# NCERT Solutions for Class 9 Science Chapter 10

## NCERT Solutions for Class 9 Science Chapter 10 - Gravitation Gravitation is an important chapter in the Class 9 Science syllabus. You have studied this chapter before but this time, the concepts will delve a little deeper. This chapter will cover the universal law, their explanations, formulas and exercises related to them. In this chapter, you will use the concepts you have studied about velocity, acceleration, force, work, energy, etc. To understand the new concepts better, you will need proper guidance from the NCERT Solutions for Class 9 Science Chapter 10 prepared by the experts at Vedantu. This chapter holds immense importance in the Science curriculum. In fact, you will need these basic concepts in future to study advanced chapters of Physics. Hence, you need to grab the concepts of Gravitation Class 9 properly and use them aptly to solve the problems. Refer to the Gravitation Class 9 solution and find out how to approach such questions to score better in the exams. Maths Students who are looking for better solutions, they can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

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Intext Exercise 1

1. State the universal law of gravitation.

Ans: Every object in the universe attracts every other object with some force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of two objects.

Let the two objects $A$ and $B$of masses $M$and $m$ lie at a distance $d$ from each other. Let the force of attraction between two objects be $F$.

(Image Will Be Updated Soon)

$F=\frac{GMm}{{{r}^{2}}}$

Where,

$G$is the universal gravitation constant which is given by:

$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

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

Ans: Let the mass of the Earth be $M$ and the mass of an object on its surface be $m$. If $R$is the radius of the Earth, then according to the universal law of gravitation, the gravitational force ($F$) that acts between the Earth and the object can be given by the relation:

$F=\frac{GMm}{{{R}^{2}}}$.

Intext Exercise 2

1. What do you mean by free fall?

Ans: Each object is drawn towards the centre of the Earth by its gravity. When any object is released from a certain height, under the impact of gravitational force, it falls to the Earth's surface. The movement of the object is said to be in free fall.

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

Ans: When any object falls freely from a certain height towards the earth's surface, its velocity changes with respect to time. This change in velocity causes acceleration. This acceleration is known as the acceleration due to gravity ($g$). The value of acceleration due to gravity is $9.8m{{s}^{-2}}$.

Intext Exercise 3

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

Ans: The difference between the mass of an object and its weight is given in the table below:

 Sr. no. Mass Weight 1. Mass can be defined as the quantity of mattercontained in the body. Weight can be defined as the force of gravity acting onthe body. 2. It is the quantity that is a measure of inertia of the body. It is the quantity that is a measure of gravity. 3. Mass is constant everywhere. The value of weight varies at different places. 4. It is a scalar quantity. Weight is a vector quantity. 5. SI unit of mass is $kg$. SI unit of weight is $N$.

2. Why is the weight of an object on the moon $\frac{1}{6}th$ its weight on the earth?

Ans: Let the mass of the Earth be ${{M}_{E}}$ and the mass of an object on the surface of earth $=m$ and the radius of earth ${{R}_{E}}$.

According to the Universal law of gravitation, weight ${{W}_{E}}$ of the object on the surface of the earth is given by,

${{W}_{E}}=\frac{G{{M}_{E}}m}{{{R}_{E}}^{2}}$

Let ${{M}_{M}}$ and ${{R}_{M}}$ be the mass and radius of the moon. Then, according to the universal law of gravitation, weight ${{W}_{M}}$ of the object on the surface of the moon is given by:

${{W}_{M}}=\frac{G{{M}_{M}}m}{{{R}_{M}}^{2}}$

So, ratio of weight of object on moon to weight on earth is

$\frac{{{W}_{M}}}{{{W}_{E}}}=\frac{{{M}_{M}}{{R}_{E}}^{2}}{{{M}_{E}}{{R}_{M}}^{2}}$

Where, ${{M}_{E}}=5.98\times {{10}^{24}}kg$

${{M}_{M}}=7.36\times {{10}^{22}}kg$

${{R}_{E}}=6.4\times {{10}^{6}}m$

${{R}_{M}}=1.74\times {{10}^{6}}m$

Substituting the values in the ratio,

$\Rightarrow \frac{{{W}_{M}}}{{{W}_{E}}}=\frac{7.36\times {{10}^{22}}\times {{\left( 6.37\times {{10}^{6}} \right)}^{2}}}{5.98\times {{10}^{24}}\times {{\left( 1.74\times {{10}^{6}} \right)}^{2}}}$

$\Rightarrow \frac{{{W}_{M}}}{{{W}_{E}}}=0.165\approx \frac{1}{6}$

Hence, the weight of an object on the moon is $\frac{1}{6}th$ of its weight on the Earth.

