Question 1: How does the force of gravitation between two objects change when the distance between them is reduced to half?
The force of gravitation between two objects is inversely proportional to the square of the distance between them; therefore, the gravity will become four times if the distance between them is reduced to half.
Question 2: Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object?
In free fall of objects, the acceleration in velocity due to gravity is independent of the mass of those objects; hence, a heavy object does not fall faster than a light object.
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 x 10^24 kg and radius of the earth is 6.4 x 10^6 m.)
= 9.81 N
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?
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.
Question 5: If the moon attracts the earth, why does the earth not move towards the moon?
Earth does not move towards the moon because the mass of the moon is very small compared to that of the earth.
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?
(i) The force between two objects will be doubled.
(ii) The force between two objects will become 1/4th and 1/9th of the present force.
(iii) The force between two objects will become four times the present force.
Question 7: What is the importance of the universal law of gravitation?
The universal law of gravitation is important due to the following:
i) This law explains well the force that binds us to the earth.
ii) This law describes the motion of planets around the sun.
iii) This law justifies the tide formation on earth due to the moon and sun.
iv) This law gives reason for the movement of the moon around the earth.
Question 8: What is the acceleration of free fall?
The acceleration of free fall is g = 9.8 m/s^2 (on earth).
Question 9: What do we call the gravitational force between the earth and an object?
Weight
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.]
Since \( W = m \times g \) and given in the question that the value of \( g \) is greater at the poles than at the equator, hence the weight of the same amount of gold will be lesser at the equator than it was at the poles. Therefore, the friend will not agree with the weight of gold bought.
Question 11: Why will a sheet of paper fall slower than one that is crumpled into a ball?
A greater surface area offers greater resistance and buoyancy, the same is true in the case of a sheet of paper that has a larger surface area as compared to paper crumpled into a ball. So, the sheet of paper falls slower.
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 newtons of a 10 kg object on the moon and on the earth?
Value of gravity on earth \( g = 9.8 \, \text{m/s}^2 \)
Value of gravity on the moon = 1/6th of earth = \( \frac{9.8}{6} = 1.63 \, \text{m/s}^2 \)
Weight of the object on the moon = \( m \times 1.63 = 10 \times 1.63 = 16.3 \, \text{N} \)
Weight of the object on earth = \( m \times 9.8 = 10 \times 9.8 = 98 \, \text{N} \)
Question 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.
(i) Use \( v = u + gt \)
Initial velocity (\( u \)) = 49 m/s
Acceleration due to gravity (\( g \)) = -9.8 m/s² (negative because it's against the motion)
At maximum height, (\( v = 0 \)), so solve for time (\( t \)):
\( 0 = 49 + (-9.8) \times t \)
\( 9.8t = 49 \)
\( t = \frac{49}{9.8} = 5 \) s
Substitute \( t \) back to find \( h \):
\( h = ut + \frac{1}{2}gt^2 = 49 \times 5 + \frac{1}{2} \times (-9.8) \times 5^2 \)
\( h = 122.5 \) m
(ii) Total time taken to return \( = 5 + 5 = 10 \) s
Question 14: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.
Height of the tower (\( h \)) = 19.6 m
Using \( v^2 = u^2 + 2gh \) to find final velocity (\( v \)):
\( v^2 = 0 + 2 \times 9.8 \times 19.6 \)
\( v^2 = 2 \times 9.8 \times 19.6 \)
\( v = \sqrt{2 \times 9.8 \times 19.6} \)
\( v = 19.6 \) m/s
Question 15: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking \( g = 10 \, \text{m/s}^2 \), find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Initial velocity (\( u \)) = 40 m/s
At maximum height, final velocity (\( v \)) = 0
Using \( v = u + gt \) to find time (\( t \)):
\( 0 = 40 + (-10) \times t \)
\( t = \frac{40}{10} = 4 \) s
Maximum height (\( h \)) = \( ut + \frac{1}{2}gt^2 = 40 \times 4 + \frac{1}{2} \times (-10) \times 4^2 \)
\( h = 160 - 80 = 80 \) m
Net displacement = 0 (stone thrown upwards then falls back to the same place)
Total distance covered = \( 80 + 80 = 160 \) m
Question 16: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = \( 6 \times 10^{24} \) kg and of the Sun = \( 2 \times 10^{30} \) kg. The average distance between the two is \( 1.5 \times 10^{11} \) m.
Mass of the earth (\( M_1 \)) = \( 6 \times 10^{24} \) kg
Mass of the Sun (\( M_2 \)) = \( 2 \times 10^{30} \) kg
Distance between the earth and the Sun (\( r \)) = \( 1.5 \times 10^{11} \) m
Using \( F = \frac{G \times M_1 \times M_2}{r^2} \) with \( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \):
\( F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2} \)
\( F \approx 35.73 \times 10^{21} \, \text{N} \)
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 m/s. Calculate when and where the two stones will meet.
Suppose both the stones will meet after \( t \) seconds.
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, and (c) its position after 4 s.
(a) Time taken by the ball to reach maximum height (\( t \)) = 6/2 = 3 s
Question 19: In what direction does the buoyant force on an object immersed in a liquid act?
The buoyant force on an object immersed in a liquid acts in the upward direction only.
Question 20: Why does a block of plastic released underwater come up to the surface of water?
Since the density of plastic is much less than that of water, the buoyant force on the plastic is greater than its weight, causing it to float or 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/cm³, will the substance float or sink?
The substance will sink since its density (2.5 g/cm³) is greater than the density of water (1 g/cm³).
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/cm³? What will be the mass of the water displaced by this packet?
The packet will sink since its density (1.428 g/cm³) is greater than the density of water (1 g/cm³).
The mass of water displaced by the packet is 350 g.