assignment of gravitation class 9

Talk to our experts

1800-120-456-456

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

CBSE Class 9 Science Chapter 10 Gravitation - Free PDF Download

Gravitation is an important chapter in the Class 9 Science syllabus . You must have studied this chapter before, but the concepts will delve deeper this time. This chapter covers the universal law, its explanations, formulas, and related exercises. 

To help you understand the new concepts, we have provided NCERT Solutions for Class 9 Science Chapter 10 prepared by the experts in the downloadable pdf format. NCERT Solutions for Class 9 Science Chapter 10   are prepared after excessive research, making it one of the best solution materials available by Vedantu. 

It provides a step-by-step explanation of every exercise question in Chapter 10 of NCERT Class 9 Science textbook. The NCERT Solutions for Chapter Gravitation can also be valuable to students in their home assignments and preparation for Class 10 boards and competitive exams.

Refer to the Gravitation Class 9 solution and find out how to attempt questions as per their marking scheme to score better in the exams. You can download NCERT Solutions for Class 9 Science Chapter 10 through the link below.

Topics Covered in Class 9 NCERT Science Chapter 10 Gravitation

Universal Law of Gravitation

Importance of The Universal Law of Gravitation

Thrust And Pressure

Archimedes Principles

Relative Density

Access NCERT Solutions For Class 9 Science Chapter 10 – Gravitation

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:

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:

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,

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

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.

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:

\[m=\]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, \[Weight\text{ }of\text{ }an\text{ }object\text{ }on\text{ }the\text{ }moon=\frac{1}{6}\times Weight\text{ }of\text{ }an\text{ }object\text{ }on\text{ }the\text{ }Earth\]

\[Weight=\,Mass\times Acceleration\]

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:

\[{{v}^{2}}-{{u}^{2}}=2gh\]

\[u=\]Initial velocity of the ball

\[v=\]Final velocity of the ball

\[h=\]Height achieved by the ball

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.

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

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

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:

Substituting the values and solving,

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

\[\Rightarrow 9.8t=49\]

\[\Rightarrow t=\frac{49}{9.8}=5s\]

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.

\[u=\]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?

\[u=\]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.

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

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

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:

\[{{M}_{Sun}}=\]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.

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 \[{s}'\].

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, \[v=0\]

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, \[g=-9.8m{{s}^{-2}}\]

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

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

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 \[4s-3s=1s\]

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

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\[=0.350g\].

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 do not float away from the earth's surface. The unit will discuss gravitational force and how it can be defined using 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 have difficulty 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 unit, which 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 acceleration. Eventually, you will understand how the equations of motion can be related to this chapter. With the help of these equations, you can easily calculate the velocity of a freely falling object.

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 significantly develop your study schedule.

Key Features of Vedantu’s NCERT Solutions for Class 9 Science Chapter 10

The subject experts curated the solutions to help you deliver easy and accurate CBSE solutions.

The solution is framed in an easy-to-understand format for better understanding.

The solutions are explained using suitable examples (wherever required) for better and quick understanding.

NCERT Solutions for Class 9 Science Chapter 10 pdf is available free of cost.

Quality explanation of all topics to provide an in-depth understanding of concepts.

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 them 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.

Important Points of NCERT Class 9 Science Chapter 10

According to the universal law of gravity, everything in the universe attracts every other object through a force known as gravitational attraction. The force acting between two objects is proportional to the product of their masses and inversely proportional to the square of their distance.

Each object is drawn to the centre of the Earth by gravity. When an object is dropped from a particular height, it begins to fall to the Earth's surface due to gravitational pull. Freefall is the term for this type of item movement.

The velocity of an object falling freely from a specific height to the earth's surface changes over time. This velocity shift causes the item to accelerate, which is known as acceleration due to gravity and is represented by the letter 'g.'

Important Questions for Practice

Define centripetal force.

Describe the importance of the universal law of gravitation.

Explain the meaning of thrust and pressure.

What is Archimedes' principle?

Define relative density.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation free pdf is designed to help students practise more and more questions of the chapter and clear all their doubts.  These Class 9 Science Chapter 10 Gravitation NCERT Solutions are one of the best tools to understand the concepts, as the solutions are presented in an easy-to-read format.  Furthermore, solutions are explained using illustration wherever possible. Hence, downloading Class 9 Science Chapter 10 Gravitation free pdf is beneficial.

Chapter-Wise NCERT Solutions for Class 9 Science

Along with the NCERT Solutions for Class 9 Science Chapter 10, students can also download a free pdf of NCERT Solutions for other chapters covered in NCERT Class 9 Science textbook through the link provided below.

Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of Atom

Chapter 5 - The Fundamental Unit of Life

Chapter 6 - Tissues

Chapter 7 - Diversity in Living Organisms

Chapter 8 - Motion

Chapter 9 - Force and Laws of Motion

Chapter 11 - Work and Energy

Chapter 12 - Sound

Chapter 13 - Why do We Fall ill

Chapter 14 - Natural Resources 

Chapter 15 - Improvement in Food Resources

Other Related Link

Ncert solutions for class 9.

Revision Notes for Class 9

CBSE Class 9 Syllabus

Vedantu's NCERT solutions for Class 9 Science Chapter 10 provide an invaluable resource for students. This chapter, on Gravitation, is made much more accessible with their comprehensive explanations and problem-solving techniques. Students can benefit greatly from the step-by-step approach and the practical examples that help simplify complex concepts. Vedantu's solutions ensure a deeper understanding of the fundamental principles of gravity, planetary motion, and the universal law of gravitation. With these resources, students are better equipped to tackle exams and gain a solid foundation in science. Vedantu's commitment to quality education shines through, making it a trusted companion in the academic journey of Class 9 students.

FAQs on NCERT Solutions for Class 9 Science Chapter 10 - Gravitation

1. How can students understand the features of Gravitational Force Properly?

Chapter 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.

2. How can I solve Gravitation problems quickly?

You must practise regularly using the NCERT Solutions Class 9 Science Chapter 10 as a reference and become more efficient. Your speed will automatically increase as you can remember the formulas properly.

3. Why do students prefer using NCERT Solutions for Class 9 Science Chapter 10?

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.

4. Why does the Earth not move towards objects due to Gravitation according to Chapter 10 Gravitation of Class 9 Science? 

Newton's third law states, “Every action has its equal and opposite reaction”. It means the force applied by an object on the Earth is equal to the force applied by the earth on the object, but we know that acceleration is inversely proportional to mass. This means when the acceleration is increased, the mass is decreased, or when the mass is increased, the acceleration is decreased. As the mass of the earth is large, the acceleration due to an object is small or negligible. Therefore, it's not noticeable. And the Earth doesn’t seem to be moving.

5. What are the different applications of Archimedes' principle?

The different applications of Archimedes' principle include the following:

It is used in designing ships and submarines.

Lactometers used to determine the purity of a milk sample and hydrometers used to determine the density of a liquid are based on this principle. 

6. Why does the Earth doesn’t move towards objects due to Gravitation according to Chapter 10 Gravitation of Class 9 Science? 

As per Newton's third law, “Every action has its equal and opposite reaction”. It means the force applied by an object on the Earth is equal to the force applied by the earth on the object, but we know that acceleration is inversely proportional to mass. This means when the acceleration is increased, the mass is decreased or when the mass is increased, the acceleration is decreased. As the mass of the earth is large, the acceleration due to an object is small or negligible. Therefore, it's not noticeable. And the Earth doesn’t seem to be moving.

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Chapter 10 Gravitation Class 9 Science NCERT Solutions

Contact form.

NCERT Solutions for Class 9 Science Chapter 9 Gravitation

• 10th June 2023

NCERT Solutions for Class 9 Science Chapter 9 Gravitation provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. Students studying in class 9 can access these solutions for free in PDF format.

All these solutions are prepared by expert teachers and updated for the current academic session. NCERT Solutions for Class 9 Science Chapter 9 Gravitation help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.

CBSE Class 9 Science Gravitation Intext Questions (Solved)

PAGE NO. 102

Question 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 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. 

For two objects of masses m 1 and m 2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:

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. 

PAGE NO. 104

Question 1: What do you mean by free fall? 

Answer: Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall. 

Question 2: What do you mean by acceleration due to gravity? 

Answer: When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s 2 . 

PAGE NO. 106

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 (1/6)th its weight on the earth? 

Answer: Let M E be the mass of the Earth and m be an object on the surface of the Earth. Let R E be the radius of the Earth. According to the universal law of gravitation, weight W E of the object on the surface of the Earth is given by,

 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:

Therefore, weight of an object on the moon is of its weight on the Earth. 

PAGE NO. 109

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 immersed in it is known as buoyancy. When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water. 

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

Answer: If the density of an object is more than the density of the liquid, then it sinks in the liquid. This is because the buoyant force acting on the object is less than the force of gravity. On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity. 

PAGE NO. 110

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 you weigh your 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 bag of cotton is heavier than an iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, a more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag lighter than its actual value. For this reason, the iron bar and the bag of cotton show the same mass on the weighing machine, but actually the mass of the cotton bag is more than that of the iron bar. 

