frac { 21 }{ 7 }\) = 3 N
Hence, F = 3 N was applied by the hammer.

Question 3.
Calculate the work done by a pair of bullocks if an object is pulled by them for 10 m which has mass equal to 20 kg and an acceleration of 20 ms^-1.
Answer:
Given, m = 20 kg, a = 20 ms^-1, s = 10 m
Using formula W = F × s
or
W = m × a × s
Putting values = 20 × 20 × 10
= 4000 kg ms^-2 m = 4000 J.

Question 4.
Give two examples of energy produced due to work or action.
Answer:
Examples showing production of kinetic energy:

1. Heat energy in body after exercise.
2. Generation of kinetic energy in a ball after falling through a point.

Question 5.
Read the following examples and tell in which case work is being done?

1. Manish is riding bicycle in a circular path.
2. John throws a pebble which comes back to initial height of ground.
3. Reading books for 2 hours.
4. Cooking pizza.

Answer:
Work is being done in none case because:

1. Case 1: In a circular path displacement is zero, hence work done is zero.
2. Case 2: Again no displacement occurs.
3. Case 3: Any kind of force is not being used, hence zero work.
4. Case 4: No force is applied here, so zero work.

Question 6.
(a) Under what conditions work is said to be done?
(b) A porter lifts a luggage of 1.5 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer:
(a) Conditions for work to be done:

1. Force should be applied.
2. Body should move in the line of action of force.
3. Angle between force and displacement should not be 90°.

(b) Given,
Mass of luggage, m = 15 kg
Displacement, s = 1.5 m
Using formula:
Work done i.e., W = F × s = mg × s
= 15 × 10 × 1.5 = 225 J.

Question 7.
Four persons jointly lift a 250 kg box to a height of 1 m and hold it.

1. Calculate the work done by the persons in lifting the box.
2. How much work is done for just holding the box?
3. Why do they get tired while holding it? (g = 10 ms²)

Answer:

1. Given, F = 250 × 10 = 2500 N
   s = 1 m
   W = F × s = 2500 × 1 = 2500 J.
2. Zero work, as there is no displacement.
3. Men are applying a force which is opposite and equal to the gravitational force acting on the box. Muscular effort is involved and therefore persons feel tried.

Question 8.
(i) Justify that “a body at a greater height has larger energy”.
(ii) A body of mass 2 kg is thrown up at a velocity of 10 m/s. Find the potential energy at the highest point.
Answer:
(i) When an object is placed at a greater height, the height increases from the reference level (Velocity remains constant i.e., zero). Hence by comparing the potential energy at two points, we can see that the P.E. at greater height will be larger.

(ii) Given,
Using formula m = 2 kg, v = 10 m/s
Initial KE = \(\frac { 1 }{ 2 }\) mv²
= \(\frac { 1 }{ 2 }\) × 2 × (10)² = 100 J
Height will be
h = \(\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2g } \) = \(\frac { { 0 }^{ 2 }-{ 10 }^{ 2 } }{ 2×10 } \)
= \(\frac { -100 }{ 20 }\) = -5 m
So, magnitude of height = 5 m and
P.E. at highest point = mgh
= 2 × 10 × 5
= 100 J.

Question 9.
A light and a heavy object have the same momentum. What is the ratio of their kinetic energies? Which one has a larger kinetic energy?
Answer:
p₁ = m₁v₁, p₂ = m₂v₂
But p₁= p₂  or  m₁v₁ = m₂v₂
and If m₁ < m₂ then v₁ > v₂
(K.E.)₁ = \(\frac { 1 }{ 2 }\) m₁v₁²
(K.E.)₂ = \(\frac { 1 }{ 2 }\) m₂v₂²
(K.E.)₁ = \(\frac { 1 }{ 2 }\) (m₁v₁)v₁ = \(\frac { 1 }{ 2 }\) p₁v₁
(K.E.)₂ = \(\frac { 1 }{ 2 }\) (m₂v₂)v₂ = \(\frac { 1 }{ 2 }\) p₂v₂.

Question 10.
(i) Define 1 kWh. Relate it to joules.
(ii) Find the energy in kWh in the month of September by four devices of power 100 W each, if each one of them is used for 10 hours daily.
Answer:
(i) 1 kWh is the energy used in 1 hour at the rate of 1000 J/s (or 1 kW)
1 kWh = 3.6 × 10⁶ J

(ii) Energy consumed by four devices in the month of September
= 4 × \(\frac { 100 }{ 1000 }\) × 10 × 30
= 120 kWh.

Work and Energy Long Answer Type Questions

Question 1.
(i) Give SI unit and commercial unit of electrical energy.
(ii) Calculate the power of an electric motor that can lift 800 kg of water to store in a tank at a height of 1500 cm in 20 s. (g = 10 m/s²)
(iii) A lamp consumes 500 J of electrical energy in 20 seconds. What is the power of the lamp?
Answer:
(i) SI unit of electrical energy is Joules (J).
Commercial unit of electrical energy is kilowatt hour (kWh)

(ii) Given,
m = 800 kg,
h = 1500 cm = 15 m,
t = 20 sec, g = 10 m/s²
Using formula,
P = \(\frac { W }{ t }\) = \(\frac { mgh }{ t }\)
= \(\frac { 800×10×15 }{ 20 }\) = 600 W.

(iii) Given,
E = 500 J, t = 20 sec
Using formula,
power P = \(\frac { W }{ E }\) = \(\frac { 500 }{ 20 }\) = 25W.
& r E 20

Question 2.
(i) Name the commercial unit of electrical energy.
(ii) Establish the relationship between the SI unit and the commercial unit of electric energy.
(iii) If 4 bulbs of 50 W for 6 hours, 3 tube lights of 40 W for 8 hours, a T.V of 100 W for 6 hours, a refrigerator of 300 W for 24 hours are used. Calculate the electricity bill amount for a month of 30 days. The cost per unit is ? ₹2.50.
Answer:
(i) The commercial unit of energy is kilowatt hour (kWh)
(ii) The SI unit of energy is joule.
Now, 1 kWh = 1 kW × 1 h
= 1000 W × 1 h
= 1000 W × 3600 s
= 3600000 J
= 3.6 × 10⁶ J
1 kWh = 3.6 × 10⁶ J

(iii)

Total energy consumed = 9960 Wh = 9.960 kWh
Electricity bill amount = 9.960 units × ₹2.50 = ₹24.90
For 30 days = 30 × 24.90 = ₹747.

Question 3.
(i) Name the physical quantity defined by rate of doing work. Define its SI unit.
(ii) Why is concept of average power useful? How is it determined?
(iii) A boy of mass 45 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power.
(g = 10 m/s²)
Answer:
(i) Power: Watt or J/s. 1 watt is the power of an object which does work at the rate of 1 J per second.

(ii) Power of an object may vary. Hence, average power is important in the case when the average power of the entire process within a given time is calculated.

(iii) P = \(\frac { mgh }{ t }\) = \(\frac { (50×10×45×0.15) }{ 9 }\) = 375 W.

Question 4.
(i) A body of mass 15 kg possesses kinetic energy of 18.75 kJ. Find the velocity.
(ii) An electric bulb of 100 W is used for 4 hours a day. Calculate the energy consumed by it in a day in joules and kilowatt hour unit.
Answer:
(i) K.E. = \(\frac { 1 }{ 2 }\) mv²
= 18.75 kJ = 18750
v² = \(\frac { 18750×2 }{ 15 }\)
v = √2500 m/s = 50 m/s

(ii) E = \(\frac { P }{ t }\) = 100 W × 4 h = 0.4 kWh
Energy consumed by it in a day = 0.4 × 3.6 × 10⁶ J
= 1.44 × 10⁶ J

Question 5.
(a) A stone is thrown upwards from a point A, as shown in the figure. After reaching the highest point B, it comes down. Explain the transformation of energy from A to B and B to A and also mention the type of energy possessed by the stone at point A, B and C of its journey.
(b) A body of mass 20 kg is dropped from a height of 100 m. Find its K.E. and P.E. after,
(i) First second
(ii) Second second
(iii) Third second
Answer:
(a) While moving upward (A to B) K.E → P.E. and while moving downward (B to A)
P.E.→ K.E.
At,
A → K.E.
B → P.E.
C → K.E. + P.E.

(b) Total Energy = mgh
= 20 × 10 × 100 = 2 × 10⁴ J

(i) After first, second: v = u + gt ms^-1
= 10 × 1 = 10 m/s
K.E. = \(\frac { 1 }{ 2 }\) mv².
= \(\frac { 1 }{ 2 }\) × 20 × 10 × 10
= 1000 J
P.E.= T.E. – K.E.
= 20,000 – 1000 = 19,000 J

(ii) After second, second: v = 20 ms^-1
K.E. = mgh + \(\frac { 1 }{ 2 }\) mv².
= 4,000 J
P.E. = T.E – K.E.
= 20,000 – 4,000 = 16,000 J

(iii) After third, second: v = 30 ms^-1
K.E. = mgh + \(\frac { 1 }{ 2 }\) mv².
= 9,000 J
P.E. = T.E.- K.E.
= 20,000 – 9,000 = 11,000 J

Work and Energy Higher Order Thinking Skills (HOTS)

Question 1.
Earth and other planets moves continuously around the sun. Do they work?
Answer:
No, Earth and other planets do not work while moving around the earth because they move in a circular path and reach the initial point after sometimes, so shows no displacement, hence no work is done by earth and other planets.

