Find here the NCERT Class 9 Science Chapter 10 – Gravitation Exercise Solutions with detailed and simple explanations. All exercise questions and answers are prepared according to the latest NCERT textbook. These solutions help students understand key concepts such as gravitation, universal law of gravitation, free fall, mass, weight, and thrust and pressure. Useful for quick revision, homework, and exam preparation.
Chapter 9 – GRAVITATION
Exercise Solutions
Ques-1:- How does the force of gravitation between two objects change when the distance between them is reduced to half ?
Ans: Let two objects (bodies) of masses m₁ and m₂ be separated by a distance r
According to the universal law of gravitation, the force of gravitation F between them is
F = G m₁ m₂ (r)²
If the distance between the two bodies is reduced to half, the new distance becomes r/2.
The new gravitational force F' is:
F′ = G m₁ m₂ (r/2)²
F′ = G m₁ m₂ r²/4
F′ = 4 G m₁ m₂ r²
∴ F′ = 4F
Therefore, the force of gravitation becomes four times when the distance between the two objects is reduced to half.
For 1 marks :- Gravitational force increases four times when distance 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: Although gravitational force acting on an object depends on its mass, the acceleration produced due to gravity during free fall is the same for all objects.
The acceleration due to gravity is given by:
F = GM R²
Where
G = Gravitational constant,
M = Mass of the earth,
R = Radius of the earth.
• This expression does not contain the mass of the falling object, so the acceleration due to gravity is the same for both heavy and light objects.
• Therefore, in the absence of air resistance, a heavy object does not fall faster than a light object.
Hence, acceleration due to gravity is independent of mass.
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, Mₑ = 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
F' = GMₑm R²
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 the universal law of gravitation, any two objects having mass attract each other with a gravitational force that is equal in magnitude and opposite in direction.
The gravitational force between the Earth and the Moon is given by:
F =
GMm
r²
where
M is the mass of the Earth,
m is the mass of the Moon and ,
r is the distance between their centres.
Hence, the Earth attracts the Moon with the same force with which the Moon attracts the Earth, but in opposite directions.
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 compared to the moon. the acceleration produced in the earth is extremely small.
Hence, the motion of the earth towards the moon is negligible, while the moon shows a noticeable motion around the earth.
Ques 6:- What happens to the force between two objects, if
(i) the mass of one object is doubled ?
Ans: By universal law of gravitation ,
F′ = G m₁ m₂ r² -----(i)
When mass of one object is doubled.
Let m₁ = 2m₁
Putting these value in Eqn. (i)
F′ = G 2m₁ m₂ r²
F′ = 2G m₁ m₂ r²
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 ,
F′ = G m₁ m₂ r² -----(i)
When distance is doubled ,
Let r = 2r
Putting these value in Eqn. (i)
F′ = G m₁ m₂ (2r)²
F′ = G m₁ m₂ 4r²
F′ = F 4
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)
F′ = G m₁ m₂ (3r)²
F′ = G m₁ m₂ 9r²
F′ = F 9
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 ,
F′ = G m₁ m₂ r² -----(i)
When masses of both the objects are doubled,
Let m₁ = 2m₁ and m₂ =2m₂
Putting these value in Eqn. (i)
F′ = G 2m₁ 2m₂ r²
F′ = 4G m₁ m₂ r²
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 the Earth.
• the motion of the Moon around the Earth.
• the motion of planets around the Sun.
• the formation of tides in oceans due to the gravitational force exerted by the Moon and the Sun on the Earth.
Ques 8 :- What is the acceleration of free fall?
Ans: The acceleration produced in a body when it falls freely under the influence of the Earth’s gravitational force is called acceleration of free fall or acceleration due to gravity.
It 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 the mass of the object and the acceleration due to gravity.
Weight (W)=mass (m)×acceleration due to gravity (g)
W = m x g
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 the gold.
• The weight of an object depends on the value of acceleration due to gravity (W=mg).
• The value of g is greater at the poles than at the equator.