Intext Exercise 4

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

Ans: Pressure can be given by the formula,

$P=\frac{F}{A}$

Pressure is inversely proportional to the surface area on which the force is acting. The smaller is the surface area, the larger will be the pressure on the surface on which the force is being acted upon. In the case of a thin strap of the school bag, the contact surface area is very less. Hence, the pressure exerted on the shoulder is very high. Therefore, it becomes difficult to hold a school bag with a thin strap.

2. What do you mean by buoyancy?

Ans: The liquid exerts an upward force on any object when it is immersed in a liquid or fluid. The tendency of the liquid to exert such an upward force on the object is called buoyancy, and the upward force which is exerted on the object by the liquid is called the buoyant force.

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

Ans: If the density of an object is greater than the density of the liquid, it will sink into the liquid. This is due to the buoyant force which is acted by the object is less than the force of gravity.

On the contrary, if the density of the object is less than the density of the liquid, it floats on the liquid's surface. This is because the force that is acting on the object is greater than the force of gravity.

Intext Exercise 5

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

Ans: An upward force acts on our body when we weigh our body while standing on a weighing machine. The buoyant force is which is a upward force that is acting. Consequently, the body is pushed up slightly, resulting in the weighing machine showing less reading than the real value.

2. You have a bag of cotton and an iron bar, each indicating a mass of $100kg$ when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

Ans: Weight measured $=$ Actual weight $-$ buoyant force

Therefore, Actual weight $=$ Weight measured $+$buoyant force

As the surface area of the cotton, the bag is greater than the iron bar, more buoyant force acts on the bag than that on the iron bar. Hence, the mass of the cotton bag is more than that of the iron bar.

NCERT Exercise

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

Ans: According to the universal law of gravitation, the gravitational force ($F$) acting between two objects of mass ${{m}_{1}}$and ${{m}_{2}}$, separated by a distance ‘$r$’ is given by

$F=\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

Where ${{m}_{1}}$and ${{m}_{2}}$are the masses of two bodies and $r$is the distance between them, $G$ is the universal gravitational constant.

When the distance is reduced to half, i.e., ${r}'=\frac{r}{2}$

$\Rightarrow F=\frac{G{{m}_{1}}{{m}_{2}}}{{{\left( \frac{r}{2} \right)}^{2}}}$

$\Rightarrow F=\frac{G{{m}_{1}}{{m}_{2}}}{\frac{{{r}^{2}}}{4}}$

$\Rightarrow F=\frac{4G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

Hence, if the distance is reduced to half, then the gravitational force becomes four times that of the previous value.

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?

Ans: All the objects fall towards the ground with constant acceleration, called acceleration due to gravity (if there is no air resistance present). It is constant and independent of the mass of the object. Hence, heavy objects do not fall faster than light objects.

3. What is the magnitude of the gravitational force between the earth and a $1kg$object on its surface? (Mass of the earth is $6\times {{10}^{24}}kg$ and radius of the earth is $6.4\times {{10}^{6}}m$).

Ans: According to the Universal law of gravitation, the gravitational force exerted on an object of mass $m$is given by:

$F=\frac{GMm}{{{r}^{2}}}$

Where,

Mass of Earth, $M=6\times {{10}^{24}}kg$

Mass of object, $m=1kg$

Universal gravitational constant, $G=6.7\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

Since the object is on the surface of the Earth, $r=$radius of the Earth ($R$)

$r=R=6.4\times {{10}^{6}}m$

Gravitational force,

$F=\frac{GMm}{{{r}^{2}}}$

$\Rightarrow F=\frac{6.7\times {{10}^{-11}}\times 6\times {{10}^{24}}\times 1}{{{\left( 6.4\times {{10}^{6}} \right)}^{2}}}=9.8N$.