Question 2.
Generally heavy objects exert more power over other objects. Give reason.
Answer:
As expression, P = wit
On expanding, P = \(\frac { m.a.s }{ t } \)
Here, P is proportional to mass in p × m
Hence, heavy object exert more pressure or power.

VI. Value-Based Question

Question 1.
Ravi saw a lady labour who carried stones on her head from one point of the construction site to the other end which was some 500 m far. He prepares a trolley for the labour to carry the stones, to make her work easier:

1. Is any work done by the labour while carrying the stones from point A to point B on head by lady labour in the construction site?
2. Is any work done by pulling the trolley of stones from point A to point B?
3. What value of Ravi is seen in the above act?

Answer:

1. No work is said to be done in carrying the stones from point A to B on head by the lady.
2. Work is said to be done by pulling the trolley of stones.
3. Ravi showed kindness, general awareness and sympathy.

MP Board Class 9th Science Solutions

MP Board Class 7th Science Solutions Chapter 7 Weather, Climate and Adaptations of Animals of Climate

Weather Climate and Adaptations of Animals of Climate  Intext Questions

Question 1.
I wonder who prepares the wheather report?
Answer:
The weather reports are prepared by the Meteorological Department of the Government.

Question 2.
I wonder why weather changes so frequently?
Answer:
Because factors affecting weather like temperature, humidity, etc. vary frequently.

Question 3.
What is the source of whether in the first place?
Answer:
All changes in the weather are caused due to sun.

Question 4.
Do fishes and butterflies also migrate like birds?
Answer:
No.

Activity
Fill all the columns according to the data in the chart that you have prepared.
Answer:
Table
Weather data of a week

(Rainfall may not be recorded for all the days since it may not rain everyday.)

Weather, Climate and Adaptations of Animals of Climate Text Book Exercises

Question 1.
Name the elements that determine the weather of a place?
Answer:
The day – to – day condition of the atmosphere at a place with respect to the temperature, rainfall, humidity, wind – speed, etc., is called the weather at that place.

Question 2.
When are the maximum and minimum temperature likely to occur during the day?
Answer:
The maximum temperature of the day occurs generally in the afternoon while the minimum temperature occurs in the early morning.

Question 3.
Fill in the blanks:

1. The average weather taken over a long time is called …………..
2. A place receives very little rainfall and the temperature is high throughout the year, the climate of that place will be ……………. and
3. The two regions of the earth with extreme climatic conditions are ………….. and ……………

Answer:

1. Climate of the place
2. Hot, dry
3. Tropical, polar regions.

Question 4.
Indicate the type of climate of the following areas:

1. Jammu and Kashmir –
2. Kerala –
3. Rajasthan –
4. North – east India –

Answer:

1. Moderately hot and moderately wet climate.
2. Very hot and wet climate.
3. Hot and dry climate.
4. Wet climate.

Question 5.
Which of the two changes frequently weather or climate?
Answer:
Weather.

Question 6.
Following are some of the characteristics of animals:

1. Diets heavy on fruits
2. White fur
3. Need to migrate
4. Loud voice
5. Sticky pads on feet
6. Layer of fat under skin
7. Wide and large paws
8. Bright colours
9. Strong tails
10. Long and Large beak.

For each characteristic indicate whether it is adaptation for tropical rainforests or polar regions. Do you think that some of there characteristics can be adapted for both regions?
Answer:

1. Tropical rainforests
2. Polar region
3. Polar region
4. Tropical rainforesl
5. Tropical rainforests
6. Polar region
7. Polar region
8. Tropical rainforets
9. Tropical rain.forests
10. Tropical rainfore’t

Question 7.
The tropical rainforest has a large population of animals. Explain why it is so?
Answer:
Tropical rain are found in Western Ghats and Assam in India, South – east Asia, central America and Central Africa. Because of continuous warmth and rain, this region supports wide variety of plants and animals. The major types of animals living in the rainforests are apes, gorillas, monkeys, tigers, lions, leopards, lizards, elephants, insects, birds and snakes.

The climate conditions in rainforests are highly suitable for supporting an enormous number and variety of animals. Thus, we can say that because of the hospitable climate conditions huge populations of plants and animals are found in the tropical rainforests.

Question 8.
Explain, with examples, why we find animals of certain kind living in particular climatic conditions?
Answer:
Animals are adapted to survive in the conditions in which they live. Animals living in very cold and hot climate must possess special features to protect themselves against the extreme cold or heat. Polar bears have white fur so that they are not easily visible in the snowy white background. It protects them from their predators. It also helps them in catching their prey. To protect them from extreme cold, they have two thick layers of fur.

They also have a layer of fat under their skin. In fact, they are so well – insulated that they have to move slowly and rest often to avoid getting overheated. Physical activities on warm days necessitate cooling. So, the polar bear goes for swimming. It is a good swimmer. Its paws are wide and large, which help it not only to swim well but also walk with ease in the snow. While swimming under water, it can close its nostrils and can remain under water for long durations.

It has a strong sense of smell so that it can catch its prey for food. Another well – known animal living in the polar regions is the penguin (Fig.). It is also white and merges well wTith the white background. It also has a thick skin and a lot of fat to protect it from cold. You may have seen pictures of penguins huddled together. This they do to keep warm.

Question 9.
How do elephant living in the tropical rainforest adapt itself?
Answer:
The elephant has adapted to the conditions of rainforests in many remarkable ways. Look at its trunk. It uses it as a nose because of whjch it has a strong sense of smell. The trunk is also used by it for picking up food. Moreover, its tusks are modified teeth. These can tear the bark of trees that elephant loves to eat.

So, the elephant is able to handle the competition for food rather well. Large ears of the elephant help it to hear even very soft sounds. They also help the elephant to 4 keep cool in the hot and humid climate of the rainforest.

Choose the correct option which answers the following question:

Question 10.
A carnivore with stripes on its body moves very fast while catching its prey. It is likely to be found in

1. Polar regions
2. Deserts
3. Oceans
4. Tropical rainforests.

Answer:
4. Tropical rainforests.

Question 11.
Which features adapt polar bears to live in extremely cold climate?

1. A white fur, fat below skin, keen sense of smell.
2. Thin skin, large eyes, a white fur.
3. A long tail, strong claws, white large paws.
4. White body, paws for swimming, gills for respiration.

Answer:
1. A white fur, fat below skin, keen sense of smell.

Question 12.
Which option best describes a tropical region?

1. Hot and humid
2. Moderate temperature, heavy rainfall
3. Cold and humid
4. Hot and dry.

Answer:
1. Hot and humid.

Extended Learning – Projects and Activities

Question 1.
Collect weather reports of seven successive days in the winter months (preferably December). Collect similar reports for the summer months (preferably June). Now prepare a table for sunrise and sunset times as shown:
Table

Try to answer the following questions:

1. Is there any difference in the time of sunrise during summer and winter?
2. When do you find that the sun rises earlier?
3. Do you also find any difference in the time of sunset during the month of June and December?
4. When are the days longer?
5. When are the nights longer?
6. Why are the days sometimes longer and sometimes shorter?
7. Plot the length of the days against the days chosen in June and December?

Answer:

1. Yes.
2. Sunrises earlier during summer.
3. Yes, sunsets earlier during winter.
4. During summer, the days are longer.
5. During winter, the nights are longer.
6. Earth revolves around the sun. During different times, the angle of earth with the sun changes. This causes the difference in length of day and night.
7. Length.

Question 2.
Collect information about the Indian Meteorological Department. If possible visit its website: http//www.imd.gov.in. Write a brief report about the things this department does.
Answer:
Do with the help of your subject teacher.

Weather, Climate and Adaptations of Animals of Climate Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (i)
The weather reports are prepared by the –
(a) Meteorological Department of the Government
(b) Agricultural Department of the Government
(c) Radio and TV Department of the Government
(d) None of these.
Answer:
(a) Meteorological Department of the Government

Question (ii)
The climate of the north – east is –
(a) Hot
(b) Wet
(c) Cold
(d) Dry.
Answer:
(b) Wet

Question (iii)
The climate of the western region is –
(a) Hot
(b) Wet
(c) Dry
(d) Hot and dry.
Answer:
(d) Hot and dry.

Question (iv)
The lion – tailed macaque lives in the rainforests of –
(a) Western Ghats
(b) Eastern Ghats
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Western Ghats

Question (v)
Which are of the following is a migratory bird –
(a) Penguin
(b) Peacock
(c) Crow
(d) Siberian crane.
Answer:
(d) Siberian crane.

Question 2.
Fill in the blanks:

1. The temperature, humidity and other factors are called the ……………. of the weather.
2. Rainfall is measured by an instrument called the …………….
3. All changes in the weather are caused by …………….
4.  ……………. Record the weather every day.
5. The polar regions are situated near the …………….
6. Musk oxen, foxes, seals, etc. are living in the ……………. regions.
7. The tropical region has generally a ……………. climate.
8. Animals are adapted to the conditions in which they …………….
9. Penguins lives in ……………. regions.
10. The beard ape lives in the rainforests of …………….

