Class 9th Science solutions Chapter 10: Gravitation

Class 9th Science solutions Chapter 10: Gravitation NCERT Solutions for Class 10 Science Chapter 10 – Gravitation solved by Expert Teachers as per NCERT (CBSE) Book guidelines. These solutions are part of NCERT Solutions for Chapter 10 Science, Gravitation class 9. Here we have given Class 9 NCERT Science Textbook Solutions for Chapter 9  Gravitation class 9

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Class 9th Science solutions Chapter 10

NCERT
Solution Class
9th Science Chapter 10:
Gravitation

Class 9th Science solutions Chapter 10: Gravitation

1: State the universal law of gravitation
Answer: The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

For two objects of masses, m1 and m2 and the distance between them r, the force
(F) of attraction acting between them is given by the universal law of gravitation
as:

F= Gm1m2r2Where, G is the universal gravitation constant and its value is: 6.67 x 10-11Nm2kg-2.

2: Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer: Let 𝑀𝐸 be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:

F=GMEmR2

1: What do you mean by free fall?
Answer: Gravity of the Earth attracts every object towards its center. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have a free fall.

2: What do you mean by the acceleration due to gravity?
Answer: When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s².

Class 9th Science solutions Chapter 10: Gravitation

See Also:  NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

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

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

2: Why is the weight of an object on the moon 1/6th its weight on the earth?
Answer: Let

ME

be the mass of the Earth and m be an object on the surface of the Earth.
Let

RE

be the radius of the Earth. According to the universal law of gravitation, weight

WE

of the object on the surface of the Earth is given by,

WE=GMEmRE2

Let

MM

and

RM

be the mass and radius of the moon. Then, according to the
the universal law of gravitation, weight

WM

of the object on the surface of the moon is given by:

WM=GMMmRM2Now,                        ⇒WMWE=MMRE2MERM2Where,ME=5.98 x 1024 kgMM=7.36 x 1022 kgRE=6.4 x 106 mRM=1.74 x 106 m                        ⇒WMWE=7.36 x 1022 x (6.4 x 106)5.98 x 1024 x (1.74 x 106)                                     =0.65                                     ≈ 16Therefore, Weight of an object on the moon is 16 of its weight on the Earth.

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.

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

3: Why does an object float or sink when placed on the surface of the 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.

NCERT Class 9th Science solutions Chapter 10: Gravitation

1: You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer: When you weigh your body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

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 others. Can you say which one is heavier and why?
Answer: The bag of cotton is heavier than an iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, a more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag lighter than its actual value. For this reason, the iron bar and the bag of cotton show the same mass on the weighing machine, but actually the mass of the cotton bag is more than that of the iron bar.

Exercises

1: How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer: According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r)
between them, i.e.,

Hence, if the distance is reduced to half, then the gravitational force becomes four
times larger than the previous value.

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: All objects fall on the ground with constant acceleration, called acceleration due to gravity (in the absence of air resistance). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

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 x 1024

kg and radius of the earth is

6.4 x 106 m

 ).
Answer: According to the universal law of gravitation, the gravitational force exerted on an object of mass m is given by

F=GMmr2

Where,
Mass of Earth, M =

6 x 1024

kg
Mass of object, m = 1 kg
Universal gravitational constant, G = 6.7 × 10−¹¹ Nm² kg−²
Since the object is on the surface of the Earth,
r = radius of the Earth (R)
r = R =

6.4 x 106 m

Therefore, the gravitational force

F=GMmr2               =6.7 x 10-11 x 6 x 1024 x 1(6.4 x 106)2               =9.8 N

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: According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth.

5: If the moon attracts the earth, why does the earth not move towards the moon?
Answer: The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon.

Ncert Class 9th Science solutions Chapter 10: Gravitation

6: What happens to the force between two objects, if
Answer: According to the universal law of gravitation, the force of gravitation between
two objects are given by

F=GMmr2

(i) the mass of one object is doubled?
Answer: F 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) the distance between the objects is doubled and tripled?
Answer: F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.

Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

(iii) the masses of both objects are doubled?
Answer: F 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 the original value.

7: What is the importance of the universal law of gravitation?
Answer: The universal law of gravitation proves that every object in the universe attracts every other object.

8: What is the acceleration of free fall?
Answer: When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms−2, which is constant for all objects (irrespective of their masses).

See Also: NCERT Solutions for Class 9 Science Chapter 10 Gravitation

9: What do we call the gravitational force between the Earth and an object?
Answer: Gravitational force between the earth and an object is known as the weight of the object.

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 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.

11: Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer: 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.

