Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to the formula for gravitational force,

$F = G \times \frac{m_1 \times m_2}{r^2}$

,
the force is inversely proportional to the square of the distance between the two objects.

If the distance

$r$

is reduced to half, then the new distance becomes

$\frac{r}{2}$

.

The new force

$F’$

will be:

$F’ = G \times \frac{m_1 \times m_2}{\left(\frac{r}{2}\right)^2} = G \times \frac{m_1 \times m_2}{\frac{r^2}{4}} = 4 \times G \times \frac{m_1 \times m_2}{r^2} = 4F$

.

Therefore, the gravitational force becomes four times as large when the distance is reduced to half.

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?

Answer: Although gravitational force is proportional to mass, the acceleration due to gravity remains the same for all objects regardless of their mass. This is because acceleration = force/mass. When mass increases, force also increases proportionally, so their ratio remains constant. From F = mg, we get g = F/m, and since F ∝ m,the acceleration g remains constant for all objects, making them fall at the same rate in the absence of air resistance.

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

$6 \times 10^{24}$

kg and radius of the earth is

$6.4 \times 10^{6}$

m.)
Answer:
Given:
Mass of earth,

$m_1 = 6 \times 10^{24}$

kg
Mass of object,

$m_2 = 1$

kg
Radius of earth (distance between centers),

$r = 6.4 \times 10^{6}$

m
Universal gravitational constant,

$G = 6.674 \times 10^{-11}$

N·m²/kg²

Formula:

$F = G \times \frac{m_1 \times m_2}{r^2}$

Substitution:

$F = 6.674 \times 10^{-11} \times \frac{6 \times 10^{24} \times 1}{(6.4 \times 10^{6})^2}$

Calculation:

$r^2 = (6.4 \times 10^{6})^2 = 40.96 \times 10^{12} = 4.096 \times 10^{13}$

So,

$F = 6.674 \times 10^{-11} \times \frac{6 \times 10^{24}}{4.096 \times 10^{13}}$

$F = 6.674 \times 10^{-11} \times 1.4648 \times 10^{11}$

$F \approx 9.77$

N

The magnitude of the gravitational force is approximately 9.77 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?

Answer: According to Newton’s third law of motion, the earth attracts the moon with exactly the same force as the moon attracts the earth. These forces are equal in magnitude but opposite in direction. The gravitational force is mutual – both objects exert equal and opposite forces on each other, regardless of their different masses.

Q5. If the moon attracts the earth, why does the earth not move towards the moon?

Answer: The earth does move towards the moon, but this motion is not noticeable because of the earth’s much larger mass compared to the moon. According to Newton’s second law, acceleration = force/mass. Since both earth and moon experience the same gravitational force, but earth’s mass is about 81 times larger than the moon’s mass, earth’s acceleration towards the moon is much smaller and practically negligible compared to the moon’s acceleration towards earth.

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?

Answer:
(i) If the mass of one object is doubled, the gravitational force between them is doubled, since force is directly proportional to mass.
(ii) If the distance is doubled, the force becomes 1/4th of the original force. If the distance is tripled, the force becomes 1/9th of the original force, since force is inversely proportional to the square of distance.
(iii) If the masses of both objects are doubled, the gravitational force becomes four times the original force, since force is directly proportional to the product of both masses.

Q7. What is the importance of universal law of gravitation?

Answer: The universal law of gravitation is important because it successfully explains several natural phenomena that were previously thought to be unconnected. It explains the force that binds us to earth, the motion of moon around earth, the motion of planets around the sun, and the occurrence of tides due to moon and sun. It helped establish that the same physical laws govern both terrestrial and celestial objects, unifying our understanding of the universe.

Q8. What is the acceleration of free fall?

Answer: The acceleration of free fall is the constant acceleration experienced by all objects when they fall freely under the influence of earth’s gravitational force alone. It is denoted by ‘g’ and has a value of approximately 9.8 m s⁻² near the earth’s surface. This acceleration is independent of the mass, size, or shape of the falling object.

Q9. What do we call the gravitational force between the earth and an object?