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

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?

Ans: According to the Universal law of gravitation, two objects attract each other and according to Newton's third law of motion, the force of attraction between two objects is the same but acts in the opposite direction. Thus, the earth attracts the moon with the same force as the moon exerts on earth but the force acts in the opposite direction.

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

Ans: The Earth and the moon experience equal gravitational forces acting towards each other.

By Newton's Second Law, $F=ma$

$\Rightarrow a=\frac{F}{m}$

For a certain force, acceleration is inversely proportional to the mass of an object.

$a\propto \frac{F}{m}$

Mass of the Earth $>>$ Mass of the moon.

Hence, the acceleration experienced by earth due to the gravitational pull of the moon is very small when compared to that experienced by the moon due to earth. That is why the Earth does not move towards the moon.

6. What happens to the force between two objects, if

a) the mass of one object is doubled?

Ans: According to the universal law of gravitation, the force of gravitation between two objects is given by: $F=\frac{GMm}{{{r}^{2}}}$

$F$is directly proportional to the product of masses of the two objects.

$F\propto Mm$

If the mass of one object is doubled, then the gravitational force will also change to double the original.

b) the distance between the objects is doubled and tripled?

Ans: $F$is inversely proportional to the square of the distance between the objects.

If the distance between the objects is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

c) the masses of both objects are doubled?

Ans: $F$is directly proportional to the product of masses of the objects.

$F\propto Mm$

If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.

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

Ans: The universal law of gravitation states that every object in the universe attracts every other object.

The force of gravitation binds us to the earth.

It is the cause for the motion of the moon around the earth and planets around the sun.

It results in the formation of tides due to the moon and the Sun. High tide occurs at the side where the moon pulls towards itself.

8. What is the acceleration of free fall?

Ans: A free-falling object is an object that is falling due to gravity without any air resistance. When it falls, there is a variation in velocity with respect to time that is associated with it.

Acceleration of free fall is denoted by $g$and its value on the surface of the earth is $9.8m{{s}^{-2}}$, which is constant for all objects (irrespective of their masses).

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

Ans: The gravitational force between the earth and an object is called the weight of that object. It is equal to the product of acceleration due to the gravity and mass of the object.

10. Amit buys a 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).

Ans: Weight of a body on the Earth is given by:

$W=mg$

Where,

$m=$Mass of the body

$g=$Acceleration due to gravity

The shape of Earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of $g$becomes greater at the poles than at the equator. Since the value of $g$ is greater at the poles than 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.

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

Ans: When a sheet of paper is crumbled into a ball, then its surface area becomes much lesser than the surface area of a plain non-crumpled sheet of paper.

Hence, the upward force exerted by air on the sheet is greater as compared to the one exerted on the ball. Hence the sheet falls slower as compared to a paper ball.

12.Gravitational force on the surface of the moon is only $\frac{1}{6}$ as strong as the gravitational force on the Earth. What is the weight in newtons of a $10kg$object on the moon and on the Earth?

Ans: It is provided that, $Weight\text{ }of\text{ }an\text{ }object\text{ }on\text{ }the\text{ }moon=\frac{1}{6}\times Weight\text{ }of\text{ }an\text{ }object\text{ }on\text{ }the\text{ }Earth$

Also,

$Weight=\,Mass\times Acceleration$

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

Therefore, the weight of a 10 kg object on the Earth $=10\times 9.8N=98N$

Weight of the same object on the moon $=\frac{1}{6}\times 9.8N=16.3N$

13. A ball is thrown vertically upwards with a velocity of $49m{{s}^{-1}}$. Calculate

a) the maximum height to which it rises.

Ans: According to the equation of motion under gravity:

${{v}^{2}}-{{u}^{2}}=2gh$

Where,

$u=$Initial velocity of the ball

$v=$Final velocity of the ball

$h=$Height achieved by the ball

$g=$Acceleration due to gravity

At maximum height, final velocity of the ball is zero, i.e., $v=0$

$u=49m{{s}^{-1}}$

During upward motion, $g=9.8m{{s}^{-2}}$

Let $h$ be the maximum height attained by the ball.

