In this post we are going to show the detailed solution for class 9th science chapter 10. All the solution are prepared by our esteemed teacher who are very well experienced in the teaching.
Chapter 10 Gravitation
Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half ?
Solution – According to the law of gravitation, the force of attraction between any two objects of mass m1 and m2 is proportional to the product of their masses and inversely proportional to the square of the distance ‘R’ between them.
⇒ F = GMm / R²
Here,
G = Gravitational constant.
M = Mass of object 1
m = Mass of object 2
r = Distance between two object
When distance is reduced to half
Distance = r / 2
And everything else remain same
Thus,
New force = GMm / (r / 2)²
= GMm / (r² / 4)
= 1 / 4 GMm / r²
= 1 / 4 × Old Force
∴ When the distance is reduced to half, force becomes one four times.
Question 2. Gravitation force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object ?
Solution – When an object falls on earth, it falls with constant acceleration known as acceleration due to gravity.
Since, acceleration is constant, therefore it does not depend upon mass.
Thus, heavy and the light object will fall at the same speed.
Question 3. What is magnitude of gravitational force between the earth and a 1 kg object on its surface ? Take mass of earth to be 6 × 1024 kg and radius of the earth is 6.4 × 106 m. G = 6.67 × 10–11 Nm2 kg–2.
Solution – Given –
Mass of earth = M = 6 × 1024 kg
Mass of object = m = 1 kg
Radius of the earth = 6.4 × 106 m.
Gravitational constant = G = 6.67 × 10–11 Nm2 kg–2
As we know F = Gm1 × m2 / r²
⇒ F = 6.67 × 10¯¹¹ × 1 × 6 × 1024 / (6.4 × 106 ) ²
⇒ F = 6.67 × 6 / (6.4)² × 10‾¹¹+24- ¹²
⇒ F = 6.67 × 6 × 10¹³ × 1 / 6.4 × 6.4 × 10¹²
⇒ F = 6.67 × 6 × 10 / (6.4 × 6.4)
⇒ F= 9.8 N (approx.)
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 ?
Solution – According to the universal law of gravitation, two objects attract each other with equal force but in opposite direction.
Therefore, the earth attract the moon with the same as the moon attract the earth.
Question 5. If the moon attracts the earth, why does the earth not move towards the moon ?
Solution – The moon attracts the earth with the same gravitational force as the earth attract the moon
But the mass of earth is to much larger than a mass of moon.
We know that,
Force = Mass × Acceleration
Hence,
Acceleration = Force / Mass
Since, mass of the earth is very large, the acceleration is produced is negligible.
Therefore, the earth does not move toward the moon.
Question 6. What happens to the force between two objects, if
- The mass of one object is doubled ?
- The distance between the object is doubled and tripled ?
- The masses of both objects are doubled ?
Solution – According to the law of gravitation, the force of attraction between two object of mass m1 and m2 is proportional to the product of their masses and inversely proportion to the square of the distance ‘R’ between them.
⇒ F = Gm1 ×m2 / R²
Here, G = Gravitational constant
R = Distance between the two object
m1 = Mass of object 1
m2 = Mass of object 2
1. When the mass of one object (m1) is doubled
⇒ F = G2m1 × m2 / R²
⇒ F = 2Gm1 × m2 / R²
⇒ F = 2 × Old Force
∴ As the mass of one object is doubled the force becomes 2 times.
2. When the distance between two bodies is doubled and tripled
(1) When the distance between two bodies is doubled
⇒ F = Gm1 × m2 / ( r / 2²)
⇒ F = Gm1 × m2 / (r / 4)
⇒ F = 1/4 × Old Force
The force is reduced to 1/4 of the original force.
(2) When the distance between two bodies is tripled
⇒ F = Gm1 × m2 / (r / 3²)
⇒ F = Gm1 × m2 / (r / 9)
⇒ F = 1/9 × Old Force
The force is reduced to 1/9 of the original force.
(3) When the masses of both the object are doubled, then
⇒ F = G2m1 × 2m2 / r²
⇒ F = 4Gm1×m2 / r²
⇒ F = 4 × Old Force
When the masses of both the objects are doubled, then the force becomes four times the original force.
Question 7. What are the importance of universal laws of gravitation ?
Solution – Universal law of gravitation is important because it explained several phenomena which were believed to be unconnected. Like
It explained how objects bound to the earth.
It explained why the moon revolve around the earth.
It explained why the planets revolve around the sun.
It explained how high tides and how low tides are formed (due to gravitational force of the moon and the sun of the surface of the water.)
Question 8. What is the acceleration of free fall ?
Solution – When a body or object fall towards the earth due to gravitational force of earth and without any other force acting on it. It is called free fall ?
The acceleration produced during free fall due to earth’s gravitational force is the acceleration of free fall.
The acceleration is acceleration due to gravity (denoted by g)
Value of acceleration due to gravity = g = 9.8 m/s².
Question 9. What do we call the gravitational force between an earth and an object ?
Solution – The gravitational force between an earth and an object is called weight of the object.
