Solutions to NCERT Class 9 Science Chapter 9 Gravitation

Welcome to Chapter 9 Gravitation! This is a crucial topic not just for your exams but also for higher classes and our solutions material is designed to help you grasp the concepts perfectly. We have answered all in-text questions, exercise questions, and in-text activities all in one place and have also provided you with a bonus set of extra questions to give you additional practice! If you study these solutions thoroughly, you will be well-prepared for your exams. Also, look at the illustrations we have created carefully to make learning enjoyable! Have fun!

Solutions to In Text Questions of NCERT Class 9 Science Chapter 9 Gravitation

Page 102:

1. State the universal law of gravitation.
Answer:
The universal law of gravitation states that every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

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:
The formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth is F = GMm/R², where G is the universal gravitation constant, M is the mass of the earth, m is the mass of the object and R is the radius of the earth.

Page 104:

1. What do you mean by free fall?
Answer:
Free fall in when an object falls towards the earth under the force of gravity alone.

2. What do you mean by acceleration due to gravity?
Answer:
Whenever an object falls towards the earth, an acceleration is involved due to the earth’s gravitational force. Therefore, this acceleration is called acceleration due to gravity, is denoted by g and has the unit of ms^-2.

Page 106:

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

Mass	Weight
(i) Mass in the amount of matter in an object.	(i) Weight is the force with which it is attracted towards the earth.
(ii) Mass of a body is constant.	(ii) Weight depends on the strength of gravity.
(iii) It only has magnitude.	(iii) It has magnitude and direction.
(iv) It is a measure of the inertia of an object.	(iv) It is the measure of the gravitational pull on an object’s mass.
(v) The SI unit of mass is kilogram (kg).	(v) The SI unit of weight is Newton (N).

2. Why is the weight of an object on the moon (1/6)^th its weight on the earth?
Answer:
The mass and radius of the moon is much smaller than that of the earth. Hence, the force of gravity on the moon is much smaller than that on earth. Hence, the weight of an object on the moon (1/6)^th its weight on the earth.

Page 109:

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 strap made of a thin and strong string because of the small area of the strap. We know, pressure = force/area. Due to the small area of the strap the pressure exerted is larger and hence the bag is difficult to hold.

2. What do you mean by buoyancy?
Answer:
Buoyancy is the upward force exerted by a liquid on an object which is floating on or immersed in the liquid.

3. Why does an object float or sink when placed on the surface of water?
Answer:
The object floats when the density of the object is less than that of water. This means that the weight of the object is lesser than the upthrust of the water on the object and hence it floats. The object sinks when the density of the object is more than that of water. This means that the weight of the object is more than the upthrust of the water on the object and hence it sinks.

Page 110:

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
Our mass will be more than 42 kg. This is because although our weight acts downwards on the machine, the buoyancy due to air acts upwards. This gives us a reduced reading on the machine. Our actual mass will therefore be more than 42 kg.

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:
In reality, the cotton bag is heavier than the iron bar. This is because the cotton bag is larger and volume and displaces a larger volume of air. By Archimedes’ principle, the buoyant force acting on the cotton bag is larger. Hence, the real weight of the cotton bag will be more than 100 kg.

Solutions to Exercises (Page No 111) of NCERT Class 9 Science Chapter 9 Gravitation

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to the universal law of gravitation, force of attraction between two bodies is F = GMm/r² , where M and m are the masses of the two objects, r is the distance between them and G is the gravitational constant.

If the distance between them becomes r/2,

Fⁱ = GMm/(r/2)²

or, Fⁱ = 4GMm/r² = 4F

Summary: The force of gravitation between two objects will become four times the initial force when the distance between them is reduced to half.

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:
A heavy object does not fall faster than a light object because the value of acceleration due to gravity ‘g’ is constant. The value of ‘g’ does not change with the mass of the object. Hence, all objects fall with the same acceleration.

3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)
Answer: According to the universal law of gravitation, force of attraction between the bodies is F = GMm/r², where M is the mass of the earth, m is the mass of the object, r is the distance between them and G is the gravitational constant.

