Class 9 is the first stepping stone for a student in the competitive world. The introduction of the CBSE Board Exam for class 10 a few years back has become an important gateway for a student. Based on the results of class 9th a student selects his future stream of Science, Commerce or Arts suiting his interest.
Takshila Learning is providing NCERT Solutions for Class 9 Science as per the latest syllabus by CBSE. Class 9 is the building block for the CBSE Class 10 Board Exams, not only for your exams but also for your higher studies and career. Science is the most essential subject and the knowledge in this field opens up wider career opportunities for the students.
Below you can find the NCERT solution for Class 9th Science. You can get a Solution for the allimportant question of Class 9 Science, Chapter 10: Gravitation
Important Formulae
F = G*M1 *m2 / r²
g = G *M / R²
v= u+gT
v² = u²+2gh.
s= ut + 1/2 gt²
W= mg.
pressure = thrust / area.
relative density = density of substance / density of water.
Pg 134
Q1. State the universal law of gravitation.
Solution:
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:
Q2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Solution:
Let M_{E} 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:
Pg 136
Q1. What do you mean by free fall?
Solution:
Gravity of earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of earth under the influence of gravity. The motion of the article is alleged to own free fall.
Q2. What do you mean by acceleration due to gravity?
Solution:
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/s^{2}.
Pg 138
Q1. What are the differences between the mass of an object and its weight?
Solution:
Differences between mass and weight are:
Mass  Weight

Mass is the amount of matter present in a body.

Weight is the force with which the earth attracts a body towards its centre.

Its SI unit is kilogram (kg).

Its SI unit of weight is newton (N).

Mass of a body is a constant quantity irrespective of the position and surroundings of the body.

Weight of a body varies from place to place.

It is a scalar quantity.

It is a vector quantity.

It is measured using a beam balance.

It is measured using a spring balance.

The mass of an object on the earth will be same everywhere as the mass is the amount of matter present in the body.
While, the weight of an object will differ on earth and the Mars because weight of an object = m x g
where, g – acceleration due to gravity.
As acceleration due to gravity is different on the earth and mars the weight of an object will vary.
Q2. Why is the weight of an object on the moon 1/6^{th} its weight on the earth?
Solution:
The weight of an object depends on the value of acceleration due to gravity g.
The value of g on earth is 6 times more than that of moon because, the mass and radius of the earth is more than the mass and radius of the moon.
We have, g = GM/R^{2} and W = mg
Weight of a body of mass m on earth is
Weight of a body of mass m on moon is
or
As mass of moo n is times the mass of earth Mg and radius of moon R_{m} is 1/4 times the radius of earth R_{e}.
Pg 141
Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Solution:
It is difficult to hold a school bag having a thin strap due to the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the area on which the force acts. The smaller is the surface area; the larger is going to be the pressure on the surface. In the case of a thin strap, the contact area is incredibly small. Hence, the pressure exerted on the shoulder is incredibly large.
Q2. What do you mean by buoyancy?
Solution:
The upward force exerted by a liquid on an object immersed in it’s called buoyancy. When you try and immerse an object in water, then you’ll feel an upward force exerted on the article, which increases as you push the article deeper into water.
Q3. Why does an object float or sink when placed on the surface of water?
Solution:
If the density of an object is more the density of the liquid, then it sinks within the liquid. This is because the buoyant force functioning on the item is a smaller amount than the force of gravity. On the opposite hand, if the density of the item is a smaller than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force functioning on the item is larger than the force of gravity.
Pg 142
Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Solution:
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.
Q2. 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?
Solution:
The bag of cotton is heavier than the iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag heavier 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 cotton bag is more that that of the iron bar.
Actual weight = Measured Weight + Buoyant Force
Pg 143
Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Solution:
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.,
F α 1 ⁄ r^{2}
If distance r becomes r/2, then the gravitational force will be proportional to
1 ⁄ (r/2)^{2 } = 4 / r^{2}
Hence, if the distance is reduced to half, then 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?
Solution:
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. 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).
Solution:
According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:
Where,
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}
Since the object is on the surface of the Earth, r = radius of the Earth (R)
r = R = 6.4 × 10^{6} m
Gravitational force,
= (6.7 × 10^{−11} X 6 × 10^{24} X 1)/ (6.4 × 10^{6})^{2}
= 9.8 N
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?
Solution:
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.
Q5. If the moon attracts the earth, why does the earth not move towards the moon?
Solution:
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.
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?
Solution:
 Doubled
 Onefourth and oneninth
 four times
According to the universal law of gravitation, the force of gravitation between two objects is given by:
(i) 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) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes onefourth of its original value.
(iii) 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.
Q7. What is the importance of universal law of gravitation?
Solution:
The universal law of gravitation proves that every object in the universe attracts every other object.
Q8. What is the acceleration of free fall?
Solution:
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 m s^{−2}, which is constant for all objects (irrespective of their masses).
Q9. What do we call the gravitational force between the Earth and an object?
Solution:
Gravitational force between the earth and an object is known as 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].
Solution:
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.
Q11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Solution:
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.
Q12. Gravitational force on the surface of the moon is only 1/6 ^{th} 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?
Solution:
Weight of an object on the moon = 1/6 X Weight of an object on the Earth
Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s^{2}
Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N
And, weight of the same object on the moon
= 1/6 X 98 = 16.3 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.
Solution:
(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
Substituting the values,
u=49 m/s
v=0
g= 9.8 m/s
2gh=v^{2}u^{2}
2*9.8*h= – 2401
h=122.5 m
v=u+gt
0=49+(9.8)t
49=9.8t
t=5 sec
Hence, Total time=5+5=10 sec
Q14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Solution:
According to the equation of motion under gravity:
v^{2} − u^{2} = 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}
∴ v^{2} − 0^{2} = 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?
Solution:
According to the equation of motion under gravity:
v^{2} − u^{2} = 2 g s
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,
0 – (40)^{2} = 2 X h X (10)
h = (40 X 40)/2 = 80 m
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.
Solution:
According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by:
Where,
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}
F = (6.7 × 10^{−11} X 2 × 10^{30} X 6 × 10^{24})/( 1.5 × 10^{11})^{2}
= 3.57 X 10 ^{22} N
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.
Solution:
Let a stone A is allowed to fall from top of a height h = 100m
Its initial velocity u_{1} = 0
a_{1} = +g = +10 ms^{2}
Another stone B is projected upwards from the ground with an initial velocity u_{2} = 25ms^{1} and for it a_{2} = g = 10 ms^{2}
Let the two stones meet at a point C at a distance y below A or (100y) above B after a time t.
Then for stone A
For stone B
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.
Solution:
 4 m/s
 1 m
 2 m above the ground
 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.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = −9.8 m s^{−2}
Equation of motion, v = u + gt will give,
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 be h.
Initial velocity during the upward journey, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s^{−2}
From the equation of motion,
s = ut + ½ at^{2}
^{ }
h = 29.3 X 3 + ½ X ( – 9.8) X 3^{2}
^{ }
= 44.1 m
(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
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, will give,
s = ut + ½ at^{2}
s = 0 X 1 + ½ X 9.8 X 1^{2} = 4.9m
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?
Solution:
An object immersed in a liquid experiences buoyant force in the upward direction.
Q20. Why does a block of plastic released under water come up to the surface of water?
Solution:
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.
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?
Solution:
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, density of the substance = (Mass of substance)/(Volume of the substance) =50/20 = 2.5 g/cm³
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 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?
Solution:
Density of the 500 g sealed packet = (Mass of the packet)/(Volume of the packet) =500/350 = 1.428 g/cm³
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.
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