CBSE Class 9 Science Gravitation Exercise Questions and Answers

Question 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, gravitational force (F) acting between two objects is inversely proportional to the square of the distance between them, i.e., 

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 the ground with constant acceleration, called acceleration due to gravity (in the absence of air resistance). 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 × 10 24 kg and radius of the earth is 6.4 × 10 6 m). 

Answer: According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:

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: According to the universal law of gravitation, the force of gravitation between two objects is given by:

(i) F is directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled. 

(ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value. Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value. 

(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value. 

Question 7: What is the importance of universal law of gravitation? 

Answer: The universal law of gravitation proves that every object in the universe attracts every other object. 

Question 8: What is the acceleration of free fall?

Answer: When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms −2 , which is constant for all objects (irrespective of their masses). 

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

Answer: When a sheet of paper is crumbled 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 (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 =  (1/6)× Weight of an object on the Earth   Also,   Weight = Mass × Acceleration  Acceleration due to gravity, g = 9.8 m/s 2   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) × 98 = 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: (i) According to the equation of motion under gravity v 2 − u 2 = 2gs  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 m/s and u = 49 m/s  During upward motion, g = − 9.8 m s −2   Let h be the maximum height attained by the ball.   Hence, using 𝑣 2 − 𝑢 2 = 2𝑔𝑠, we have,  

(ii) Let t be the time taken by the ball to reach the height 122.5 then according to the equation of motion v = u + at, we get

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 v 2 − u 2 = 2gs  Where,  u = Initial velocity of the stone = 0 m/s  v = Final velocity of the stone  s = Height of the stone = 19.6 m  g = Acceleration due to gravity = 9.8 ms −2  

Now using v 2 − u 2 = 2gs, we get v 2 − 0 2 = 2 × 9.8 × 19.6  ⇒ v 2 = 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 ms −1 .  

Question 15: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s 2 , 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 v 2 − u 2 = 2gs  Where,  u = Initial velocity of the stone = 40 m/s  v = Final velocity of the stone = 0 m/s  s = Height of the stone   g = Acceleration due to gravity = −10 ms −2   Let h be the maximum height attained by the stone.  

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m  Net displacement 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 × 10 24 kg and of the Sun = 2 × 10 30 kg. The average distance between the two is 1.5 × 10 11 m. 

Answer: According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by

Hence, the force of gravitation between the Earth and the Sun is 3.57 × 10 22 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 the two stones meet after a time t. 

When the stone dropped from the tower  Initial velocity, u = 0 m/s  Let the displacement of the stone in time t from the top of the tower be s.   Acceleration due to gravity, g = 9.8 ms −2   From the equation of motion, 

Initial velocity, u = 25 ms −1   Let the displacement of the stone from the ground in time t be 𝑠′.   Acceleration due to gravity, g = −9.8 ms −2   Equation of motion, 

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

In 4 s, the falling stone has covered a distance given by (1) as 𝑠 = 4.9 × 4 2 = 78.4 𝑚 Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from the ground.

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 m/s  Acceleration due to gravity, g = −9.8 ms −2 Using equation of motion, v = u + at, we have  0 = u + (−9.8 × 3)  ⇒ u = 9.8 × 3 ⇒ u = 29.4 m/s  Hence, the ball was thrown upwards with a velocity of 29.4 m/s. 

(b) Let the maximum height attained by the ball be h.  Initial velocity during the upward journey, u = 29.4 m/s  Final velocity, v = 0 m/s Acceleration due to gravity, g = −9.8 ms −2   Using the equation of motion, 

Hence, the maximum height is 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 m/s 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. 

Now, total height = 44.1 m  This means 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: Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within the water. Due to this reason, a block of plastic released under water comes up to the surface of the water. 

Question 21: The volume of 50 g of a substance is 20 cm 3 . 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. 

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 cm 3 . 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

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. 

Leave a Reply Cancel Reply

Your email address will not be published. Required fields are marked *

Name  *

Email  *

Add Comment  *

Post Comment

• Class 6 Maths
• Class 6 Science
• Class 6 Social Science
• Class 6 English
• Class 7 Maths
• Class 7 Science
• Class 7 Social Science
• Class 7 English
• Class 8 Maths
• Class 8 Science
• Class 8 Social Science
• Class 8 English
• Class 9 Maths
• Class 9 Science
• Class 9 Social Science
• Class 9 English
• Class 10 Maths
• Class 10 Science
• Class 10 Social Science
• Class 10 English
• Class 11 Maths
• Class 11 Computer Science (Python)
• Class 11 English
• Class 12 Maths
• Class 12 English
• Class 12 Economics
• Class 12 Accountancy
• Class 12 Physics
• Class 12 Chemistry
• Class 12 Biology
• Class 12 Computer Science (Python)
• Class 12 Physical Education
• GST and Accounting Course
• Excel Course
• Tally Course
• Finance and CMA Data Course
• Payroll Course

Interesting

• Learn English
• Learn Excel
• Learn Tally
• Learn GST (Goods and Services Tax)
• Learn Accounting and Finance
• GST Tax Invoice Format
• Accounts Tax Practical
• Tally Ledger List
• GSTR 2A - JSON to Excel

Are you in school ? Do you love Teachoo?

We would love to talk to you! Please fill this form so that we can contact you

You are learning...

Chapter 10 Class 9 - Gravitation

Click on any of the links below to start learning from Teachoo ...

Get NCERT Solutions, Notes, Solutions to Intext Questions, Examples of Chapter 10 Class 9 Gravitation free at Teachoo. 

In this chapter, we will learn

What is Gravity ?

What is Universal Law of Gravitation

Important Natural Phenomena Occurring Due to Gravitation

What is Free Fall ?

What is Acceleration Due To Gravity

Deriving value of Acceleration due to Gravity

Different Equations of Motion for Free Falling Object

What is the Difference between Mass and Weight 

What is thrust ?

What is Pressure

What is buoyancy ?

Density And Relative Density Of An Object

Why do Objects float or sink in Water

Archimedes Principle

Different Factors Affecting Buoyancy

Why is gravity maximum at poles and minimum at equator ?

What is the SI unit of Thrust and Pressur e?

What is the SI Unit of g and G ?

What is the difference between gravity and gravitation ?

Click on a link to open the first post.

To open any other post, like NCERT Question 18, there is a list with arrows at the bottom. You can click on a link to open it.

Or you can also click Next

NCERT Questions

Questions from inside the chapter, examples from ncert book, teachoo questions.

What's in it?

Hi, it looks like you're using AdBlock :(

Please login to view more pages. it's free :), solve all your doubts with teachoo black.

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• DK Goel Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• ML Aggarwal Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Sandeep Garg Textbook Solution
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• HOTS Question
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• JEE Crash Course
• JEE Previous Year Paper
• Important Info
• JEE Mock Test
• JEE Sample Papers
• SRM-JEEE Mock Test
• VITEEE Mock Test
• BITSAT Mock Test
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• COMEDK Previous Year Paper
• GUJCET Previous Year Paper
• KCET Previous Year Paper
• KEAM Previous Year Paper
• Manipal Previous Year Paper
• MHT CET Previous Year Paper
• WBJEE Previous Year Paper
• AMU Previous Year Paper
• TS EAMCET Previous Year Paper
• SRM-JEEE Previous Year Paper
• VITEEE Previous Year Paper
• BITSAT Previous Year Paper
• UPSEE Previous Year Paper
• CGPET Previous Year Paper
• CUSAT Previous Year Paper
• AEEE Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

Gravitation Worksheet for Class 9 PDF with Answers

In Class 9 Science there is a chapter “Gravitation”, it is a crucial lesson for the students as they get to know all the basic topics of Gravitation. Since it is an important lesson, students should take Gravitation Worksheet Class 9 to better develop an understanding of the concepts explained.

It is very important for Class 9 students to practise questions of the chapter Gravitation because it will help them create their own exam strategy to score well in the upcoming final examination. 

In addition to that, Gravitation Worksheet Class 9 has quite interactive and creative tasks which boosts student’s creativity level.

Gravitation Worksheet With Solutions

The questions in the chapter Gravitation worksheet for Class 9 are provided with solutions. Through these solutions, students can easily solve all their doubts. By clearing the doubts of Gravitation, students can build a strong foundation. Accordingly students can also score good marks in questions related to the chapter Gravitation. The subject matter experts at Selfstudys has prepared Gravitation Worksheet With Solutions in a way that helps students answer all types of questions regardless of its difficulty.

Gravitation Worksheet Class 9 PDF

Class 9 students can easily download the Portable Document Format (PDF) of Gravitation worksheet with the help of Selfstudys website. This can help Class 9 students to understand all the topics and concepts of the chapter Gravitation which will help students increase their self-confidence level. A perfect level of self- confidence can easily decrease students' level of stress to prepare for the chapter Gravitation.

How to Download Gravitation Worksheet Class 9?

To look through the questions included in Gravitation worksheet Class 9, students can follow the given steps. These steps are the easiest one that one can follow to download Gravitation Worksheet Class 9. 

• Open the Selfstudys website. 
• Bring the arrow towards the CBSE which can be seen in the navigation bar. 
• A drop down menu will appear, select KVS NCERT CBSE Worksheet.

• A new page will appear, Class 9 from the list of classes.

• Click Science from the list of subjects.

• Again a new page will appear, now select Gravitation Worksheet from the list of chapters. 