M = 6 × 10²⁴ kg

m = 1 kg

r = 6.4 × 10⁶ m

G = 6.7 × 10^-11 N m² kg^-2

F = (6.7 × 10^-11 × 6 × 10²⁴ × 1)/(6.4 × 10⁶ )²

or, F = 9.8 N

Summary:

The magnitude of the gravitational force between the earth and a 1 kg object on its surface is 9.8 N.

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:

The earth and the moon are attracted to each other by gravitational force. The earth attracts the moon with a force which is same as the force with which the moon attracts the earth. The force of attraction is same in magnitude but opposite in direction.

According to the universal law of gravitation, force of attraction between earth and the moon is F = GMm/r², where M and m are the masses of the earth and the moon respectively, r is the distance between them and G is the gravitational constant.

5. If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The earth does not move towards the moon because the mass of the earth is very large and hence the acceleration of the earth is very small.

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:

According to the universal law of gravitation, force of attraction between two bodies is F = GMm/r² , where M and m are the masses of the two objects, r is the distance between them and G is the gravitational constant.

(i) The mass of one object is doubled, i.e. M becomes 2M.

Fⁱ = (G(2M)m)/r² = 2GMm/r² = 2F

Hence, the force between the two objects gets doubled.

(ii) The distance between the objects is doubled, that is r becomes 2r.

Fⁱ = GMm/(2r)² = GMm/4r² = F/4

Hence, the force between the two objects becomes one-fourth.

The distance between the objects is tripled.

Fⁱ = GMm/(3r)² = GMm/9r² = F/9

Hence, the force thus becomes one-ninth of its initial force.

(iii) The masses of both objects are doubled. The new masses are 2M and 2m.

Fⁱ = (G(2M)(2m))/r² = 4GMm/r² = 4F

The force becomes four times the initial force.

7. What is the importance of universal law of gravitation?
Answer:
The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:

(i) the force that binds us to the earth;

(ii) the motion of the moon around the earth;

(iii) the motion of planets around the Sun; and (iv) the tides due to the moon and the Sun.

8. What is the acceleration of free fall?
Answer:
Acceleration due to gravity, denoted by g, is the rate at which an object falls under Earth’s gravitational force alone. On Earth’s surface, its value is 9.8 ms^-2.

9. What do we call the gravitational force between the earth and an object?
Answer:
We call the gravitational force between the earth and an object the weight of the object.

10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]
Answer:
The weight of a body = mg, where m is the mass of the object and g is the acceleration due to gravity. Since the value of g is greater at the poles than at the equator, the gold will weigh less at the equator than at the poles. Hence, the friend at the equator will not agree with the weight of gold bought.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
A flat sheet of paper has a greater surface area than a crumpled paper ball. Due to its larger surface area, the sheet of paper experiences more air resistance, causing it to fall more slowly than the crumpled ball.

12. Gravitational force on the surface of the moon is only  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:

Let the gravitational force on the surface of the moon be F_m and the gravitational force on the surface of the earth be F_e.

It is given that F_m = (1/6)F_e.

Acceleration due to earth’s gravity = g_e = 9.8 ms^-2.

Acceleration due to moon’s gravity = g_m.

F_m = (1/6)F_e

or, mg_m = (1/6)mg_e

or, g_m = (1/6)g_e = (1/6) × 9.8 ms^-2 = 1.63 ms^-2

Therefore, weight of the object on earth = mg_e = 10 × 9.8 = 98 N.

Weight of the object on moon = mg_m = 10 × 1.63 ms^-2 = 16.3 N.

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

(i) the maximum height to which it rises,

(ii) the total time it takes to return to the surface of the earth.

Answer:

(i) Initial velocity (u) = 49 m/s

Final speed (v) = 0 (Since at maximum height the ball comes to rest)

Acceleration due to earth gravity g = -9.8 m/s².

Let maximum height be h.