Question 10. Amit buys a few gram of gold at the poles as per instruction of one of his friend. He hands over the same when he meets him at equator. Will the friend agree with the weight of gold bought ? If not, why ?
Solution – Weight of an object is given by,
Weight = mass × gravity
Where,
M = mass of the object
g = acceleration due to gravity at that place.
Since the acceleration due to gravity is less at equator as compared to that at poles, the weight of the will be less at the equator than the poles.
Hence, Amit’s friends won’t agree with the weight of the gold bought since it would be less than what he expecting.
Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball ?
Solution – A sheet of paper has more surface area than a crumpled ball, it faces more air resistance.
Because it faces more air resistance, it’s speed decreases
Therefore, it will fall slower than a crumpled ball.
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’s of a 10 kg object on the moon and on the earth ?
Solution – Given –
Gravitational force on the surface of moon = 1/6 × Gravitational force on the surface of earth
Therefore,
Weight on moon = 1/6 × weight on earth
Finding weight on earth
Mass of the object = m =10 kg
Acceleration due to gravity = g = 9.8
Weight = mass × gravity
= 10 × 9.8
= 98 N
And,
Weight on moon = 1/6 × weight on earth
= 1/6 × 98
= 16.33 N
Thus, the weight of object on earth = 98 N
and on moon = 16.33 N respectively.
Question 13. A ball is thrown vertically upward with a velocity of 49 m s–1. Calculate:
- the maximum height to which it rises.
- the total time it takes to return to the surface of the earth.
Solution – 1. As per the statement given –
Initial velocity of the ball (u) = 49 ms¯¹
Final velocity of the ball (v) = 0 ms¯¹
Downward gravity (g) = 9.8 m s¯²
Upward gravity (g) = -9.8 m s¯²
Maximum height attained by the ball (s) = ?
We need to find maximum height (s)
We know u, v, and a
Finding s using 3rd equation of motion
⇒ v² – u² = 2as
⇒ 0² – 49² = 2 × (-9.8)s
⇒ -2401 = -19.6s
⇒ -19.6 × s = -2401
⇒ s = -2401 / -19.6
⇒ s = 24010 / 196
⇒ s = 122.5 m
Therefore, maximum height attained by the ball will be 122.5 m
Now, we also need to find time taken to reach the ground
Let’s find Time Taken to reach the highest point
We know v, u and a
Finding time using 1st equation of motion
⇒ v = u + at
⇒ 0 = 49 + (-9.8) t
⇒ 0 = 49 – 9.8 t
⇒ -9.8 = -49 t
⇒ t = 49 / 9.8
⇒ t = 490 / 98
⇒ t = 5 second.
∴ Ball takes 5 s to reach the highest point
Now,
Time taken to reach the ground
⇒ 2 × Time taken to reach the highest point
⇒ 2 × 5
⇒ 10 s
Thus, it takes 10 seconds to reach the ground.
Question 14. A stone is released from a top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Solution – Since the stone is released from top,
Initial velocity = u = 0 m/s
Distance travelled by the stone = height of the tower
= 19.6 m
And,
Acceleration = a = 9.8 m/s²
Finding final velocity (v)
We know u, a, and s,
Finding v using 3rd equation of motion
⇒ v² -u² = 2as
⇒ v² – 0 = 2 × 9.8 × 19.6
⇒ v² = 19.6 × 19.6
⇒ v = 19.6 m/s
Final velocity of the ball is 19.6 m/s.
Question 15. A stone is thrown vertically upward with an initial velocity with an 40 m/s. Taking g = 10 m/s². Find the maximum height reached by the stone. What is the net displacement and total distance covered by the stone ?
Solution – Given –
Initial velocity = u = 49 m/s
At the highest point of motion,
velocity becomes = 0
Therefore,
final velocity = v= 0 m/s
and,
Acceleration = -g = -10 m/s²
We need to find maximum height (s)
We know u, v, and a
Finding s using 3rd equation of motion
⇒ v² – u² = 2as
⇒ 0² – 40² = 2 × (-10) s
⇒ -1600 = -20 s
⇒ s = -1600 / -20
⇒ s = 80 m
Therefore, maximum height attained by the ball is 80 m
Now, when we through stone upward
It reaches highest point and comes down
So, it’s initial and final point is same
∴ Net displacement = 0 m
And,
Total distance covered = 2 × Maximum height covered by the stone
= 2 × 80 m
= 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.
Solution – As given in the statement, Me = 6 × 1024 kg, Ms = 2 × 1030
r = 1.5 × 10¹¹ m
As we know that, F = G m1m2 / r²
Therefore, F = 6.67 × 10¯¹¹ × 6 × 1024 × 2 × 1030 / (1.5 × 10¹¹)²
⇒ F = 6.67 × 12 × 10²¹ / 1.5 × 1.5
⇒ Therefore, F = 3.56 × 10²² N
Question 17. A stone is allowed to fall from the top of a tower 100m 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 ?