Answer:

1. Elements
2. Rain gauge
3. Sun
4. Meteorologists
5. Poles
6. Polar
7. Hot
8. Live
9. Very cold
10. Western Ghats.

Question 3.
Which of the following statements are true (T) or false (F):

1. The weather is generally not the same on any two days and week after week.
2. The camel is called the ship of the desert.
3. The times of sunrise and sunset also change during the year.
4. The climate of the Kerala is very cold for most part of the year
5. Penguins lives in very cold.
6. Animals are adapted to the conditions in which they live.
7. All the changes in the weather are driven by the sun.
8. Migration is another means to escape the harsh, cold conditions.
9. Polar bears are found in Indian tropical rainforests.
10. The winter sleep of animals is called migration.

Answer:

1. True (T)
2. True (T)
3. True (T)
4. False (F)
5. True (T)
6. True (T)
7. True (T)
8. True (T)
9. False (F)
10. False (F).

Question 4.
Match the items in Column A with Column B:

Answer:

(i) (b)
(ii) (d)
(iii) (a)
(iv) (c)

Weather, Climate and Adaptations of Animals of Climate vert short  Answer Type Questions

Question 1.
Who prepares the weather report?
Answer:
The weather reports are prepared by the Metreological Department of the Government.

Question 2.
What do you mean by weather?
Answer:
The day – to – day condition of the atmosphere at a place with respect to the temperature, rainfall, wind speed, humidity, is called the weather of that place.

Question 3.
What do you mean by climate?
Answer:
The average weather pattern taken over a long time, say 25 years, is called the climate of the place.

Question 4.
When is the climate of a place called hot and wet?
Answer:
If there is heavy rainfall on most of the days as well as the temperature is high in the same place, then we can say that the climate of that place is hot and wet.

Question 5.
When is the climate of a place called hot?
Answer:
If the temperature at a place is high most of the time, then we say that the climate of that place is hot.

Question 6.
Why do some places have hotter climate than others?
Answer:
The places nearer to the equator are usually hotter. This is because the sim’s rays are more concentrated near the equator than they are farther North or South.

Question 7.
Which causes the changes in weather?
Answer:
Sun.

Question 8.
Name the location in India where climate is hot and dry?
Answer:
Rajasthan.

Question 9.
Name the location in India where climate is wet.
Answer:
North – east.

Question 10.
Define raingauge.
Answer:
Rainfall is measured by an instrument called the raingauge.

Question 11.
Name two animals found in cold climate?
Answer:
Polar bear and Penguins.

Question 12.
Name two animals found in hot and humid climate.
Answer:
Beard ape and Red – eyed frog.

Question 13.
Name two deserts animal.
Answer:
Camel and snake.

Question 14.
Where does penguin live?
Answer:
Penguin lives in very cold places.

Question 15.
What makes penguins good swimmers?
Answer:
Penguin’s bodies are streamlined and their feet have webs, making them good swimmers.

Question 16.
How are the paws of a polar bear?
Answer:
Wide and large.

Question 17.
Can a polar bear live happily on land?
Answer:
No, it lives happily where the land is fully covered with snow.

Question 18.
Where do the elephant live ?
Answer:
Elephant lives in forest.

Question 19.
Name two countries where the tropical rainforests are found?
Answer:
India and Brazil.

Question 20.
Name two countries where polar regions are found?
Answer:
Sweden and Canada.

Question 21.
Name four countries in polar region?
Answer:
Norway, Iceland, Canada and Greenland.

Question 22.
Name the major types of animals living in rainforest?
Answer:
The major types of animals living in the rainforests are apes, lions, tigers, monkeys, gorillas, elephants, leopards, snakes, birds and lizards.

Question 23.
Where do the following animals live?

1. Fish
2. Monkeys
3. Snakes

Answer:

1. In water
2. On land and trees
3. On land and water.

Question 24.
Name any two animals which are active during night?
Answer:
Owl and Bat.

Question 25.
Name the bird from Siberia that comes to India?
Answer:
Siberian Crane.

Weather, Climate and Adaptations of Animals of Climate Short Answer Type Questions

Question 1.
Differentiate between weather and climate?
Answer:
Difference between weather and climate:

Weather:
The day – to – day condition of the atmosphere at a place with respect to the temperature, rainfall, windspeed, humidity, etc. is called the weather at that place.

Climate:
The average weather pattern taken over a long time, say 25 years, is called the climate of that place.

Question 2.
Write a short note on the sun?
Answer:
All changes in the weather are caused by the sun. The sun is a huge sphere of hot gases at a very high temperature. The distance of the sun from us is very large. Even then the energy sent out by the sun is so huge that it is the source of all heat and light on the earth. So, the sun is the primary source of energy that causes changes in the weather.

Energy absorbed and reflected by the earth’s surface, oceans and the atmosphere play important roles in determining the weather at any place. If you live near the sea, you would have realised that the weather at your place is different from that of a place in a desert, or near a mountain.

Question 3.
How do penguins adapt with polar climate?
Answer: Penguins are white and merge well with the white background. They also have a thick skin and a lot of fat to protect it from cold. Penguins huddle together. This they do to keep warm. Further, their bodies are streamlined and their feet have webs, making them good swimmers.

Question 4.
How is camel adapted to live in desert?
Answer:
Camel lives in desert. It has long legs which help it to lift its body above the ground. So, camel is able to avoid direct contact with the hot ground. Camel drinks more than 50 litres of water at a time. Camel store the water in his body. So that it lives without water for a longer time. Due to its thick skin, transpiration of water is also prevents. That’s why camel is suited to live in desert.

Question 5.
How are fishes adapted to live in water?
Answer:
Fishes are best suited to live in water. They have structure like a boat, which help them in swimming in water. They have gills from which they get food and oxygen. The body of fishes contain different types of fins which help them for swimming in water.

Question 6.
What is the climate of polar regions?
Answer:
The polar regions present an extreme climate. These regions are covered with snow and it is very cold for most part of the year. For six months the sun does not set at the poles while for the other six months the sun does not rise. In winters, the temperature can be as low as _–37°C. Animals living there have adapted to these severe conditions.

Question 7.
Where the polar regions are situated? Name some of the countries that belong to the polar regions. Also name the some countries where the tropical rainforests are found.
Answer:
The polar regions are situated near the poles, i.e., north pole and south pole. Some well – known countries that belong to the polar regions are Canada, Greenland, Iceland, Norway, Sweden, Finland, Alaska in U.S.A. and Siberian region of Russia. Some countries where the tropical rainforests are found are India, Malaysia, Indonesia, Brazil, Republic of Congo, Kenya, Uganda, and Nigeria.

Question 8.
Write the features of lion – tailed macaque?
Answer:
The lion – tailed macaque (also called Beard ape) lives in the rainforests of Western Ghats. Its most outstanding feature is the silver – white mane, which surrounds the head from the cheeks down to its chin. It is a good climber and spends a major part of its life on the tree. It feeds mainly on fruits. It also eats seeds, young leaves, stems, flowers and buds. This beard ape also searches for insects under the bark of the trees. Since it is able to get sufficient food on the trees, it rarely comes down on the ground.

Weather, Climate and Adaptations of Animals of Climate Long Answer Type Questions

Question 1.
Answer:

Question 2.
Write a note about “climate of the tropical rainforests”?
Answer:
The tropical region has generally a hot climate because of its location around the equator. Even in the coldest month the temperature is generally higher than about 15°C. During hot summers, the temperature may cross 40°C. Days and nights are almost equal in length throughout the year. These regions get plenty of rainfall. An important feature of this region is the tropical rainforests. Tropical rainforests are found in Western Ghats and Assam in India, Southeast Asia, Central America and Central Africa. Because of continuous warmth and rain, this region supports wide variety of plants and animals. The major types of animals living in the rainforests are monkeys, apes, gorillas, lions, tigers, elephants, leopards, lizards, snakes, birds and insects.

Question 3.
For which animals the climatic conditions in rainforests are highly suitable?
Answer:
The climatic conditions in rainforests are highly suitable for supporting an enormous number and variety of animals. Since the numbers are large, there is intense competition for food and shelter. Many animals are adapted to living on the trees. Red – eyed frog (Fig. (a)) has developed sticky pads on its feet to help it climb trees on which it lives. To help them live on the trees, monkeys (Fig. (b)) have long tails for grasping branches. Their hands and feet are such that they can easily hold on to the branches.

MP Board Class 7th Science Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.5

Question 1.
In the figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:
We have a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30°

∴ ∠AOB + ∠BOC = ∠AOC
∠AOB = 60° + 30° = 90°
Now, tne arc ABC subtends ∠AOC = 90° at the centre and ∠ADC at a point D on the circle other than the arc ABC.
∴ ∠ADC = \(\frac{1}{2}\) [∠AOC]
∠ADC = \(\frac{1}{2}\) (90°) = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle = 60
∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ABC at a point on the minor arc of the circle.

