Class 9 Science Chapter 9 – Gravitation Exercise Solutions | NCERT Q&A

Find here the NCERT Class 9 Science Chapter 10 – Gravitation Exercise Solutions with detailed and simple explanations. All intext questions and answers are prepared according to the latest NCERT textbook. These Class 9 Science solutions help students understand key concepts like work, energy, power, and their relationship. Useful for quick revision, homework, and exam preparation.
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Class 9 Science Ch 9 :- Gravitation Exercise question Answer :- PDF

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GRAVITATION 

EXERCISE QUESTION ANSWER

 

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

Ans :- Let there be two bodies of masses m_₁ and m₂ and let the distance between them be r. 

The gravitational force "F" between them would be given by

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

\ F' = 4F

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

Ques-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 :- An object experiences acceleration during free fall denoted by g.

It is given by :- 

 

Where G = Gravitational constant, 

M = Mass of the earth,

R = Radius of the earth.

Thus, it is independent of mass.

Hence a heavy body does not fall faster than a light body. 

Ques-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 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶ m)

Ans:- Gravitational constant

==> G= 6.67 × 10⁻¹¹ N·m²/kg²

Mass of the earth, M_E = 6 x 10²⁴ kg   

==> Mass of the body, m = 1 kg

Radius of the earth (r) = 6.4 x 10⁶ m

By universal law of gravitation,

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. 

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

This force is given by: 

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

Ques 5 :- If the moon attracts the earth, why does the earth not move towards the moon?

Ans :- 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. 

Ques 6:- What happens to the force between two objects, if

(i) the mass of one object is doubled ?

F = force of attraction

G = constant of proportionality

m_₁ = mass of first object

m_₂ = mass of second object

 r = distance between them

Ans :- By universal law of gravitation ,

When mass of one object is doubled.

Let m' = 2m_₁

==> Putting these value in Eqn. (i)

==>  F' = 2F

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

(ii) the distance between the objects is doubled and tripled?

Ans :- By universal law of gravitation ,

* When distance is doubled ,

Let r' = 2r 

==> Putting these value in Eqn. (i)

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

* Now, When distance is Tripled,

Let r' = 3r 

==>  Putting these value in Eqn. (i)

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

(iii) the masses of both objects are doubled ?

Ans :- By universal law of gravitation ,

* When masses of both the objects are doubled, 

Let m₁' = 2 m_₁  and m₂' =2m₂  

==> Putting these value in Eqn. (i)

F'= 4F

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

Ques 7 :- What is the importance of universal law of gravitation?

Ans :- Universal law of Gravitation is important because :-

-- the force that is responsible for binding us to Earth.

-- the motion of moon around the earth.

-- the motion of planets around the sun.

-- the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both sun and moon on the earth.

Ques 8 :- What is the acceleration of free fall?

Ans:- Whenever an object falls towards the earth, acceleration is involved This acceleration is due to earth's gravitational force. Therefore, the acceleration is called as acceleration due to "gravitational force of earth".

The acceleration of free fall is denoted by "g" and its value on the surface of the earth is 9.8 m/s².

Ques 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 the object. Weight is equal to the product of acceleration due to the gravity and mass of the object.

Weight (W) = acceleration(a /g) x mass(m)

Ques 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.]

 

Ans:- No, the friend will not agree with the weight of gold because weight of an object depends on the value of 'g'. The value of 'g' is greater at poles than at equator. Therefore, weight of gold will be less at the equator than at the poles.

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

Ans :- The surface area of a crumpled paper ball is smaller than that of a sheet of paper. There will be a lot of air resistance for a sheet of paper.

A piece of paper falls slower than a ball that has crumpled as a result. 

Ques 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?

Ans :- Mass of an object remains constant. It does not change on moon or on the earth. Hence mass of the object on moon and on the earth is 10 kg.

* On earth

W = m x g  

==> W = 10 kg x 9.8 m /s²  

\ W = 98 N

* On moon

Gravitational force on moon,

g' = 1/6 x g  ==>  g' = 1/6 x 9.8  

==> g' = 1.63 m/s²

Weight, W₁ = m x g'

W₁ = 10 kg x 1.63 m/s²

==> W₁ =  16.3 N

Thus, weight of the object on the earth is 98 N and its weight on the moon will be 16.3.

Ques 13 :- A ball is thrown vertically upwards with a velocity of 49 m/s. calculate: -

(i) the maximum height to which it rises,

Ans :- Initial velocity (u) = 49 m/s

Final velocity v at maximum height = 0 m/s

Acceleration due to earth gravity " g"  = -9.8 m/s²

(negative as ball is thrown up).

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

By third equation of motion,

2gH = v² − u²   

==> Substituting we get,

2 x (-9.8) x H = 0 - (49)² 

-19.6 x H = - 2401

==> H = 122.5 m

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

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

Ans :- Total time T = Time to ascend (T_a) + Time to descend (T_d)

(Time to ascend (T_a) = Time to descend (T_d)

According to the first equation of motion,

 

Hence, total time it takes to return to the surface of the earth = 10 s.

