NCERT Solutions for Class 9 Science Chapter 9 Gravitation

NCERT
SOLUTIONS
CLASS - 9th

aglase .co

Book : Science Ncert Solutions | Chapter-10 Science

Class : 9th
Subject : Science
Chapter : 10
Chapter Name : GRAVITATION

Q1 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 m1 and m2 and the distance between them r, the force (F) of
attraction acting between them is given by the universal law of gravitation as:
Gm1 m2
F= r2
Where, G is the universal gravitation constant given by:
G = 6.67 × 10−11 Nm2 kg−2

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

Answer. Let ME 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:
Gm1 m2
F= r2

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Q1 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 in�uence of gravitational
force. The motion of the object is said to have free fall.

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Q2 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

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known as acceleration due to gravity (g). Its value is given by 9.8 m/s2.

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

Answer.

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

Answer. Let ME be the mass of the Earth and m be an object on the surface of the Earth. Let RE
be the radius of the Earth. According to the universal law of gravitation, weight WE of the
object on the surface of the Earth is given by,
GME m
WE = Let MM and RM the mass and radius of the moon. Then, according to the
R2E
universal law of gravitation, weight WN of the object on the surface of the moon is given by:
GMM m
WM =
R2M
WM MM R2E
=
WE ME R2M
ME = 5.98 × 1024 kg, MM = 7.36 × 1022 kg
RE = 6.4 × 106 m, RM = 1.74 × 106 m
2
W 7.36×1022 ×(6.37×106 )
∴ WM = 24 6 2
= 0.165 ≈ 16
E 5.98×10 ×(1.74×10 )

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Therefore, weight of an object on the moon is ⅙ of its weight on the Earth.

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Q1 Why is it dif�cult to hold a school bag having a strap made of a thin and strong string?

Answer. It is dif�cult 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.

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Q2 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.

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Q3 Why does an object �oat 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
�oats on the surface of the liquid. This is because the buoyant force acting on the object is
greater than the force of gravity.

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Q1 You �nd 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.

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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?

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Answer. The iron bar is heavier than the bag of cotton. 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 iron bar is more that that of the cotton bag.

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Q1 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 (r) between them, i.e.,
F ∝ r12
If distance r becomes r/2, then the gravitational force will be proportional to
1
2
= r42
( 2r )

Hence, if the distance is reduced to half, then the gravitational force becomes four times larger
than the previous value.

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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?

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.

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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 × 106 m.)

Answer. According to the universal law of gravitation, gravitational force exerted on an object
of mass m is given by:
F = GMm
r2
Where,
Mass of Earth, M = 6 × 1024 kg
Mass of object, m = 1kg
Universal gravitational constant, G = 6.7 × 10−11 Nm2 kg−2
since the object is on the surface of the Earth, r = radius of the Earth (R)
r = R = 6.4 × 106 m

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Book : Science Ncert Solutions | Chapter-10 Science

F = GMm
R2
Gravitational force,
−11
×6×1024 ×1
= 6.7×10 2
= 9.8N
(6.4×106 )

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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?

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 mcn)n attracts the earth.

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Q5 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.

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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?

Answer. (i) Doubled (ii) One-fourth and one-ninth (iii) four times
According to the universal law of gravitation, the force of gravitation between two objects is
given by:
Gm1 m2
F= r2
(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.

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Q7 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.

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Q8 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 m s-2, which is constant for all
objects (irrespective of their masses).

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Q9 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.

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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.]

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.

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Q11 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.

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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 newtons of a 10 kg object on the moon and on the earth?