• Therefore, when the same gold is taken to the equator, its weight becomes less, even though its mass remains the same.
Ques 11 :- Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans: A sheet of paper has a larger surface area than a crumpled paper ball.
• Due to its larger surface area, it experiences greater air resistance while falling.
• The crumpled paper ball has a smaller surface area and hence experiences less air resistance.
• Therefore, a sheet of paper falls slower than a crumpled paper ball.
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: The mass of an object remains constant. It does not change on the moon or on the earth.
Hence, the mass of the object on both the moon and the earth is 10 kg.
On earth
W = m x g
==> W = 10 kg x 9.8 m s⁻²
==> W = 98 N
On the moon
Acceleration due to gravity on the moon,
g' = 1/6 x g
==> g' = 1/6 x 9.8
==> g' = 1.63 m s⁻²
Weight, W1 = m x g'
W1 = 10 kg x 1.63 ms⁻²
==> W1 = 16.3 N
Thus, the weight of the object on the earth is 98 N and its weight on the moon is 16.3 N.
Ques 13 :- A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate :-
(i) the maximum height to which it rises,
Given
Initial velocity (u) = 49 m/s.
Final velocity v at maximum height = 0 m/s
Acceleration due to earth gravity " g" = -9.8 m s⁻²
(negative as ball is thrown up).
Let the maximum height be H.
Using the third equation of motion:
v² = u² + 2gH
2gH = v² - u²
2(-9.8)H = (0)² - (49)² +
2(-9.8)H = - 2401
-19.6 H = - 2401
H = -2401 -19.6 = 122.5 m
∴ Maximum height = 122.5 m
(ii) the total time it takes to return to the surface of the earth
Ans: Time to ascent = Time to descent
using the first equation of motion,
v = u + gt
0 = 49 + (-9.8 t)
0 = 49 -9.8 t
9.8 t = 49
t = 49 9.8 = 5 s
Time to ascent = Time to descent = 5 s
Total time (T) = Time to ascent + Time to descent
Total time (T) = 5 + 5 = 10 s
∴ Total time = 10 s
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.
Initial velocity, u = 0 m s⁻¹
Acceleration due to gravity,
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,
v² − u² = 2gs
v² − 0 = 2 × 9.8 × 19.6
v² = 19.6 × 19.6
v = √ 19.6 × 19.6
v = 19.6 m s⁻¹
∴ The velocity of the stone just before touching the ground is 19.6 m s⁻¹ (downward).
Ques 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?
Ans: (i) Maximum height
Initial velocity, u = 40 m s⁻¹
Acceleration due to gravity,
g = −10 m s⁻²
Final velocity at the highest point, v = 0 m s⁻¹
Let the maximum height be s.
Using the third equation of motion,
v² − u² = 2gs
0 − (40)² = 2 × (−10) × s
-1600 = −20s
s = -1600 -20 = 80 m
Maximum height reached = 80 m
(ii) Net displacement
Since the stone returns to the point from where it was thrown,
Net displacement = 0
(iii) Total distance covered
Distance going up = 80 m
Distance coming down = 80 m
Total distance = 2s = 2 × 80
Total distance covered = 160 m
Ques 16 :- Calculate the force of gravitation between the Earth and the Sun.
Ans: Given:
Mass of the Earth,
mₑ = 6 × 10²⁴ kg
Mass of the Sun,
mₛ = 2 × 10³⁰ kg
Distance between Earth and Sun,
r = 1.5 × 10¹¹ m
Gravitational constant,
G = 6.67 × 10⁻¹¹ N m² kg⁻²
By universal law of gravitation, F = G mₑ mₛ r²
F = 6.67×10⁻¹¹×6×10²⁴×2×10³⁰ (1.5×10¹¹)²
F = 6.67× 6 x 2 x 10⁻¹¹⁺²⁴⁺³⁰ (1.5×10¹¹)²
F = 80.84 x 10⁴³ 2.25 x 10²²
F = 35.57 x 10⁴³⁻²²
F = 35.57 x 10²¹ N
F = 3.56 x 10²² N
∴ The gravitational force between the Earth and the Sun is = 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⁻¹. Calculate when and where the two stones will meet.