Features of Gravitation Worksheet Class 9

Before solving questions from Gravitation worksheet Class 9, students should understand what makes Gravitation Worksheet Class 9 PDF special. Features of the worksheet are discussed below: 

• All Concepts are Covered: In Gravitation worksheet Class 9, all concepts and topics are covered in an elaborate manner in the questions format. Through this elaboration, students can understand all the topics of the chapter Earth in a better way.  
• Explained in an Easy Language: Answers of Gravitation Worksheet Class 9 is explained in an easy language which helps students easily understand the process of answering questions.  
• Varieties of Questions are Included: In the Class 9 Gravitation worksheet, varieties of questions are included. Through this students can solve all kinds of questions of the chapter Gravitation. 
• Eye Catching Format: Gravitation Worksheet of Class 9 is considered to be an eye catching one. This eye-catching format can attract many students to solve the questions of the chapter Gravitation.
• Solutions are Provided: For all the questions in the worksheet of Class 9 Gravitation, solutions are provided. Through the solutions, Class 9 students will be able to solve challenging questions which will help them develop a critical thinking capability.
• According to the Class 9 Syllabus: The questions in Class 9 Science Gravitation Worksheet are as per the Class 9 Syllabus and prescribed NCERT books. With the help of this, kids will be able to make their foundational understanding stronger.

How to Know If You're Ready for Gravitation Worksheet Class 9? 

First of all Class 9 students need to complete the chapter Gravitation from the main Science book. After covering all the topics, definitions and concepts from the chapter, students are totally ready to solve the questions from Gravitation Worksheet Class 9. Solving the questions from the Class 9 worksheet can help students to increase their conceptual understanding of Gravitation. 

What Is Gravitation Worksheet Class 9 and How to Use It?

Gravitation worksheet Class 9 is mainly given to students to practise a variety of questions. After practising questions from Gravitation worksheet, students can also look through the answers. Answers for all questions can help students to improve their practising skills. These skills can help Class 9 students to increase their level of understanding. 

Parents are advised to tell their Class 9rd kids to begin solving Gravitation Worksheet Class 9 the moment they finish their study of the chapter. Doing this, will help students brush up their all learning as well as be completed for upcoming annual exams or tests.

Advantages of Gravitation Worksheet Class 9

Solving questions from Gravitation worksheet Class 9, students can be benefited a lot. Those important advantages are: 

• Boosts Confidence Level: To solve questions from the Class 9 Gravitation worksheet can help students to boost their confidence. A perfect level of confidence can help students to improve the study process. 
• Assist the Preparation Process: Solving questions from the Class 9 Gravitation worksheet can help students in preparing for the chapter. 
• Helps in Self Evaluation: Through Gravitation worksheet, students can easily evaluate themselves. According to the self evaluation process, students can easily improve their preparation. 
• Builds a Strong Foundation: It is important for all students to solve questions from the chapter Gravitation. Regular solving questions from the worksheet can help students to build a strong foundation for the chapter Gravitation.  
• Enhances the Learning Process: Constant solving of questions from Gravitation Worksheet can enhance a student's learning process so that students can understand all topics easily.  
• Quick Revision: By solving questions from the Class 9 Gravitation worksheet, students can easily revise all the topics and concepts included in the chapter. 

Why Gravitation Worksheet Class 9 Is Right for You?

It is a must for students to exercise questions from Gravitation worksheet Class 9 as perfectly right for them. With the help of Gravitation worksheet Class 9 questions, students can increase their capability of solving questions in a different and creative manner.

Tips to Understand All Questions of Gravitation Worksheet Class 9 in a Better Way: 

Students should understand all questions of Gravitation worksheet Class 9 in a better way. Better understanding of questions can help students to score good marks in questions which are related to the chapter Gravitation. Those important tips are: 

• Finish Off The Chapter: First and foremost tip is to finish off the chapter Gravitation. Students need to complete each and every topic included in the lesson. 
• Practise Questions: After completing the chapter Gravitation, students need to practise questions from the Class 9 worksheet. Routine practice of questions can help students in identifying their strengths and weaknesses. 
• Note Down the Mistakes: While practising questions, it is very important that students note down their mistakes. Noting down the mistakes is very important as accordingly students can improve their preparation strategies. 
•  Correction of Mistakes: After noting down the mistakes, it is a must to correct all the mistakes made. Correction of mistakes can help students to solve worksheet questions in a better way. 
• Maintain a Positive Attitude: While solving questions from the Class 9 Gravitation worksheet, students need to maintain a positive attitude. A positive attitude can help students to remove stress and anxiety while preparing for the chapter Gravitation. 
• Remain Focused: To understand questions of Gravitation worksheet, students need to remain focused while preparing for it. 

Why Should Students Start Solving Gravitation Worksheet Class 9 From the PDF?

Gravitation Class 9 Worksheet is provided in the PDF so that students don’t need to search for them here and there. By solving the questions from Gravitation worksheet Class 9, students can understand the chapter in a fine way. Routine solving of questions from the Class 9 Science Worksheet can help students increase their comprehension skills. Comprehension skills will help in performing outstanding in the final examination. 

What are Included in Gravitation Worksheet Class 9?

In Gravitation worksheet Class 9, questions from the chapter are included. After solving questions of the chapter Gravitation, students can also go through the answers included in the worksheet. Answers to these questions of Gravitation are explained in a detailed manner. As it can also help teachers to make students understand in a better and elaborate way. Through this students can easily identify their skills and flaws for the Science chapter Gravitation.

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences

One Last Step...

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.

Gravitation

Class 9 - ncert science solutions, intext questions 1.

State the universal law of gravitation.

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 them. The force is along the line joining the centres of two objects.

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

F = G Mm d 2 \dfrac{\text{Mm}}{\text{d}^2} d 2 Mm ​

F = force of attraction

G = constant of proportionality

M = mass of earth

m = mass of the object

d = distance between the earth's centre and object's centre

Intext Questions 2

What do you mean by free fall?

Earth attracts objects towards itself due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

What do you mean by acceleration due to gravity?

Whenever an object falls towards the earth, an acceleration is involved due to the earth's gravitational force. So, this acceleration is called the acceleration due to the gravitational force of the earth (or acceleration due to gravity). It is denoted by g and its value is 9.8 ms -2 .

Intext Questions 3

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

Differences between the mass of an object and its weight:

Why is the weight of an object on the moon 1 6 \dfrac{1}{6} 6 1 ​ th its weight on the earth?

The moon's gravitation force is determined by the mass and the size of the moon.

The mass of moon is 1 100 \dfrac{1}{100} 100 1 ​ times and its radius 1 4 \dfrac{1}{4} 4 1 ​ 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 1 6 \dfrac{1}{6} 6 1 ​ th its weight on the earth.

Intext Questions 4

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

Pressure is inversely proportional to the surface area on which the force acts. Incase of a strap made of a thin and strong string, the surface area of the strap in contact with the shoulder is less, hence, pressure on the shoulders is more which makes it difficult to hold the school bag.

What do you mean by buoyancy?

The property of liquid to exert an upward force on a body immersed in it, is called buoyancy.

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

An object floats or sinks in water because of the difference in densities of the substance and water. When the density of the substance is less than the density of water, then the upthrust of water is greater than the weight of the substance. Hence it floats.

When the density of substance is more than water, then the upthrust of water on the substance is less than the weight of the substance. Hence it sinks.

Intext Questions 5

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

When we stand on a weighing machine, the weight acts downwards while the upthrust due to air acts upwards.

As, true weight = (apparent weight + up thrust)

So our apparent weight becomes less than the true weight. This apparent weight is measured by the weighing machine and hence, the mass indicated is less than the actual mass. So our actual mass will be more than 42 kg.

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?

The cotton bag is heavier than an iron bar as it experiences a larger air thrust than the iron bar.

The cotton bag's density is less than that of the iron bar, so, the volume of the cotton bag is more compared to the iron bar. Hence, the cotton bag experience more upthrust due to the presence of air. So, in the presence of air, the cotton bag's true weight is more compared to the true weight of the iron bar.

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

F = G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​

m 1 = mass of first object

m 2 = mass of second object

d = distance between the two objects.

When distance is reduced to half then, d' = d 2 \dfrac{\text{d}}{2} 2 d ​

So, substituting we get,

F' = G m 1 m 2 ( d 2 ) 2 \dfrac{\text{m}_1\text{m}_2}{\Big(\dfrac{\text{d}}{2}\Big)^2} ( 2 d ​ ) 2 m 1 ​ m 2 ​ ​ = G m 1 m 2 d 2 4 \dfrac{\text{m}_1\text{m}_2}{\dfrac{\text{d}^2}{4}} 4 d 2 ​ m 1 ​ m 2 ​ ​ = 4G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​ = 4F

Hence, when the distance between the objects is reduced to half, the gravitational force increases four times.

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

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). This acceleration remains constant and is independent of an object's mass. Hence, heavy objects do not fall faster than light objects.

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 × 10 24 kg and radius of the earth is 6.4 × 10 6 m.)