Answer. Weight of an object on the moon = 16 × Weight of an object on the Earth
Also,
Weight = Mass x Acceleration
Acceleration due to gravity, g = 9.8 m/s2
Therefore, weight of a 10 kg object on the Earth = 10 x 9.8 = 98 N
And, weight of the same object on the moon = 16 × 98 = 16.3N

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Q13 . 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) 122.5 m (ii) 10 s
According to the equation of motion under gravity:
v2 − u2 = 2gs
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, �nal velocity of the ball is zero, I.e. V = 0
u = 49 m/s
During upward motion, g = - 9.8m s-2
Let h be the maximum height attained by the ball.
Hence,
0 − (49)2 = 2 × (−9.8) × h
h = 49×49
2×9.8
= 122.5m
Let t be the time taken by the ball to reach the height 122.5 m, then according to
the equation Of motion:
v = u + gt
We get,
0 = 49 + t x (-9.8)
9.8t = 49
49
t = 9.8 = 5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

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

Answer. According to the equation of motion under gravity:
v2 − u2 = 2gs
Where,
Initial velocity Of the Stone = 0
v = Final velocity Of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity 9.8 m s-2
∴ v2 − 02 = 2 × 9.8 × 19.6
v2 = 2 × 9.8 × 19.6 = (19.6)2
v = 19.6ms−1
Hence, the velocity of the stone just before touching the ground is 19.6 m s-1.

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Q15 A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2 ,
�nd 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:
v2 − u2 = 2gs
Where,
u = Initial velocity of the stone = 40 m/s
v = �nal velocity of the stone = 0
S = Height of the stone
g = Acceleration due to gravity = -10 m s-2
Let h be the maximum height attained by the stone.
Therefore,
0 − (40)2 = 2 × h × (−10)
h = 40×40
20
= 80m
Therefore, total distance covered by the stone during its upward and downward Journey = 80 +
80 = 160 m
Net displacement of the stone during its upward and downward journey = 80+(-80) = 0

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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.

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Answer. According to the universal law Of gravitation, the force Of attraction between the
Earth and the Sun is given by:
GMsun MEarth
F= R2
Where,
M sun = Mass of the Sun = 2 × 1030 kg
MEarth = Mass of the Earth = 6 × 1024 kg
R = Average distance between the Earth and the Sun = 1.5 × 1011 m
G = Universal gravitational constant = 6.7 × 10−11 Nm2 kg−2
−11
×2×1030 ×6×1024
F = 6.7×10 11 2
= 3.57 × 1022 N
(1.5×10 )

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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.

Answer. Let the two stones meet after a time t.
(i) 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.
Acceleration due to gravity, g = 9.8 m s-2
From the equation of motion,
s = ut + 12 gt2
= 0 × t + 12 × 9.8 × t2
∴ s = 4.9t2
1

(ii) For 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

Equation of motion,
s′ = ut + 12 gt2
= 25t − 12 × 9.8 × t2
∴ s′ = 25t − 4.9t2
2

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

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∴ s + s′ = 100
1 2
2
gt + 25t − 12 gt2 = 100
∴ t = 100
25
= 4s
In 4 s, the falling stone has covered a distance given by equation (1) as
s = 12 × 10 × 42 = 80m
Therefore, the stones will meet after 4 s at a height (100 - 80) = 20 m from the ground.

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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 4 s.

Answer. (a) 29.4m/s (b) 44.1m (c) 39.2 m above the ground
(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
Acceleration due to gravity, g = -9.8 m s -2
Equation of motion, v = u + gt will give,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4ms−1
Hence, the ball was thrown upwards with a velocity of 29.4 m s-1
(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m s1
Final velocity, v = 0
Acceleration due to gravity, g = -9.8 m s-2
From the equation of motion, s = ut + 12 at2
h = 29.4 × 3 + 12 × −9.8 × (3)2 = 44.1m
(c) dall attains the maximum height after 3 s. After attaining this height, it will start falling
downwards.
In this case,
Initial velocity, u = 0
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.
Equation of motion, s = 0 × t + 12 × 9.8 × 12 = 4.9m
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m - 4.9 m) above the ground after 4 seconds.

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Q19 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.

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Q20 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.

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

Answer. If the densitdy 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
�oats on the surface of the liquid.
Mass of the substance
Here, density Of the substance = Volume of the substance
= 50
20
= 2.5gcm−3
The density of the substance is more than the density of water (1 g cm-3). Hence, the substance
will sink in water.

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Q22 The volume of a 500 g sealed packet is 350 cm3 . Will the packet �oat 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?

Mass of the packet
Answer. Density Of the 500 g sealed packet = Volume of the packet
= 500
350
= 1.428gcm−3
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.,
350 g.

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