∠ACB = \(\frac{1}{2}\) [reflex ∠AOB]
= \(\frac{1}{2}\) [300°] = 150°
Similarly, ∠ADB = \(\frac{1}{2}\) [∠AOB]
= \(\frac{1}{2}\) x [60°] = 30°
Thus, the angle subtended by the chord on the minor arc = 150° and on the major arc = 30°.

Question 3.
In the figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Solution:
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

∴ reflex ∠POR = 2∠PQR,
But ∠PQR = 100°
reflex ∠POR = 2 x 100° = 200°
Since, ∠POR + reflex ∠POR = 360°
∠POR + 200° = 360°
∠POR = 360° – 200°
∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In ∆POR, ∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180° [Sum of the angles of a triangle = 180°]
∠OPR + ∠OPR + 160° = 180° [∴ ∠OPR = ∠ORP]
2∠OPR = 180° – 160° = 20°
∠OPR = \(\frac{20^{\circ}}{2}\) =10°

Question 4.
In the Figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Solution:
We have, in ∆ABC,
∠ABC = 69° and
∠ACB = 31°

But ∠ABC + ∠ACB + ∠BAC = 180°
∴ 69° +31° + ∠BAC = 180°
∠BAC = 180° – 69° – 31°
= 80°
Since, angles in the same segment are equal.
∴ ∠BDC = ∠BAC
∠BDC =80°

Question 5.
In the figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ∆CDE,
Exterior ∠BEC – ∠AED = 130°

130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
∠EDC = 130° – 20° = 110°
∠BDC = 110°
Since, angles in the same segment are equal.
∠BAC = ∠BDC
∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
Given:
∠BAC = 30°, ∠DBC = 70°
To find:
∠BCD and ∠ECD.
∠BDC = ∠BAC = 30° (∠s on the same segment of a circle are equal)

In ∆BCD, 70° + 30° + ∠C = 180° (ASP)
100° + ∠C = 180°
∠C = 80°
AB = BC (Given)
∠BAC = ∠BCA = 30°
∠BCD = ∠BCA + ∠ECD
80° = 30° + ∠ECD
∠ECD – 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Given
ABCD is a cyclic quadrilateral in which AC and BD are diagonals.
To prove:
ABCD is a rectangle.

Proof:
BD is the diameter of the circle.
∠BAD = 90°(angle in a semicircle)
Similarly ∠BCD = 90°
AC is the diameter of the circle
∠ABC = 90° (angle in a semicircle)
Similarly ∠ADC = 90°
In quadrilateral ABCD, ∠A = ∠B = ∠C = ∠D = 90°
ABCD is a rectangle.

Question 8.
If the non-parallel sides of h trapezium are equal, prove that it is cyclic.
Solution:
Given:
AD = BC, AB ∥ DC.
To prove:
ABCD is cyclic quadrilateral.
Construction:
Draw AE and BF As on DC.

Proof:
In ∆ADE and ∆BCF
∴ AE = BF
(∴ Distance between two parallel lines are equal)
∠E = ∠F (Each 90°)
AD = BC (Given)
∆ADE = ∆BCF (By RHS)
and so ∠D = ∠C (By CPCT)
AB ∥ DC and AD is the transversal.
∴ ∠BAD + ∠ADC =180° (CIA’s)
⇒ ∠BAD + ∠BCD = 180° (∠ADC = ∠BCD)
∴ ABCD is cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are J drawn to intersect the circles at A, D and Q respectively (see Fig.). Prove that ∠ACP = ∠QCD.

Solution:
Given
C (O, r) and C (O₁, r₁) are two circles. Two lines ABD and PBQ are drawn which intersect at B.
To prove:
∠1 = ∠3

Proof:
∠1 = ∠2 …(1) (Z s on the same segment of a circle are equal.)
∠3 = ∠4 …..(2) (Z s on the same segment of a circle are equal)
∠2 = ∠4 …(3) (OA’s)
From (1), (2) and (3), we get
∠1 = ∠3 …(2)

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Given
C (O, r) and C (O₁, r₁) are two Circles in whichAB andAC are diameter. These circles intersect at point A and D.
To prove:
BDC is a line.
Construction: JoinAD.
Proof:
∠ADB = 90° (∠s in a semicircle)
∠ADC) = 90° (∠s in a semicircle)

Adding (1) and (2), we get
∠ADB + ∠ADC = 90° + 90°
∠BDC = 180°
∴ BDC is a line and hence D lies on the third side.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD – ∠CBD.
Solution:
Given:
ABC and ADC are two right ∆’s on common base AC. ∠B – 90° and ∠D = 90°.
To prove: ∠CAD = ∠CBD
Proof:
∠ABC + ∠ADC = 90° + 90°
= 180°

ABCD is a cyclic quadrilateral.
∠CAD and ∠CBD are angles on the same segment of a circle.
∴ ∠CAD = ∠CBD.

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Given: ABCD is a ∥^gm
To prove: ABCD is a rectangle.

Proof:
∠A = ∠C
∠A + ∠C =180°
∠A + ∠A = 180°
∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ ABCD is a rectangle.
∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ ABCD is a rectangle.

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 6 Physical and Chemical Changes

Activities

Activity 1
Make a list of eight changes you have noticed from your surroundings.
Answer:
The changes are as:

1. Motion of fan
2. Lighting of a tubelight
3. Vaporisation of water
4. Melting of ice
5. Sound produced by radio
6. Blooming of flower
7. Changing of day and night
8. Changing of the shape of sun.

Activity 2
Try to indentify changes that you observe around you as physical or chemical changes.
Answer:

Physical and Chemical Changes Text book Exercises

Question 1.
Classify the changes involved in the following processes as physical or chemical changes?

1. Photosynthesis
2. Dissolving sugar in water
3. Burning of coal
4. Melting of wax
5. Beating aluminium to make aluminium foil
6. Digestion of food

Answer:

1. Chemical change
2. Physical change
3. Chemical change
4. Physical change
5. Physical change
6. Chemical change.

Question 2.
State whether the following statements are true or false. In case a statement is false, write the corrected statement in you notebook?

1. Cutting a log food into pieces is a chemical change. (T/F)
2. Formation are from leaves is a physical change. (T/F)
3. Iron pipes coated with zinc do not get rusted easily. (T/F)
4. Iron and rust are the same substances. (T/F)
5. Condensation of steam is not a chemical change. (T/F)

Answer:

1. False (F)
2. False (F)
3. True (T)
4. True (T)
5. True (T)

Question 3.
Fill in the blanks in the following statements:

1. When carbon dioxide is passed through lime water, it turns milky due to the formation of ……………
2. The chemical name of baking soda is ……………
3. Two methods by which rusting of iron can be prevented are …………… and ……………
4. Changes in which only …………… properties of a substance change are called physical changes.
5. Changes in which new substances are formed are called …………… changes.

Answer:

1. calcium carbonate
2. Sodium hydrogen carbonate
3. coating, galvanization
4. physical
5. chemical

Question 4.
When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of changes is it? Explain.
Answer:
It is a chemical change. When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas carbaon – dioxide.
Lemon juice + Baking soda Carbon dioxide + Lime water.

Question 5.
When a candle burns, both physical and chemical changes take place. Identify these changes. Give another example of a familiar process in which both the chemical and physical changes take place?
Answer:
Melting of wax is a physical change while burning of candle is a chemical change. Lightning torch bulb using dry cell is another example where both physical and chemical changes takes place. The lighting of the bulb is physical change while current from the dry cell is obtained by the chemical substances inside it.

This is chemical change because the chemicals in the cell get converted into new substances and hence the cell ultimately becomes useless.

Question 6.
How would you show that setting of curd is a chemical change?
Answer:
The conversion of milk into curd, i.e., setting of curd is a permanent as well as irreversible and lead to the production of a new substance. The curd is not be converted into milk. Thus, the formation of curd is a chemical change.

Question 7.
Explain why burning of wood and cutting it into small pieces are considered as two different types of changes.
Answer:
Burning of a wood is a chemical change because in addition to new products burning is always accompanied by production of heat. Cutting of wood into small pieces is a physical change because pieces of wood under went changes in size and no new substances is formed.

Question 8.
Describe how crystals of copper sulphate are prepared?
Answer:
Take a cupful of water in a beaker and add a few drops of dilute sulphuric acid. Heat the water. When it starts boiling add copper sulphate powder slowly while stirring continuiously (fig.). Continue adding copper sulphate powder, till no more powder can be dissolved. Filter the solution. Allow it to cool.

Do not disturb the solution when it is cooling. Look at the solution after some time. Can you see the crystals of copper sulphate? If not, wait for some more time. Crystals of copper sulphate slowly from at the bottom of the beaker.

Question 9.
Explain how painting of an iron gate prevents it from rusting?
Answer:
For rusting, iron must be in contact with both air and moisture. When iron gate is painteel the layer of paint cuts the contact between air, moisture and iron. Thus, it prevents rusting.

Question 10.
Explain why rusting of iron objects is faster in coastal areas than in deserts?
Answer:
The water of coastal areas containing many salts. The salt water makes the process of rust formation faster. Thus, rusting of iron objects is faster in coastal areas than deserts.