Hence,

$\Rightarrow {{\left( 0 \right)}^{2}}-{{\left( 49 \right)}^{2}}=2\times \left( -9.8 \right)\times h$

$\Rightarrow h=\frac{49\times 49}{2\times 9.8}=122.5$

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

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

$v=u+gt$

Substituting the values and solving,

$\Rightarrow 0=49+t\times \left( -9.8 \right)$

$\Rightarrow 9.8t=49$

$\Rightarrow t=\frac{49}{9.8}=5s$

However,

Time of ascent = Time of descent

Therefore, the total time taken by the ball to return is $5+5=10s$.

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

Ans: According to the equation of motion under gravity:

${{v}^{2}}-{{u}^{2}}=2gh$

Where,

$u=$Initial velocity of the stone $=0$

$v=$Final velocity of the stone

$s=$Height of the stone $=9.6m$

g = Acceleration due to gravity $=9.8m{{s}^{-2}}$

$\Rightarrow {{v}^{2}}-{{0}^{2}}=2\times 9.8\times 19.6$2

$\Rightarrow {{v}^{2}}=2\times 9.8\times 19.6={{\left( 19.6 \right)}^{2}}$

$\Rightarrow v=19.6m{{s}^{-1}}$

Hence, the velocity of the stone just before touching the ground is $19.6m{{s}^{-1}}$.

15. A stone is thrown vertically upward with an initial velocity of $40m{{s}^{-1}}$. Taking $g=10m{{s}^{-2}}$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Ans: According to the equation of motion under gravity:

${{v}^{2}}-{{u}^{2}}=2gh$

Where,

$u=$Initial velocity of the stone $=40m{{s}^{-1}}$

$v=$Final velocity of the stone$=0$

$s=$Height of the stone

g = Acceleration due to gravity $=-10m{{s}^{-2}}$

Let $h$ be the maximum height attained by the stone.

Therefore,

$\Rightarrow 0-{{\left( 40 \right)}^{2}}=2\times h\times \left( -10 \right)$

$\Rightarrow h=\frac{40\times 40}{20}=80m$

Therefore, the total distance covered by the stone during its upward and downward journey is $80+80=160m$.

The net displacement of the stone during its upward and downward

journey is $80+\left( -80 \right)=0m$.

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth $=6\times {{10}^{24}}kg$ and of the Sun $=2\times {{10}^{30}}kg$. The average distance between the two is $1.5\times {{10}^{11}}m$.

Ans: According to the Universal l law of gravitation, the force of attraction between the Earth and the Sun is given by:

Given,

${{M}_{Sun}}=$Mass of the Sun $=2\times {{10}^{30}}kg$

${{M}_{Earth}}=$Mass of the Earth $=6\times {{10}^{24}}kg$

$R=$ Average distance between the Earth and the Sun $=1.5\times {{10}^{11}}m$

$G=$Universal gravitational constant $=6.7\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$.

$F=\frac{G{{M}_{Sun}}{{M}_{Earth}}}{{{R}^{2}}}$

$\Rightarrow F=\frac{6.7\times {{10}^{-11}}\times 2\times {{10}^{30}}\times 6\times {{10}^{24}}}{{{\left( 1.5\times {{10}^{11}} \right)}^{2}}}$

$\Rightarrow F=3.57\times {{10}^{22}}N$

Hence, the force of gravitation between the earth and the sun is $3.57\times {{10}^{22}}N$.

17. A stone is allowed to fall from the top of a tower $100m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25m{{s}^{-1}}$. Calculate when and where the two stones will meet.

Ans: Let the two stones meet after time $t$from the start.

a) For the stone dropped from the tower:

Initial velocity, $u=0$.

Let the displacement of the stone in time t from the top of the tower be s.

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

From the equation of motion,

$s=ut+\frac{1}{2}g{{t}^{2}}$

$\Rightarrow s=0\times t+\frac{1}{2}\times 9.8\times {{t}^{2}}$

$\Rightarrow s=4.9{{t}^{2}}$……. (1)

b) For the stone thrown upwards:

Initial velocity, $u=25m{{s}^{-1}}$

Let the displacement of the stone from the ground in time $t$be ${s}'$.