Therefore,

v² = u² + 2gh

or, 2gh = v² – u²

2 × (- 9.8) × H = 0² – (49)²

– 19.6h = – 2401

h = 122.5 m

(ii) Let the time taken to reach maximum height be t.

v = u + gt

or, 0 = 49 – 9.8t

or, t = 49/9.8 = 5 s

Time required to descend to the surface of the earth = time taken to reach maximum height = 5 s.

Therefore, total time it takes to return to the surface of the earth = 5s + 5s = 10 s.

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

Initial velocity (u) = 49 m/s

Height (h) = 19.6 m.

Let final velocity be v.

Acceleration due to earth gravity g = 9.8 m/s².

v² = u² + 2gh

or, v² = 0² + 2 × 9.8 × 19.6

or, v² = 384.16

or, v = √384.16

or, v = 19.6 m/s

Its final velocity just before touching the ground is 19.6 m/s.

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:

Initial velocity (u) = 40 m/s

Final speed (v) = 0 (Since at maximum height the ball comes to rest)

Acceleration due to earth gravity g = 10 m/s².

Let maximum height be h.

v² = u² + 2gh

or, 2gh = v² – u²

2 × (- 10) × H = 0² – (40)²

or, – 20h = – 1600

or, h = 80 m

The stone descends after its ascent.

Therefore, total distance covered by the stone = 80 m + 80 m = 160 m.

Net displacement = 0 (since the stone returns to its initial point)

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

According to the universal law of gravitation, force of attraction between earth and the Sun is F = GMm/r², where M and m are the masses of the Sun and the earth respectively, r is the distance between them and G is the gravitational constant.

M = 2 × 10³⁰ kg

m = 6 × 10²⁴ kg

r = 1.5 × 10¹¹ m

G = 6.7 × 10^-11 N m² kg^-2

F = (6.7 × 10^-11 × 2 × 10³⁰ × 6 × 10²⁴)/(1.5 × 10¹¹ )²

or, F = 3.57 × 10²² N

Force of gravitation between the earth and the Sun = 3.57 × 10²² N.

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

Let the falling stone cover a height of h.

Initial velocity = 0.

Time taken = t.

h = ut + (1/2)gt²

or, h = 0 + (1/2) × 10 ×t²

or, h = 5t²

Let the height covered by the stone projected vertically upwards be hⁱ.

Initial velocity = 25 m/s.

Time taken = t.

hⁱ = ut – (1/2)gt²

or, hⁱ = 25t – (1/2) × 10 × t²

or, hⁱ = 25t – 5t²

Now, h + h¹ = 100

or, 5t² + 25t – 5t² = 100

or, 25t = 100

or, t = 100/25 = 4 s

The stones meet after 4 seconds.

h = 5t² = 5 × 4² = 80 m.

hⁱ = 100 – 80 = 20 m. The two stones will meet at a height of 20 m from the ground.

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 taken to ascend = time taken to descend = 3 s.

Let initial velocity be u.

g = 10 m/s²

Final velocity (v) = 0 (Since the ball attains a state of rest at maximum height)

v = u – gt

or, 0 = u – 10 × 3

or, u = 30 m/s

The velocity with which the stone was thrown up was 30 m/s.

(b) Let maximum height be h.

h = ut – (1/2)gt²

or, h = 30 × 3 – (1/2) × 10 × 3²

or, h = 45 m

Maximum height = 45 m.

(c) In 3 seconds it reaches maximum height. Distance travelled in another 1 second be h¹.

h = ut + (1/2)gt²

or, h = 0 + (1/2) 10 × 1²

or, h = 5 m

Therefore, its position after 4 s is (45 – 5) = 40 m from the ground.

19. In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
The buoyant force on an object immersed in a liquid acts in the vertically upwards direction.

20. Why does a block of plastic released under water come up to the surface of water?
Answer:
Due to its lower density compared to water, a plastic block experiences a stronger buoyant force than its own weight. Consequently, the block accelerates upwards, leading it to rise to the surface of the water.

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

Density of the substance = mass/volume = 50 g/20 cm³ = 2.5 g/cm³.

Density of water is 1 g cm^–3.

Since the density of the substance is greater than the density of water, the substance will sink.

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 substance = mass/volume = 500 g/350 cm³ = 1.43 g/cm³.