Question 11.
The gas we use in the kitchen is called liquified petroleum gas (LPG). In the cylinder it exist as a liquid. When it comes out from the cylinder it becomes a gas (Change A) then it burns (Change B). The following statements pertain to these changes. Choose the correct one.

1. Process – A is a chemical change.
2. Process – B is a chemical change.
3. Both processes – A and B are chemical changes.
4. None of these processes is a chemical change.

Answer:
3. Both processes – A and B are chemical changes.

Question 12.
Anaerobic bacteria digest animal waste and produce biogas (Change A). The biogas is then burnt as fuel (Change B). The following statements pertain to these changes. Choose the correct one.

1. Process – A is a chemical change.
2. Process – B is a chemical change.
3. Both processes – A and B are chemical changes.
4. None of these processes is a chemical change.

Answer:
3. Both processes – A and B are chemical changes.

Extended Learning – Activities and Projects

Question 1.
Describe two changes that are harmful. Explain why you consider them harmful. How can you prevent them?
Answer:
1. Spoilage of Food:
Food items when kept carelessly, get spoiled. This is a chemical change and obviously harmful for our health. Basically, the food is spoiled by microorganisms.

Preventation of Food Spoilage:
Microorganisms do not survive high or low temperature. So, food items stored in refrigerator do not spoil. Also, we should keep them covered, so that microorganism do not get any change to spoil them.

2. Rusting:
If a piece of iron is left in open for some time, it acquires a film of brownish substance. This substance is called rust and the process is called rusting.

Iron benches kept in parks, gardens and lawns, iron gates of parks, farm houses, houses; almost every article of iron, kept in open gets rusted.

The process of rusting can be represented by the following equation:
Iron (Fe) + Oxygen (O₂, from the air) + water (H₂O) → Rust (ironoxide, Fe₂O₃)
For rusting, the presence of both water (or water vapour) and oxygen is essential. Rusting is harmful because, it destorys the iron objects. Iron is the most widely used metal and so rusting is a serious problem.

Preventation of rusting:
It can be prevented by preventing iron things from coming in contact with water, oxygen, or both. One simple way is to apply a coat of paint or grease. In fact, these coats should be applied regularly to prevent rusting.

Another way is to deposit a layer of a metal like chromium or zinc on iron. This process is called galvanisation. Generally the iron pipes, which are used in our homes for water supply are galvanised to prevent rusting.

Question 2.
Take three glass bottles with wide mouths. Label them, A, B and C. Fill about half of bottle A with ordinary tap water. Fill bottle B with water which has been boiled for several minutes, to the same level as in A. In bottle C, take the same boiled water and of the same amount as in other bottles. In each bottle put a few similar iron nails so that they are completely under water. Add a teaspoonful of cooking oil to the water in bottle C so that it forms a film on its surface. Put the bottles away for a few days. Take out nails from each bottle and observe them. Explain your observations?
Answer:
The nails in bottle B rust a little, nails in bottle A are the most rusted and that is bottle C remain unchanged. For rusting both oxygen and water are necessary. Both of these factors are present in the bottle A, since oxygen is dissolved in water. In bottle B, water is boiled and hence dissolved air is removed. Due to lack of oxygen, iron nail does not rust much. In bottle C, the layer of oil present dissolution of air in the water and hence no rusting occurs.

Question 3.
Prepare crystals of alum.
Answer:
Take a cupful of water in a beaker and add a few drops of dilute sulphuric acid. Heat the water. When it starts boiling add copper sulphate powder slowly while stirring continuiously (fig.). Continue adding copper sulphate powder, till no more powder can be dissolved. Filter the solution. Allow it to cool.

Do not disturb the solution when it is cooling. Look at the solution after some time. Can you see the crystals of copper sulphate? If not, wait for some more time. Crystals of copper sulphate slowly from at the bottom of the beaker.

Question 4.
Collect information about the types of fuels used for cooking in you area. Discuss with your teachers/ parents/ others which fuels are less polluting and why?
Answer:
The different fuels used for cooking are wood, cow – dung cake, kerosene, biogas and LPG. Among all of these, biogas and LPG are least polluting. Both of these burn completely and do not give smoke. Also there is no residue i.e. ash after burning. Thus, these fuels are less polluting.

Physical and Chemical Changes Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative

Question (i)
Which of the following is not a chemical change –
(a) Digestion of food
(t) Burning of oaal
(c) Curdling of milk
(d) Melting Of ice.
Answer:
(d) Melting Of ice.

Question (ii)
Properties such as shape, size, colour and state of a sub-stance are called its –
(a) Chemical properties
(b) Physical properties
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Physical properties

Question (iii)
The substances formed as a result of chemical reaction are called –
(a) Materials
(b) Products
(c) Reactants
(d) ingredients.
Answer:
(b) Products

Question (iv)
Rusting of iron is a
(a) Slow reaction
(b) Fast reaction
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Slow reaction

Question (v)
Qutub minar was built more than years ago –
(a) 1600
(b) 1650
(c) 1700
(d) 1800.
Answer:
(a) 1600

Question 2.
Fill in the blanks:

1. A physical change is generally ……………..
2. A chemical change is also called a …………….. reaction.
3. Chemical changes are very important in our ……………..
4. Ozone layer protects us from the harmful …………….. radiation which come from the sun.
5. Oxygen is …………….. from ozone.
6. If the content of moisture in air is high, which means if it is more humid, rusting becomes ……………..
7. The process of depositing a layer of zinc on iron is called ……………..
8. Stainless steel does …………….. rust.
9. In chemical changes …………….. substances are produced.
10. Burning of coal, wood or leaves is a …………….. change.

Answer:

1. Reversible
2. Chemical
3. Lives
4. Ultraviolet
5. Different
6. Faster
7. Galvanisation
8. Hot
9. New
10. Chemical.

Question 3.
Which of the following statements are true (T) or false (F):

1. Conversion of milk into curd is a physical change.
2. A physical change is generally reversible.
3. On dissolving the ash in water it forms a new substance.
4. When carbon dioxide is passed throuth lime water, calcium carbonate is formed, which makes lime water milky.
5. Chemical changes are not important in our lives.
6. All new substances are formed as a result of chemical changes.
7. A medicine is the end product of a chain of chemical reactions.
8. Stainless steel is made by mixing iron with carbon and metals.

Answer:

1. False (F)
2. True (T)
3. True (T)
4. True (T)
5. False (F)
6. True (T)
7. True (T)
8. True (T).

Question 4.
Match the items in Column A with Column B:

Answer:

(i) (b)
(ii) (d)
(iii) (a)
(iv) (c)

Physical and Chemical Changes Very Short  Answer Type Question

Question 1.
Define a physical change.
Answer:
The change in which the identity of the substance does not change is called a physical change.

Question 2.
Give any one characteristic of a physical change.
Answer:
No new substance is formed as a result of physical change.

Question 3.
List two physical changes.
Answer:

1. Change of an iron bar to a magnet.
2. Melting of ice.

Question 4.
Complete the following:

1. In a physical change the state of the substance is ………….. if cause of the change is removed.
2. In a physical change only ……………… physical properties of the substance are

Answer:

1. Restored
2. Changed.

Question 5.
Define a chemical change.
Answer:
It is the change in which identity of the substance is changed and a new substance is formed-

Question 6.
Which of the following is true in case of a chemical change:

1. Identity of the substance does not change.
2. No, new substance is formed.
3. Only physical properties of the substance are changed.
4. Chemical properties of the substance are changed.

Answer:
4. Chemical properties of the substance are changed.

Question 7.
Give any one important characteristic of a chemical change?
Answer:
In this change new substance is formed.

Question 8.
Give any two examples of a physical change?
Answer:

1. Dissolution of common salt (or sugar) in water.
2. Magnetisation of an iron piece and demagnetisation of a magnet.

Question 9.
Is glowing of an electric bulb is a physical change or chemical ?
Answer:
Physical change.

Question 10.
List any two chemical changes with which we come across in our daily life.
Answer:

1. Durdling of milk, and
2. Digestion of food.

Question 11.
Classify the following into physical change and chemical change:

(a) Souring of kneaded flour
(b) Change of sugarcane juice to vinegar
(c) Evaporation of water
(d) Change of an iron rod to a magnet
(e) Pickling
(f) Dissolving common salt in water
(g) Change of cattle dung to biogas
(h) Glowing of an electric bulb.

Answer:

Physical change: (c), (d), (f), (h).
Chemical change: (a), (b), (e), (g).

Question 12.
Give one difference between physical and chemical change.
Answer:
In a physical change new substance is not formed, whereas in a chemical change new substance is formed.

Question 13.
Are the following changes physical or chemical:

1. Burning of candle
2. Preparation of soap from oil and caustic soda
3. Dissolving sugar in cone, sulphuric acid, and
4. conversion of grape juice to wine.

Answer:
Chemical change.

Question 14.
Pick up chemical changes out of the following:

(a) Souring of milk
(b) Making of ice – cream
(c) Burning of a candle
(d) Lighting of an electric bulb
(e) Respiration
(f) Heating of ammonium chloride
(g) Melting of wax.
(h) Breaking a chalk.