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 newtons of a 10 kg object on the moon and on the Earth?
Answer:
Weight of an object on the moon =1/6 × Weight of 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, the weight of the same object on the moon = 1/6 × 98 = 16.3 N

13: 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:
(i) According to the equation of motion under gravity v²− u² = 2gs
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 m/s and u = 49
m/s
During upward motion, g = − 9.8 m s−²
Let h be the maximum height attained by the ball.
Hence, using 𝑣² − 𝑢² = 2𝑔𝑠

We have, 02-492=2(-9.8)h                        ⇒h=49 x 492 x 9.8 m                                =122.5 m

(ii)

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,
0 = 49 + (−9.8)𝑡 ⇒ 9.8𝑡 = 49 ⇒ 𝑡 =

499.8=5s

But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

See Also: NCERT Class 9th Science solutions Chapter 10: Gravitation

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: According to the equation of motion under gravity v² − u² = 2gs
Where,
u = Initial velocity of the stone = 0 m/s
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 ms−²
∴ v² − 02 = 2 × 9.8 × 19.6
⇒ v² = 2 × 9.8 × 19.6 = (19.6)²
⇒ v = 19.6 ms−¹
Hence, the velocity of the stone just before touching the ground is 19.6 ms−²

.
15: 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 are the net displacement and the total distance covered by the stone?

Answer: According to the equation of motion under gravity v² − u²= 2gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0 m/s
s = Height of the stone
g = Acceleration due to gravity = −10 ms−²
Let h be the maximum height attained by the stone.
Therefore, 0² − 402 = 2(−10)ℎ ⇒ ℎ =

40 x 4020

= 80 𝑚

Therefore, total distance covered by the stone during its upward and downward
journey = 80 + 80 = 160 m
Net displacement during its upward and downward journey = 80 + (−80) = 0.

16: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth =

6 x 1024

kg and of the Sun =

2 x 1030

kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer: According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by

F=G x Msum x MEarthR2

Where,

Museum

= Mass of the Sun =

2 x 1030

kg

MEarth

= Mass of the Earth =

6 x 1024

kg
R = Average distance between the Earth and the Sun = 1.5 × 10¹¹m
G = Universal gravitational constant = 6.7 × 10−¹¹ Nm2 kg−²

F=6.7 x 10-11 x 2 x 1030 x 6 x 1024(1.5 x 1011)2 N   =3.57 x 1022 N

Hence, the force of gravitation between the Earth and the Sun is 3.57 × 10²² 𝑁.

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 m/s. Calculate when and where the two stones will meet.
Answer: Let the two stones meet after a time t.
When the stone dropped from the tower
Initial velocity, u = 0 m/s
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 ms−²
From the equation of motion,

s=ut+12at2s=0 x t +12 x 9.8 x t2⇒s=4.9t2 …………………….(1)

When the stone is thrown upwards
Initial velocity, u = 25 ms−¹
Let the displacement of the stone from the ground in time t be 𝑠′.
Acceleration due to gravity, g = −9.8 ms−²
Equation of motion,

s=ut+12at2s’=25 x t -12 x 9.8 x t2⇒s’=25t-4.9t2 …………………….(2)

The combined displacement of both the stones at the meeting point is equal to the
height of the tower 100 m.

𝑠′ + 𝑠 = 100
⇒ 25𝑡 − 4.9𝑡² + 4.9𝑡² = 100
⇒ 𝑡 = 100/25𝑠 = 4𝑠
In 4 s,

The falling stone has covered a distance given by (1) as 𝑠 = 4.9 × 42 = 78.4 𝑚
Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from
the ground.

18: A ball thrown up vertically returns to the thrower after 6 s. Find
(a)the velocity with which it was thrown up,

Answer: Time of ascent is equal to the time of descent. The ball takes a total of 6 s for
its upward and downward journey.

Hence, it has taken 3 s to attain the maximum height.
The final velocity of the ball at the maximum height, v = 0 m/s
Acceleration due to gravity, g = −9.8 ms−²
Using the equation of motion, v = u + at, we have
0 = u + (−9.8 × 3)
⇒ u = 9.8 × 3 = 29.4 m/s
Hence, the ball was thrown upwards with a velocity of 29.4 m/s.

(b)the maximum height it reaches, and
Answer: Let the maximum height attained by the ball be 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 ms−²
10
Using the equation of motion,

s=ut + 12at2     h=29.4 x 3-12 x 9.8 x 32⇒h=44.1 m

Hence, the maximum height is 44.1 m.

Class 9th Science solutions Chapter 10: Gravitation

(c)its position after 4 s.
Answer: 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 traveled by it
during its downward journey in 4 s − 3 s = 1 s.
Using the equation of motion,

s=ut + 12at2     s=0 x 1+12 x 9.8 x 12⇒s=4.9 m

Now, total height = 44.1 m
This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.

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.

20: Why does a block of plastic released under water come up to the surface of the 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 the water. Due to this reason, a block of plastic released under water comes up to the surface of the water.

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.
Here, the density of the substance

=Mass of the substance volume of the substance=5020=2.5g/cm3

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

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−²? What will be the mass of the water displaced by this packet?
Answer: Density of the 500 g sealed packet

=Mass of the packet volume of the packet=500350=1.428 g/cm3

The density of the substance is more than the density of water (1 𝑔/𝑐𝑚3). 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.

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