Mass of the body (m) = 1 kg

Mass of the earth (M) = 6 × 10 24 kg

Radius of the Earth (R) = 6.4 × 10 6 m

G = 6.67 x 10 -11 Nm 2 kg -2

According to the formula for gravitational force we know,

F = G Mm R 2 \dfrac{\text{M}\text{m}}{\text{R}^2} R 2 M m ​

Substituting we get,

F = 6.67 × 1 0 − 11 × 6 × 1 0 24 × 1 ( 6.4 × 1 0 6 ) 2 = 6.67 × 1 0 13 6.4 × 6.4 × 1 0 12 = 9.77  N ≈ 9.8  N \text{F} = \dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} \\[1em] = \dfrac{6.67 \times 10^{13}}{6.4 \times 6.4 \times10^{12}} \\[1em] = 9.77 \text{ N} \approx 9.8 \text{ N} F = ( 6.4 × 1 0 6 ) 2 6.67 × 1 0 − 11 × 6 × 1 0 24 × 1 ​ = 6.4 × 6.4 × 1 0 12 6.67 × 1 0 13 ​ = 9.77  N ≈ 9.8  N

Hence, magnitude of the gravitational force = 9.8 N

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?

According to universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions.

This force is given by: F = G Mm R 2 \dfrac{\text{M}\text{m}}{\text{R}^2} R 2 M m ​

Hence, the earth attracts the moon with the same force with which the moon attracts the earth.

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

According to the universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions. Hence, moon and earth both attract each other with the same force. Therefore, the earth does not move towards the moon.

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?

(i) We know,

When mass of one object is doubled.

F' = G 2m 1 m 2 d 2 \dfrac{\text{2m}_1\text{m}_2}{\text{d}^2} d 2 2m 1 ​ m 2 ​ ​ = 2G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​ = 2F

Hence, when the mass of one object is doubled, the gravitational force increases two times.

(ii) When distance is doubled then, d' = 2r

F' = G m 1 m 2 ( 2 d ) 2 \dfrac{\text{m}_1\text{m}_2}{(2d)^2} ( 2 d ) 2 m 1 ​ m 2 ​ ​ = G m 1 m 2 4 d 2 \dfrac{\text{m}_1\text{m}_2}{4d^2} 4 d 2 m 1 ​ m 2 ​ ​ = F 4 \dfrac{\text{F}}{4} 4 F ​

Hence, when the distance between the objects is doubled, the gravitational force becomes one fourth of its original force.

Now, if it's tripled

F' = G m 1 m 2 ( 3 d ) 2 \dfrac{\text{m}_1\text{m}_2}{(3d)^2} ( 3 d ) 2 m 1 ​ m 2 ​ ​ =G m 1 m 2 9 d 2 \dfrac{\text{m}_1\text{m}_2}{9d^2} 9 d 2 m 1 ​ m 2 ​ ​ = F 9 \dfrac{\text{F}}{9} 9 F ​

Hence, when the distance between the objects is tripled, the gravitational force becomes one ninth of its original force.

(iii) If masses of both the objects are doubled, then

m 1 ' = 2m 1

m 2 ' = 2m 2

F' = G 2m 1 2m 2 d 2 \dfrac{\text{2m}_1\text{2m}_2}{\text{d}^2} d 2 2m 1 ​ 2m 2 ​ ​ = 4G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​ = 4F

Hence, when the masses of both the objects are doubled, the gravitational force becomes four times the original force.

What is the importance of universal law of gravitation?

The universal law of gravitation successfully explained several phenomena that were believed to be unconnected:

• The force that binds us to the earth
• The motion of the moon around the earth
• The motion of planets around the sun
• The tides due to the moon and the sun.

What is the acceleration of free fall?

The acceleration of free fall is denoted by g and its value on the surface of the earth is 9.8 ms -2 .

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

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.]

The weight of a body on the earth's surface (W) = mg (where m = mass of the body and g = acceleration due to gravity)

Given, value of g is larger at poles when compared to the equator. So, for the same mass of gold, weight will be less at the equator as compared to the poles. Therefore, Amit's friend will not agree with the weight of gold bought.

Question 11

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

A sheet of paper has a larger surface area in comparison to a crumpled paper ball. The larger surface area will suffer greater air resistance. Hence, a sheet of paper falls slower than the crumpled ball.

Question 12

Gravitational force on the surface of the moon is only 1 6 \dfrac{1}{6} 6 1 ​ 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?

Acceleration due to earth's gravity g = 9.8 ms -2

Object's mass, m = 10 kg

Acceleration due to moon gravity = g m

Weight on the earth = W e

Weight on the moon = W m

Weight = mass x gravity

g m = 1 6 \dfrac{1}{6} 6 1 ​ g e (given)

So, W m = m g m = m x 1 6 \dfrac{1}{6} 6 1 ​ x g e

W m = 10 x 1 6 \dfrac{1}{6} 6 1 ​ x 9.8 = 16.34 N

W e = m x g e = 10 x 9.8

Hence, weight on moon and earth = 16.34 and 98 N respectively.

Question 13

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

(i) The maximum height to which it rises,

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

Initial velocity (u) = 49 ms -1

Final velocity v at maximum height = 0

Acceleration due to earth gravity g = -9.8 ms -2 (negative as ball is thrown up).

Let H be the maximum height to which the ball rises.

By third equation of motion,

2gH = v 2 - u 2

2 × (-9.8) × H = 0 - (49) 2

-19.6 H = -2401

H = − 2401 − 19.6 \dfrac{-2401}{-19.6} − 19.6 − 2401 ​ = 122.5 m

Hence, the maximum height to which it rises = 122.5 m

(ii) Total time T = Time to ascend (T a ) + Time to descend (T d )

According to the first equation of motion,

0 = 49 + (-9.8) x T a

T a = 49 9.8 \dfrac{49}{9.8} 9.8 49 ​ = 5 s

Also, T d = 5 s

∴ T = T a + T d = 5 + 5 = 10 s

Hence, total time it takes to return to the surface of the earth = 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.

Initial velocity (u) = 0

Tower height = total distance = 19.6 m

g = 9.8 ms -2

Final velocity (v) = ?

According to the third equation of motion,

v 2 = 2gs + u 2

v 2 = (2 x 9.8 x 19.6) + 0

v = 384.16 \sqrt{384.16} 384.16 ​

v = 19.6 ms -1

Hence, final velocity = 19.6 ms -1

Question 15

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

Initial velocity (u) = 40 ms -1

Final velocity (v) = 0

g = 10 ms -2

Max height = ?

v 2 = u 2 - 2gs

Note: [negative g as the object goes up]

0 = (40) 2 - 2 x 10 x s

⇒ 1600 = 20s

⇒ s = 1600 20 \dfrac{1600}{20} 20 1600 ​ = 80 m

Total Distance = s a + s d = 80 + 80 = 160 m

where, s a and s d are distance going up and coming down, respectively.

Displacement = 0 [as the stone comes to the same point from where it was thrown ].

Hence, total distance covered = 160 m and displacement = 0, as the first point is the same as the last point

Question 16

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10 24 kg and of the Sun = 2 × 10 30 kg. The average distance between the two is 1.5 × 10 11 m.

Mass of the sun (m) = 2 × 10 30 kg

Average distance R = 1.5 × 10 11 m

F = 6.67 × 1 0 − 11 × 6 × 1 0 24 × 2 × 1 0 30 ( 1.5 × 1 0 11 ) 2 \dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 2 \times 10^{30} }{(1.5 \times 10^{11})^2} ( 1.5 × 1 0 11 ) 2 6.67 × 1 0 − 11 × 6 × 1 0 24 × 2 × 1 0 30 ​ = 3.56 x 10 22 N

Hence, magnitude of the gravitational force = 3.56 x 10 22 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 ms -1 . Calculate when and where the two stones will meet.

(a) In the case, when the stone falls from the top of the tower,

Initial velocity u = 0

Distance travelled = x

Time taken = t

According to the second equation of motion,

S = ut + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

⇒ S = 0 x t + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

⇒ S = 1 2 \dfrac{1}{2} 2 1 ​ gt 2    [Eq 1]

(b) When another stone is projected vertically upwards,

Initial velocity u = 25 ms -1

Distance travelled = (100 - x)

Using S = ut + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

S' = 25t - 1 2 \dfrac{1}{2} 2 1 ​ gt 2    [Eq 2]

From equations (1) and (2)

S + S' = 1 2 \dfrac{1}{2} 2 1 ​ gt 2 + 25t - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

100 = 1 2 \dfrac{1}{2} 2 1 ​ gt 2 + 25t - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

t = 100 25 \dfrac{100}{25} 25 100 ​ = 4 sec.

Hence, after 4 secs, the two stones will meet.

S = 1 2 \dfrac{1}{2} 2 1 ​ x 10 x 4 2 = 5 x 16 = 80m.

Hence, after 4 sec, 2 stones meet a distance of 80 m from the top.

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 4s.

Ball returns to thrower in 6s. So, time up + time down = 6 s hence, time up = 3 s.

(a) Final velocity (v) = 0

From first equation of motion

v = u – gt up

⇒ u = v + gt up

⇒ u = 0 + (9.8 x 3)

⇒ u = 29.4 ms -1

Hence, the velocity with which it was thrown up = 29.4 ms -1 .