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

Equation of motion,

${s}'=ut+\frac{1}{2}g{{t}^{2}}$

$\Rightarrow {s}'=25t-\frac{1}{2}\times 9.8\times {{t}^{2}}$

$\Rightarrow {s}'=25t-4.9{{t}^{2}}$…… (2)

The combined displacement of both the stones at the meeting point is equal to the height of the tower $100$m.

$s+{s}'=100$…… (3)

Substituting equation (1) and (2) in (3),

$4.9{{t}^{2}}+25t-4.9{{t}^{2}}=100$

$\Rightarrow 25t=100$

$\Rightarrow t=\frac{100}{25}=4s$

In $4s$, the falling stone has covered a distance given by equation (1) as

$s=\frac{1}{2}\times 9.8\times {{4}^{2}}=78.4m$

Therefore, the stones will meet after $4s$ at a height $\left( 100-78.4 \right)=21.6m$ from the ground.

18. A ball thrown up vertically returns to the thrower after $6s$. Find

a) the velocity with which it was thrown up,

Ans: Time of ascent is equal to the time of descent. The ball takes a total of $6s$for its upward and downward journey.

Hence, time taken for upward journey, $t=\frac{6}{2}=3s$

Final velocity of the ball at the maximum height, $v=0$

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

Equation of motion, $v=u+gt$will give,

$\Rightarrow 0=u+\left( -9.8\times 3 \right)$

$\Rightarrow u=9.8\times 3=29.4m{{s}^{-1}}$

Hence, the ball was thrown upwards with a velocity of $29.4m{{s}^{-1}}$.

b) the maximum height it reaches

Ans: Let the maximum height attained by the ball be $h$.

Initial velocity during the upward journey, $u=29.4m{{s}^{-1}}$

Final velocity, $v=0$

Acceleration due to gravity, $g=-9.8m{{s}^{-2}}$

$s=ut+\frac{1}{2}a{{t}^{2}}$

From the equation of motion,

$h=29.4\times 3+\frac{1}{2}\times \left( -9.8 \right)\times {{\left( 3 \right)}^{2}}=44.1m$

c) its position after $4s$.

Ans: Ball attains the maximum height after $3s$. After attaining this height, it will start falling downwards.

In this case, Initial velocity, $u=0$

Position of the ball after $4s$ of the throw is given by the distance

travelled by it during its downward journey in $4s-3s=1s$

Equation of motion,

$s=ut+\frac{1}{2}a{{t}^{2}}$

will give,

$s=0\times t+\frac{1}{2}\times 9.8\times {{1}^{2}}=4.9m$

Total height $=44.1m$

This means that the ball is $44.1m-4.9m=39.2m$ above the ground

after $4$seconds.

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

Ans: An object immersed in a liquid is acted upon by the buoyant force in the vertically upward direction.

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

Ans: The number of forces acting on a certain item in water are two. The first one is the gravitational force pulling down the object, and the other is the buoyant force pushing up the object. If the buoyant force acting in the upward direction is higher than the gravitational force that is acting downward, then the object goes up to the water's surface as quickly as it is released into water. That is why a block of plastic released under the water comes up to the surface of the water.

21. The volume of $50g$of a substance is $20c{{m}^{3}}$ . If the density of water is $1gc{{m}^{-3}}$, will the substance float or sink?

Ans: If the density of an object is more than the density of a liquid, then it sinks in the liquid. If the density of an object is less than the density of a liquid, then it floats

$Density\text{ }of\text{ }the\text{ }substance=\frac{Mass\text{ }of\text{ }the\text{ }substance}{Volume\text{ }of\text{ }the\text{ }substance}$

$\Rightarrow Density\text{ }of\text{ }the\text{ }substance=\frac{50}{20}$

$\Rightarrow Density\text{ }of\text{ }the\text{ }substance=2.5gc{{m}^{-3}}$.

The density of the substance $>$ The density of water $\left( 1gc{{m}^{-3}} \right)$.

Hence, the substance will sink in water.