Density of water is 1 g cm^–3.

Since density of the substance is greater than the density of water, the substance will sink.

Volume of water displaced = volume of substance = 350 cm³.

Mass of water displaced = density of water × volume of water displaced = 1 g cm^–3 × 350 cm³ = 350 g.

Solutions to All Activities of NCERT Class 9 Science Chapter 9 Gravitation

1. Complete Activity 9.1 (Page 100).

• Take a piece of thread.

• Tie a small stone at one end. Hold the other end of the thread and whirl it round, as shown in Fig. 9.1.

• Note the motion of the stone.

• Release the thread.

• Again, note the direction of motion of the stone.

Answer:

Solution to Activity 9.1

2. Complete Activity 9.2 (Page 102).

• Take a stone.

• Throw it upwards.

• It reaches a certain height and then it starts falling down.

Answer:

Solution to Activity 9.2

3. Complete Activity 9.3 (Page 103).

• Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously.

Answer:

Solution to Activity 9.3

4. Complete Activity 9.4 (Page 108).

• Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You see that the bottle floats.

• Push the bottle into the water. You feel an upward push. Try to push it further down. You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed.

• Now, release the bottle. It bounces back to the surface.

• Does the force due to the gravitational attraction of the earth act on this bottle? If so, why doesn’t the bottle stay immersed in water after it is released? How can you immerse the bottle in water?

Answer:

Solution to Activity 9.4

5. Complete Activity 9.5 (Page 108).

• Take a beaker filled with water.

• Take an iron nail and place it on the surface of the water.

• Observe what happens.

Answer:

Solution to Activity 9.5

6. Complete Activity 9.6 (Page 109).

• Take a beaker filled with water.

• Take a piece of cork and an iron nail of equal mass.

• Place them on the surface of water.

• Observe what happens.

Answer:

Solution to Activity 9.6

7. Complete Activity 9.7 (Page 109).

• Take a piece of stone and tie it to one end of a rubber string or a spring balance.

• Suspend the stone by holding the balance or the string as shown in Fig. 9.6 (a).

• Note the elongation of the string or the reading on the spring balance due to the weight of the stone.

• Now, slowly dip the stone in the water in a container as shown in Fig. 9.6 (b).

• Observe what happens to elongation of the string or the reading on the balance.

Answer:

Solution to Activity 9.7

Extra Questions to Complement Solutions to NCERT Class 9 Science Chapter 9 Gravitation

Very Short Answer Type Questions:

1. What is the weight of an object of mass 1 kg on the surface of the earth?
Answer:
Weight = mg = 1 kg × 9.8 m/s² = 9.8 N.

2. Due to which force does the moon orbit around the earth?
Answer:
Gravitational force of attraction of the earth.

3. What is the shape of the orbit of a planet?
Answer:
Ellipse.

4. The distance between two objects is tripled. The magnitude of the new force is what fraction of the original force?
Answer:
1/9.

5. Does the universal gravitational constant change from place to place?
Answer:
No, it has a constant value of G = 6.7 × 10^-11 N m² kg^-2.

6. What happens to the value of acceleration due to gravity as we go upwards from the surface of the earth?
Answer:
It decreases.

7. If the buoyant force exerted by a liquid on a body is equal to the weight of the body, what will happen?
Answer:
The body will float inside the liquid.

8. Which law did Newton use to calculate gravitational force of attraction?
Answer:
Kepler’s third law.

9. The motion of the moon around the earth is due to which force?
Answer:
Centripetal force.

10. Every object in the universe attracts every other object with a force. What is the force?
Answer:
Gravitational force.

Multiple Choice Questions (MCQ):

1. The weight of an object positioned at infinite radius from the centre of the earth is:

(A) 0
(B) Infinite
(C) 9.8 kg
(D) 98 kg

Answer: (A) 0

We know, F = GMm/r². As the distance from the Earth’s centre increases, the gravitational force decreases. At an infinite distance, the gravitational force approaches zero, resulting in negligible weight for the object.

