NCERT Solutions for Class 9 Science Chapter 10 Gravitation

The acceleration of free fall is the acceleration experienced by any object when it falls freely under the influence of Earth’s gravity, without any resistance. It is denoted by g, and its value on Earth is 9.8 m/s². This acceleration is constant for all objects (irrespective of their masses). This means that for every second an object is in free fall, its velocity (speed) increases by 9.8 metres per second towards the Earth’s surface.

9. What do we call the gravitational force between the earth and an object?

Answer:

The gravitational force between the Earth and an object is called its weight. Weight is equal to the product of the object’s mass and the acceleration due to gravity $)$.

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 an object is given by \( W = m \times g \), where \( m\) is the mass and \( g \) is the acceleration due to gravity.

Again, gravity depends on the radius of the Earth. \(g = \frac{GM}{R^2}\)​

The Earth is not a perfect sphere. The radius of the Earth is smaller at the poles and larger at the equator. So, the value of  \( g \)  is greater at the poles than at the equator. So, gold at the equator weighs less than at the poles. That’s why Amit’s friend will not agree with the weight of gold bought at the poles.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer:

A sheet of paper falls slower than a crumpled paper ball because of air resistance.

When an object falls, gravity pulls it downward. At the same time, air pushes upward against it. A flat sheet of paper has a large surface area, so it experiences more air resistance. This slows down its fall.

When we crumple the paper into a small ball, its surface area becomes smaller. Less air resistance acts on it, so it falls faster than the flat sheet.
(N.B. If there were no air (like in a vacuum), both would fall at the same speed because only gravity would act on them. )

12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton’s of a 10 kg object on the moon and on the Earth?

Answer:

Given:

• Mass of the object, \( m= 10 kg \)
• Acceleration due to gravity on Earth \( g = 9.8  \text{m/s}^2 \)
• \( W_{Moon} = \frac{1}{6} \times W_{Earth} \)

1. Weight on Earth:

We know that, Weight \( W = m \times g \)

\( W_{Earth} = 10 \times 9.8 = 98  \text{N} \)

So, the weight of the 10 kg object on the Earth is 98 N

2. Weight on Moon:

\( W_{Moon} = \frac{1}{6} \times W_{Earth} \)

\( W_{Moon} = \frac{1}{6} \times 98 N \)

\( W_{Moon} ≈16.33 N \)

The weight of the 10 kg object on the Moon is approximately 16.33 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.

Solution:

(i) Given that:

• Initial velocity, \( u = 49  \text{m/s} \)
• Final velocity, \( v = 0  \text{m/s} \) (at the maximum height, the ball stops)
• Acceleration due to gravity \( g = – 9.8 \text{m/s}^2 \) (it’s negative because the ball is moving against gravity)

We use the third equation of motion,

\( v^2 = u^2 + 2 gs \)

⇒ \( 0^2 = 49^2 – 2 \times 9.8 \times s \)
⇒ \( 0=2401−19.6s \)
⇒ \( 19.6s=2401 \)
⇒ \( s= \frac{2401}{19.6} \)
⇒ \( s= 122.5 m \)
∴ The maximum height to which the ball rises is 122.5 meters.

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

Solution:-

Let the time taken by the ball to reach the height \(122.5 m \) be \( t \), then according to the equation of motion:

\( v = u + gt \)

Substituting the values we get,

⇒ \( 0 = 49 + ( -9.8)t \)

⇒ \( 9.8 t = 49  \)

⇒ \( t = \frac{49}{9.8}  \)

⇒ \( t = 5 s  \)

So, the time taken to reach the maximum height is 5 seconds.

However, Time to ascent = Time to descent

∴ Total time, T = Time to ascend + Time to descend

∴ Total time, T = 5 s + 5 s = 10 seconds 

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.

Solution:

Given that:

• Height of tower = Distance, \( s = 19.6  \text{m} \)
• Initial velocity, \( u = o m/s \)
• Acceleration due to gravity, \( g = 9.8  \text{m/s}^2 \)
• Final velocity, \( v = ? \)

We know that,

\( v^2 = u^2 + 2 gs \)

⇒ \( v^2 = 0^2 + 2 \times 9.8 \times 19.6 \)

⇒ \( v^2 = 2 \times 192.08 \)

⇒ \( v^2 = 384.16 \)

\( ⇒ v = \sqrt{384.16} \)

⇒ \( v = 19.6  \text{m/s} \)

∴ The final velocity of the stone just before it touches the ground is \( 19.6  \text{m/s} \)

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?

