NCERT Solutions for Class 9 Science Chapter 10 - Gravitation

NCERT Solutions For Class 9 Science Chapter 10 PDF Free Download

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

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

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

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Class 9 Science NCERT Solutions for Gravitation part 1
Class 9 Science NCERT Solutions for Gravitation part 2
Class 9 Science NCERT Solutions for Gravitation part 3
Class 9 Science NCERT Solutions for Gravitation part 4
Class 9 Science NCERT Solutions for Gravitation part 5
Class 9 Science NCERT Solutions for Gravitation part 6
Class 9 Science NCERT Solutions for Gravitation part 7
Class 9 Science NCERT Solutions for Gravitation part 8
Class 9 Science NCERT Solutions for Gravitation part 9
Class 9 Science NCERT Solutions for Gravitation part 10
Class 9 Science NCERT Solutions for Gravitation part 11
Class 9 Science NCERT Solutions for Gravitation part 12
Class 9 Science NCERT Solutions for Gravitation part 13

 

Access Answers of Science NCERT Class 9 Chapter 10 – Gravitation ( All In text and Exercise Questions Solved)

In text Questions-10.1 Page number 134

Q1. State the universal law of gravitation.

Ans :

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.

Q2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Ans :

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

Also consider ‘m’ is the mass of the object on the surface of earth and ‘Me’ is the mass of earth.

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

G = \(\frac{GMm}{r^{2}}\)

In text Questions-10.2 Page Number 136

Q1. What do you mean by free fall?

Ans :

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.

Q2. What do you mean by acceleration Q due to gravity?

Ans :

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

The gravity-based acceleration value is g = \(\frac{9.8m}{s^{2}}\)

 

In text Questions -10.4 Page Number 138

Q1. What are the differences between the mass of an object and its weight?

Ans :

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

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

Q2. Why is the weight of an object on the moon 1/6th its weight on the earth?

Ans :

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

Intext Questions -10.5 Page: 141

Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Ans :

It is tough to carry a college bag having a skinny strap as a result of the pressure on the shoulders is sort of giant. This can be as a result of the pressure is reciprocally proportional to the expanse on that the force acts. The smaller is that the surface area; the larger are going to be the pressure on the surface. Within the case of a skinny strap, the contact expanse is extremely tiny. Hence, the pressure exerted on the shoulder is extremely giant.

Q2. What do you mean by buoyancy?

Ans :

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

Q3. Why does an object float or sink when placed on the surface of water?

Ans :

An object float or sink when placed on the surface of water because of two reasons.

  • If its density is greater than that of water, an object sinks in water.
  • If its density is less than that of water, an object floats in water.

 

In text Questions-10.6 Page 142

Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Ans :

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

Q2. 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?

Ans :

The bag of cotton is heavier than the bar of iron. The cotton bag has a larger air thrust than the iron bar. The weighing machine therefore indicates a smaller cotton bag weight than its actual weight.

Exercise Questions Page 143

Q3. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Ans :

Consider the Universal law of gravitation,

According to that law, the force of attraction between two bodies is F = \(\frac{Gm_{1}m_{2}}{r^{2}}\)

Where

  • m1 and m2 are the masses of the two bodies.
  • G is the gravitational constant.
  • r is the distance between the two bodies.

Given that the distance is reduced to half then

r = 1/2 r

Therefore,
F = \(\frac{Gm_{1}m_{2}}{r^{2}}\)

F = \(\frac{Gm_{1}m_{2}}{r/2^{2}}\)

F = \(\frac{4Gm_{1}m_{2}}{r^{2}}\)

F = 4F

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

Q2. 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 objects fall on the bottom with constant acceleration called acceleration thanks to gravity (g). It’s constant and therefore the price of ‘g’ doesn’t depend on the mass of associate object. So serious objects don’t fall quicker than light-weight objects provided there’s no air resistance.

Q3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106m.)