2. Acceleration due to gravity

(A) depends on mass and volume of the object
(B) depends on mass but not volume of the object
(C) does not depend on mass but depends on volume of the object
(D) does not depend on mass or volume of the object

Answer: (D) does not depend on mass or volume of the object

3. The mean distance of planet A from Sun is twice the mean distance of planet B from the Sun. The orbital period of A:

(A) Is equal to the orbital period of B
(B) Is twice the orbital period of B
(C) Is 2√2 times the orbital period of B
(D) Is 8 times the orbital period of B

Answer: (C) Is 2√2 times the orbital period of B

From Kepler’s third law we know:

r³/T² = constant where r is the mean distance of the planet from the sun and T is the square of the orbital period T.

Therefore,

(2r)³/T_A² = r³/(T_B²)

or, T_A² = 8T_B²

or, TA = 2√2TB

4. Which celestial body exerts the strongest gravitational force on objects at its surface?

(A) The Moon
(B) Earth
(C) Mars
(D) Jupiter

Answer: (D) Jupiter

The reason is that the mass of Jupiter is by far the largest. Gravitational force depends on the mass of the object and the distance from its centre, and Jupiter’s immense mass results in a stronger gravitational pull compared to the other options listed.

5. Which of the following factors does NOT affect the buoyant force acting on an object submerged in a fluid?

(A) Volume of the object
(B) Density of the fluid
(C) Depth of immersion
(D) Material with which the object is made

Answer: (D) Material with which the object is made

Short Answer Type Questions:

1. Will an object weigh more on the surface of the earth or at a certain height above the earth?
Answer:
An object will weigh slightly less at a height h above the surface of the earth compared to its weight on the surface of the earth. This is due to the decrease in gravitational force with increasing distance from the centre of the earth.

F = GMm/r², where M is the mass of the earth and m is the mass of the object, d is the distance between the earth and the object and G is the universal gravitational constant. As we can see, F decreases with increasing d. Hence, the weight of the object decreases with increasing height above the earth.

2. An iron nail sinks in water while a ship of the same density floats. Why?
Answer:
The iron nail sinks in water because it displaces less water and hence, according to Archimedes’ principle the buoyant force is lesser than its weight. However, the ship, despite being made of iron, floats because its shape and volume allow it to displace a greater amount of water, exerting an upward buoyant force that exceeds its weight, thus keeping it afloat.

3. Why do astronauts experience weightlessness in space even through the force of gravity acts on them?
Answer:
Astronauts are in a state of free fall in space. Although the force of gravity is present, the normal reaction force which we feel when standing on solid ground is not present in space. Hence, there is a feeling of weightlessness.

4. What factors affect the acceleration due to gravity on a planet’s surface?
Answer:
On the planet’s surface, g = GM/R² where M is the mass of the planet, r is the radius and G is the universal gravitational constant. Thus, we can see that the value of ‘g’ on the planet’s surface depends on mass of the planet and radius of the planet.

5. How are tides caused?
Answer:
The gravitational force of the moon attracts the sea water towards it, causing a bulge. As the moon orbits around the Earth, these bulges result in the periodic rise and fall of sea levels, known as high tides and low tides, respectively.

6. Is the buoyant force experienced by an object in oil less or more than the buoyant force experienced by the object in water?
Answer:
The buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object, according to Archimedes’ principle. The buoyant force depends on the density of the fluid and the volume of the displaced fluid.

The volume of the displaced fluid will be the same in both cases. The density of oil is generally lower than that of water, which means that the buoyant force experienced by an object submerged in oil is usually less than the buoyant force experienced in water for the same volume of fluid displaced.

7. You are given two bags of the same mass. One bag has a broad strap and the other bag has a thin strap for carrying. Why bag should you choose and why?
Answer:
We know, Pressure = Force/Area. Since area of the broad strap is more, pressure exerted by it will be less. Since area of the thin strap is less, pressure exerted by it will be more. Therefore, we should choose the bag with the broad strap.

8. How can density of a substance determine its purity?
Answer:
When a substance is impure, it may contain other materials or substances that alter its density compared to the pure form. Therefore, by measuring the density of a substance and comparing it to the known density of the pure substance, one can assess the level of purity.

