# NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

NCERT Solutions for Class 9 Science Chapter 10 Gravitation provides you with necessary insights on the concepts involved in the chapter. Detailed answers and explanations provided by us will help you in understanding the concepts clearly.

Gravity is a fascinating topic that explains many things. From how our planet stays in orbit to why things fall down. Explore NCERT Solutions for Class 9 Science Chapter 10 – Gravitation to learn everything you need to know about gravity. Content is crafted by highly qualified teachers and industry professionals with decades of relevant knowledge. Moreover, the solutions have been updated to include the latest content prescribed by the CBSE board.

Furthermore, we ensure that relevant content on NCERT Solutions Class 9 is regularly updated as per the norms and prerequisites that examiners often look for in an exam. This ensures that the content is tailored to be class relevant, but without sacrificing the informational quotient. CoolGyan’S also strives to impart maximum informational value without increasing the complexity of topics. This is achieved by ensuring that the language is simple and all technical jargons are explained at the required school level.

## Exercise-10.1 Page: 134

1. State the universal law of gravitation.

Solution:

The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Solution:

Consider F as the force of attraction between an object on the surface of earth and the earth

Also, consider ‘m’ as the mass of the object on the surface of earth and ‘M’ as the mass of earth

The distance between the earth’s centre and object = Radius of the earth = R

Therefore, the formula for the magnitude of the gravitational force between the earth and an object on the surface is given as

F = G Mm/R2

## Exercise-10.2 Page: 136

1. What do you mean by free fall?

Solution:

Earth’s gravity attracts each object to its center. When an object is dropped from a certain height, under the influence of gravitational force it begins to fall to the surface of Earth. Such an object movement is called free fall.

2. What do you mean by acceleration due to gravity?

Solution:

When an object falls freely from a certain height towards the earth’s surface, its velocity keeps changing. This velocity change produces acceleration in the object known as acceleration due to gravity and denoted by ‘g’.

The value of the acceleration due to gravity on Earth is,

## Exercise-10.3 Page: 138

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

Solution:

The differences between the mass of an object and its weight are tabulated below.

 Mass Weight Mass is the quantity of matter contained in the body. Weight is the force of gravity acting on the body. It is the measure of inertia of the body. It is the measure of gravity. It only has magnitude. It has magnitude as well as direction. Mass is a constant quantity. Weight is not a constant quantity. It is different at different places. Its SI unit is kilogram (kg). Its SI unit is the same as the SI unit of force, i.e., Newton (N).

2. Why is the weight of an object on the moon 1/6th its weight on the earth?

Solution:

The moon’s mass is 1/100 times and 1/4 times the earth’s radius. As a result, when compared to earth, the gravitational attraction on the moon is about one sixth. Thus, an object’s weight on the moon is 1/6th its weight on earth.

## Exercise-10.4 Page: 141

1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Solution:

It is tough to carry a school bag having a skinny strap because of the pressure that is being applied on the shoulders. The pressure is reciprocally proportional to the expanse on which the force acts. So, the smaller the surface area, the larger is going to be the pressure on the surface. In the case of a skinny strap, the contact expanse is quite small. Hence, the pressure exerted on the shoulder is extremely huge.

2. What do you mean by buoyancy?

Solution:

The upward force possessed by a liquid on an object that’s immersed in it is referred to as buoyancy.

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

Solution:

An object floats or sinks when placed on the surface of water because of two reasons.

(i) If its density is greater than that of water, an object sinks in water.

(ii) If its density is less than that of water, an object floats in water.

## Exercise-10.5 Page: 142

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Solution:

When weighing our body, it is acting by an upward force. The buoyant force is this upward force. As a result, the body is pushed up slightly, resulting in the weighing machine showing less reading than the actual value.

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?

Solution:

The bag of cotton is heavier than the bar of iron. The cotton bag experiences a larger air thrust than the iron bar. Therefore, the weighing machine indicates less weight than its actual weight for the cotton bag.

## Exercises-10.6 Page: 143

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Solution:

Consider the Universal law of gravitation,

According to that law, the force of attraction between two bodies is

Where,

m1 and m2 are the masses of the two bodies.

G is the gravitational constant.

r is the distance between the two bodies.

Given that the distance is reduced to half then,

r = 1/2 r

Therefore,

F = 4F

Therefore once the space between the objects is reduced to half, then the force of gravitation will increase by fourfold the first force.

2. Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object?

Solution:

All objects fall from the top with a constant acceleration called acceleration due to gravity (g). This is constant on earth and therefore the value of ‘g’ doesn’t depend on the mass of an object. Hence, heavier objects don’t fall quicker than light-weight objects provided there’s no air resistance.

