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GRAVITATION
10
TEXTBOOK QUESTIONS AND THEIR ANSWERS
Q. 1. State the Universal law of Gravitation.
Ans. Newton’s Universal law of Gravitation states that every object
in the universe attracts every other object with a force which is
proportional to the square of the distance between them. This
force acts along the line joining the centre of the two objects. If
m1 and m2 are the masses of two objects lying at distance ‘d’
apart, them force and between them is given by :
g =
Gm1m2
d2
{Where ‘G’ is the
universal
gravitational constant}
Q. 2. Write the formula to find the magnitude of gravitational
force between the earth and an object on the surface of the
earth.
Ans.
Fgravitation =
G . M earth × mobject
d2
When ‘G’ is the gravitational constant, ‘Mearth’ the mass of the
earth, ‘mobject’ the mass of the object and ‘d’ the distance
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between the centre of gravity of the object and the centre of the
earth.
Q. 3. What do you mean by free fall?
Ans. When an object falls with a constant acceleration, under the
influence of force of gravitation of the earth, object is said to
have free fall.
Q. 4. What do you mean by acceleration due to gravity?
Ans. The constant acceleration experienced by a freely falling object
towards the earth is called acceleration due to gravity. Its
average value on the surface of the earth is 9.81 ms–2.
Q. 5. What are the differences between the mass of an object and
its weight?
Ans.
Mass
Weight
(a) It is the amount of matter in (a) It is the force of gravity acting
on a body.
a body.
(b) It is a constant quantity at all (b) It is variable quantity and
changes with the change in
the places in universe.
acceleration due to gravity.
(c) It is measured by a physical (c) It is measured by a spring
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balance.
(d) Its SI unit is kilogram.
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balance.
(d) Its SI unit is Newton.
Q. 6 Why is the weight of an object on the moon
1
th its
6
weight on the earth?
Ans.Mass of the Moon (M)= 7.4 × 1022 kg.
Radius of the Moon (R) = 1.74 × 106 m
Gravitational constant (G) = 6.7 × 10–11 Nm2/kg2
∴ Acceleration due to gravity on the Moon (gMoon) =
GM
R2
6.7 × 10−11 Nm 2 kg −2 × 7.4 × 1022 kg
gMoon =
(1.74 × 106 m) 2
6.7 × 7.4 10−11 + 22
×
N /kg
gMoon =
1.74 × 1.74
1012
gMoon = 1.63 ms–2.
g Moon 1 ⋅ 63 ms −2 1
∴
=
= approx .
g Earth 9.81 ms −2 6
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∴ gMoon =
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1
g
.
6 Earth
Q. 7 Why is it difficult to hold a school bag having a strap made
of thin and strong string?
Ans.For the constant force, the pressure experienced is inversely
proportional to area of cross-section. Now, when the string is
thin, it has less area of cross-section, and hence, exerts greater
pressure on the hand for the given weight of school bag. Thus, it
becomes difficult to hold the school bag.
Q. 8
What do you mean by buoyancy?
Ans. Buoyancy is the tendency of a fluid to exert an upward force on
a body immersed partially or wholly in it.
Q. 9
Why does an object float or sink, when placed on the
surface of water?
Ans. When the object has density less than 1 gcm–3 i.e. density of
water, then it floats on the surface of water, because, it always
displaces more weight of water than its own weight. As buoyant
force is more than its own weight, therefore, it floats.
When the object has density more than 1 gcm–3 i.e. density of
water, then it sinks in water, because, it always displaces less
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weight of water than its own weight. As buoyant force is less
than its own weight, therefore, it sinks.
Q. 10. You find your mass to be 42 kg on a weighing machine. Is
your mass more or less than 42 kg?
Ans. Mass is always a constant quantity. So, it cannot be more or less
than 42 kg.
Q. 11. 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. Iron is heavier.
Reason : Heaviness of an object is determined by its density.
As the iron has far more density than cotton, therefore, in reality
iron is heavier than cotton.
Q.12 How does the force of gravitation between two objects
change when the distance between them is reduced to half?
Ans. All other variables remaining constant, the force of gravitation
is inversely proportional to the square of distance between the
two objects.
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F ∝
F ∝
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1
d2
1
(1 2 )
2
= 4.