9. An unknown liquid has a relative density of 1.2. If a 500 cm³ container is filled with this liquid, find the mass of the liquid.
Answer:
Density of liquid = Relative density × Density of water = 1.2 × 1 g/cm³ = 1.2 g/cm³.

Now, Density of water = Mass/volume

or, Mass = Density of water × volume

or, Mass = 1.2 g/cm³ × 500 cm³

or, Mass = 600 g

Mass of liquid = 600 g.

10. A ball is dropped from a height of 20 m. After 2 seconds the ball will be at what height above the ground?
Answer:
v = u + gt

After 2 s:

v = 0 + 9.8 × 2 = 19.6 m s^-1

v² = u² + 2gh

or, (19.6)² = 0² + 2 × 9.8 × h

or, h = 19.6 m

Therefore, the ball will be at a height of (20 – 19.6) = 0.4 m above the ground.

Long Answer Type Questions:

1. Two solid spheres each of radius r made of the same material are kept in contact with each other. Show that the gravitational force of attraction between them is proportional to r⁴.

Answer:

Let d be the density of each sphere.

Mass of each sphere = Volume × density = (4/3)πr³d.

Distance between their centres = r + r = 2r.

F = G((4/3) πr³ d)((4/3) πr³ d)/(2r)²

or, F = (4/9)π²Gd²r⁴

or, F is proportional to r⁴.

2. A metal block weighs 500 N in air. When submerged in water, it weighs 450 N. Calculate the buoyant force acting on the block and determine its density.

Answer:

Weight of the block in air = 500 N.

Weight of the block in water = 450 N.

Therefore, buoyant force acting upwards = 500 N – 450 N = 50 N.

We know that the buoyant force in water is equal to the weight of the water displaced by the block.

Buoyant force = Weight of water displaced = Mass of water displaced × g = Density of water × Volume of water displaced × g

The density of water = 1000 kg/m³.

Therefore,

Volume of water displaced = Buoyant force / (Density of water × g)

g ≈ 9.8 m/s²,

Volume of water displaced = 50 N / (1000 kg/m³ × 9.8 m/s²) ≈ 0.0051 m³.

Therefore, volume of the block also = 0.0051 m³.

Now,

Mass of the block = Weight of the block/g = 500 N/9.8 ms^-2 = 51.02 kg.

Density of the block = Mass/Volume = 51.02 kg/0.0051 m³ = 10000 kg/m³.

Fill in the Blanks:

(a) The value of g at the centre of the earth is __________.

(b) All objects experience a force of _________ when immersed in a liquid.

(c) The value of the acceleration due to gravity varies with _________, but it is approximately constant near the surface of the Earth.

(d) Gravitational force is responsible for keeping celestial bodies like planets and stars in _________.

(e) The buoyant force acting on an object submerged in a fluid depends on the object’s __________ and the density of the fluid.

Answers:

(a) The value of g at the centre of the earth is 0.

(b) All objects experience a force of buoyancy when immersed in a liquid.

(c) The value of the acceleration due to gravity varies with altitude, but it is approximately constant near the surface of the Earth.

(d) Gravitational force is responsible for keeping celestial bodies like planets and stars in orbit.

(e) The buoyant force acting on an object submerged in a fluid depends on the object’s volume and the density of the fluid.

Match and Pair:

Column A	Column B
(i) Isaac Newton	(a) 980 cms-1
(ii) Kepler	(b) Gravity
(iii) g	(c) Unitless
(iv) G	(d) Elliptical orbit
(v) Relative density	(e) 6.67 × 1011 N m2 kg-2

Answer:

Column A	Column B
(i) Isaac Newton	(b) Gravity
(ii) Kepler	(d) Elliptical orbit
(iii) g	(a) 980 cms-1
(iv) G	(e) 6.67 × 1011 N m2 kg-2
(v) Relative density	(c) Unitless

++++++++++++++

Frequently Asked Questions (FAQs) on NCERT Solutions to Class 9 Science Chapter 9 Gravitation