NCERT Solutions for Class 9 Science Chapter 10 Gravitation

# NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Edited By Vishal kumar | Updated on Apr 03, 2024 09:44 AM IST

## NCERT Solutions Class 9 Science Chapter 10 – CBSE Free PDF Download

NCERT Solutions for Class 9 Science Chapter 10 Gravitation: This article aims to give you NCERT Solutions that make it easier for you to grasp the concepts in the Class 9 Gravitation chapter. Through CBSE NCERT class 9 science chapter 10 exercise solutions, you will get the necessary insights into the chapter. Have you ever noticed that when you shake a tree leaf, it falls down? When we drop a pen, it falls. A leaf & pen fall down because of gravitational force due to the earth. The solutions for NCERT Gravitation Class 9 Science Chapter 10 are going to provide you with the solutions for some important topics of the chapter such as gravitation, free fall, weight, thrust, and pressure, etc. Class 9 Science Chapter 10 exercise question answers are also available class-wise and subject-wise.

Detailed class 9th gravitation NCERT solutions are prepared by highly qualified subject matter experts of careers360. Students can download the solution in the form of PDF by visiting the link below for free. Apart from this team of Careers360 has also prepared the NCERT Notes. Students can access the NCERT Class 9th Notes Subject wise notes to prepare well in the exam which is free of cost.

### NCERT Solutions for Class 9 Science: Important Formulas and Diagrams + eBook link

Gravitation is an important chapter in the Class 9 NCERT syllabus. In order to solve questions from the NCERT book or any other related books, it is necessary to have a solid understanding of the concepts and formulas. The experts at Careers360 have compiled the following important formulas from Chapter 10 of NCERT Class 9 Science textbook which are listed below.

where:

F is the gravitational force between two objects, G is the gravitational constant (approximately 6.67430 × 10-11 m³/(kg·s²)), m1 and m2 are the masses of the two objects, and r is the distance between the centres of the two objects.

• T² = k × R³

Where:

T represents the orbital period, R represents the average distance from the Sun (also known as the semi-major axis), and k is a constant.

Students can download chapter-wise important formulas in PDF format by clicking on the link below. These formulas can be extremely helpful for quick revision during exams and tests.

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

The important topics of ncert class 9 science chapter 10 exercise solutions are listed below:

• Gravitation: TGravitation is the force that attracts any two objects in the universe
• Universal Law of Gravitation: The law that states the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
• Free Fall: The motion of an object under the influence of only the gravitational force.
• Acceleration Due to Gravity (g): The acceleration acquired by an object in free fall.
• Difference between G and g: The gravitational constant (G) is a universal constant, while the acceleration due to gravity (g) varies from place to place.

## NCERT Solutions for Class 9 Science Chapter 10 Gravitation: Solved Exercise Questions

Gravitation class 9 ncert solutions - Topic 10.1 Gravitation

The universal law of gravitation states that everybody in the universe attracts every other body by the virtue of its mass. This force is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between them.

Let there be two bodies of masses m 1 and m 2 and let the distance between them be r. The gravitational force F g between them would be given by

$F_{g}=G\frac{m_{1}m_{2}}{r^{2}}$

where G is the universal gravitational constant and is equal to $6.67\times 10^{-11}N\ m^{2}\ kg^{-2}$

Let M E and m be the masses of the earth and the body and let the distance between the centre of Earth and the body be d. The magnitude of the gravitational force between the earth and the object would be given by the relation.

Mass of Earth, $M_{E}=5.972\times 10^{24}kg$

d would be approximately equal to the radius of the earth.

Radius of Earth $=6.378\times 10^{6}m$

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

$\\F=6.67\times 10^{-11}\times \frac{5.972\times 10^{24}\times m}{(6.378\times 10^{6})^{2}}\\ F=9.8m\ N$

Chapter 10 Science Class 9 exercise answers - 10.2 Free fall

We say an object is freely falling when it is dropped from some height and is attracted by the gravitational force of earth only and is under the influence of no other considerable force.

Each object on the Earth is under the influence of the gravitational force of the earth. The acceleration due to the earth's gravitational force is known as acceleration due to gravity.