Answer: The gravitational force between the earth and an object is called the weight of the object. Weight is the force with which the earth attracts any object towards its center. It is calculated as W = mg, where m is the mass of the object and g is the acceleration due to gravity.

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?

Answer: The friend will not agree with the weight of gold bought. The weight of gold will be less at the equator compared to the poles because the value of acceleration due to gravity ‘g’ is greater at the poles than at the equator. This happens because the earth is not a perfect sphere – it is flattened at the poles and bulges at the equator, making the radius larger at the equator. Since weight = mg, the gold will weigh less at the equator.

Q11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer: A sheet of paper falls slower than a crumpled paper ball because of air resistance. The flat sheet has a larger surface area exposed to air, which offers more resistance to its motion through air. The crumpled ball has a smaller surface area and more compact shape, so it experiences less air resistance and falls faster. In vacuum, both would fall at the same rate.

Q12. Gravitational force on the surface of the moon is only

$\frac{1}{6}$

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?
Answer:
Given:
Mass of object,

$m = 10$

kg
Acceleration due to gravity on Earth,

$g_{\text{earth}} = 9.8$

m/s²
Acceleration due to gravity on Moon,

$g_{\text{moon}} = \frac{1}{6} \times 9.8 = 1.63$

m/s²

Formula:
Weight,

$W = m \times g$

Weight on Earth:

$W_{\text{earth}} = 10 \times 9.8 = 98$

N

Weight on Moon:

$W_{\text{moon}} = 10 \times 1.63 = 16.3$

N

Weight of the object on Earth is 98 N and on the Moon is 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.
Answer:
Given:
Initial velocity,

$u = 49$

m/s
Final velocity at maximum height,

$v = 0$

m/s
Acceleration due to gravity,

$g = 9.8$

m/s²
Maximum height,

$h = ?$

Total time to return,

$T = ?$

(i) Maximum height:
Formula:

$v^2 = u^2 – 2 g h$

Substitution:

$0 = (49)^2 – 2 \times 9.8 \times h$

Calculation:

$0 = 2401 – 19.6 h$

$19.6 h = 2401$

$h = \frac{2401}{19.6} = 122.5$

m

(ii) Total time to return:
Time to reach maximum height,

$t = \frac{u}{g} = \frac{49}{9.8} = 5$

s

Total time for upward and downward journey:

$T = 2 \times t = 2 \times 5 = 10$

s

Maximum height = 122.5 meters
Total time to return = 10 seconds

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.

Answer:
Given: u = 0, h = 19.6 m, g = 9.8 m s⁻²
Formula: v² = u² + 2gh
Substitution: v² = 0² + 2 × 9.8 × 19.6
v² = 384.16
v = 19.6 m/s
The final velocity just before touching the ground is 19.6 m/s.

Q15. 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 is the net displacement and the total distance covered by the stone?
Answer:
Given:
Initial velocity,

$u = 40$

m/s
Acceleration due to gravity,

$g = 10$

m/s²
Maximum height,

$h = ?$

Net displacement,

$d = ?$

Total distance covered,

$s = ?$

Maximum height:
Formula:

$v^2 = u^2 – 2 g h$

where

$v = 0$

at maximum height.

Substitution:

$0 = 40^2 – 2 \times 10 \times h$

Calculation:

$0 = 1600 – 20 h$

$20 h = 1600$

$h = \frac{1600}{20} = 80$

m

Net displacement:
The stone returns to the thrower’s hand, so net displacement is zero.