(b) According to the second equation of motion,

S = ut - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

S = (29.4 x 3) - ( 1 2 \dfrac{1}{2} 2 1 ​ x 9.8 x 3 x 3) = 88.2 - 44.1 = 44.1 m

Hence, maximum height stone reaches is 44.1 m

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

Distance travelled in another 1 sec = s'

g is +ve as ball is going down

S = 0 + ( 1 2 \dfrac{1}{2} 2 1 ​ x 9.8 x 1 x 1) = 4.9 m

Hence, in 4 sec the ball will be 4.9 m from the top or 39.2 m (i.e., 44.1 - 4.9) from the bottom.

Question 19

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

The buoyant force acts vertically upwards on the object that is immersed in a liquid.

Question 20

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

When plastic is immersed in water, two forces act on it:

• Gravitational force or its weight in vertically downward direction
• Buoyant force in vertically upward direction.

As the density of plastic is less than that of water, the buoyant force is greater than its weight. Hence, the plastic comes up to the surface.

Question 21

The volume of 50 g of a substance is 20 cm 3 . If the density of water is 1 gcm -3 , will the substance float or sink?

Volume of substance = 20 cm 3

Mass of substance = 50 g

Density of water = 1 gcm -3

Density of the substance = ?

Density = Mass Volume \dfrac{\text{Mass}}{\text{Volume}} Volume Mass ​ = 50 20 \dfrac{50}{20} 20 50 ​ = 2.5 gcm -3

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

Question 22

The volume of a 500 g 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 mass of the water displaced by this packet?

Volume of sealed packet = 350 cm 3

Mass of sealed packet = 500 g

Density of the sealed packet = ?

Density = Mass Volume \dfrac{\text{Mass}}{\text{Volume}} Volume Mass ​ = 500 350 \dfrac{500}{350} 350 500 ​ = 1.42 gcm -3

As density of the packet is greater than density of water. Hence, the sealed packet will sink.

According to Archimedes Principle,

Displaced water volume = Volume of packet

Volume of water displaced = 350 cm 3

Mass of water displaced = ρ x V = 1 × 350 = 350 g

Hence, mass of displaced water = 350 g

• CBSE Class 9 Science Chapter 10 – Gravitation Class 9 Notes

Gravitation Class 9 Revision Notes

The chapter of Gravitation in Class 9 Science is an essential chapter which deals in everything related to gravity. Why do you come down when you jump or why does the fruit fall on the ground? This chapter will help you solve these queries with ease. It explores the concept of gravitation in an easy manner. Students are already familiar with the motion of objects and force, the gravitation class 9 notes just further explain the gravitational force.

Other than that, it discusses the motion of objects under the influence of gravitational force on our earth. Similarly, it will help in understanding how the weight of anybody might vary from one place to another. Most importantly, you learn how objects float in liquids. Thus, the Gravitation class 9 notes will come in handy in doing so.

Download Toppr app for Android and iOS or signup for free.

Sub-topics covered under Gravitation –

• Newton’s Universal Law of Gravitation – Students will learn about the Universal Law of Gravitation that Newton gave.
• Thrust, Pressure, and Buoyancy – This sub-topic elucidates about thrust, pressure and buoyancy in detail.
• Acceleration Due to Gravity – You often see acceleration is caused due to gravity, this unit explains how it occurs.
• Earth Satellites – It explains satellites orbiting the earth in detail over here.
• Escape Velocity – What happens when something escapes out of gravitational velocity, this unit will help you understand it.
• Gravitational Potential Energy – This explains the complex topic of gravitational potential energy easily with this section.
• Kepler’s Law – The three different Kepler’s laws will be discussed through this sub-topic.
• Weightlessness – What causes weightlessness and what is zero gravity? You will find these answers here.

You can download CBSE Class 9 Science Chapter 10 Revision Notes by clicking on the download button below

Customize your course in 30 seconds

Which class are you in.

CBSE Class 9 Science Revision Notes

• CBSE Class 9 Science Chapter 11 – Work and Energy Class 9 Notes
• CBSE Class 9 Science Chapter 9 – Force and Laws of Motion Class 9 Notes
• CBSE Class 9 Science Chapter 6 – Tissues Class 9 Notes
• CBSE Class 9 Science Chapter 7 Diversity in Living Organisms Class 9 Notes
• CBSE Class 9 Science Chapter 8 Motion Class 9 Notes
• CBSE Class 9 Science Chapter 12 – Sound Class 9 Notes
• CBSE Class 9 Science Chapter 13 – Why Do We Fall ill Class 9 Notes
• CBSE Class 9 Science Chapter 15 – Improvement in Food Resources Class 9 Notes
• CBSE Class 9 Science Chapter 14 – Natural Resources Class 9 Notes

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

NCERT Solutions for Class 9 Science (physics) Chapter 10 Gravitation are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 10 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 10 Textbook Questions and Answers

INTEXT QUESTION

PAGE NO 134

Question 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 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. 

For two objects of masses m 1 and m 2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:

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. If 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:

PAGE NO 136

Question 1: What do you mean by free fall? 

Answer: 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.

Question 2: What do you mean by acceleration due to gravity? 

Answer: When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s 2 . 

PAGE NO 138

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

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

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 immersed in it is known as buoyancy. When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water. 

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

Answer: An object float or sink 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.

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 you weigh your 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 bag of cotton is heavier than iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag lighter than its actual value. For this reason, the iron bar and the bag of cotton show the same mass on the weighing machine, but actually the mass of the cotton bag is more than that of the iron bar. 

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

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 × 10 24 kg and radius of the earth is 6.4 × 10 6 m). 

Answer:  Given, Mass of Earth, M = 6 × 10 24 kg  Mass of object, m = 1 kg  Universal gravitational constant, G = 6.7 × 10 −11 Nm 2 kg −2   Radius of the Earth, R = 6.4 × 10 6 m 

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: According to the universal law of gravitation, the force of gravitation between two objects is given by:

Question 7: What is the importance of universal law of gravitation? 

Answer: 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

Question 8: What is the acceleration of free fall?

Answer: When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms −2 , which is constant for all objects (irrespective of their masses). 

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

Answer: Surface area of a Sheet which is crumpled into a ball, is much smaller than the surface area of a plain or flat sheet. Therefore, despite both experience same force of gravity, the plain or flat sheet of paper will have to face more air resistance than the crumpled ball, so it will fall slower than the sheet crumpled into a ball.

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: (i) According to the equation of motion under gravity v 2 − u 2 = 2gs  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 m/s and u = 49 m/s

During upward motion, g = − 9.8 m s −2   Max. Height attained by the ball (s) = ?

∴ Max. Height attained by the ball (s) = 122.5 m

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

Time for upward journey of the ball will be the same as time for downward journey i.e., t = 5 s.

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 v 2 − u 2 = 2gs  Where,  u = Initial velocity of the stone = 0 m/s  v = Final velocity of the stone  s = Height of the stone = 19.6 m  g = Acceleration due to gravity = 9.8 ms −2  

Now, v 2 = u 2 + 2as ⇒ v 2 − u 2 = 2as ⇒ v 2 − 0 2 = 2 × 9.8 × 19.6  ⇒ v 2 = 2 × 9.8 × 19.6 ⇒ v 2 = 19.6 × 19.6 ⇒ v 2 = (19.6) 2 ⇒ v = 19.6 ms −1  

Hence, the velocity of the stone just before touching the ground is 19.6 ms −1 .  

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

Answer:  Given u = Initial velocity of the stone = 40 m/s  v = Final velocity of the stone = 0 m/s  s = Height of the stone   g = Acceleration due to gravity = −10 ms −2   the maximum height attained by the stone (s) = ? 

According to the equation of motion under gravity, v 2 − u 2 = 2gs 

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

Net displacement 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 × 10 24 kg and of the Sun = 2 × 10 30 kg. The average distance between the two is 1.5 × 10 11 m. 

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 the two stones meet after a time t. 

When the stone dropped from the tower  Initial velocity, u = 0 m/s  Let the displacement of the stone in time t from the top of the tower be s.   Acceleration due to gravity, g = 9.8 ms −2  

When the stone thrown upwards  Initial velocity, u = 25 ms −1   Let the displacement of the stone from the ground in time t be 𝑠′.   Acceleration due to gravity, g = −9.8 ms −2  

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

In 4 s, the falling stone has covered a distance given by (1) as 𝑠 = 4.9 × 4 2 = 78.4 𝑚

Therefore, the stones will meet after 4s at a height (100 – 78.4) = 20.6 m from the ground.

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 4s.

Answer: (a) The velocity with which ball was thrown up : Acceleration due to gravity, g = – 9.8 ms –2   As the total time taken in upward and return journey by the ball is 6 s. Therefore, The upward journey, t = 6/2 s = 3 s Final velocity, v = 0 ms –1 Initial velocity, u = ?

Using equation of motion, v = u + at, we have 

0 = u + (−9.8 × 3)  ⇒ u = 9.8 × 3 ⇒ u = 29.4 m/s 

Hence, the ball was thrown upwards with a velocity of 29.4 m/s. 

(b) Let the maximum height attained by the ball be s.  Initial velocity during the upward journey, u = 29.4 m/s  Final velocity, v = 0 m/s Acceleration due to gravity, g = −9.8 ms −2  

Hence, the maximum height is 44.1 m.

(c) Ball attains the maximum height after 3s. After attaining this height, it will start falling downwards.  

