RBSE Solutions for Class 9 Science Chapter 10 Gravitation

Page No. 138

Question 1.
What are the differences between the mass of an object and its weight?
Answer:

Mass	Weight
1. The total quantity of matter contained in an object is called mass of an object.	1. The gravitational force by which Earth attracts an object is called weight of the object.
2. Mass of an object can be measured by its inertia.	2. Weight = Mass × acceleration (m × g)
3. Mass of the object remains constant at all the places.	3. Weight of the object is different at different places.
4. Mass can be measured by using a pan or beam balance.	4. Weight can be measured by using a spring balance.
5. Mass does not change even if value of g is zero at any place.	5. Weight of the object becomes zero if g is zero.

Question 2.
Why is the weight of an object on the Moon 1/6th its weight on the Earth?
Answer:
The mass of the Moon of 1/100 times and its radius 1/4 times that of the Earth. As a result, the gravitational attraction on the Moon is about one sixth when compared to that on the Earth. Hence, the weight of an object on the Moon is 1/6th of the weight on the Earth.

Page No. 141

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult 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.

Question 2.
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 we try to immerse an object in water, then we can feel an upward force exerted on the object, which increases as we push the object deeper into water.

Question 3.
Why does an object float 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 floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity.

Page No. 142

Question 1.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
Using a weighing machine, we actually measure our weight and from this, we infer about our mass. While measuring weight, the weight recorded is slightly less than the true weight on account of force of buoyancy due to air acting on us. As a result, the true weight and consequently, the true mass must be greater than. 42 kg.

Question 2.
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?
Answer:
We know that density of cotton is lesser than that of an iron bar. Hence, for same mass of 100 kg, the volume of cotton is greater than that of iron bar. Due to this, while weight in air using a weighing machine, the upthrust due to air is more on cotton and less on iron bar. As a result, true weight and consequently, the true mass of cotton must be greater than that of iron bar. Thus, cotton is heavier than the iron

RBSE Class 9 Science Chapter 10 Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
When all other variables remain constant, the force of gravitation is inversely proportional to the square of distance between the two objects.
F ∝ \(\frac{1}{r^{2}}\)
If distance r becomes \(\frac{r}{2}\), then the gravitational force will be proportional to
F₁ = \(\frac{1}{(r / 2)^{2}}=\frac{4}{r^{2}}\)
Hence, if the distance is reduced to half then the gravitational force becomes four times larger than the previous value.

Question 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?
Answer:
A freely falling object of any mass falls under the action of gravity given by
g = \(\mathrm{G} \frac{\mathrm{M}}{\mathrm{R}^{2}}\)
where, ‘G’ = constant to gravitation
g = gravitational acceleration
M = Mass of Earth
r = The distance between the object and the centre of Earth.
As, the acceleration due to gravity is independent of the mass of the objects. All objects fall at the same speed towards the Earth.

Question 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 × 10²⁴ kg and radius of the Earth is 6.4 × 10⁶ m.)
Answer:
Given : m = 1 kg, Me = 6 × 10²⁴ kg, r = 6.4 × 10⁶ m, G = 6.67 × 10^-11 Nm²/kg²

Question 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?
Answer:
The Earth attracts the Moon with the same force with which the Moon attracts the Earth because, the gravitational force between any two bodies is mutual and equal according to Newton’s universal law of gravitation.

Question 5.
If the Moon attracts the Earth, why does the Earth not move towards the Moon?
Answer:
The Earth does not move towards the Moon because the force exerted by the Earth or the Moon on each other is insufficient to move the Earth on account of its huge mass.

Question 6.
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:
The gravitational force between two objects is given as :
F = \(\mathrm{G} \frac{m_{1} m_{2}}{r^{2}}\)
(a) If the mass of one object is doubled to 2mj, then new force is given as :
F' = \(\frac{\mathrm{G}\left(2 m_{1}\right)\left(m_{2}\right)}{r^{2}}\) = 2F
∴ Force is doubled.

(b) If the distance between the objects is doubled, then r' = 2r and hence new force is given as
F' = \(\frac{\mathrm{G}_{m_{1}} m_{2}}{(2 r)^{2}}=\frac{F}{4}\)
∴ The force is reduced to \(\frac{1}{4}\)th of its original value. 
If distance is tripled then r'' = 3r
F'' = \(\frac{G m_{1} m_{2}}{(3 r)^{2}}=\frac{F}{9}\)
∴ The force is reduced to \(\frac{1}{9}\)th of its original value.

(c) If masses of both the objects are doubled i.e.
m'₁ = 2m₁ and m'₂ = 2m₂
∴ F' = \(\frac{\mathrm{G}\left(2 m_{1}\right)\left(2 m_{2}\right)}{r^{2}}\) = 4F
∴ The force becomes 4 times of its original value.

Question 7.
What is the importance of universal law of gravitation?
Answer:
Importance of universal law of gravitation is as follows :

• It is the gravitational force between the Sim and the Earth, which makes the Earth to move around the Sun with a uniform speed.
• The tides formed in sea are because of gravitational pull exerted by the Sun and the Moon on the surface of water.
• It is the gravitational pull of Earth, which keeps us and other bodies firmly on the ground.
• It is the gravitational pull of the Earth, which holds our atmosphere in place.

Question 8.
What is the acceleration of free fall?
Answer:
The average acceleration of free fall on the surface of Earth is 9.81 ms^-2.