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

Ans: If the density of an object is greater than the density of a liquid, then the object will sink in the liquid. If the density of an object is less than the density of a liquid, then it will float on the surface of the liquid.

$Density\text{ }of\text{ }the\text{ }500\text{ }g\text{ }sealed\text{ }packet=\frac{Mass\text{ }of\text{ }the\text{ }packet}{Volume\text{ }of\text{ }the\text{ }packet}$

$\Rightarrow Density\text{ }of\text{ }the\text{ }500\text{ }g\text{ }sealed\text{ }packet=\frac{500}{350}$

$\Rightarrow Density\text{ }of\text{ }the\text{ }500\text{ }g\text{ }sealed\text{ }packet=1.428gc{{m}^{-3}}$

The density of the substance is more than the density of water $\left( 1gc{{m}^{-3}} \right)$.

Hence, the object will sink in water.

Clearly, the mass of water displaced by the packet can be considered equal to the volume of the packet$=0.350g$.

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

You can opt for Chapter 10 - Gravitation NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Science

### Gravitation: Class 9 Science Chapter 10 Summary

The chapter will focus on new terms and definitions related to gravitation. It will discuss why the objects fall on the ground and not float away from the surface of the earth. The unit will discuss what gravitational force means and how it can be defined using the specific terms. You will also learn the factors deciding the value of the gravitational force between two objects in space. Considering the universal law of gravitation, a formula will be prepared. It will also tell you how to identify the unit of gravitational force and how to identify it. If you face any difficulty regarding understanding these new concepts, you can refer to the NCERT Solution of 9th Class Science Chapter 10 Gravitation.

The chapter will proceed to the next units where it will teach you the prime characteristics of gravitational force and its importance. The prime concept of free fall will also become clear. You will also study the changes in the velocity of a freely falling body. This will also be discussed with the proper explanation of gravitational equations. Chapter 10 Science Class 9 will also define the gravitational constant and gravitational acceleration. Eventually, you will proceed to understand how the equations of motion can be related to this chapter. With the help of these equations, you can calculate the velocity of a freely falling object easily.

Eventually, you will reach the exercise of the respective units and can test your level of understanding. For this, refer to the NCERT Science Book Class 9 Solutions so that you can develop the skill to approach all kinds of questions without any hassle. Your preparation time will reduce and you can make significant development in your study schedule.

### Why Should You Use NCERT Solutions For Class 9 Science Chapter Gravitation?

The main reason for using NCERT Solution for Class 9 Science Chapter 10 is to focus on the exercise problems and solve them appropriately. Even if the answers are correct, the marks are allotted for other parameters too. This is why the experts always follow the latest CBSE format to frame the solutions so that students can follow it and make a good impression on the examiner. Hence, when you are done with the chapter, you can proceed to the exercise and use the solution as a reference to follow the CBSE answering rules perfectly.

Learning new concepts can be intimidating when you do not have the right platform or study material. There is no need to wait for your doubts to be cleared when you have the ideal solution for this chapter. Download the Gravitation Class 9 numerical problems with solutions in PDF format on your computer, use them to solve the problems and practice. This will help you grab the concepts well and use them to solve problems on your own later.

Q1. How can Students Understand the Features of Gravitational Force Properly?

Ans. Ch 10 Science Class 9 is an important part of the Science syllabus. Focus on the classroom sessions and concentrate on what the teachers are explaining. Study the chapter unit-wise and clear your doubts by using the Science Class 9 NCERT Solutions provided by Vedantu. You will surely understand these newfound concepts well.

Q2. How can I Solve Gravitation Problems Quickly?

Ans. You need to practice regularly using the NCERT Solutions Class 9 Science Chapter 10 as a reference and become more efficient. Your speed will automatically increase as you will be able to remember the formulas properly.

Q3. Why Students Prefer Using NCERT Solutions for Class 9 Science Ch 10?

Ans. By using the NCERT Solutions for Class 9 Science Gravitation, a student can save time in finding the right answers. They can focus better in preparing the chapter and score higher in the exams by following the ideal answering format recommended by the experts. SHARE TWEET SHARE SUBSCRIBE