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 × 106m.)

Solution:

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by

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?

Solution:

The earth attracts the moon with a force same as the force with which the moon attracts the earth. However, these forces are in opposite directions. By universal law of gravitation, the force between moon and also the sun can be

Where,

d = distance between the earth and moon.

m1 and m2 = masses of earth and moon respectively.

5. If the moon attracts the earth, why does the earth not move towards the moon?

Solution:

According to the universal law of gravitation and Newton’s third law, we all know that the force of attraction between two objects is the same, however in the opposite directions. So the earth attracts the moon with a force same as the moon attracts the earth but in opposite directions. Since earth is larger in mass compared to that of the moon, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. Therefore, for this reason the earth does not move towards the moon.

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?

Solution:

(i)

According to universal law of gravitation, the force between 2 objects (m1 and m2) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

If the mass is doubled for one object.

F = 2F, so the force is also doubled.

(ii)

If the distance between the objects is doubled and tripled

If it’s doubled

Hence,

F = (Gm1m2)/(2R)2

F = 1/4 (Gm1m2)/R2

F = F/4

Force thus becomes one-fourth of its initial force.

Now, if it’s tripled

Hence,

F = (Gm1m2)/(3R)2

F = 1/9 (Gm1m2)/R2

F = F/9

Force thus becomes one-ninth of its initial force.

(iii)

If masses of both the objects are doubled, then

F = 4F, Force will therefore be four times greater than its actual value.

7. What is the importance of universal law of gravitation?

Solution:

The universal law of gravitation explains many phenomena that were believed to be unconnected:

(i) The motion of the moon round the earth

(ii) The force that binds North American nation to the world

(iii) The tides because of the moon and therefore the Sun

(iv) The motion of planets round the Sun

8. What is the acceleration of free fall?

Solution:

When a body is in free fall, the only force acting on the body is that of the earth’s gravitational force. By Newton’s second law of motion, all the forces produce acceleration and thus all the objects accelerate toward the surface of the earth due gravitational attraction of the earth.

This acceleration is known as the acceleration due to gravity on the earth’s surface. It’s denoted by ‘g’ and its value is 9.8m/s2 and it’s constant for all objects close to earth’s surface (irrespective of their masses).

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

Solution:

The gravitational force between the earth and an object is known as the object’s weight.

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.]

Solution:

The weight of a body on the earth’s surface;

W = mg (where m = mass of the body and g = acceleration due to gravity)

The value of g is larger at poles when compared to the equator. So gold can weigh less at the equator as compared to the poles.

Therefore, Amit’s friend won’t believe the load of the gold bought.

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

Solution:

A sheet of paper has a larger surface area when compared to a crumpled paper ball. A sheet of paper will face a lot of air resistance. Thus, a sheet of paper falls slower than the crumpled ball.

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?

Solution:

Given data:

Acceleration due to earth’s gravity = ge or g = 9.8 m/s2

Object’s mass, m = 10 kg

Acceleration due to moon gravity = gm

Weight on the earth= We

Weight on the moon = Wm

Weight = mass x gravity

gm = (1/6) ge (given)

So Wm = m gm = m x (1/6) ge

Wm = 10 x (1/6) x 9.8 = 16.34 N

We = m x ge = 10 x 9.8

We = 98N

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:

Given data:

Initial velocity u = 49 m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/s2 (thus negative as ball is thrown up).

By third equation of motion,

v2 = u2 – 2gs

Substitute all the values in the above equation

Total time T = Time to ascend (Ta) + Time to descend (Td)

v = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5 s

Also, Td = 5 s

Therefore T = Ta + Td

T = 5 + 5

T = 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.

Solution:

Given data:

Initial velocity

u = 0

Tower height = total distance = 19.6m

g = 9.8 m/s2

Consider third equation of motion

v2 = u2 + 2gs

v2 = 0 + 2 × 9.8 × 19.6

v2 = 384.16

v = √(384.16)

v = 19.6m/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 data:

Initial velocity u = 40m/s

g = 10 m/s2

Max height final velocity = 0

Consider third equation of motion

v2 = u2 – 2gs [negative as the object goes up]

0 = (40)2 – 2 x 10 x s

s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

Total displacement = 0 (The first point is the same as the last 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.

