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Q1. State the universal law of gravitation.
Ans: According to Newton’s universal law of gravitation, everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of distance between their centres.
Universal Law of Gravitation
Q1. What do you mean by free fall?
Ans: When an object falls with a constant acceleration, under the influence of the force of gravitation of the earth, the object is said to have a free fall.
Q2. What do you mean by acceleration due to gravity?
Ans: During free fall, any object that has mass experiences force towards the center of the earth and hence acceleration works as well. “acceleration experienced by an object in its free fall is called acceleration due to gravity.” It is denoted by g.
Q1. What are the differences between the mass of an object and its weight?
Ans:
Q2. Why is the weight of an object on the moon 1/6th its weight on the earth?
Ans: The mass of an object remains the same whether on Earth or moon but the value of acceleration on the moon is 1/6th of the value of acceleration on Earth. Because of this weight of an object on the moon is 1/6th its weight on the Earth.
Mass of the moon (M) = 7.4 × 10^{22} kg
Radius of the moon (R) = 1.74 × 10^{6} m
Gravitational constant (G) = 6.7 × 10^{11} Nm^{2}kg^{2}
We know that acceleration due to gravity(g) = GM/R^{2}
Also, weight , W = mg
W = GMm/R^{2}
Let the mass of the object be m, Let's its weight of moon be M_{m }and_{ }its radius be R_{m}
By applying the universal law of gravitation, the weight of the object on the moon will be: (1)
Let the weight of the same object on the earth be We. The mass of earth is M and its radius is R.
(2)
Substituting the values in equations (1) and (2) we get:
W_{m}= 2.431 x 10^{10}G x m
and W_{e} = 1.474 x 10^{11} G x m
On dividing W_{m} and W_{e} we get:
∴ Weight of the object on the moon = its weight on earth
Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Ans: 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.
Q2. What do you mean by buoyancy?
Ans:
Q3. Why does an object float or sink when placed on the surface of water?
Ans:
Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Ans: Mass is always a constant quantity. Therefore, it cannot be more or less than 42 kg.
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?
Ans: The correct answer is cotton bag is heavier than iron bar.
The reason is:
True weight = (apparent weight + up thrust)
Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Ans: The force of gravitation between two objects of masses m_{1} and m_{2} at separation r is given by:
F = (G m_{1}m_{2})/r^{2}
When the distance between them is reduced to half, the gravitational force.
F' = (G m_{1}m_{2})/r^{2}
= (G m_{1}m_{2})/ (r/2)^{2}
= (4 G m_{1}m_{2})/(r^{2}) = 4F
The gravitational force becomes four times.
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: The acceleration due to gravity of a freely falling body is independent of the mass of falling body. Thus, both heavy and light object fall with same acceleration.
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: Here, mass of the earth, M = 6 x 10^{24 }kg
Mass of the object, m = 1 kg, radius of the earth, R = 6.4 x 10^{6 }m, G = 6.7 x 10^{11} Nm^{2}kg^{2}
The magnitude of gravitational force between the earth and object is given by:
Force of gravitation between them,
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 force or the same as the force with which the moon attracts the earth? Why?
Ans:
Q5. If the moon attracts the Earth, why does the earth not move towards the moon?
Ans:
Q6. What happens to the force between two objects, if
(a) the mass of one object is doubled?
(b) the distance between the objects is doubled and tripled?
(c) the masses of both objects are doubled?
Ans: Let masses = m_{1} and m_{2}, distance between masses = r.
According to universal law of gravitation F ∝ m_{1}m_{2} and F ∝ 1/r_{2}.
(a) If one mass, say m_{1} is doubled, and F ∝ m_{1}, then F gets doubled.
(b) If distance is doubled and F ∝ 1/r^{2}, force becomes onefourth, i.e. 1/(2)^{2}.
If distance is tripled and F ∝ 1/r^{2}, force becomes oneninth, i.e. 1/(3)^{2}.
(c) If both masses are doubled and F ∝ m_{1}m_{2}, force becomes 4 times, i.e. (2m_{1})(2m_{2}) = 4m_{1}m_{2}
Q7. What is the importance of universal law of gravitation?
Ans: The universal law of gravitation has successfully explained the phenomena that were earlier considered to be separate:
(a) It explains the force which holds us on earth.
(b) It explains the motion of planets around the sun.
(c) It explains the motion of the moon around the Earth.
(d) It explains the occurrence of tides in the ocean.
Q8. What is the acceleration of free fall?
Ans: When a body falls freely and accelerates at every point of its motion due to gravitational force alone. This is called the acceleration of free fall. This acceleration is known as the acceleration due to gravity on the earth’s surface. It’s denoted by ‘g’, and its value is 9.8m/s2, and it’s constant for all objects close to the earth’s surface (irrespective of their masses).