Topic 10.4 Weight

The following are the differences between the mass of a body and its weight

 Mass Weight (i) Mass is the amount of matter contained in the body. (i) Weight is the gravitational force experienced by the body. (ii) Mass of a body is always constant. (ii) Weight of a body depends on the place where it is at. (iii) Mass is the measure of the inertia of the body. (iii) Weight is the measure of the gravitational force acting on the body. (iv) Mass only has magnitude. (iv) Since weight is a force it naturally has a direction associated with it as well (v) It's SI unit is the kilogram (kg). (v) It's SI unit is Newton (kg m s -2 )

The weight of an object on the moon would be given by

$W_{M}=G\frac{M_{M}m}{r_{_{M}}^{2}}$

where M M is the mass of the moon, m is mass of the body, r M is the radius of the moon and G is the gravitational constant.

The weight of an object on the Earth would be given by

$W_{E}=G\frac{M_{E}m}{r_{_{E}}^{2}}$

where M E and r E are the mass and radius of the earth respectively.

$\frac{W_{M}}{W_{E}}=\frac{M_{M}r_{E}^{2}}{M_{E}r_{M}^{2}}$

The above ratio is approximately equal to 1/6 and this is why the weight of an object on the moon (1/6) th of its weight on the earth .

NCERT textbook Solutions for Class 9 Science Chapter 10 Gravitation

Topic 10.5 Thrust and Pressure

It is difficult to hold a school bag having a strap made of a thin and strong string because as the surface area of the string in contact with the shoulders is very less and due to this, its weight applies a large pressure on the shoulders.

When an object is placed in a fluid it displaces a volume of liquid equal to its own volume. Due to this, the liquid exerts an upward force on the body called the Buoyant force. This tendency of a liquid to exert the upward buoyant force is called buoyancy.

When an object is placed on the surface of the water it displaces a certain volume of water. If the density of the object is less than that of water the buoyant force due to this displacement of water is equal to the weight of the object and it floats on the surface of the water. If the density of the object is more than the density of water, the volume of water displaced would be equal to the volume of the object itself and the buoyant force acting upwards due to this displacement of water would be less than the weight of the object and the object would sink.

Class 9 Science Chapter 10 Question Answer: Gravitation

Topic 10.6 Archimedes' Principle

The weighing scale shows the reading according to the weight applied to it. The weight on the weighing scale would be slightly less than our weight because there is a small upward force acting on us due to the buoyancy of the atmosphere and the reading would be slightly lower than our actual weight, therefore, our mass must be more than 42 kg.

The volume of the bag of cotton would be much more than the iron bar and the upward buoyant force on the bag of cotton would be more than that acting on the iron bar and therefore the value is shown in case of the bag of cotton must be lesser than the actual value by a larger amount and therefore the bag of cotton is heavier than the iron bar.

## NCERT Solutions for Class 9 Science Chapter 10 Gravitation: Solved Exercise Questions

Let there be two bodies of masses m 1 and m 2 and let the distance between them be r. The gravitational force F between them would be given by

$F_{}=G\frac{m_{1}m_{2}}{r^{2}}$

Let the distance between them be halved by two. The gravitational force between them would now be given by

$\\F'=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}}\\ F'=4G\frac{m_{1}m_{2}}{r^{2}}\\ F'=4F$

The force of gravitation between two objects would increase by 4 times if the distance between them is halved.

The gravitational force acting is definitely more in the case of a heavier object than a light object but the acceleration depends on the ratio of the force acting on the body to the mass of the body and since gravitational force acting on a body is proportional to its mass the acceleration due to gravity is the same for all bodies irrespective of their masses.

Mass of Earth M E is $6\times 10^{24}\; kg$

Mass of the body m is 1 kg

The radius of the earth R E is $6.4\times 10^{6}\; m.$

Universal gravitational constant, G = $6.67\times 10^{-11}N\ m^{2}\ kg^{-2}$

The magnitude of the gravitational force between the earth and the body would be

$\\F=G\frac{M_{E}m}{r_{E}^{2}}\\ F=6.67\times 10^{-11}\times \frac{6\times 10^{24}\times 1}{(6.4\times 10^{6})^{2}}\\ F=9.77N$

The earth attracts the moon with the same force as the moon attracts the earth.

We know from the third law of motion that each force has an equal and opposite force and the universal gravitational law also states the same i.e. the gravitational force of attraction between two bodies is the same.

The moon and the earth attract each other with the same gravitational force. It is because of the much larger mass of the earth than the mass of the moon the earth does not move towards the moon.