Solution:

Given that:

• Initial velocity, \( u = 40  \text{m/s} \)
• Final velocity, \( v = 0  \text{m/s} \) (at the maximum height, the ball stops)
• Acceleration due to gravity \( g = – 10 \text{m/s}^2 \) (negative because the stone is moving against gravity)

(i) Maximum height

We know that,

\( v^2 = u^2 + 2 gs \)

\( ⇒ 0^2 = 40^2 + 2 \times (-10) \times s \)

\( ⇒ 0 = 1600 – 20 s \)

\( ⇒ 20 s = 1600 \)

\( ⇒ s = \frac{1600}{20} = 80 m \)

∴ The maximum height reached by the stone is 80 meters.

(ii) Net displacement

• The stone is thrown vertically upward from the ground and eventually returns to the ground.
• So, initial and final positions are the same.
• Net displacement = 0

(iii) Total distance covered

• Distance traveled upwards  \( = 80 m \)
• Distance traveled downwards \( = 80 m \)
• Total distance \( = 80 + 80 = 160 m \)

Total distance covered = 160 m

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.

Solution:

Given that:

• Mass of Earth, \( M_e = 6 \times 10^{24} \text{kg} \)

• Mass of Sun, \( M_s = 2 \times 10^{30} \text{kg} \)

• Distance between Earth and Sun, \( d = 1.5 \times 10^{11} \text{m} \)

• Gravitational constant, \( G = 6.67 \times 10^{-11} \text{Nm}^2 \)

We know the gravitational force:

\( F = \frac{G M_e M_s}{d^2} \)

\( ⇒ F = \frac{(6.67 \times 10^{-11} ) \times ( 6 \times 10^{24}) \times (2 \times 10^{30})} {(1.5 \times 10^{11})^2} \)

\( ⇒ F = \frac{(6.67 \times 10^{-11} ) \times (12 \times 10^{54}) } {(2.25 \times 10^{22})} \)

\( ⇒ F = (6.67 \times 10^{-11} ) \times (5.33 \times 10^{32}) \)

\( ⇒ F ≈ 3.57\times 10^{22} N \)

∴ The gravitational force between the Earth and the Sun is approximately \(  3.57\times 10^{22} N \)

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.

Solution:

Let the two stones meet after time \( t \) from the start.

(i) When the stone from the top of the tower is thrown,

Initial velocity, \( u = 0 \)

Let the distance covered by the first stone in time \( t \) from the top of the tower be \( s_1 \)

Acceleration due to gravity, \( g = 9.8  \text{m/s}^2 \)

From the equation of motion, \( s_1 = ut + \frac{1}{2} gt^2 \)

\( ⇒ s_1 = 0\times t + \frac{1}{2} \times 9.8\times t^2 \)

\( ⇒ s_1 = 4.9 t^2 \)————-(i)

(ii) When the stone is thrown upwards,

Initial velocity, \( u = 25 m/s \)

Let the distance covered by the stone from the ground in time \( t \) be \( s_2 \)

Acceleration due to gravity, \( g = – 9.8  \text{m/s}^2 \)

From the equation of motion, \( s_2 = ut + \frac{1}{2} gt^2 \)

\( ⇒ s_2 = 25\times t + \frac{1}{2} \times (-9.8)\times t^2 \)

\( ⇒ s_2 = 25t – 4.9 t^2 \)———(ii)

The combined distance covered by both the stones at the meeting point is equal to the height of the tower \( 100 m\).

∴ \( s_1 + s_2 = 100 \)———–(iii)

Substituting equation (i) and (ii) in (iii), we get

\( ⇒ 4.9 t^2 + 25t – 4.9 t^2 = 100 \)

\( ⇒ 25t = 100 \)

\( ⇒ t = \frac{100}{25} \)

\( ⇒ t = 4 s \)

So, substituting the value of \( t \) in equation (ii), we get;

\( ⇒ s_2 = 25\times 4 – 4.9 \times4^2 \)

\( ⇒ s_2 = 100 – 4.9 \times16 \)

\( ⇒ s_2 = 100 – 78.4 \)

\( ⇒ s_2 = 21.6 m \)

∴ The two stones will meet at a height of \( 21.6 m \)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,