Q9. What do we call the gravitational force between the earth and an object?
Ans: Gravity is the name of the gravitational force between the earth and 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: Earth is not a perfect sphere. It is flattened at the poles. Thus, the value of ‘g’ is greater at the poles than at the equator (as g ∝ 1/r^{2}). This means the weight of gold bought at the poles becomes lesser at the equator, and the friend does not agree with the weight.
Q:11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans:
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: Here, mass of object, m = 10 kg, g on the earth’s surface = 9.8 ms^{2}
Weight of the object on the earth’s surface = mg
= 10 kg x 9.8 ms^{2}
= 98.0 kg ms^{2} = 98 N
The weight of object on moon is 1/6th that of object on earth, so the weight of the object on the moon:
= 1/6 x 98 N= 16.34 N.
Q13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(a) the maximum height to which it rises.
(b) the total time it takes to return to the surface of the earth.
Ans:
(a) Initial velocity, u = 49 ms^{1}, Final velocity, v = 0
Maximum height, H = ?
Acceleration due to gravity, g = + 9.8 ms^{2} (downward)
As object is moving upward, so a = gFrom relation v^{2} = u^{2 }+ 2as , we have a = g , s = H
v^{2 }= u^{2 } 2gH
⇒ H = 122.5 m
(b) When ball returns on the surface of earth, displacement, s = 0
∴ Relation, s = ut + at^{2} gives s = ut gt^{2}
0 = (49 ms^{1}) t x 9.8 t^{2}
⇒ t (49 4.9 t) = 0
t = 0 or 49  4.9 t = 0 ⇒ t =
As t ≠ 0, so t = 10s
The maximum height is 122.5 m and total time is 10 s.
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.
Ans: Initial velocity, u = 0, height, h = 19.6 m and g = 9.8 ms^{2}
Final velocity,
= 19.6 ms^{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 are the net displacement and the total distance covered by the stone?
Ans: Initial velocity, u = 40 ms^{1}
g = 10 ms^{2} (upward motion)
Final velocity, v = 0
During upward motion, g = 10 ms^{2}
Net displacement on returning back = zero
Total distance = 80 m + 80 m = 160 m
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: Mass of earth, M_{e} = 6 x 10^{24} kg
Mass of sun M_{s} = 2 x 10^{30} kg and
Distance, r = 1.5 x 10^{11} m
Gravitational constant, G = 6.67 x 10^{11} Nm^{2}/kg^{2}
Force,
= 3.56 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.
Ans: u_{1} = 0
Let the stones meet at a height h above the ground. Stone 1 covers a distance (100  h); g = 10 m/s^{2}.
(100  h) = 0 +
⇒ 100  h = 5 t^{2} ...(i)
u_{2} = 25 m/s
Stone 2 covers a distance h
g =  10m/s^{2}
h = u_{2}t 
⇒ h = 25 t  5 t^{2} ...(ii)
From (i) and (ii), we get
100  25 t + 5 t^{2} = 5 t^{2}
100  25 t = 0
or t = 4 s (Stones meet 4 s after they are thrown)
h = 25(4)  5(4)^{2 }= 100  80 = 20 m above the ground.
Q18. A ball thrown up vertically returns to the thrower after 6s. Find
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches.
(c) its position after 4s.
Ans:
Time of ascend = Time of descending
⇒ Time taken to reach the top, t = 3 s.
⇒ Final velocity, v = 0, g = 9.8 ms^{2}.
(a) Initial velocity, u = v  gt = 0  (9.8) (3)
= 29.4 ms^{1}
(b) Maximum height reached, h = (v^{2}  u^{2})/g = (29.4)^{2}/(2 x 9.8)
= 44.1 m
(c) After 4s, ball has started falling and has fallen by some distance h' for 1s.
Here, initial velocity u' = 0, t = 1s
g = + 9.8 ms^{1}
h' = u't + = 0 +
The ball is at a height, (44.1  4.9) = 39.2 m above the ground.
Q19. In what direction does the buoyant force on an object immersed in a liquid act?
Ans: 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?
Ans:
Q21. The volume of 50g of a substance is 20 cm^{3}. If the density of water is 1g cm^{3}, will the substance float or sink?
Ans:
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:
80 videos363 docs97 tests

NCERT Textbook  Gravitation Doc  15 pages 
Gravitation Video  11:28 min 
Numericals Practice Questions  Gravitation Doc  21 pages 
1. What is the formula for gravitational force? 
2. How does the gravitational force between two objects change with distance? 
3. What is the difference between mass and weight? 
4. What is free fall? 
5. How does the gravitational force affect the motion of planets? 
NCERT Textbook  Gravitation Doc  15 pages 
Gravitation Video  11:28 min 
Numericals Practice Questions  Gravitation Doc  21 pages 

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