Q1 
How does the force of gravitation between two objects change when the distance between them is reduced to half ? 
Ans: 
We know that the universal law of gravitation,
gravitational force acting between two objects is inversely proportional to the square of the distance (r) between them, i.e.,
If distance r becomes r/2, then the gravitational force will be proportional to
When the distance is reduced to half the gravitational force becomes four times larger than the previous value. 

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? 
Ans: 
Gravitational force acts on all objects in proportion to their masses. But a heavy object does not fall faster than a light object. This is because force is directly proportional to mass, acceleration is constant for a body of any mass. Hence, heavy objects do not fall faster than light objects. 

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 × 10^{24} kg and radius of the earth is 6.4 × 10^{6} m.) 
Ans: 
Given; Mass of Earth, M = 6 × 10^{24} kg
Mass of object, m = 1 kg
Universal gravitational constant, G = 6.7 × 10^{−11} Nm^{2} kg^{−2}
radius of the Earth (R) = R = 6.4 × 10^{6} m
Gravitational force,


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? 
Ans: 
The Earth attracts the moon with an equal force with which the moon attracts the earth but these forces are in opposite directions.
According to universal law of gravitation,
Where, m_{1}m_{2} is mass of earth and the moon respectively and r is distance between the earth and the moon. 

Q5 
If the moon attracts the earth, why does the earth not move towards the moon? 
Ans: 
Apply Newton’s third law which states that every action has an equal and opposite reaction. Therefore, the Earth and the moon experience the same amount of gravitational forces from each other. Both bodies revolve around their common centre of mass and centrifugal force balances the gravitational force. However, the mass of the Earth is much larger than the mass of the moon.
For this reason, the Earth does not move towards the moon. 

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? 
Ans: 
According to the universal law of gravitation,
(i) Force 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) Force is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes onefourth.
Similarly, if the distance is tripled, then the gravitational force becomes oneninth.
(iii) Force 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. 

Q7 
What is the importance of universal law of gravitation? 
Ans: 
The importance of universal law of gravitation ;
1. It helps us bind together with the Earth.
2. A body which goes up will not come down if there is no gravitational force. 

Q8 
What is the acceleration of free fall? 
Ans: 
When the body falls due to Earth’s gravitational pull, its velocity changes and is said to be accelerated due to Earth’s gravity and it falls freely called free fall. Acceleration of free fall is 9.8 ms^{−2}, which is constant for all objects. 

Q9 
What do we call the gravitational force between the earth and an object? 
Ans: 
Gravitational force between the earth and an object is called the weight of the object. 

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.] 
Ans: 
Weight of an object on the Earth is given by:
W = mg
Where,
m = Mass of the object
g = Acceleration due to gravity
Since the acceleration due to gravity is less at the equator as compared to that at poles, the weight of the gold will be less at the equator than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought. 

Q11 
Why will a sheet of paper fall slower than one that is crumpled into a ball? 
Ans: 
A sheet of paper will fall slower than one that is crumpled into a ball because of the different drag force. Crumpled ball has a smaller surface area so the resistance offered by air is less than in the case of a sheet of paper which has a larger surface area. 

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? 
Ans: 
Given; gravitational force on surface of the moon = 1/6 x gravitational force on surface of the Earth
Also,
Weight = Mass × Acceleration
Mass of the object = 10 kg
Acceleration due to gravity, g = 9.8 m/s^{2}
Therefore, weight of object on the Earth = 10 × 9.8 = 98 N
And, weight of the same object on the moon = 98 / 6 = 16.33 N 

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. 
Ans: 
We know that the equation of motion under gravity:
v_{2} − u_{2} = 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}
(i) Let; h = the maximum height attained by the ball.
Hence,
(ii) Let; t = time taken by the ball to reach the height 122.5 m
then using 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


Q14 
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity. 
Ans: 
Initial velocity of the stone = u = 0
Final velocity of the stone = v
Height of the stone = s = 19.6 m
Acceleration due to gravity = g = 9.8 m s^{−2}
According to the equation of motion under gravity:
v_{2} − u_{2} = 2 gs
v_{2} − 02 = 2 × 9.8 × 19.6
v_{2} = 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}. 

Q15 
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s^{2}, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? 
Ans: 
Initial velocity of the stone = u = 40 m/s
Final velocity of the stone = v = 0
Height of the stone = s
Acceleration due to gravity = g = −10 m s^{−2}
Let h = maximum height attained by the stone.
According to the equation of motion under gravity:
v_{2} − u_{2} = 2 gs
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 

Q16 
Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10^{24} kg and of the Sun = 2 × 10^{30} kg. The average distance between the two is 1.5 × 10^{11} m. 
Ans: 
M_{Sun} = Mass of the Sun = 2 × 10^{30} kg
M_{Earth} = Mass of the Earth = 6 × 10^{24} kg
R = Average distance between the Earth and the Sun = 1.5 × 10^{11} m
G = Universal gravitational constant = 6.7 × 10^{−11} Nm^{2} kg^{−2}
Force of gravitation = F =


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. 
Ans: 
Let the two stones meet after a time t.
(i) For the stone dropped from the tower:
Initial velocity, u = 0 m/s
Let the displacement = s
Acceleration due to gravity, g = 9.8 m s^{−2}
From the equation of motion,
...(1)
(ii) For the stone thrown upwards:
Initial velocity, u = 25 m/s
Let the displacement = s'.
Acceleration due to gravity, g = −9.8 m s^{−2}
Equation of motion,
...(2)
The combined displacement is ;
In 4 s, the falling stone has covered a distance given by equation (1) as
Therefore, the stones will meet after 4 s and the distance is 80 m from the top.


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. 
Ans: 
The ball takes a total of 6 s for its upward and downward journey.
Time taken to reach maximum height = 6 / 2 = 3 s
Hence, it has taken 3 s to attain the maximum height
(a)
Let initial velocity = u m/s
Final velocity of the ball at the maximum height, v = 0 m/s
Acceleration due to gravity, g = −9.8 m s^{−2}
Equation of motion
v = u + gt
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms^{−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 = h
Initial velocity during the upward journey, u = 29.4 m/s
Final velocity, v = 0 m/s
Acceleration due to gravity, g = −9.8 m s^{−2}
From the equation of motion,
(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0 m/s
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 x 1 + ½ x 9.8 x 1^{2}
s = 4.9 m
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.


Q19 
In what direction does the buoyant force on an object immersed in a liquid act? 
Ans: 
The buoyant force acts in the upward direction on the object which is immersed in a liquid. 

Q20 
Why does a block of plastic released under water come up to the surface of water? 
Ans: 
When a plastic is immersed in water, there are two forces acting upon it. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. In this case, the buoyant force on plastic is more than the gravitational force. Thus, the plastic will come up to the surface of water. 

Q21 
The volume of 50 g of a substance is 20 cm^{3}. If the density of water is 1 g cm^{3}, will the substance float or sink? 
Ans: 
An object sinks in liquid when its density is greater than the liquid. On the other hand, an object floats on liquid when its density is less than the liquid.
Here, density of the substance =
The density of the substance is more than the density of water (1 g cm^{−3}). Hence, the substance will sink in water. 

Q22 
The volume of a 500 g sealed packet is 350 cm^{3}. 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: 
Density of the 500 g sealed packet
Density of packet > density of water
Hence, it will sink in water.
Mass of water displaced by packet = Volume of the packet = 350 g
The mass of water displaced is 350 g. 
