Chapter 10 Gravitation Solutions
Question - 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 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 - 21 : -
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.
Question - 22 : - Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer - 22 : - 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.
Question - 23 : - Gravitational force on the surface of the moon is only
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 - 23 : - Weight of an object on the moon 
Weightof an object on the Earth
Also,
Weight= Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s2
Therefore,weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N
And,weight of the same object on the moon 
Question - 24 : - 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 - 24 : -
(i) 122.5 m (ii) 10 s
According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
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, final velocity of the ball is zero, i.e., v = 0
u = 49 m/s
During upward motion, g = − 9.8 m s−2
Let h be the maximum height attained by the ball.
Hence,
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,
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s
Question - 25 : - A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer - 25 : -
Accordingto the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = 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 = 19.6 m s−1
Hence,the velocity of the stone just before touching the ground is 19.6 m s−1.
Question - 26 : - 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?
Answer - 26 : -
According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final 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,
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
Question - 27 : - Calculate the force of gravitation betweenthe 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 thetwo is 1.5 × 1011 m.
Answer - 27 : -
Accordingto the universal law of gravitation, the force of attraction between the Earthand the Sun is given by:
Where,
MSun = Mass of the Sun =2 × 1030 kg
MEarth = Mass of the Earth= 6 × 1024 kg
R = Average distance between the Earthand the Sun = 1.5 × 1011 m
G = Universal gravitational constant = 6.7 ×10−11 Nm2 kg−2
Question - 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 upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer - 28 : -
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,
(ii) For the stone thrown upwards:
Initial velocity, u = 25 ms−1
Let the displacement of the stone from theground in time t be s‘.
Acceleration due to gravity, g = −9.8 m s−2
Equationof motion,
Thecombined displacement of both the stones at the meeting point is equal to theheight of the tower 100 m.
In4 s, the falling stone has covered a distance given by equation (1) as
Therefore,the stones will meet after 4 s at a height (100 − 80) = 20 m from the ground
Question - 29 : - 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 - 29 : -
(a)Time of ascent is equal to the time of descent. The ball takes a total of 6 sfor its upward and downward journey.
Hence,it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximumheight, v = 0
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion, v = u + gt willgive,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms−1
Hence, the ball was thrown upwards with avelocity of 29.4 m s−1.
(b) Let the maximum height attained by theball be h.
Initial velocity during the upwardjourney, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s−2
Fromthe equation of motion,
(c)Ball attains the maximum height after 3 s. After attaining this height, it willstart falling downwards.
Inthis case,
Initial velocity, u = 0
Positionof the ball after 4 s of the throw is given by the distance travelled by itduring its downward journey in 4 s − 3 s = 1 s.
Equationof motion, 
willgive,Totalheight = 44.1 m
Thismeans that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4seconds.
Question - 30 : - In what direction does the buoyant force on an object immersed in a liquid act?
Answer - 30 : - An object immersed in a liquid experiences buoyant force in the upward direction.