Thus, force of gravitation increases 4 times.
Q. 13 Gravitational force acts on all objects in proportion to their
masses. Why then, a heavy object does not fall faster than
the light object?
Ans. A freely falling object of any mass falls under the action of
gravity which is given by the expression
g =
G.M
d2
Where ‘G’ is the constant of gravitation, ‘M’ the mass of earth
and ‘d’ the distance between the object and the centre of earth.
As the acceleration due to gravity is independent of the mass of
the objects, therefore, all objects fall at the same speed towards
the earth.
Q. 14 What is the magnitude of gravitational force between the
earth and 1 kg object on its surface? [Mass of the earth is
6 × 1024 kg and radius of the earth is 6.4 × 106 m.]
Ans. Fgravitation=
G × M earth × mobject
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d2
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=
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6.67 × 10−11 Nm 2 kg −2 × 6 × 2424 kg × 1 kg
(
6.4 × 106
)
2
m2
6.67 × 6 × 10−11+ 24
=
6.4 × 6.4 × 1012
=
6.67 × 6
× 10−11+ 24 −12 N
6.4 × 6.4
= 0.9770 × 101 N = 9.770 N.
Q. 15 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 the same force as the moon
attracts the earth.
It is because, the gravitational force between any two bodies is
mutual and equal according to Newton’s universal law of
gravitation.
Q. 16 If the moon attracts the earth, why does the earth not move
towards the moon.
Ans. The force exerted by the earth or the moon on each other is
insufficient to move the earth on account of its huge mass.
Q. 17 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?
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(iii) the masses of objects are doubled?
We know that F =
Ans. (i)
GmM
R2
If m = 2m
Then F =
G 2mM
GmM
=
=
2
R2
R2
Hence, when mass of one object is doubled, force of gravitation
doubles.
(a) We know that F =
GmM
R2
When the distance between the objects is doubled we get
F=
GmM GmM
=
(2 R) 2
4R2
1 ⎛ GmM ⎞
⇒F= ⎜
⎟
4 ⎝ R2 ⎠
Hence, when R is doubled, force of gravitation decreases by 4
times.
b) We know that F =
GmM
R2
When the distance between the object is tripled
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We get F =
GmM
( 3R )
2
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1 ⎛ mM ⎞
= ⎜ 2 ⎟
9⎝ R ⎠
Hence, when R is tripled, force of gravitation decreases by 9
times.
(c) We know that F =
GmM
R2
When masses of both objects are doubled we get,
F=
G (2m)(2 M )
R2
⎛ mM ⎞
⇒ F = 4⎜ 2 ⎟
⎝ R ⎠
Hence, masses of objects are doubled, force of gravitation
increases four times.
Q. 18 What is the importance of Universal Law of Gravitation?
Ans. The following are the importance of Universal Law of
Gravitation:
1. It is the gravitational force between the sun and the earth,
which makes the earth to move around the sun with a uniform
speed. Similarly, it is the gravitational force between the earth
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and the moon, which makes the moon to move around the earth
with uniform speed. In fact, it is the gravitational force which is
responsible for the existence of our solar system.
2. The tides (rising and falling of water level in sea) are formed in
sea due to the gravitational pull exerted by the sun and the moon
on the surface of water.
3. It is the gravitational pull of the earth, which holds our
atmosphere in place. It is the gravitational pull of the earth, due
to which the rain drops fall towards the earth and the rivers flow
towards the sea.
4. It is the gravitational pull of earth, which keeps us and other
bodies firmly on the ground.
5. It is the gravitational pull of the earth, which keeps the
enormous amount of water in oceans in place and prevents it
from flowing.
Q. 19 What is the acceleration of free fall?
Ans. The average acceleration of free fall on the surface of earth is
9.81 ms–2.
Q. 20 What do we call the gravitational force between the earth
and the object?
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Ans. It is called force of gravity.
Q. 21 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 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 an object is given by the expression ‘mg’ where
‘m’ is the mass of the object and ‘g’ the acceleration due to
gravity.
Thus, the weight of gold is likely to be less at the equator than at
the poles, because, the magnitude of ‘g’ is less at the equator
than at the poles.
So, his friend will not agree with weight of the gold at the poles
when measured at equator.