• the mass of one object is doubled?

As the force between the objects is directly proportional to the product of masses of the objects if the mass of one object is doubled the force between them will also double.

The distance between the objects is doubled and tripled?

The force between two objects is inversely proportional to the square of the distance between them. Therefore if the distance between the objects is doubled and tripled the force between them would become one fourth and one-ninth of the initial value respectively.

• the masses of both objects are doubled?

As the force between the objects is directly proportional to the product of masses of the objects if the masses of both objects are doubled the force will become four times the initial value.

The importance of universal law of gravitation lies in the fact that it proves that every object in the universe is attracted by every other object in the universe by the virtue of their masses.

The acceleration of free fall on earth is 9.8 m s -2 and is the same for all objects i.e. its value is independent of their masses.

The gravitational force between the earth and an object is called the weight of the object.

As the value of g is more at the poles than that at the equator the weight of the same amount of gold would be more at the poles than that at the equator and therefore the friend will not agree with the weight of gold bought.

A sheet of paper has much more area than the same paper crumbled and due to this the sheet experiences more air resistance and thus falls at a speed slower than when it is crumbled.

The weight of an object on earth is given by w = mg where m is the mass of the object and g is the gravitational acceleration.

g = 9.8 m s -2

Weight of a 10 kg object on earth = 10 X 9.8 = 98 N

As the gravitational force on the surface of the moon is only 1/6th of that on earth the gravitational accelerations on the moon would be equal to g/6.

Weight of an object of mass 10 kg on the moon is therefore given as follows

$\\w=10\times \frac{g}{6}\\ w=10\times \frac{9.8}{6}\\ w=16.33N$

• the maximum height to which it rises,

The initial velocity of the ball u = 49 m s -1

Final velocity at the highest point would be v = 0

The magnitude of the acceleration is equal to the acceleration due to gravity g

Acceleration, a = -g = -9.8 m s -2

Let the maximum height to which it rises be s

Using the third equation of motion we have

$\\v^{2}-u^{2}=2as\\ s=\frac{v^{2}-u^{2}}{2a}\\ s=\frac{0^{2}-49^{2}}{2\times -9.8}\\ s=122.5m$

The ball would rise to a maximum value of 112.5 m.

• The total time it takes to return to the surface of the earth.

Let the time taken by the ball to reach the highest point be t

$\\v=u+at\\ 0=49+(-9.8)\times t\\ t=5\ s$

At the same time, t would be taken to come back to the ground from the highest point.

Therefore the total time it takes to return to the surface of the earth = 2t = 10 s.

Initial velocity u = 0

Acceleration, a = g = 9.8 m s -2

Distance travelled, s = 19.6 m

Let the final velocity be v

According to the third equation of motion

$\\v^{2}-u^{2}=2as\\ v^{2}-0^{2}=2\times 9.8\times 19.6\\ v=\sqrt{2\times 9.8\times 19.6}\\ v=19.6\ m\ s^{-1}$

Its final velocity just before touching the ground will be 19.6 m s -1 .

Initial velocity u = 40 m s -1

Acceleration a = -g = -10 m s -2

Final velocity at the highest point would be v = 0

Let the maximum height reached be s

As per the third equation of motion

$\\v^{2}-u^{2}=2as\\ s=\frac{v^{2}-u^{2}}{2a}\\ s=\frac{0^{2}-40^{2}}{2\times -10}\\ s=80m$

The net displacement would be zero as the stone will return to the point from where it was thrown.

The total distance covered by the stone = 2s = 160 m

Mass of the earth, M E $=6\times 10^{24}kg$

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

Distance between the earth and the sun, $d=1.5\times 10^{11}m$

Universal gravitational constant, $G=6.67\times 10^{-11}N\ kg^{-2}\ m^{2}$

The force of gravitation between the earth and the Sun would be given as

$\\F=G\frac{M_{E}M_{E}}{d^{2}}\\ F=6.67\times 10^{-11}\times \frac{6\times 10^{24}\times 2\times 10^{30}}{(1.5\times 10^{11})^{2}}\\ F=3.56\times 10^{22}N$

Let the distance travelled by the stone which is dropped from the top upto the instant when the two stones meet be x

Initial velocity u = 0

Acceleration a = g = 9.8 m s -2

Using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ x=0\times t+\frac{1}{2}\times 9.8\times t^{2}\\ x=4.9t^{2}$ $(i)$