Question 9.
What do we call the gravitational force between the Earth and an object?
Answer:
It is called weight of the object.

Question 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.]
Answer:
Weight of an object = mg
where ‘m’ is mass of the object at the equator than at the poles as 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.

Question 11.
Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
When a sheet of paper is crumpled 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 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 newton of a 10 kg object on the Moon and on the Earth?
Answer:
Mass of object (m) = 10 kg
Acceleration due to gravity on Earth (g_e) = 9.81 ms^-2
Acceleration due to gravity on Moon (g_m) = \(\frac{9.81}{6}\)ms^-2
Weight of the object on the Earth = mg_e = 10 × 9.81 = 98.1 N
Weight of the object on the Moon = mg_m = 10 × 9.81/6 = 16.35 N

Question 13.
A ball is thrown vertically upwards with a velocity of 49 ms^-1. 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) Initial velocity of the ball (u) = 49 ms^-1
Final velocity of the ball (v) = 0
Acceleration due to gravity (g) = 9.8 ms^-2
[In upward direction g is -ve.]
Height attained by the ball (s) = ?
Time for rising up (t) = ?
We know; v² - u² = 2gs
(0)² - (49)² = 2 × (9.8) × s
s = \(\frac{49 \times 49}{2 \times 9.8}\) = 122.5 m
We know; v = u + gt
0 = 49 - 9.8 × t
t = \(\frac{49}{9.8}\) = 5 s
(ii) Now, time for upward journey of the ball = 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 = 10s

Question 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.
Answer:
Initial velocity of stone (u) = 0
Final velocity of stone (v) = ?
Height attained (h) = 19.6 m
Acceleration due to gravity (g) = 9.8 ms²
We know, v² = u² + 2gh
v² = (0)² + 2 × 9.8 × 19.6
v² = 19.6 × 19.6
v = \(\sqrt{19.6 \times 19.6}\) = 19.6 ms^-1.

Question 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?
Answer:
Initial velocity of stone (u) = 40 ms^-1
Final velocity of stone (v) = 0
Acceleration due to gravity (g) = 10 ms^-2
(For upward direction g is -ve.)
Height attained by stone (h) = ?
We know, v² - u² = 2gh
(0)² - (40)² = 2 × (10) × h
h = 1600/20 = 80 m
Maximum height attained by stone = 80 m
Net displacement of stone = 0
(Because the stone returns back to the same point)
Total distance covered by the stone = 2 × height attained
= 2 × 80 = 160 m

Question 16.
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.
Answer:
Given : G = 6.7 × 10^-11 Nm²/kg², M_S = 2 × 10³⁰ kg, M_E = 6 × 10²⁴ kg, r = 1.5 × 10¹¹ m

Question 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 ms^-1. Calculate when and where the two stones will meet.
Answer:
Let the stone meet at a height of x m from the ground.
For stone 1
Acceleration due to gravity (g) = 10 ms^-2
Initial velocity (u) = 0 ms^-1
Distance (s) = 100 - x
Time (t) = ?
s = ut + \(\frac{1}{2}\)at²
(100 - x) = 0 × t + \(\frac{1}{2}\) × 10 × t²
100 - x = 5t²
x = 100 - 5t² ............(1)

For stone 2
Acceleration due to gravity (g) = - 10 ms^-2
Initial velocity (u) = 25 ms^-1
Distance (s) = x
Time (t) = ?
s = ut + \(\frac{1}{2}\)gt²
= 25 × t + \(\frac{1}{2}\) × (-10) × t²
x = 25t - 5t²
Substituting the value of x from (2) in (1), we get
100 - 5t² = 25t - 5t²
100 = 25t
t = 4 sec.
Putting the value of t in equation (1), we get
x = 100 - 5 × (4)²
x = 100 - 5 × 16
x = 20 m
The stones will meet at a height of 20 m from ground after 4 sec.

Question 18.
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.
(c) its position after 4 s.
Answer:
The ball returns back to the thrower in 6 s, the time for its upward journey = \(\frac{6}{2}\) = 3s
(a) For the upward motion of ball
Initial velocity (u) = ?
Final velocity (v) = 0 (Ball comes to rest)
Time (t) = 3 s
Acceleration due to gravity (g) = 10 ms^-2 (In upward direction g is -ve.)
We know
v = u + gt
0 = u + (- 10) × 3
u = 30 ms^-1

(b) We know
s = ut + \(\frac{1}{2}\) gt²
s = 30 × 3 + \(\frac{1}{2}\) (-10) × (3)²
s = 90 - 45
s = 45 m

(c) For the downward motion of ball
Initial velocity (u) = 0
Time for downward fall (t) = 4 - 3 = 1 s
Acceleration due to gravity (g) = 10 ms^-2
Distance covered in downward direction (s) = ?
We know
s = ut + \(\frac{1}{2}\) gt²
s = 0 × t + \(\frac{1}{2}\)× 10 × (1)²
s = 5 m
Position of ball after 4 s from ground = 40 m.

Question 19.
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.

Question 20.
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.

Question 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?
Answer:
If the density 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 floats on the surface of the liquid.

The density of the substance is more than the density of water (1 g cm^-3).
Hence, the substance will sink in water.

Question 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?
Answer:

The density of the substance is more than the density of water (1 g 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.