In this case, 

Initial velocity, u = 0 m/s

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. 

Now, total height = 44.1 m 

This means, 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: Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within water. Due to this reason, a block of plastic released under water comes up to the surface of the water. 

Question 21: The volume of 50 g of a substance is 20 cm 3 . 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. 

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 cm 3 . 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

The density of the substance is more than the density of water (1𝑔/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., 350g. 

Class 9 Science NCERT Solutions Chapter 10 Gravitation

CBSE Class 9 Science NCERT Solutions Chapter 10 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 10 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 10. The list gives you a quick look at the different topics and subtopics of this chapter.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

• Last modified on: 3 years ago
• Reading Time: 28 Minutes

Here we have given NCERT Solutions for Class 9 Science Chapter 10 Gravitation.

INTEXT Questions

Question 1. State the universal law of gravitation. Solution: According to universal law of gravitation, every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of distance between them. The direction of force is along the line joining the two particles.

Question 3. What do you mean by free fall? Solution: All objects falling towards earth under the action of gravitational force of earth alone are said to be in free fall.

Question 4. What do you mean by acceleration due to gravity? Solution: The acceleration with which an object fall freely towards the earth is known as acceleration due to gravity. It is denoted by g and its value is 9.8 m s -2 .

Question 5. What are the differences between the mass of an object and its weight? Solution:

Question 8. What do you mean by buoyancy? Solution: Whenever an object is immersed in a liquid, either partially or fully, an upward force is exerted on the object by the liquid. This upward force is called upthrust or buoyant force or force of buoyancy and the property of the fluid due to which this upthrust is exerted on the object is called buoyancy.

Question 9. Why does an object float or sink when placed on the surface of water? Solution: When the object has density less than 1 g cm -3  (density of water), then it floats on the surface of water, because, it always displaces more weight of water than its own weight. As buoyant force is more than its own weight, therefore, it floats. When the object has density more than 1 g cm-3, then it sinks in water, because it always displaces less weight of water than its own weight. As buoyant force is less than its own weight, therefore, it sinks.

Question 10. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg? Solution: A weighing machine is a sort of spring balance which measure the weight and not the mass of a body. When we stand on the weighing machine, our weight which is due to gravitational attraction of the earth acts vertically downwards. But the buoyancy due to air on our body acts vertically upwards. As a result of this, our apparent weight true weight – buoyant force is less than the true weight. Since the weighing machine measures the apparent weight, our true weight is more, i.e., more than 42 kg.

Question 11. 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: We know that true weight = apparent weight + upthrust. The cotton bag is heavier than the iron bar. This is due to the reason, that the bag of cotton which has more volume (as it has less density) than the iron bar (which has more density), experiences more upthrust due to air.

NCERT Exercises

Question 5. If the moon attracts the earth, why does the earth not move towards the, moon? Solution: Both the earth and the moon attract each other with the same force. But according to Newton’s second law of motion, acceleration produced in a body by any force is inversely proportional to the mass of the body. Since, mass of the earth is much more than that of the moon, the acceleration produced in the earth is negligible. As a result, it appears as if the earth does not move towards the moon.

Question 7. What is the importance of universal law of gravitation? Solution: Universal law of gravitation is important as it accounts, (a) for the existence of the solar system, i.e., motion of planets around the sun. (b) for holding the atmosphere near the surface of the earth. (c) for the flow of water in rivers. (d) for rainfall and snowfall. (e) for occurrence of tides.

Question 8. What is the acceleration of free fall? Solution: All objects moving towards earth on account of gravitational force of earth on them are said to be in free fall. This force produces a uniform acceleration in the object. This is acceleration of free fall and its value is 9.8 m s -2 .

Question 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 called the force of gravity or simply earth’s gravity.

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? Solution: We know that the value of g is greater at the poles than at the equator. So the weight of gold at the equator will be less than the weight of gold at the poles. So it is obvious that the friend at equator will not agree with the weight of gold bought at poles.

Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball? Solution: A sheet of paper will fall slower than the one that is crumpled into a ball. This is because the air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the sheet of paper is more than the resistance offered by air to the paper ball because the sheet has larger area.

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. Solution:

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. Solution: Question 15. A stone is- thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s -2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? Solution: As final position of the stone coincides with its initial position, net displacement = 0 Total distance covered by the stone = h + h = 80m + 80m = 160 m

Question 16. Calculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 × 10 24  kg and of the sun = 2 × 10 30  kg. The average distance between the two is 1.5 × 10 11  m. Solution: Thus, the earth and the sun attract each other by a gravitational force of 3.56 × 10 22  N.

Question 20. Why does a block of plastic released under water come up to the surface of water? Solution: The buoyant force acting on the block of plastic is more than its weight. As a result of this, it comes up when released under water.

The cause of this larger buoyant force on the block of plastic is due to its density being lesser than that of water.

Question 21. The volume of 50 g of a substance is 20 cm 3 . If the density of water is 1 g cm -3 , will the substance float or sink? Solution: As the density of the substance is greater than that of water, the given substance will sink in water.,

Download CBSE Books

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online JEE Test Series for 499/- Only (Web + App) for 1 Year

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

1 thought on “ NCERT Solutions for Class 9 Science Chapter 10 Gravitation ”

• Pingback: NCERT Solutions for Class 9 Science – Gurukul For JEE & NEET

Leave a Reply Cancel reply

Join our Online Test Series for CBSE, ICSE, JEE, NEET and Other Exams

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Unit 3: Gravity

About this unit.

In this chapter, we will explore what gravity is and how it keeps all of us stuck to earth, and our solar system together. We will also explore pressure in fluids and what makes things float.

Universal law of gravity

• Introduction to gravity (Opens a modal)
• Gravity for astronauts in orbit (Opens a modal)
• Universal law of gravity example (Opens a modal)

Free fall & Acceleration due to gravity:

• Would a brick or feather fall faster? (Opens a modal)
• Mass & weight (Opens a modal)

Motion of objects in the influence of gravitational force of earth

• Free fall 1 body - solved example (Opens a modal)
• Free fall - 2 body solved numerical (Opens a modal)
• Free fall - total time up & down solved example (Opens a modal)
• Solving freefall problems using kinematic formulas 4 questions Practice

Thrust & pressure

• Pressure & thrust (Opens a modal)
• Calculating pressure due to weight (Opens a modal)

Pressure in liquids & Archimedes principle

• Pressure in liquids (Opens a modal)
• Archimedes principle & buoyancy (Opens a modal)
• What is buoyant force? (Opens a modal)

Density and condition for floating

• Condition for floating (Opens a modal)
• Calculating mass of displaced liquid (Opens a modal)

Relative density

• Relative density (Specific gravity) (Opens a modal)

Assignment - Gravitation, Class 9, Science PDF Download

MATCH THE COLUMN SINGLE COLUMN (MATRIX)

FILL IN THE BLANKS :

1. .................... is the force of attraction between any two bodies in the universe.

2. .................... is the force of attraction between a body and a planet. 3. Acceleration due to gravity .................... with height from the surface of the earth. 4. Acceleration due to gravity .................... with depth from the surface of the earth. 5. Acceleration due to gravity is a maximum at the .................... 6. .................... of a body changes from place to place but its .................... remains constant. 7. The upward force experienced by a body immersed partially or fully in a fluid is called .................... 8. Density of a substance is defined as the ratio of the mass of a body to its .................... 9. Relative density of a substance is defined as the ratio of the density of the substance to the density of water at .................... 10. In cgs the relative density of a substance is .................... equal to its density in gcm–3. 11. The combined weight of the sinker and cork is .................... than the weight of the sinker alone. 12. For a body to float the density of the floating object should be .................... than or equal to the density of the liquid in which it is to float.

13. The centre of ..........is a point, where the total upthrust, due to fluid displaced by part ...........of body acts. 14. A fish .................... by squeezing out air, from its .................... 15. When a body is partly or wholly immersed in a .................... , it experience an .................... 16. An iceberg floats with .................... of its volume below .................... 17. The unit of upthrust in SI is .................... 18. Lactometer is used to measure .................... of .................... 19. A hydrometer sinks .................... in water than in pure milk. 20. The density of hot air is .................... that of cold air.