Answer:

Physical change: (b), (d), (f), (g), (h).
Chemical change: (a), (c), (e).

Question 15.
What are main points under which physical changes can be classified?
Answer:
The three main points viz. change in state, dispersion in solution, magnetisation and electrical changes.

Question 16.
Define rate of a chemical reaction.
Answer:
The rate of reaction is the quantity of products obtained per unit time from the reactant.

Question 17.
List a fast reaction.
Answer:
2Na + 2H₂O → 2NaOH + H₂ ↑

Question 18.
List the factors which influence the rate of a chemical reaction.
Answer:

1. Nature and state of reactants
2. Temperature
3. Concentration of the reactants
4. Catalyst.

Question 19.
What happens when temperature of a chemical reaction increases?
Answer:
Rate of reaction increases.

Physical and Chemical Changes Short Answer Type Question

Question 1.
How can you say that burning of coal and passing of electric current through water are chemical changes?
Answer:
Burning of coal:
When coal is burnt carbon dioxide gas is formed and ash is left behind as residue. Thus, new substances with new properties are formed. It is because the molecular structure of carbon-dioxide and coal are quite different from that of coal. Further more coal cannot be obtained back, hence it is a permanent change. So, it is a chemical change.

Passing of electric currnet through water:
When electric current is passed throuth water it decomposes to hydrogen and oxygen. The molecular structure of the later viz. H₂ and O₂ is different from that of H₂O, therefore the change is a chemical one.

Question 2.
Define a physical change?
Answer:
Physical change:
A physical change is a change in the physical properties (colour, state, density etc.) and does not involve a change in molecular structure. No new substance is formed and the change can be reversed by ordinary physical means.

Question 3.
Prove that evaporation of water and heating of iron to redness are physical changes?
Answer:
When water is heated, it changes into steam. No new substance has been formed because the molecular structures of water and steam are the same. Water has merely undergone a change in form (liquid → gas).

Further steam can be converted back into water just by the process of cooling therefore, this is a temporary change. Similary when iron is heated to redness, it is a temporary change in colour because on cooling iron is obtained back in its original form.

Question 4.
How can you argue that the following changes are physical changes:

1. Dissolution of sugar in water
2. Melting of wax
3. Lighting of bulb
4. Breaking of a chalk stick.

Answer:
1. Dissolution of sugar in water:
Is a physical change because no new substance is formed. On boiling off water, sugar can be obtained back.

2. Melting of wax:
When wax is heated in a dish, it melts but its composition does not change. On cooling we get back wax.

3. Lighting of bulb:
When an electric bulb is lighted by-passing electric current, the filament glows and emits light. But no new substance is formed. So it is a temporary change.

4. Breaking of a chalk stick:
When a chalk breaks, no new substance is formed. The identity of chalk aslo does not change. Therefore, it is a physical change.

Question 5.
Dispersion of a substance is a physical change or chemical one.
Answer:
Dispersion in solution:
This is also a physical change in many cases. The solute molecules get dispersed amongst the solvent molecules. They do not suffer a change in molecular structure and the solute can be recovered unchanged by cation of the solvent.

Question 6.
Give some chemical reactions which occurs in our daily life.
Answer:
Some important chemical reactions that occur in our daily life are:

1. Respiration (during respiration oxidation of glucose takes place).
2. Digestion of food.
3. Combustion of fuels.
4. Ripening of fruits.
5. Cooking of food.
6. Rusting of iron.

Question 7.
Explain that change in state is nothing but a physical change?
Answer:
Change in state:
All the substances can be transformed to gases, liquids or solids by changing temperature or the other physical conditions. In the solid state the constituent particles are ordred. In the liquid state they are less ordered; in the gaseous state they are most disordered. So, the physical state of a substance can be changed by application of energy.

Thus, changes of a solid to liquid i.e., melting of a solid, change of vapour to liquid etc. can be reversed by application of enerty i.e., either by supplying heat or by cooling. This leads to the conclusion that change of state is nothing but a physical change.

Question 8.
Why heating of a metal say platinum, to redness is a physical change?
Answer:
When platinum is heated to redness it glows. This is only due to temporary excitement of electrons in its atoms. When we stop heating in electrons get unexcited and consequently it ceases glowing. Therefore, change under consideration is definitely a physical change.

Question 9.
Explain as to why the following changes are chemical changes?

1. Rusting of iron
2. Souring of milk
3. Burning of magnesium
4. Pickling.

Answer:
1. Rusting of iron:
During the phenomenon new substances viz. iron oxide etc. are formed on the surface of iron. Iron cannot be obtained back from rust. So, rusting of iron is a chemical change.

2. Souring of milk:
As a result of souring of milk new sub¬stances are formed which cannot be converted back to milk, hence souring of milk is a chemical change.

3. Burning of magnesium:
When Mg burns new substance viz. magnesium oxide is formed, which cannot be converted back to magnesium and oxygen. Therefore burning of magnesium is a chemical change.
2Mg + O₂ → 2MgO

4. Pickling:
Pickling is nothing but the fermentation of the vegetable. Pickled vegetables cannot be changed back to fresh vegetables. Hence, it is also a chemical changed.

Question 10.
Give brief distinction between physical change and chemical change.
Answer:
Distinction between physical and chemical changes:
Physical Change:

• Change in physical properties.
• Temporary change.
• Composition of matter does not change.
• No new substance is formed.
• Can be reversed by physical means.
• Not accompanied by large energy changes.
• No change in weight.

Chemical Change:

• Change in chemical properties.
• More or less a permanent change.
• Composition of matter changes.
• New substance is formed.
• Cannot be reversed easily.
• Accompanied by large energy changes.
• Though total mass remains constant, mass of individual substance changes.

Question 11.
Is the solution of hydrogen chloride in water a physical changes?
Answer:
Hydrogen chloride is a covalent molecule. In solution it becomes electrovalent. The hydrogen ion combines with a water molecule to form hydronium ion. The properties of the solution are entirely different. There is a change in molecular structure and so as per our definition it is a chemical change.

Question 12.
Is an allotropic change a chemical change?
Answer:
The constituent atoms are identical in allotropes but allotropy is caused by change in molecular structure. Oxygen gas contains two atoms in each molecule but ozone contains three. In many cases the chemical as well as physical properties are different. So we have to conclude that allotropic change is a chemical change.

Question 13.
Is the cooking of rice a physical change?
Answer:
In cooking the fast molecules of boiling water pierce the walls of the cell in which strach is enclosed and release the starch. Sight hydrolysis also takes place. But the change is physical to a major extent since most of the starch remains unchanged.

Question 14.
What type of change is sublimation of ammonium chloride?
Answer:
Ammonium chloride on heating splits up into gases ammonia and hydrogen chloride. These gaseous products recombine on cooling to give the original salt. The apparent physical change is the result of two distinct chemical changes decomposition and combination. Sublimation can also be a physical change when the sub – stance is below its triple point and the change to vapour and back to solid state takes place directly, but in the case of ammonium chloride and many other compounds dissociation occurs.

Question 15.
Magnetisation of an iron bar is a physical change. Explain.
Answer:
Magnetisation of an iron bar:
Magnetisation caused by alignment of constituent atoms [figs, (a) and (b)] and as such it is to be considered only as a physical change. Charging and discharging of a condenser a temporary and physical change.

Physical and Chemical Changes Long Answer Type Question

Question 1.
Give in detail the distinction between physical and chemical changes?
Answer:
Distinction between physical and chemical changes:
As already stated a chemical change involves a change in the structure of the molecules while a physical change does not involve a change in molecular structure. This definition is comprehensive and all other characteristic of a chemical change can be deduced easily as floows.

1. During a chemical change new substances with new properties are produced:
Change in molecular structure means a new substance with new properties. On the other hand, a physical change does not involve a change in the molecular structure.

2. A chemical change is a permanent change and the original substance is not got back by reversing the conditions. A physical change is temporary and exist only as long as the imposed conditions exist. A physical change can be reversed by reversing the conditions.

3. No change, physical or chemical can take place without energy change. A large amount of heat energy has to be given to convert water into steam. But the energy involved in chemical changes is generally far greater than that involved in physical changes.

4. A chemical change means a physical change also, because won news substances with new properties are produced the physical state and physical properties are altered.

Question 2.
Give a detailed account of modem approach of physical and chemical changes?
Answer:
Modern approach of physical and chemical changes:
As your studies in science progress you will appreciate the point of view that all changes are chemical changes. Chemical changes are changes in structure. Minor electronic changes do occure even when a physical change takes place. In solids, subjected to heat, the ions oscillate with increasing amplitude and electrons move with greater speeds in expanded orbitals. Breaking a piece of diamond is an accepted chemical change because covalent chemical bonds are broken in the process. The same is happening when any substance is broken.

Thedivision into physical and chemical changes is, therefore, purely a matter of convenience. The three types of changes are classified as fallows:

1. Physical change:
Slight changes in outer electron states. Comparatively less energy change. No change in molecular patterns. No change in nucleonic pattern.

2. Chemical change:
Change in molecular pattern. Large scale energy change. Rearrangement, of electronic orbitals. No change in nucleonic pattern.