The distance travelled by the stone which is projected vertically upwards from the ground up to the instant when the two stones meet would be equal to 100 - x

Initial velocity = 25 m s -1

Acceleration a = -g = -9.8 m s -2

Using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ 100-x=25\times t+\frac{1}{2}\times (-9.8)\times t^{2}\\ 100-x=25t-4.9t^{2}\\ x=4.9t^{2}-25t+100$ $(ii)$

Equating x from (i) and (ii) we get

4.9t 2 = 4.9t 2 - 25t + 100

25t = 100

t = 4s

x = 4.9t 2

x = 4.9 X 4 2

x = 78.4 m

100 - x = 100 - 78.4 = 21.6 m

The stones meet after a time of 4 seconds at a height of 21.6 meters from the ground.

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4 s.

(a) Let the ball be thrown with initial velocity u

Time taken to get back to the thrower = 6 s

Time taken to reach the highest point is t = 6/2 = 3 s

Final velocity at the highest point is v = 0

Acceleration a = -g = -9.8 m s -2

Using the first equation of motion

$\\v=u+at\\ 0=u-9.8\times 3\\ u=29.4\ m\ s^{-1}$

(b) Let the maximum height it reaches be s

Using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ s=29.4\times 3-4.9\times 3^{2}\\ s=44.1m$

(c) Out of the 4 seconds, 3 have been spent in reaching the highest point

The distance travelled by the ball in the next 1 second is s' given by

$\\s'=ut+\frac{1}{2}gt^{2}\\ s=0\times +4.9\times 1^{2}\\ s=4.9m$

Distance from the ground after 4 seconds = s - s' = 44.1 - 4.9 = 39.2 m

The position of the ball after 4 seconds is 39.2 m from the ground.

The buoyant force acts on an object in the vertically upward direction opposite to that of gravitational force.

The density of plastic is less than that of water and due to which the upwards acting buoyant force is more than the downwards acting gravitational force. Due to this, a block of plastic released under water comes up to the surface of the water.

Mass of the given amount of substance = 50 g

Volume of the given amount of substance = 20 cm 3

Density of the given substance is $\rho$

$\\\rho =\frac{50}{20}\\ \rho =2.5\ cm^{3}$

As the given substance has a higher density than that of water it will be sinking in water.

Mass of the packet = 500 g

Volume of the packet = 350 cm 3

Density of the packet is given by

$\\\rho =\frac{500}{350}\\ \rho =1.428\ cm^{3}$

As the density of the packet is less than that of water it will sink in water.

Volume of the water displaced by the packet = volume of the packet = 350 cm 3

Mass of the water displaced by the packet = Volume of the water displaced by the packet X Density of water

= 350 X 1

= 350 g

Mass of water displaced is less than the mass of packet, so the packet will sink.

### Gravitation NCERT Science Class 9 Chapter 10 Topics

Topics for class 9 science chapter 10 question answer are listed below:

10.1 Gravitation

10.1.1 Universal Law of Gravitation

10.1.2 Importance of The Universal Law of Gravitation

10.2 Free Fall

10.2.1 To Calculate The Value Of g

10.2.2 Motion of Objects Under the Influence of Gravitational Force of the Earth

10.3 Mass

10.4 Weight

10.4.1 Weight of an Object on the Moon

10.5 Thrust and Pressure

10.5.1 Pressure in Fluids

10.5.2 BUOYANCY

10.5.3 Why Objects Float Or Sink When Placed on The Surface of Water?

10.6 Archimedes’ Principle

10.7 Relative Density

## Gravitation Class 9 Some Important Points

• We study the gravitational force between two bodies, acceleration due to gravity, free fall, buoyancy force, and relative density.
• The gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between the objects. That is if two masses M and m are separated by a distance d apart then the gravitational force

$\\F\propto\frac{Mm}{d^2}\\F= G \frac{Mm}{d^2}$

Where G is called universal gravitational constant whose value is

$G= 6.7\times10^{-11}Nm^2Kg^{-2}$

• The value of acceleration due to the gravity of earth (g) on an object is given by the equation

$g= \frac{GM}{d^2}$

Where M is the mass of the earth and d is the distance between the object and the earth. If the object is on the surface of earth or near to earth then the radius of the earth is approximately = distance between the object and the earth. Therefore

$g= \frac{GM}{R^2}$

Where R is the radius of the earth.