UNDERSTANDING BASE QUESTIONS :-

1. In what sense does the moon fall towards the earth ? Why does it not actually fall on the surface of earth ? 2. If the earth attracts an apple. Does the apple also attract the earth? If so, why does the earth not move towards the apple ? 3. If the force of gravity somehow vanished away, why would we sent flying in space ? 4. There are two kinds of balances, that is, a beam balance and a spring balance. If both the balances give same measure of a given body on the surface of the earth, will they give same measure on the surface of moon ? Explain your answer. 5. A bag of sugar weighs W at some place on the equator. If this bag is taken to Antarctica, will it weigh same, more or less ? Give a reason for your answer. 6. Which is greater : the attraction of Earth for 1 kg of lead or attraction of 1 kg lead for earth ? 7. The weight of a man on the earth is 100 kg. Does this weight on the moon increase or decrease ? 8. When dropped from the same height a body reaches the ground quicker at poles than at the equator. Why ? 9. What would be the effect on the weight of a body if the earth stopped rotating ? 10. A man at the top of a tower throws and object horizontally whereas he simply drops another. Will these two objects reach the earth at the same time ? 11. If gravitation is the power by which all bodies tend to towards each other, then why do all bodies tend towards the centre of the earth ? 12. What is weightlessness ? 13. Why is the bottom part of the foundation of a building made wider ? 14. Why cutting instruments are sharpened ? 15. Explain why it is difficult to push a tin can into water keeping its mouth upwards than when its mouth is kept downwards towards the water. 16. A piece of ice is gently placed on the surface of water filled in glass tumbler, so the water rises to the brim. What will happen to the level of water when the ice melts ? Will the water overflow ? If not, explain with reason. 17. You are provided with a hollow iron ball of volume 20 cm3, a mass of 15 g and a solid iron ball of mass 20 g. Both are placed on the surface of water contained in a large tub. Which will float ? Give reason for your answer. 18. Explain the following : (i) Icebergs floating in the sea are dangerous for the ship. (ii) An egg sinks in fresh water but floats in a strong solution of salt. (iii) An iron nail sinks in water but a ship made of iron floats. (iv) A ship sinks to a great depth in river water than in sea water. (v) It is easier for a man to swim in sea water than in river water. (vi) A dead body floats, with its head immersed in water. 19. A solid brass cylinder tied to a thread is hanging from the hook of a spring balance. The cylinder is gradually dipped into the water contained in a jar. What change do you expect in the reading of the spring balance ? Give reason for your answer. 20. If a solid of the same density as that of a liquid is placed in it, what will happen to the solid ?

HIGH ORDER THINKING SKILL QUESTIONS

1. Newton's law of gravitation is called universal law. Why ? 2. According to Newton's law of gravitation, every, particle of matter attracts every other particle. But bodies on the surface of earth never move towards each other on account of this force of attraction, why ? 3. What is the nature of motion of an object falling freely under the action of gravity ? 4. Will a body weight more on equator or on pole ? 5. Earth is continuously pulling the moon towards its centre, still it does not fall to the earth, why ? 6. How does acceleration due to gravity depend on the mass of planet ? 7. If the diameter of the earth becomes two times its present value and its mass remains unchanged, then how would the weight of an object on the surface of the earth be affected ? 8. Choose the correct alternative : (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth. (c) Acceleration due to gravity is independent of the mass of the earth/mass of the body. 9. A person can jump higher on the surface of the moon than on the earth. Why ? 10. A glass jar contains a liquid of density 'd' upto a height of 'h' at a place where acceleration due to gravity is 'g', the atmospheric pressure is PA : (i) What is the pressure at the free surface of the liquid ? (ii) Write an expansion for the total pressure at the base of the jar. 11. Explain why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of the lake. 12. A body dipped into a liquid experiences an upthrust. State the factors on which the upthrust on the body depends. 13. Explain briefly why a balloon filled with helium gas rises in air. Why does the balloon rise to a particular height above the ground and does not rise further ? 14. Will a body weigh more in air or in vacuum when weighed with a spring balance ? Give a reason for your answer.

NUMERICAL PROBLEMS

1. What is the gravitational acceleration of a spaceship at a distance equal to two Earth's radius from the centre of the Earth? 2. A boy on a cliff 49 m high drops a stone. One second later, he throws a second stone after the ftrst. They both hit the ground at the same time. With what speed did he throw the second stone ? 3. A stone drops from the edge of the roof. It passes a window 2 m high in 0·1 s. How far is the roof above the top of the window? 4. A particle is dropped from a tower 180 m high. How long does it take to reach the ground? What is the velocity when it tourhes the ground ? Take g = 10 m/s 2 . 5. To estimate the height of a bridge over a river, a stone is dropped freely on the river from the bridge. The stone takes 2 s to touch the water surface in the river. Calculate the height of the bridge from the water level. Take g = 9·8 m/s 2 . 6. How much would a 70 kg man weigh on moon? What will be his mass on Earth and Moon? Given g on Moon = 1·7 m/s 2 .

Top Courses for Class 9

Faqs on assignment - gravitation, class 9, science, past year papers, assignment - gravitation, previous year questions with solutions, extra questions, video lectures, viva questions, study material, objective type questions, semester notes, practice quizzes, mock tests for examination, sample paper, important questions, shortcuts and tricks.

Assignment - Gravitation, Class 9, Science Free PDF Download

Importance of assignment - gravitation, class 9, science, assignment - gravitation, class 9, science notes, assignment - gravitation, class 9, science class 9 questions, study assignment - gravitation, class 9, science on the app, welcome back, create your account for free.

Forgot Password

Unattempted tests, change country.

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Science

NCERT Solutions for Class 9 Science

Ncert solutions for class 9 science updated for 2023-24 free chapterwise pdf download.

NCERT Solutions for Class 9 Science help students to clear any doubts instantly and efficiently. These NCERT Solutions guide students to learn the important concepts which are included in the CBSE Class 9 Science syllabus. Students are required to solve the exercise questions included in the textbook to create a proper understanding of the topics.

While solving the textbook questions, doubts arise among students. The NCERT Solutions for Class 9 come in handy at such times, as they include precise explanations and detailed answers to those questions. These CBSE Science NCERT Solutions for Class 9 cover solutions to all the important chapters included in the textbook, like Matter, Atoms, Tissues, Living Organisms, Motion, Force, Laws of Motion, Gravitation, Energy and Work, Sound, Natural Resources, etc.

Along with answers to the textbook questions, these solutions provide you with extra questions, exemplar problems, the important questions from previous year question papers, sample papers , worksheets, MCQs, short answering questions, descriptive type questions, their solutions, as well as tips and tricks.

NCERT Solutions for Class 9 Science Chapter-Wise PDFs

NCERT Solutions for Class 9 Science are provided in a format for better accessibility to the students. These solutions serve as an important studying tool for students who are preparing for their board examinations and assignments.

NCERT Solutions of Class 9 Science Book Chapter Brief

Chapter 1: matter in our surrounding.

The Class 9 science textbook starts with the chapter “Matter in Our Surroundings”. Everything that we see around us is made up of material which is called matter. These things occupy space and have mass. Earlier, Indian philosophers had classified matter into 5 basic elements called Panch Tatva – air, water, earth, sky and fire. Now, modern scientists have come up with 2 types of classification, i.e., based on physical property and chemical nature. In this chapter, students will learn about the physical properties of matter .

The matter is made up of particles and these particulars are very small. The particles of matter have space between them, they are continuously moving and attract each other. The matter around us exists in 3 different states – solids, liquids and gases. These states of matter arise due to variation in the characteristic of the particles of matter. All the 3 states of matters have been explained in-depth with the help of activities. Further, the textbook explains that the state of matter is inter-convertible. The state of matter can be changed by changing temperature or pressure. The phenomenon of change of a liquid into vapour at any temperature below its boiling point is called evaporation. The rate of evaporation depends upon the surface area exposed to the atmosphere, the temperature, the humidity and the wind speed. Evaporation also causes cooling.

Topics Covered in Class 9 Science Chapter 1 Matter in Our Surrounding :

Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state –  melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation.

Also access the following resources for Class 9 Chapter 1 Matter in Our Surrounding at BYJU’S:

• Matter in Our Surroundings Class 9 Notes – Chapter 1
• CBSE Class 9 Science Chapter 1 Matter in Our Surroundings Important Questions
• NCERT Exemplar Class 9 Science Solutions for Chapter 1 – Matter in Our Surroundings

Chapter 2: Is Matter Around Us Pure?

This chapter of NCERT Class 9 Science students will teach about mixtures, solutions, properties of solutions, separation of mixtures, and physical and chemical changes. Along with this, they will also learn about compounds and their properties, the difference between mixtures and compounds, the classification of matter, etc.

Topics Covered in Class 9 Science Chapter 2 Is Matter Around Us Pure?  :

Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture).

Also, access the following resources for Class 9 Chapter 2 Is Matter Around Us Pure? at BYJU’S:

• CBSE Class 9 Science Chapter 2 Is Matter Around Us Pure? Notes
• Important Questions for CBSE Class 9 Science Chapter 2 – Is Matter Around Us Pure?
• NCERT Exemplar Class 9 Science Solutions for Chapter 2 – Is Matter Around Us Pure?

Chapter 3: Atoms and Molecules

In NCERT Class 9 Science Chapter 3, students will learn the laws of chemical combination , atoms and molecules. They will get to know how to write a chemical formula, molecular mass and mole concepts and some numerical problems related to these concepts.

Topics Covered in Class 9 Science Chapter 3 Atoms and Molecules  :

Particle nature and their basic units: Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses.

Also, access the following resources for Class 9 Chapter 3 Atoms and Molecules, at BYJU’S:

• Atoms And Molecules Class 9 Notes – Chapter 3
• Important Questions for Class 9 Science Chapter 3 – Atoms and Molecules
• NCERT Exemplar Class 9 Science Solutions for Chapter 3 – Atoms And Molecules

Chapter 4: Structure of the Atom

This chapter of NCERT Class 9 Science deals with the various atomic models of atoms that were proposed by different scientists. In addition to it, this chapter also covers electrons’ distribution in different orbits, calculation of valency , atomic number and mass number.