Question 3.
What are slow and fast reactions? Give example.
Answer:
A reaction can be called slow or fast depending upon the speed rate of reaction.
Whenever elements and compounds react their reaction is said to be slow if the rate of the reaction, i.e., the quantity of products obtained per unit time from the reactants, is slow.
Example:

1. Rusting of iron is a slow reaction.
2. When a copper vessel kept in moist air it gets greenished due to the formation of basic copper carbonate. To the contrary if the speed or the rate of the reaction is fast, than the reaction is said to be the fast reaction.

Example:
By adding sodium chloride in silver nitrate solution, immediately a white ppt. appears. It is a fast reaction i.e.,

Question 4.
What is a chemical reaction? Give an example.
Answer:
A reaction in which the following are being occurred is called a chemical reaction:

1. Absorption or liberation of heat or any other form of energy.
2. Change in the identity of the combining substances i.e., the reactants.
3. Collision among the reactant molecules.
4. Use of some catalyst or provision of some form of energy to the reactants for initiation of reaction.

Example:
1. Sodium reacts with water to form sodium hydroxide alongwith the liberation of hydrogen gas.
The chemical equation of the reaction is as follows:
2Na + 2H₂O → NaOH + H₂

2. Lime stone, on heating, liberates carbon dioxide.
The equation for the reaction is as follows:
CaCO₃ → CaO + CO₂.

MP Board Class 7th Science Solutions

MP Board Class 9th Science Solutions Chapter 10 Gravitation

Gravitation Intext Questions

Gravitation Intext Questions Page No. 134

Question 1.
State the universal law of gravitation.
Answer:
Suppose there are two objects having mass M and m respectively.
The distance between their centres is equal to d.
The force of attraction is F.
Thus, F ∝ M . m … (i)
and, F ∝ 1/d² … (ii)
Joining equation (i) and (ii)
we get F ∝ M . m/d²
⇒ F = G . M . m/d² … (iii)
where, G is the proportionality constant and called Universal Gravitation Constant.
The expression (iii) is called expression for Universal Law of Gravitation.
The universal law of gravitation is represented as:
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
Where, G is the universal gravitation constant given by:
G = 6.67 × 10^-11 Nm² kg^-2.

Question 2.
Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the earth.
Answer:
Let M_e be the mass of the Earth and m be the mass of an object on its surface.
And say R is the radius of the Earth,
then according to the universal law of gravitation,
the gravitational force (F) acting between the Earth and the object is given by the relation:
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
F = GM_em/R

Gravitation Intext Questions Page No. 136

Question 1.
What do you mean by free fall?
Answer:
Its a phenomenon of gravity. When an object falls from any height under the influence of gravitational force only, it is said to have a free fall. In the case of free fall, no change in direction takes place but the magnitude of velocity changes because of acceleration.

Question 2.
What do you mean by acceleration due to gravity?
Answer:
Change in velocity due to variation in height produces acceleration which is due to gravity in the object and is known as acceleration due to gravity denoted by letter g. The value of acceleration due to gravity is g = 9.8 m/s².

Gravitation Intext Questions Page No. 138

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

Question 2.
Why is the weight of an object on the moon \(\frac { 1 }{ 6 }\)th its weight on the earth?
Answer:
The mass of moon is \(\frac { 1 }{ 100 }\) times and its radius \(\frac { 1 }{ 4 }\) times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is \(\frac { 1 }{ 6 }\)th of its weight on the earth.

Gravitation Intext Questions Page No. 141

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Question 2.
What do you mean by buoyancy?
Answer:
The upward force exerted by a liquid on an object that is immersed in it is known as buoyancy.

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

1. An object sinks in water if its density is greater than that of water.
2. An object floats in water if its density is less than that of water.

Gravitation Intext Questions Page No. 142

Question 1.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
When we weigh our body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

Question 2.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?
Answer:
The cotton bag is heavier than the iron bar. The cotton bag experiences larger up – thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

Gravitation NCERT Textbook Exercises

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to Universal Law of gravitation, the gravitational force of attraction between any two objects of mass M and m is proportional to the product of their masses and inversely proportional to the square of distance r between them. So, force F is given by
F = G\(\frac { M\times m }{ { r }^{ 2 } } \)
Now, when the distance ‘r’ is reduced to half then force between two masses becomes
F’ = G\(\frac { M\times m }{ { (\frac { r }{ 2 } ) }^{ 2 } } \)
Or
F’ = 4F
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 2.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m).
Answer:
Given that,
Mass of the body, m = 1 kg
Mass of the Earth, M = 6 × 10²⁴ kg
Radius of the Earth, R = 6.4 × 10⁶ m
Now, magnitude of the gravitational force (F) between the Earth and the body can be given as,
F = G\(\frac { M\times m }{ { r }^{ 2 } } \) = \(\frac { 6.67 × 10 × 6 × 10 × 1 }{ (6.4 × 6.4) }\)
= \(\frac { 6.67 × 6 × 10 }{ (6.4 × 6.4) }\) = 9.8N (approx.)

Question 4.
The Earth and the Moon are attracted to each other by gravitational force. Does the Earth attract the Moon with a force that is greater or smaller or the same as the force with which the Moon attracts the Earth? Why?
Answer:
According to the Universal Law of Gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the Moon with an equal force with which the Moon attracts the Earth.

Question 5.
If the Moon attracts the Earth, why does the Earth not move towards the Moon?
Answer:
The Earth and the Moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the Moon. Hence, it accelerates at a rate lesser than the acceleration rate of the Moon towards the Earth. For this reason, the Earth does not move towards the Moon.

Question 6.
What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Answer:
(i) From Universal Law of Gravitation, force exerted on an object of mass at by Earth is given by
F = G\(\frac { M\times m }{ { R }^{ 2 } } \) ….1
When nws of the object say ne is doubled than
F’ = G\(\frac { M\times 2m }{ { R }^{ 2 } } \)  = 2F
So as the mass of any one of the object is doubled the force is also doubled,

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

(iii) Again from Universal Law of Attraction, from equation 1, force ‘F’ is directly proportional to the product of both the masses, So, if both the masses are doubled then, the gravitational force of attraction become four times the original value.

Question 7.
What in the importance of Universal Law of Gravitation?
Answer:
Universal Law of Gravitation is important because it tells us about:

1. the force that is responsible for binding us to Earth.
2. the motion of Moon around the Earth.
3. the motion of planets around the Sun.
4. the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both Sun and Moon on the Earth.

Question 8.
What in the acceleration of free fall?
Answer:
Acceleration of free tall is the acceleration produced when a body falls under the influence of the force of gravitation of the Earth alone. It is denoted by ‘g’ and its value on the surface of the Earth is 9.8 ms^-2.

Question 9.
What do we call the gravitational force between the Earth and an object?
Answer:
Gravitational force between the Earth and an object is known as the weight of the object.

Question 10.
Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator}.
Answer:
Weight of a body on the Earth is given by:
W = mg
Where,
m = Mass of the body
g = Acceleration due to gravity
The value of g is greater at poles than at the equator.
Therefore, gold at the equator weighs less than at the poles.
Hence, Amit’s friend will not agree with the weight of the gold bought.

Question 11.
Why does a sheet of paper fell slower than one that is crumpled into a ball?
Answer:
When a sheet of paper is crumpled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Question 12.
Gravitational force on the surface of the Moon is only \(\frac { 1 }{ 6 }\) as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Answer:
Weight of an object on the Moon = \(\frac { 1 }{ 6 }\) × Weight of an object on the Earth.
Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s²
Therefore, weight of a 10 kg object on the earth = 10 × 9.8 = 98 N
And, weight of the same object on the Moon = 1.6 × 9.8 = 16.3 N.

Question 13.
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Answer:
According to the equation of motion under gravity:
-u² = 2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s.
During upward motion, g = – 9.8 ms^-2.
(i) Let ‘h’ be the maximum height attained by the ball.
Hence,
0 – 49² = 2 × 9.8 × h
h = \(\frac { 49 × 49 }{ 2 × 9.8}\) = 122.5 m

(ii) Let ‘t’ be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
v = u + gt
We get,
= 49 + t × (- 9.8)
9.8 t = 49
t = \(\frac { 49 }{ 9.8}\) = 5 s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
According to the equation of motion under gravity: v² – u² = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 ms^-2
∴ v² – 0² = 2 × 9.8 × 19.6
v² = 2 × 9.8 × 19.6 = (19.6)²
v = 19.6 ms^-1
Hence, the velocity of the stone just before touching the ground is 19.6 m s^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
According to the equation of motion under gravity:
v² – u² = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = -10 ms^-2
Let h be the maximum height attained by the stone.
Therefore,
0 – (40)² = 2 × h × (-10)
h = \(\frac { 40 × 40 }{ 20 }\) = 80 m
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (-80) = 0.

Question 16.
Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer:
According to question,
M_Sun = Mass of the Sun = 2 × 10³⁰ kg
M_Earth = Mass of the Earth = 6 × 10²⁴ kg
R = Average distance between the Earth and the Sun = 1.5 × 10¹¹ m.
From Universal Law of Gravitation,
F = G\(\frac { M\times m }{ { R }^{ 2 } } \)
Therefore, putting all the values given in question in above equation we get
F = 6.67 × 10^-11 \(\frac { (6\times { 10 }^{ 24 })\times (2\times { 10 }^{ 30 }) }{ { (1.5\times { 10 }^{ 11 }) }^{ 2 } } \) = 3.56 × 10²² N.