Ans :

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by F = \(\frac{Gm_{1}m_{2}}{r^{2}}\)

Here,

m1 = mass of Earth = 6.0 x 1024kg

m2 = mass of the body = 1kg

r = distance between the two bodies

Radius of Earth = 6.4 x 106m

G = Universal gravitational constant = 6.67 x 10-11 Nm2kg-2

By substituting all the values in the equation

F = \(\frac{Gm_{1}m_{2}}{r^{2}}\)

F = 6.67 x 1011 (6 x 1024 x 1 ) / (6.4 x 106)2

F = 9.8 N

This shows that Earth exerts a force of 9.8 N on a body of mass 1 kg. The body will exert an equal force of attraction of 9.8 N on the Earth.

Q4. 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 :

The earth attracts the moon with associate degree equal force with that the moon attracts the planet however these forces are in opposite directions. By universal law of gravitation, the force between moon and also the sun can be,
F = \(\frac{Gm_{1}m_{2}}{d^{2}}\)

Where

d = distance between the earth and moon.

m1 m2 = masses of earth and moon respectively.

Q5. If the moon attracts the earth, why does the earth not move towards the moon?

Ans :

According to universal law of gravitation and Newton third law, we all know that the force of attraction between 2 objects is same however in wrong way. So the planet attracts the moon with the identical force because the moon exerts on earth however in opposite directions. Since earth is far larger in size than moon, that the acceleration cannot be detected on earth surface.

Q6. 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?

Ans :

(i)

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

F = \(\frac{Gm_{1}m_{2}}{R^{2}}\)

If the mass is doubled for one object.

F = 2F, so force is also doubled.

(ii)

If the distance between the objects is doubled and tripled

If it’s doubled

Hence,

F = \(\frac{Gm_{1}m_{2}}{2R^{2}}\)

F = 4F, Force thus becomes one-fourth of its initial force.

If it’s tripled

Hence,

F = \(\frac{Gm_{1}m_{2}}{3R^{2}}\)

F = 9F, Force thus becomes one-ninth of its initial force.

(iii)

If masses of both the objects are doubled, then

F = \(\frac{G2m_{1}2m_{2}}{R^{2}}\)

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

Q7. What is the importance of universal law of gravitation?

Ans :

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

  • The motion of the moon round the earth
  • The force that binds North American nation to the world
  • The tides because of the moon and therefore the Sun
  • The motion of planets round the Sun

Q8. What is the acceleration of free fall?

Ans :

When anybody is in free fall, the sole force functioning on the article is that the earth’s field of force. By Newton’s second law of motion all the forces manufacture acceleration therefore all the objects accelerate toward the world’s surface thanks to attraction of the earth.

This acceleration is thought as acceleration thanks to gravity close to earth’s surface. It’s denoted by ‘g’ and its worth is 9.8ms-2 and it’s constant for all objects close to earth’s surface (irrespective of their masses).

Q9. What do we call the gravitational force between the earth and an object?

Ans :

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

Q10. 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.]

Ans :

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

W = mg wherever (m = mass of the body and g = acceleration thanks to gravity)

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

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

Q11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Ans :

A sheet of paper has a lot of area as compared to a crumpled paper ball. A sheet of paper must face a lot of air resistance. As result a sheet of paper falls slower than the crumpled ball.

Q12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton’s of a 10 kg object on the moon and on the earth?

Ans :

Given data:

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

Object weight m = 10 kg

Acceleration due to moon gravity = gm

Weight on the earth= We

Weight on the moon = Wm

Weight = mass x gravity

gm = (1/6) ge (given)

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

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

We = m x ge = 10 x 9.8

We = 98N

Q13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate The maximum height to which it rises, The total time it takes to return to the surface of the earth.

Ans :

Given data:

Initial velocity u = 49m/s

Final speed v at maximum height = 0

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

By third equation of motion

v2 = u2 – 2gs

Substitute all the values in the above equation

0 = (49)2-2 x 9.8 x s

s = 122.5m

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

V = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5s

Also Td = 5s

Therefore T = Ta + Td

T = 5 + 5

T = 10s

Q14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Ans :

Given data:

Initial velocity

u = 0

Tower height = total distance = 19.6m

g = 9.8 m/s2

Consider third equation of motion

v2 = u2 + 2gs

v2 = 0 + 2 x 9.8 x 19.6

v2 = 384.16

v = 19.6m/s

Q26. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Ans :

Given data:

Initial velocity u = 40m/s

g = 10 m/s2

Max height final velocity = 0

Consider third equation of motion

v2 = u2 – 2gs [negative as the object goes up]
⇒ 0 = (40)2 – 2 x 10 x s
⇒ s = (40 x 40) / 20
Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

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

Q16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.