Solution:

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, the time taken for the upward journey, \( t= \frac{6}{2} = 3s \)

At the maximum height, the final velocity (\( v \)) of the ball is \( 0 m/s \)

Acceleration due to gravity \( g = – 9.8  \text{m/s}^2 \) ((negative because the ball is moving against gravity)

From first equation of motion:- \( v = u + gt \)

\( ⇒ 0 = u + ( – 9.8) (3) \)

\( ⇒ 0 = u – 29.4 \)

\( ⇒ 0 = 29.4 m/s \)

∴ The ball was thrown upwards with a velocity of \( 29.4 m/s \)

b) The maximum height it reaches 

Solution:

Let the maximum height attained by the ball be \( s \).

Initial velocity, \( u = 29.4 m/s \)

Time to reach max height \( t = 3 s \)

Acceleration due to gravity \( g = – 9.8  \text{m/s}^2 \)

We know that, \( s = ut + \frac{1}{2} gt^2 \)

\( ⇒ s = 29.4\times 3 + \frac{1}{2} \times(-9.8)\times (3)^2 \)

\( ⇒ s = 88.2 – \frac{1}{2} \times(9.8)\times 9 \)

\( ⇒ s = 88.2 – 44.1 \)

\( ⇒ s = 44.1 m \)

∴ The maximum height the ball reaches is \(  44.1 m \)

c) Its position after \( 4s \).

Solution:

Since, in \( 3s \), the ball attains its maximum height. So, after attaining this height, the ball starts falling downwards.

Position of the ball after \( 4s \) of the throw is given by the distance travelled by it during its downward journey in ( \( 4s – 3s = 1s \))

In this case, Initial velocity, \( u = 0 m/s \)

Time, \( t = 1s \)

Acceleration due to gravity \( g = 9.8  \text{m/s}^2 \)

We have,  \( s = ut + \frac{1}{2} gt^2 \)

\( s = 0 + \frac{1}{2} \times 9.8 \times (1)^2 \)

\( ⇒ s = 4.9 m \)

Since, total height \( = 44.1 m \)

This means that the ball is \( 44.1m – 4.9m = 39.2 m \) above the ground after \( 4 \) seconds.

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 upward direction.

20. Why does a block of plastic released under water come up to the surface of water?

Answer:- 

When a plastic block is released under water, two forces act on it:

• Gravity pulls it downward.
• Buoyant force pushes it upward.

Since the density of plastic is less than water, the buoyant force is greater than the weight of the block. As a result, the block moves upward and comes to the surface.

21. The volume of \( 50g \) of a substance is \( 20 cm^3 \) . If the density of water is \( 1g cm^3 \), will the substance float or sink?

Solution:

Given:
Mass of the substance \( m = 50 \text{g} \)
Volume of the substance \( V = 20 \text{cm}^3 \)
Density of water \( = 1 \text{g/cm}^3 \)

We know, Density of the substance, \( D = \frac{m}{V} \)

\( ⇒ D = \frac{50}{20} \)

\( ⇒ D= 2.5 \text{g/cm}^3 \)

Since the density of the substance ( \( 2.5 \text{g/cm}^3 \))is more than the density of water (\( 1 \text{g/cm}^3 \)). So the substance will sink.

22. The volume of a \( 500g \) sealed packet is \( 350 cm^3 \). Will the packet float or sink in water if the density of water is \( 1 g cm^{-3} \)? What will be the volume of the water displaced by this packet?

Solution:

Given:

• Mass of packet, \( m = 500 \text{g} \)
• Volume of packet, \( V = 350 \text{cm}^3 \)
• Density of water, \(  D = 1 \text{g/cm}^3 \)

We know, Density of the packet, \( D = \frac{m}{V} \)

\( ⇒ D = \frac{500}{350} \)

\( ⇒ D = \frac{10}{7} \) g/cm³

\( ⇒ D = 1.43 \) g/cm³

Since the density of the packet (1.43 g/cm³) is greater than the density of water (1 g/cm³), the packet will sink.

Since the packet sinks completely, the volume of water it displaces is equal to its own volume. So, Volume of the packet: 350 cm³

Mass of water displaced by the packet = volume of the packet × density of water

⇒ Mass of water displaced by the packet = 350 cm³ × 1 g/cm³

∴ Mass of water displaced by the packet = 350 g