Question 17.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let be the point at which two stones meet and let ‘h’ be their height from the ground. It is given in the question that height of the tower is H = 100 m
Now, first consider the stone which falls from the top of the tower.
So, distance covered by this stone at time ‘t’ can be calculated using equation of motion:
x – x₀ = u₀t + \(\frac { 1 }{ 2 }\)gt²
Since, initial velocity u = 0,
so we get
100 =  x \(\frac { 1 }{ 2 }\)gt² ………… (1)
The distance covered by the same stone that is thrown in upward direction from ground is
x = 25t –\(\frac { 1 }{ 2 }\)gt²
In this case intitial velocity is 25 m/s.
So, x = 25t – \(\frac { 1 }{ 2 }\)gt²  ………… (2)
Adding equations (1) and (2) we get,
100=25t
or,
t = 4s
Putting value in equation (2).
x = 25 × 4 – \(\frac { 1 }{ 2 }\) × 9.8 × (4)²
= 100 – 78.4
= 21.6 m.

Question 18.
A ball thrown up vertically returns to the thrower after 6 s. Find:
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answer:
(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0 Acceleration due to gravity, g = -9.8 m s^-2
Equation of motion, v = u + gt
will give,
0 = u + (-9.8 × 3)
u = 9.8 × 3
= 29.4 ms^-1
Hence, the ball was thrown upwards with a velocity of 29.4 ms^-1.

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

(c) Ball attains the maximum height after 3 s.
After attaining this height, it will start falling downwards.
In this case, Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in
4 s – 3 s = 1 s.
Equation of motion, s = ut + \(\frac { 1 }{ 2 }\) g²
will give,
s = 0 x t + \(\frac { 1 }{ 2 }\) x 9.8 x 12 = 4.9 m
Total height  = 44.1 m.
This means that the ball is 39.2 m (44.1 m – 4.9 m) above the ground after 4 seconds.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
An object immersed in a liquid experiences buoyant force in the upward direction.

Question 20.
Why does a block of plastic released under water come up to the surface of water?
Answer:
For an object immersed in water two forces act on it:

1. Gravitational force which tends to pull object in downward direction
2. Buoyant force that pushes the object in upward direction.

Here, in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

Question 21.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^-3, will the substance float or sink?
Answer:
If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.

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

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

= \(\frac { 500 }{ 350 }\)
= 1.428 g cm^-3
The density of the substance is more than the density of water (1 g cm^-3). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

Gravitation Additional Questions

Gravitation Multiple Choice Questions

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

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

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

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

Question 5.
An object is put one by one in three liquids having different densities. The object floats with 1 2 3, and 9 11 7 parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₂₁ > d₂ < d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃.
Answer:
(d) d₁ < d₂ < d₃.

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

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

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

Question 9.
Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be __________ .
(a) 14 times
(b) 4 times
(c) 12 times
(d) unchanged.
Answer:
(b) 4 times

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

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

Question 12.
The weight of an object at the centre of the Earth of radius R is __________ .
(a) zero.
(b) infinite.
(c) R times the weight at the surface of the Earth.
(d) \(\frac { 1 }{ { R }^{ 2 } } \) times the weight at surface of the Earth.
Answer
(a) zero.

Gravitation Very Short Answer Type Questions

Question 1.
Why Moon revolves around the Earth?
Answer:
Gravitational force of Earth.

Question 2.
What is the SI unit of gravitational force?
Answer:
Newton (N).

Question 3.
Which law of physics is represented by the statement every object attract other object in universe towards itself’?
Answer:
Universal Law of Gravitation.

Question 4.
State the relation between gravitational force and distance among objects.
Answer:
Inversely proportional.

Question 5.
In which conditions free fall of an object occur?
Answer:
When an object falls from a height under the influence of gravity and no other force, it is said to have a free fall.

Question 6.
What is the SI unit of gravitational constant?
Answer:
Nm²kg^-2.

Question 7.
Express the relation between thrust and pressure.
Answer:
Pressure = thrust / area.

Question 8.
What kind of force is exerted by a liquid?
Answer:
Equal and unidirectional.

Question 9.
In what condition an object sinks?
Answer:
If the weight of the object is more than 9.8 N, then the object will sink.

Question 10.
Which material is taken as standard to calculate any object’s relative density?
Answer:
Water.

Gravitation Short Answer Type Questions

Question 1.
What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?
Answer:
Gravitational force. This force depends on the product of the masses of the planet and Sun, and the distance between them.

Question 2.
On the Earth, a stone is thrown from a height in a direction parallel to the Earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?
Answer:
Both the stones will take the same time to reach the ground because the two stones fall from the same height.

Question 3.
Suppose gravity of Earth suddenly becomes zero, then in which direction will the Moon begin to move if no other celestial body affects it?
Answer:
The Moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of Moon is due to centripetal force provided by the gravitational force of Earth.

Question 4.
Two identical packets are dropped from two Aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming all conditions are identical, will those packets take same time to reach the surface of Earth. Justify your answer.
Answer:
The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Question 5.
The weight of any person on the Moon is about \(\frac { 1 }{ 6 }\) times that on the Earth. He can lift a mass of 15 kg on the Earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon?
Answer:
The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Gravitation Long Answer Type Questions

Question 1.
State ‘Archimedes’ Principle and write its two applications.
Answer:
‘Archimedes’ Principle:
Force exerted by liquid on wholly or partly immersed object is equal to the weight of the fluid displaced by the object. Applications based on ‘Archimedes’ principle are:

1. designing of water transport vehicles.
2. hydrometers used for determining the density of liquids

Question 2.
What is relative density? What is the density of water?
Answer:
Relative Density (RD) or Specific Gravity (SG) is the ratio of either densities or weights. Hence, when we compare or divide value of an objects’ density with water’s density, it is called Relative Density of a substance to water.

1. In SI units, the density of water is (approximately) 1000 kg/m³ Or 1 g/cm³.

Question 3.
Mass of a rectangular copper solid piece is 300 g. With dimensions 5 × 2 × 5 cm³, what should be its specific gravity, calculate? Will the bar float or sink in water?
Answer:

Given:
Mass of copper = 300 g
5 × 2 × 5 = 50 cm³
Density of copper = mass / volume
\(\frac { 300 }{ 50 }\) = 6 g / cm³
Density of water, = 1 g/cm³
Specific gravity of iron = \(\frac { 6 }{ 1 }\) = 6.
Hence the bar will sink.

Question 4.
How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?
Answer:
We know, weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth, i.e.,.
Weight of a body ∝ \(\frac { M }{ { R }^{ 2 } } \)
Original weight, W₀  = mg = mG\(\frac { M }{ { R }^{ 2 } } \)
When hypothetically M becomes 4 M and R becomes \(\frac { R }{ 2 }\) then weight becomes
W₀ = mG \(\frac { 4M }{ { (\frac { R }{ 2 } ) }^{ 2 } } \) = (16 m G) M
R² = 16 × W₀
The weight will be 16 times heavier.

Question 5.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water? Give reason for each case.
(b) A ball weighing 4 kg of density 4000 kg m^-3 is completely immersed in water of density 10³ kg m^-3. Find the force of buoyancy on it. (Given: g = 10 ms^-2.)
Answer:
(a)

1. The cube will experience a greater buoyant force in the saturated salt solution because the density of the salt solution is greater than that of water.
2. The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube.

(b) Buoyant force = weight of the liquid displaced = density of water x volume of water displaced xg 4
= 1000 × \(\frac { 4 }{ 4000 }\) × 10 = 10N.
4000

Gravitation Higher Order Thinking Skills (HOTS)

Question 1.
How will the weight of a body of mass 250 g of changes, if it is taken from equator to the poles? Give reasons.
Answer:
As we move from equator to poles, acceleration due to gravity increases. It is because radius of earth is less at poles than at equator. Therefore, its weight will increase.

Question 2.
Aman tried to immerse an empty plastic bottle in a bucket of water. But each time he fails. Why does this happen?
Answer:
When Aman tried to immerse an empty plastic bottle in a bucket of water, it comes above the surface of water. It is due to the upward force (upthrust or buoyant force). The upthrust exerted by water on the bottle is greater than its own weight.

Gravitation Value Based Question

Question 1.
Rashmi was wearing a high heel shoes for a beach party. Her friend told her to wear flat shoes as she will be tired soon with high heel and will not feel comfortable.

1. What is the reason of one’s feeling tired with high heel shoes on a beach?
2. Name the unit of pressure.
3. What value of Rashmi’s friend is reflected in the above act?

Answer:

1. Because the high heel shoes would exert lot of pressure on the loose sand of beach and will sink more in the soil as compared to flat shoes. Therefore, large amount of force will be required to walk with high heels.
2. Pascal.
3. Rashmi’s friend showed the value of being intelligent, concerned and helpful.

MP Board Class 9th Science Solutions