Topics Covered in Class 9 Science Chapter 4 Structure of the Atom  :

Structure of atoms: Electrons, protons and neutrons, Valency, Atomic Number and Mass Number, Isotopes and Isobars.

Also, access the following resources for Class 9 Chapter 4 Structure of the Atom at BYJU’S:

• Structure of the Atom Class 9 CBSE Notes – Chapter 4
• Important Questions of CBSE Class 9 Science Chapter 4 – Structure of the Atom
• NCERT Exemplar Class 9 Science Solutions for Chapter 4 – Structure of the Atom

Chapter 5: The Fundamental Unit of Life

This chapter is related to Biology. In this chapter, students will get to know that the cell is the fundamental unit of life. The whole chapter revolves around the cell and its structural organisation, in which students will learn about the plasma membrane , cell wall, nucleus, cytoplasm and structure of an animal cell.

Topics Covered in Class 9 Science Chapter 5 The Fundamental Unit of Life  :

Cell – Basic Unit of life: Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number.

Also, access the following resources for Class 9 Chapter 5 The Fundamental Unit of Life at BYJU’S:

• The Fundamental Unit of Life Class 9 CBSE Notes – Chapter 5
• Revision Notes For Class 9 Science Chapter 5 The Fundamental Unit of Life
• NCERT Exemplar Class 9 Science Solutions for Chapter 5 – The Fundamental Unit of Life

Chapter 6: Tissues

This chapter pertains to the basic definition of tissue and then elaborates on Plant and Animal tissue with proper diagrams. Students will get to know the different types of plant and animal tissues with a detailed explanation of each.

Topics Covered in Class 9 Science Chapter 6 Tissues  :

Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Also, access the following resources for Class 9 Chapter 6 Tissues at BYJU’S:

• Tissues Class 9 CBSE Notes – Chapter 6
• Revision Notes For Class 9 Science Chapter 6 Tissues
• NCERT Exemplar Class 9 Science Solutions for Chapter 6 – Tissues

Chapter 7: Diversity in Living Organism

This chapter deals with the classification of plants and animals. Students get to know that all living organisms are divided into 5 kingdoms namely Monera, Protista, Fungi, Plantae and Animalia. It also describes the classification and evolution, the hierarchy of classification.

Also access the following resources for Class 9 Chapter 7 Diversity in Living Organism at BYJU’S:

• Diversity in Living Organisms Class 9 CBSE Notes – Chapter 7
• Important Questions for CBSE Class 9 Science Chapter 7 – Diversity in Living Organisms
• NCERT Exemplar Class 9 Science Solutions for Chapter 7 – Diversity In Living Organisms

Chapter 8: Motion

NCERT Class 9 Science Chapter 8 describes motion, the speed with direction, rate of change of velocity , and graphical representation of motion. Also, students will find the 3 equations of motion and numerical problems related to them. Overall, this chapter will be a mix of theory as well as the numerical part.

Topics Covered in Class 9 Science Chapter 8 Motion :

Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion.

Also, access the following resources for Class 9 Chapter 8 Motion at BYJU’S:

• CBSE Class 9 Physics Motion Notes
• NCERT Exemplar Class 9 Science Solutions for Chapter 8 – Motion

Chapter 9: Force and Law of Motion

This chapter explains the 3 laws of motion with the help of diagrams and examples. Below are the 3 laws of motion:

• First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.
• Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force .
• Third law of motion: To every action, there is an equal and opposite reaction, and they act on two different bodies.

Topics Covered in Class 9 Science Chapter 9 Force and Law of Motion  :

Force and Newton’s laws: Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration.

Also, access the following resources for Class 9 Chapter 9 Force and Law of Motion at BYJU’S:

• Force and Laws Of Motion Class 9 CBSE Notes – Chapter 9
• Revision Notes For Class 9 Science Chapter 9 Force and Laws of Motion
• NCERT Exemplar Class 9 Science Solutions for Chapter 9 – Force and Laws Of Motion

Chapter 10: Gravitation

In chapter 10 of NCERT Class 9 Science, students will learn the universal law of gravitation and its importance, free fall, mass, weight, thrust and pressure, Archimedes’ principle and relative density. Students will also find numerical problems related to these topics.

Topics Covered in Class 9 Science Chapter 10 Gravitation :

Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall.

Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy.

Also, access the following resources for Class 9 Chapter 10 Gravitation at BYJU’S:

• Gravitation Class 9 CBSE Notes – Chapter 10
• Revision Notes For Class 9 Science Chapter 10 Gravitation
• NCERT Exemplar Class 9 Science Solutions for Chapter 10 – Gravitation

Chapter 11: Work and Energy

In this chapter, the concept of work is defined with different activities, numerical and examples. The chapter also deals with energy and its different forms. Examples are given to explain all types of energy. The chapter ends with the topic “rate of doing work”.

Topics Covered in Class 9 Science Chapter 11 Work and Energy  :

Work, Energy and Power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy).

Also, access the following resources for Class 9 Chapter 11 Work and Energy at BYJU’S:

• Work And Energy Class 9 CBSE Notes – Chapter 11
• Revision Notes For Class 9 Science Chapter 11 Work and Energy
• NCERT Exemplar Class 9 Science Solutions for Chapter 11 – Work And Energy

Chapter 12: Sound

Chapter 12 of CBSE Class 9 Science, deals with concepts such as the production of sound, propagation of sound, the reflection of sound, range of hearing, applications of ultrasound, and the structure of the human ear . A few numericals are also there in the chapter which can only be solved after understanding the concepts. So, students must grasp and study the chapter carefully.

Topics Covered in Class 9 Science Chapter 12 Sound :

Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Also access the following resources for Class 9 Chapter 12 Sound at BYJU’S:

• CBSE Class 9 Science Chapter 12 Sound Notes
• Important Questions for CBSE Class 9 Science Chapter 12 – Sound
• NCERT Exemplar Class 9 Science Solutions for Chapter 12 – Sound

Chapter 13: Why Do We Fall Ill?

NCERT Class 9 Science chapter 13 deals with health issues and different types of diseases. It covers topics like; Health and its failure , disease and its cause, infectious diseases. This chapter is added to make students aware of different types of diseases so that they take care of their health and be fit and healthy.

Also access the following resources for Class 9 Chapter 13 Why Do We Fall Ill? at BYJU’S:

• Why Do We Fall Ill? Class 9 CBSE Notes – Chapter 13
• Important Questions for CBSE Class 9 Science Chapter 13 – Why Do We Fall Ill?
• NCERT Exemplar Class 9 Science Solutions for Chapter 13 – Why Do We Fall Ill?

Chapter 14: Natural Resources

We are blessed with natural resources. These are essential to meet the basic requirements of all forms of life on Earth. So, this chapter has been included by NCERT to provide knowledge on the types of resources available on the Earth. This chapter deals with the topics like air, water, soil, the biogeochemical cycle and the ozone layer, its importance and how humans are polluting them.

Also access the following resources for Class 9 Chapter 14 Natural Resources at BYJU’S:

• Natural Resources Class 9 CBSE Notes – Chapter 14
• Revision Notes For Class 9 Science Chapter 14 Natural Resources
• NCERT Exemplar Class 9 Science Solutions for Chapter 14 – Natural Resources

Chapter 15: Improvement in Food Resources

The last chapter of NCERT Class 9 Science covers topics like Improvement in crop yields, manure, fertilizer, storage of grains, and animal husbandry . This chapter provides knowledge regarding agriculture, farming and dairy.

Also access the following resources for Class 9 Chapter 15 Improvement in Food Resources at BYJU’S:

• Improvement in Food Resources Class 9 CBSE Notes – Chapter 15
• Important Questions for CBSE Class 9 Science Chapter 15 – Improvement in Food Resources
• Revision Notes For Class 9 Science Chapter 15 Improvement in Food Resources
• NCERT Exemplar Class 9 Science Solutions for Chapter 15 – Improvement In Food Resources

To download the complete book in PDF format, visit NCERT Science Book Class 9 . For more practice tests and mock tests, sign up on BYJU’S the learning app. Apart from Science, get NCERT Solutions for all the subjects. Students can download worksheets, assignments, NCERT Books , notes and study materials for exam preparation and for a better understanding of the topics.

CBSE Class 9 Science Evaluation Scheme (Theory) –

Features of ncert 9 th class science book solutions.

The book has a wide variety of features which are listed below:

• CBSE Class 9 Science NCERT Solutions are available to everyone for free.
• Covers all the exercise problems from the Class 9 textbook.
• Consists of extra questions, exemplar problems, the important questions from previous year question papers and sample papers, worksheets, MCQs, short answering questions, descriptive type questions, their solutions and tips and tricks.
• CBSE Class 9 Science Solutions files are available for download in PDF format for easy access.
• Diagrams are included to help students visualize the topics.
• Most effective solutions are given which can help to score well in the exams.

Frequently Asked Questions on NCERT Solutions for Class 9 Science

How many chapters are there in the ncert solutions for class 9 science , can i expect the questions from ncert solutions for class 9 science to be asked in the annual exam, is the ncert solutions for class 9 science sufficient for the exam preparation, why should students refer to the ncert solutions for class 9 science , leave a comment cancel reply.

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Very good app

I am a subscriber and my feedback is that it really helped me to understand how the question’s answers came. Thank you Byju’s

Very nice app

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Counselling