Q. 22 Why will a sheet of paper fall slower than one that is
crumpled into a ball?
Ans. Unfolded sheet has large surface area as compared to the similar
sheet crumpled into a ball. Thus, unfolded sheet will experience
more friction due to air as compared to the one crumpled into a
ball, no matter same force of gravity acts upon them.
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It is the larger friction of air which slows down the unfolded
sheet, and hence, it falls slowly as compared to sheet crumpled
into a ball.
Q. 23 Gravitational force on the surface of moon is only 1/6 as
strong as gravitational force on the earth. What is weight in
newtons of 10 kg object on the moon and on the earth?
Ans. Mass of object (m) = 10 kg
Acceleration due to gravity on earth (gearth) = 9.81 ms–2
∴ Acceleration due to gravity on moon (gmoon) =
9.81
ms–2
6
Weight of the object on the earth= mgearth = 10 kg × 9.81 ms–2
= 98.1 N.
Weight of the object on the moon = mgmoon = 10 kg × ms–2
= 16.35 N.
Q. 24 A ball is thrown vertically upward with a velocity of 49 ms–
1. Calculate :
(i) the maximum height to which it rises.
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(ii) the total time it takes to return to the surface of the earth.
Ans. (i) Initial velocity of the ball (u)
Final velocity of the ball (v)
= 49 ms–1
= 0
[The ball will come to rest before moving downward]
Height attained by the ball (S)= ?
[To
be
calculated]
Acceleration
due
to
gravity
(g) = – 9.8 ms–2 [In upward direction g is negative]
Time for rising up (t)
= ? [To be calculated]
Applying, v2 – u2
⇒ (0)2 – (49 ms–1)2
= 2gS
= 2 × – 9.8 ms–2 × S
−49 × 49m 2 s −2
S=
−2 × 98ms −2
⇒
⇒
S = 122.5 m
Applying,
v = u + gt
⇒
0 = 49 ms–1 – 9.8 ms–2 × t
⇒
49ms −1
t =
= 5 s.
−2
9.8ms
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(ii) Now, time for upward journey of the ball is equal to the time
for downward journey of the ball.
∴ Total time taken by the ball to return to the surface of earth
= 2 × t = 2 × 5 s = 10 s.
Q. 25 A stone is released from the top of a tower of height 19.6 m.
Calculate its final velocity.
Ans. Initial velocity of stone (u)
=0
Final velocity of stone (v)= ? [To be calculated]
Height (S)
= 19.6 m
Acceleration due to gravity (g)= + 9.8 ms–2.
v2 = u2 + 2gS
Applying,
⇒
v2= (0)2 + 2 × 9.8 ms–2 × 19.6 m
⇒
v2
= 19.6 × 19.6 m2s–2
∴
v
= 19.6 × 19.6m s s −2
= 19.6 ms–1.
Q. 26 A stone is thrown vertically upward with 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 total
distance covered by the stone?
Ans. Initial velocity of stone (u)= 40 ms–1
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Final velocity of stone (v) = 0
Acceleration due to gravity (g) = –10 ms–2
[For upward direction g is negative]
Height attained by stone (S)= ? [To be calculated]
v2 – u2 = 2gS
Applying,
⇒
(0)2 – (40 ms–1)2 = 2 × – 10 ms–2 × S
−1600m 2 s −2
S=
= 80 m
−20ms −2
⇒
∴
Maximum height attained by stone = 80 m.
Net displacement of stone is zero, as the stone returns back to
the same point.
Total distance covered by the stone = 2 × height attained
= 2 × 80 m = 160 m.
Q. 27 Calculate the force of gravitation between the earth and the
sun, given that mass of earth = 6 × 1024 kg and mass of the
sun = 2 × 1030 kg. The average distance between the two is
1.5 × 1011 m.
Ans. Fgravitation=
G × M earth × M sun
(Distance of the earth from the sun)2
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6.67×10-11 Nm 2 kg -2 ×6×1024 kg×2×1030 kg
=
1.5×1011m×1.5×1011m
6.67 × 6 × 2 × 10−11+ 24+30−11−11
=
N
2.25
= 35.57 × 1021 N.
Q. 28 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
upward from the ground with a velocity of 25 ms–1.
Calculate, when and where the two stones will meet. [Take g
= 10 ms–2].