Ans :

Given data:

Mass of the sun ms = 2 × 1030 kg

Mass of the earth me = 6 × 1024 kg

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

Average distance r = 1.5 × 1011 m

Consider Universal law of Gravitation

F = \(\frac{Gm_{1}m_{2}}{d^{2}}\)

F = (6.67 × 10-11 × 6 × 1024 × 2 × 1030 ) / (1.5 × 101)2

F = 3.56 x 1023 N

Q17. 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.

Ans :

Given data:

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

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,
⇒ x = 0 + (1/2) gt2
⇒ x = 5t2 ———(a)
(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t
⇒ (100 – x) = 25t + (1/2) x 10 x t2
x = 100 -25t + 5t2 ——- (b)

From equations (a) and (b)

5t2 = 100 -25t + 5t2

⇒ t = (100/25) = 4sec.

After 4sec, two stones will meet

From (a)

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

Putting the value of x in (100-x)

= (100-80) = 20m.

This means that after 4sec, 2 stones meet a distance of 20 m from the ground.

Q18. 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.

Ans :

Given data:

g = 10m/s2

Total time T = 6sec

⇒ Ta = Td = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u – gta

⇒ u = v + gta

= 0 + 10 x 3

= 30m/s

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

(b) From second equation of motion

s = uta – 1/2(g (ta)2

= 30 x 3 – (1/2) x 10 x (3)2

= 90-45

= 45m

The maximum height stone reaches is 45m.

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

Distance travelled in another 1sec = s’

s = uta – 1/2(g (ta)2

s = 0 + 10 x 1 x 1

s = 5m.

The distance travelled in another 1sec = 5m.

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

= 40m from the ground.

Q19. In what direction does the buoyant force on an object immersed in a liquid act?

Ans :

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

Q20. Why a block of plastic does released under water come up to the surface of water?

Ans :

The density of plastic is a smaller amount than that of water, therefore the force of buoyancy on plastic block are going to be bigger than the load of plastic block displaced. Hence, the acceleration of plastic block are going to be in upward direction, and comes up to the surface of water.

Q21. The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm–3, will the substance float or sink?

Ans :

To find the Density of the substance the formula is

Density = (Mass/Volume)

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

Density of water = 1g/cm3

Density of the substance is greater than density of water. So the substance will float.

Q22. The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm–3? What will be the mass of the water displaced by this packet?

Ans :

Density of sealed packet = 500/300 = 1.42 g/cm3

Density of sealed packet is greater than density of water

Therefore the packet will sink.

Considering Archimedes Principle,

Displaced water volume = Force exerted on the sealed packet.

Volume of water displaced = 350cm3

Therefore displaced water mass = ρ x V

= 1 x 350

Mass of displaced water = 350g.

NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

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

The topics usually covered under this chapter are:

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

NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

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

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

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

  1. Solutions provided in an easy-to-understand language
  2. Qualified teachers and their vast experience helps formulate the solutions
  3. Questions updated to the latest prescribed syllabus
  4. A detailed breakdown of the most important exam questions
  5. Access to a learning resources like sample papers and previous year question papers

Further Reading: NCERT Solutions Class 9 Science

Frequently Asked Questions on chapter 10 Gravitation

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 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.

What do you mean by free fall?

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.

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

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

What do you mean by buoyancy?

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

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

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

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 the bottom with constant acceleration called acceleration thanks to gravity (g). It’s constant and therefore the price of ‘g’ doesn’t depend on the mass of associate object. So serious objects don’t fall quicker than light-weight objects provided there’s no air resistance.

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

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

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

A sheet of paper has a lot of area as compared to a crumpled paper ball. A sheet of paper must face a lot of air resistance. As result a sheet of paper falls slower than the crumpled ball.

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

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

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

The density of plastic is a smaller amount than that of water, therefore the force of buoyancy on plastic block are going to be bigger than the load of plastic block displaced. Hence, the acceleration of plastic block are going to be in upward direction, and comes up to the surface of water.

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