Ques 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.

Ans :- Initial velocity u = 0 m/ s

==> Final Velocity v = ?

Acceleration due to gravity (g) = 9.8 m/s²

Height of tower = Distance travelled (s) = 19.6 m

Let the final velocity be "v"

According to the third equation of motion

v² − u² = 2gs    

==> v² - (0 m/s)² = 2 x 9.8 x 19.6

v² = 2 x 9.8 x 19.6  

\ V = 19.6

Hence, the velocity of the stone just before touching the ground is 19.6 m/s.

Ques 15 :- A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s² find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Ans :- (i) Initial velocity u = 40 m/s

Acceleration due to gravity "g" = -10 m/s²

(negative as ball is thrown up)

Final velocity at the highest point "V" = 0 m/s

Let the maximum height reached be "s"

As per the third equation of motion 

\ S = 80 m

(ii) Net displacement = 0 ( as the stone will return to the point from where it was thrown).

(iii) Total distance covered by the stone = 2s

=  2 x 80

Therefore total distance covered by the stone = 160 m.

Ques 16 :- Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Ans:- Mass of the earth (m_E) = 6 x 10²⁴ kg

Mass of the sum (m_s) = 2 x 10³⁰ kg

Distance between earth and sun (r) = 1.5 x 10¹¹ m

Gravitational constant (G) = 6.67 × 10⁻¹¹ N·m²·kg⁻².

By universal law of gravitation,

==> F = 3.56 x 10²² N

Ques 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.

 

Ans :-  (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 + ½ at²

 ==> x = 0 x t + ½  x 9.8 x t² 

==> x =  ½  x 9.8 x t²

x = 4.9 t²  …………. (i)

(b) When another stone is projected vertically upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 - x)  ....... (ii)

Time taken = t

According to the second equation of motion,

S = ut + ½ at²

 ==> 100 - x = 25 x t + ½  x (-9.8) x t²

=> 100 - x  = 25t + (-4.9 t²)

=> 100 - x  = 25t - 4.9 t² …………. (iii)

Putting value of x from Eqn. (i) to Eqn. (iii) we get

100 - 4.9 t²  = 25t - 4.9 t² 

==> 100 = 25t - 4.9 t² + 4.9 t²

100 = 25t  ===> t = 100/25  

\ t = 4s

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

Putting value of t= 4s in Eqn. (i) we get

X = 4.9 x (4)²   => 4.9 x 16 => 78.4 m

Putting value of x = 80m in Eqn. (ii) we get

Distance travelled = (100 - 78.4) =21.6 m

Hence, after 4 sec, 2 stones meet a distance of 21.6 m from the ground.

Ques 18 :- A ball thrown up vertically returns to the thrower after 6 s. Find

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

 

(i) Acceleration due to gravity, g =9.8

Time (t) = 3 s (since the time to go up and return is 6s The time to go up = 3s

Final velocity v = 0  
==> Initial velocity, u

By first equation of motion

V = u + gt    ==> 0 = u + (-9.8 m/s²) x 3s
==> 0 m/s = u - 29.4 m/s

u = 29.4 m/s

Thus, the velocity with which it was thrown up =29.4

Distance, s = Height, h = ?

 

(ii) By second equation of motion,

S = ut + ½ gt²    ==> S = 29.4 m/s x 3S + ½ x 9.8 m/s² x(t)²

S = 88.2 m - 44.1 m  ==> S = 44.1 m

Hence, maximum height stone reaches is 44.1 m

 

(iii) Time (t) = 4s  ==> Distance, s ?

By second equation of motion,

S = ut + ½ gt²           
==> S = 29.4 m/s x 4S + ½ x 9.8 m/s²  x(4)²                                                                    

S = 117.6 m - 78.4 m  ==> S = 39.2 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.

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

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

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

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

1. Gravitational force or its weight in vertically downward direction

2. 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. 

Ques 21:- The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/ cm³, will the substance float or sink?

Ans :- Given,  Volume of substance = 20 cm³

Mass of substance = 50 g 

==> Density of water = 1 g/cm³

Density of the substance ?

==> Density = 50/20  
==> Density = 2.5 g/cm³

Relative Density = 2.5/1  

==> Relative Density =  2.5

 

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

Ques 22:- The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g /cm³ ? What will be the mass of the water displaced by this packet?

Ans :- Given, Volume of sealed packet 350 cm³

Mass of sealed packet = 500 g    

==> Density of water = 1 g/cm³

Density of the sealed packet ?

Mass = 500 

==> Density =500/350  ==> Density = 1.42 g/cm³

 

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 cm³

Mass of water displaced p x V = 1 x 350 =350 g

Hence, mass of displaced water = 350 g