Ans. Let x be the distance from the ground where the two stones
meet after time t.
For the stone coming down :
Distance (S) = 100 – x
Time (t)
=?
Acceleration due to gravity (g) = 10 ms–2
Initial velocity (u) = 0
Applying,
S = ut +
1 2
gt
2
(100 – x) = 0 × t + × 10t2
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100 – x = 5t2 ....(1)
For the stone moving vertically upward :
Initial velocity (u) = 25 ms–1
Time (t) = ?
Acceleration due to gravity (g) = –10 ms–2
[g is negative in upward direction]
Distance (S) = x
Applying,
S = ut +
1 2
gt
2
x = 25 × t +
1
× – 10t2
2
x = 25t – 5t2 ....(2)
Substituting the value of x from (2) in (1)
⇒
100 – (25t – 5t2) = 5t2
⇒
100 – 25t + 5t2 = 5t2
⇒
25t = 100
t = 4 s.
Substituting the value of ‘t’ in (1)
⇒
100 – x
= 5 (4)2
⇒
100 – x
= 80
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⇒
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100 – 80 = x
⇒
x
= 20 m.
Thus, the stones will meet at a height of 20 m from ground, after
4 s.
Q. 29 A ball thrown up vertically returns to the thrower after 6 s.
Find :
(a) velocity with which it was thrown up,
(b) maximum height it reaches, and
[Take g = 10 ms–2]
(c) its position after 4 s.
Ans. As the ball returns back to thrower in 6 s, the time for its
upward journey = 6 s ÷ 2 = 3 s.
For the upward motion of ball
Initial velocity (u)= ? [To be calculated]
Final velocity (v) = 0 [Ball comes to rest]
Time (t) = 3 s
Acceleration due to gravity (g) = – 10 ms–2
[g is negative in upward direction]
Applying,
v = u + gt
⇒
0 = u – 10 ms–2 × 3 s
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⇒
– u = – 30 ms–1
⇒
u = 30 ms–1
Applying,
S = ut +
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1 2
gt
2
S = 30 ms–1 × 3 s –
1
× 10 ms–2 × (3 s)2
2
= 90 m – 45 m = 45 m.
For the downward motion of ball
Time for downward fall (t)= 4 s – 3 s = 1 s
Acceleration due to gravity (g)= 10 ms–2
Initial velocity (u)
=0
Distance covered in downward direction (S) = ?
(To be calculated)
S = ut +
Applying,
S =0×t+
1 2
gt
2
1
× 10 ms–2 × (1 s)2
2
= 0 + 5 m = 5 m.
∴ Position of ball after 4 s from ground = 45 m – 5 m = 40 m.
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Q. 30 In what direction does the buoyant force of an object
immersed in a liquid act?
Ans. The buoyant force acts in the vertically upward direction.
Q. 31 Why does a block of plastic released under water come up to
the surface of water?
Ans. The piece of plastic displaces more weight of water than its own
weight, and hence, experiences buoyant force more than its own
weight. Thus, on being released the buoyant force acting on it,
pushes it up the surface of water.
Q. 32 The volume of 50 g of a substance is 20 cm3. If the density of
water is 1 gcm–3, will the substance float or sink?
Ans. Mass of substance (m)= 50 g
Volume of substance (v)= 20 cm3
∴ Density of substance (D) =
m
500 g
–3
=
=
2.5
gcm
.
=
2
V 350cm
The substance will sink in water, as its density (2.5 gcm–3) is
more than the density of water (1 gcm–3).
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Q. 33 The volume of a 500 g sealed packed is 350 cm3. Will the
packet float or sink in water, if density of water is 1 gcm–3?
What will be the mass of water displaced by this packet?
Ans. Mass of packet (m)= 500 g
Volume of packet (V)= 350 cm3
∴ Density of packet (D)=
m
500 g
–3
=
1.43
gcm
.
=
V 350cm 2
The packet will sink in water, as its density (1.43 gcm–3) is more
than the density of water (1 gcm–3).
Volume of the water displaced = Volume of the packet
(V)
= 350 cm3
Density of water (D) = 1 gcm–3
∴ Mass of water displaced (M)
= V×D = 350 cm3 × 1 gcm–3
= 350 g.
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