• We have experienced that it is easy to lift a bucket of water from a well till the surface of the water and above the water weight increase. The less weight inside the water is due to a force exerted by water against the weight of a bucket of water. This force exerted by the liquid is known as upthrust or buoyant force.
• All the objects experience a force of buoyancy when they are immersed in a fluid. The magnitude of the upthrust (buoyant force) depends on the density of water. Objects of density less than that of a liquid float on the liquid and the objects of density greater than that of a liquid sink in the liquid.

## Key features NCERT solutions for class 9 science chapter 10 Gravitation

• Easy to Use: The class 9 science chapter 10 exercise question answer Gravitation are presented in a user-friendly manner, allowing for easy scrolling to find the desired solutions or students can download them and use them offline anytime without charge.
• Comprehensive Coverage: In addition to the NCERT solutions for Gravitation, Class 9 Science, the article also provides NCERT Solutions for Class 9 Maths.
• Exercise Solutions: The NCERT solutions for Class 9 Science Chapter 10 Gravitation cover the exercises in the chapter, ensuring that the concepts and questions are thoroughly understood.
• Topic-wise Solutions: The Gravitation Class 9 NCERT Science Chapter 10 is organized in a topic-wise manner, making it easier to locate specific solutions for better comprehension.
• Exam-oriented Preparation: The NCERT Solution for Class 9 Science Chapter 10 are designed to help students understand the questions and concepts frequently asked in exams. This can aid in scoring well on science exams.
• Free Access: These chapter 10 science class 9 exercise answers are freely accessible, making them available to all students seeking assistance with gravitation class 9.

Also Check -

NCERT Books and NCERT Syllabus here:

## NCERT Solutions for Class 9 Science- Chapter Wise

 Chapter No. Chapter Name Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Gravitation Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15

## NCERT Science Exemplar Solutions Class 9 - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Chapter 1 Matter in our Surroundings Chapter 2 Chapter 2 Is Matter Around Us Pure? Chapter 3 Chapter 3 Atoms and Molecules Chapter 4 Chapter 4 Structure of the Atom Chapter 5 Chapter 5 The Fundamental Unit of Life Chapter 6 Chapter 6 Tissues Chapter 7 Chapter 7 Diversity in Living Organisms Chapter 8 Chapter 8 Motion Chapter 9 Chapter 9 Forces and Laws of Motion Chapter 10 Chapter 10 Gravitation Chapter 11 Chapter 11 Work and Energy Chapter 12 Chapter 12 Sound Chapter 13 Chapter 13 Why do We Fall ill? Chapter 14 Chapter 14 Natural Resources Chapter 15 Chapter 15 Improvement in Food Resources

1. What is the difference between weight and mass?

Mass of an object is a measure of inertia. It is the measure of the amount of matter in a body and is constant everywhere. But weight is the force due to gravity. If m is the mass of the object and g is the acceleration due to gravity then weight w=mg. Weight varies according to the distance since g is dependent on distance. The weight of an object in the earth is different from the weight of the same object in the moon since the value of g is different in the moon and Earth. But the mass of the object will be the same in the moon and earth.

2. What are the important topics covered in NCERT Class 9 chapter Gravitation?

The important topics covered in Gravitation chapter of NCERT book for Class 9 Science are-

•  Universal Law of Gravitation
•  Free Fall
•  How To Calculate The Value Of g
•  Motion of Objects Under the Influence of Gravitational Force of the Earth
•  Concepts Of Mass And Weight
•  Thrust and Pressure
3. What is the SI unit of gravitation?

The SI unit of gravitation is also Newtons (N).

4. What is the value of g?

The value of acceleration due to gravity (g) on the Earth's surface is approximately 9.8 m/s². However, it can vary slightly depending on the location and altitude. For example, it is slightly less at the equator and slightly more at the poles.

5. What is gravitation Class 9 and formula?

The formula for the gravitational force between two objects is given by:

F = G * (m1 * m2) / r^2

6. Is class 9 science chapter gravitation question answer is helpful for higher class?

The Class 9 Science chapter on Gravitation, with its question-and-answer format, establishes a fundamental understanding of gravity that proves beneficial for higher-level studies in physics and related disciplines.

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