Ans: (a) Stone falling from the top
Initial velocity,
u = 0
Distance travelled = x
Time taken = t
Using the second equation of motion,
s = ut + (1/2)at²
x = 0 + ½ × 9.8 × t²
x = 4.9t² ----(i)
Ans: (b) Stone projected upwards from the ground
Initial velocity,
u = 25 m s⁻¹
Distance travelled = (100 − x)
Acceleration,
a = −9.8 m s⁻²
Time taken = t
Using the second equation of motion,
100 − x = 25t + ½(−9.8)t²
100 − x = 25t − 4.9t² ----(ii)
Putting x = 4.9t² in equation (ii)
100 − 4.9t² = 25t − 4.9t²
100 = 25t
100 / 25 = t
t = 4
Position where they meet
Putting t = 4 s in Eqn. (i):
x = 4.9 X (4)²
x = 4.9 X 16 = 78.4 m
Distance from the ground :
100 - 78 = 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
Ans: Given
Total time of flight = 6 s
Time to reach maximum height = 3 s
Acceleration due to gravity:
g = −9.8 m s⁻² (negative because motion is upward)
(a) Velocity with which the ball was thrown up
At maximum height:
Final velocity v = 0
Using first equation of motion
v = u + gt
0 = u + (-9.8)(3)
0 = u - 29.4 m s⁻¹
u = 29.4 m s⁻¹
(ii) Maximum height reached
Using second equation of motion
s = ut + ½ at²
s = (29.4)(3) + ½ (-9.8)(3)²
s = 88.2 - ½ (9.8)(3 x 3)
s = 88.2 - 4.9 x 9
s = 88.2 - 44.1
s = 44.1 m
Hence, maximum height stone reaches is 44.1 m
(iii) Position of the ball after 4 s
s = ut + ½ gt²
s = (29.4)(4) + ½ (-9.8)(4)²
s = 117.6 - ½ (9.8)(4 x 4)
s = 117.6 - 4.9 x 16
s = 117.6 - 78.4
s = 39.2 m
So after 4 s, the ball is:
• 39.2 m above the point of projection
• Maximum height = 44.1 m
Distance below top: 44.1−39.2= 4.9 m
Ball is 4.9 m below the highest point
Ques 19 :- In what direction does the buoyant force on an object immersed in a liquid act?
Ans: The buoyant force on an object immersed in a liquid acts vertically upwards.
Because the pressure exerted by the liquid increases with depth, the force on the lower surface of the object is greater than the force on the upper surface, resulting in a net upward force.
Ques 20:- Why does a block of plastic released under water come up to the surface of water?
Ans: When a block of plastic is released under water, two forces act on it:
1. Gravitational force (weight) acting vertically downward. 2. Buoyant force acting vertically upward.
Since the density of plastic is less than the density of water, the buoyant force acting on the plastic is greater than its weight. Therefore, the block of plastic moves upward and comes to the surface of water.
Ques 21:- 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: Given:
Volume of substance = 20 cm³
Mass of substance = 50 g
Density of water = 1 g cm⁻³
Density of substance = Mass / Volume
Density = 50 / 20
Density = 2.5 g cm⁻³
Relative Density = (Density of substance) / (Density of water)
Relative Density = 2.5 / 1 = 2.5
Since the density of the substance is greater than the density of water, 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 cm⁻³ ? 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 cm⁻³
Density of the sealed packet:
Density=Mass/Volume=500/350
Density =1.42 g cm⁻³
Since the density of the sealed packet is greater than the density of water, the packet will sink.
According to Archimedes’ Principle:
When a body sinks completely in water, the volume of water displaced = volume of the body.
Volume of water displaced = 350 cm³
Mass of water displaced = ρ × V
= 1 × 350
= 350 g
• The packet will sink.
• Mass of water displaced = 350 g.
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