Solution – Let the stones meet at time = t seconds
Finding Distance moved by stones both stones in time t second
Distance moved by stone 1
Initial velocity = u = 0
Acceleration = g = 10 m s²
We know u, a and t
Finding distance (s) using 2nd equation of motion
⇒ s = ut + ½ at²
⇒ s = 0 × t + ½ × 10 t²
⇒ s = 5 t²
Distance moved by stone 2
Initial velocity = u = 0
Acceleration = g = -10 m s²
We know v, u and t
Finding distance (s) using 2nd equation of motion
⇒ s = ut + ½ at²
⇒ s = 25 × t + ½ (-10) t²
⇒ s = 25t – 5 t²
Now,
Total distance covered by both stones = 100 m
⇒ Distance 1 + Distance 2 = 100 m
⇒ 5t² + 25 t – 5t² = 100
⇒ 25 t = 100
⇒ t = 100 / 25
⇒ t = 4 s
At t = 4 s ,
⇒ Distance 1 = 5t²
⇒ 5 × 4²
⇒ 5 × 16
⇒ 80 m
⇒ Distance 2 = 25t – 5t²
⇒ 25 × 4 – 5 × 4²
⇒ 100 – 5 × 16
⇒ 100 – 80
⇒ 20 m
Therefore, stones will meet when time is 4 second,
And Distance is 80 m from the top.
Question 18. A ball thrown vertically returns to the thrower after 6s Find:
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches, and
(c) its position after 4s.
Solution – Since, it takes 6 s to return to the thrower
Therefore,
Time taken to reach maximum height = 6 /2
= 3 s
Now solving all parts,
(a) the velocity with which it was thrown up.
Solution – We need to find initial velocity
Let initial velocity = u
At the highest point of motion velocity becomes 0
Therefore,
⇒ Final velocity = 0 m/s
⇒ Acceleration = -9.8 m/s²
⇒ Time = 3 seconds
⇒ We know v, a, and t
Finding u by using 1st equation of motion
⇒ v = u + at
⇒ 0 = u + (-9.8) × 3
⇒ -u = -29.4
⇒ u = 29.4 m/s
Thus, initial velocity is 29.4 m/s.
(b) the maximum height it reaches, and
Solution – Calculating maximum height
We know u, t, and a
Finding distance (s) using 2nd equation of motion
⇒ s = ut + ½ at²
⇒ s = 29.4 × 3 + ½ (-9.8) × 3²
⇒ s = 88.2 – ½ × 9.8 × 9
⇒ s = 88.2 – 44.1
⇒ s = 44.1 m
Thus, maximum height it reaches is 44.1 m.
(c) its position after 4 second
Solution – The ball reaches highest point in 3 seconds
At 4 s it will be moving downward,
This is same as,
A ball is dropped from highest point
We need to calculate distance it moves in 1 second.
When ball is dropped,
⇒ Initial velocity = 0 m/s
⇒ Time = 1 second
⇒ Acceleration = g = 9.8 m/s²
⇒ We need to find distance (s)
⇒ We know u, a, and t
Finding distance (s) using 2nd equation of motion
⇒ s = ut + ½ at²
⇒ s = 0 × 1 + ½ × 9.8 × 1²
⇒ s = ½ × 9.8
⇒ s = 4.9 m
So, position of ball is 4.9 from the top.
Therefore,
Initial velocity of the ball = 29.4 m/s
The maximum height it reaches = 44.1 m
It’s position of ball at time 4s = 4.9 (from top)
Question 19. In what direction does the buoyant force on an object immersed in a liquid act ?
Answer – Buoyant force acts in vertically upward direction
The direction of the buoyant force is always opposite to the direction of the weight of the object.
Question 20. Why does a block of plastic released under water come up to the surface of the water ?
Answer – When a plastic is immersed in water, there are two forces acting upon it
- Gravitational force in downward direction
- Buoyant force in upward direction
In this case,
the buoyant force on plastic is more than the gravitational force.
Thus, plastic will come up to the surface of 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 ?
Solution – As given in the statement:
The mass of substance = 50 g
The volume of substance = 20 cm³
Density of water is 1 g cm¯³
Density of substance = Mass of substance volume of substance
= 50 / 20
= 2.5 g cm¯³
Clearly, As the density of substance (2.5 g cm‾³) is more than that of water (1 g cm¯³ ), so it will sink.
Question 22. The volume of a 500 g sealed packed 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 water displaced by this packet ?
Solution – As per the statement of question :
Mass of sealed packed = M = 500 g
Volume of sealed packed = V = 350 cm³
Density of water = 1 g cm‾³
Density of sealed packet = m / v
⇒ 500 / 350
⇒ 10 / 7
⇒ 1.43 g cm‾³
As density of packet which is 1.43 g cm‾³ is more than that of water (1 g cm‾³),
Therefore, the packet will sink in water
We also need to find mass of the water displaced by this packet
Mass of water displaced by packet = Volume of the packet
= 350 g
∴ Mass of the water displaced is 350 g.
You can see other posts here:
- NCERT Class 9th Science Chapter 8 – Motion Summary
- NCERT Science Class 9th Chapter 8 Motion
- NCERT Science Class 9th Chapter 9 Force and Laws of Motion Summary
- NCERT Science Class 9th Chapter 9 Force and Laws of Motion
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