Solution:

Given data:

Mass of the sun ms = 2 × 1030 kg

Mass of the earth me = 6 × 1024 kg

Gravitation constant G = 6.67 x 10-11 N m2/ kg2

Average distance r = 1.5 × 1011 m

Consider Universal law of Gravitation

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:

Given data:

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

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

From equations (a) and (b)

5t2 = 100 -25t + 5t2

t = (100/25) = 4sec.

After 4sec, two stones will meet

From (a)

x = 5t2 = 5 x 4 x 4 = 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

This means that after 4sec, 2 stones meet a distance 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 4s.

Solution:

Given data:

g = 10m/s2

Total time T = 6sec

Ta = Td = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u – gta

u = v + gta

= 0 + 10 x 3

= 30m/s

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

(b) From second equation of motion

The maximum height stone reaches is 45m.

(c) In 3sec, it reaches the maximum height.

Distance travelled in another 1sec = s’

The distance travelled in another 1sec = 5m.

Therefore in 4sec, the position of point p (45 – 5)

= 40m from the ground.

19. In what direction does the buoyant force on an object immersed in a liquid act?

Solution:

The buoyant force on an object that is immersed in a liquid will be in a vertically upward direction.

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

Solution:

The density of plastic is lesser than that of water. Therefore, the force of buoyancy on plastic block will be greater than the weight of plastic block. Hence, the acceleration of plastic block is going to be in the upward direction. So, the plastic block comes up to the surface of 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?

Solution:

To find the Density of the substance the formula is

Density = (Mass/Volume)

Density = (50/20) = 2.5g/cm3

Density of water = 1g/cm3

Density of the substance is greater than density of water. So 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?

Solution:

Density of sealed packet = 500/350 = 1.42 g/cm3

Density of sealed packet is greater than density of water

Therefore the packet will sink.

Considering Archimedes Principle,

Displaced water volume = Force exerted on the sealed packet.

Volume of water displaced = 350cm3

Therefore displaced water mass = ρ x V

= 1 × 350

Mass of displaced water = 350g.

 Also Access NCERT Exemplar Solutions for class 9 Science Chapter 10 CBSE Notes for Class 9 Science Chapter 10

## NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

Chapter 10 – Gravitation is a part of Unit 3 – Motion, Force and Work, which carries a total of 27 out of 100. Usually, 2 or 3 questions do appear from this chapter every year, as previous trends have shown.

The topics usually covered under this chapter are:

• Universal Law of Gravitation and its Importance
• Characteristics of Gravitational Forces
• Concept of Free Fall
• Difference between Gravitation Constant and Gravitational Acceleration

## NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

Often times, the term gravity and gravitation are used interchangeably and this is wrong. However, these two terms are related to each other but their implications are quite different. Academically, Chapter 10 – Gravitation is an important concept as it carries a considerable weightage in the exam. Therefore, ensure that all relevant concepts, formulas and diagrams are studied thoroughly.

Explore how gravitation works at the molecular level, discover its applications and learn other related important concepts by exploring our NCERT Solutions.

### Key Features of NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

1. Solutions provided in an easy-to-understand language
2. Qualified teachers and their vast experience helps formulate the solutions
3. Questions updated to the latest prescribed syllabus
4. A detailed breakdown of the most important exam questions

Further Reading: NCERT Solutions Class 9 Science

## Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 10

### Does the NCERT Solutions for Class 9 Science Chapter 10 help students to grasp the features of gravitational force?

The chapter 10 of NCERT Solutions for Class 9 Science is an important part in the syllabus. Students need to focus more on the classroom sessions and try to understand what the teacher is explaining during the class hours. The chapters must be studied unit wise and the students must clear their doubts instantly using the solutions available on CoolGyan’S. The new concepts are also explained in an interactive manner to help students grasp them without any difficulty.

### How to solve the problems based on gravitation quickly in the Chapter 10 of NCERT Solutions for Class 9 Science?

Regular practice is the main key to remember the concepts efficiently. Students are advised to solve the problems present in the textbook and understand the method of answering them. If they possess any doubts regarding the problems, they can refer to the NCERT Solutions for Class 9 Science Chapter 10 from CoolGyan’S. The problems are solved in the most systematic way by keeping in mind the marks weightage allotted for each step in the CBSE exam.

### Why should I use the NCERT Solutions for Class 9 Science Chapter 10 PDF from CoolGyan’S?

Science is one of the important subjects for Class 9 students as most of the concepts are continued in higher levels of education. For this purpose, obtaining a strong foundation of the fundamental concepts is important. Students are recommended to answer the textbook questions using the solutions PDF available on CoolGyan’S to gain a grip on the important concepts. The PDF of solutions can be downloaded and referred to understand the method of answering complex questions.