$d = 0$

m

Total distance covered:
The stone travels up to maximum height and then comes down the same distance.
Total distance,

$s = h + h = 2 \times 80 = 160$

m

Maximum height = 80 meters
Net displacement = 0 meters
Total distance covered = 160 meters

Q16. 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.
Answer:
Given:
Mass of Earth,

$m_1 = 6 \times 10^{24}$

kg
Mass of Sun,

$m_2 = 2 \times 10^{30}$

kg
Distance between Earth and Sun,

$r = 1.5 \times 10^{11}$

m
Universal gravitational constant,

$G = 6.674 \times 10^{-11}$

N·m²/kg²

Formula:

$$F = G \times \frac{m_1 \times m_2}{r^2}$$

Substitution:

$$F = 6.674 \times 10^{-11} \times \frac{6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}$$

Calculation:

$$r^2 = (1.5 \times 10^{11})^2 = 2.25 \times 10^{22}$$

$$F = 6.674 \times 10^{-11} \times \frac{12 \times 10^{54}}{2.25 \times 10^{22}} = 6.674 \times 10^{-11} \times 5.333 \times 10^{32}$$

$$F \approx 3.56 \times 10^{22} \, \text{N}$$

The gravitational force between the Earth and the Sun is approximately

$3.56 \times 10^{22}$

newtons.

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.
Answer:
Given:
Height of tower,

$h = 100$

m
Initial velocity of stone thrown upward,

$u = 25$

m/s
Acceleration due to gravity,

$g = 9.8$

m/s²
Time when stones meet,

$t = ?$

Distance from ground where they meet,

$s = ?$

Let the time taken for the stones to meet be

$t$

seconds.

For stone falling down from the tower:
Initial velocity,

$u_1 = 0$

(released from rest)
Distance fallen in time

$t$

,

$s_1 = \frac{1}{2} g t^2 = \frac{1}{2} \times 9.8 \times t^2 = 4.9 t^2$

m

Height of stone from the ground at time

$t$

,

$h_1 = 100 – s_1 = 100 – 4.9 t^2$

m

For stone moving upward from the ground:
Initial velocity,

$u_2 = 25$

m/s
Distance risen in time

$t$

,

$s_2 = u_2 t – \frac{1}{2} g t^2 = 25 t – 4.9 t^2$

m

At meeting point:
Heights of both stones from ground are equal,

$h_1 = s_2$

So,

$100 – 4.9 t^2 = 25 t – 4.9 t^2$

Simplify:

$100 = 25 t$

$t = \frac{100}{25} = 4$

seconds

Distance from ground where they meet:

$s = s_2 = 25 \times 4 – 4.9 \times 4^2 = 100 – 4.9 \times 16 = 100 – 78.4 = 21.6$

m

The two stones will meet after 4 seconds at 21.6 meters above the ground.

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.
Answer:
Given:
Total time of flight,

$T = 6$

s
Acceleration due to gravity,

$g = 9.8$

m/s²
Initial velocity,

$u = ?$

Maximum height,

$h = ?$

Position after 4 s,

$s = ?$

(a) Velocity with which it was thrown up:
Time to reach maximum height,

$t = \frac{T}{2} = \frac{6}{2} = 3$

s

Using,

$v = u – g t$

At maximum height, final velocity

$v = 0$

, so:

$0 = u – 9.8 \times 3$

$u = 9.8 \times 3 = 29.4$

m/s

(b) Maximum height:
Using,

$v^2 = u^2 – 2 g h$

At max height,

$v = 0$

, so:

$0 = u^2 – 2 g h$

$h = \frac{u^2}{2 g} = \frac{(29.4)^2}{2 \times 9.8} = \frac{864.36}{19.6} = 44.1$

m

(c) Position after 4 seconds:
Using,

$s = u t – \frac{1}{2} g t^2$

$s = 29.4 \times 4 – \frac{1}{2} \times 9.8 \times 4^2 = 117.6 – 78.4 = 39.2$

m

(a) Initial velocity = 29.4 m/s
(b) Maximum height = 44.1 meters
(c) Position after 4 seconds = 39.2 meters above the thrower

Q19. In what direction does the buoyant force on an object immersed in a liquid act?

Answer: The buoyant force on an object immersed in a liquid always acts vertically upward, opposite to the direction of gravitational force. This upward force is exerted by the liquid on the immersed object and opposes the weight of the object.

Q20. Why does a block of plastic released under water come up to the surface of water?

Answer: A block of plastic comes up to the surface because its density is less than the density of water. When the plastic block is immersed in water, the buoyant force (upward force exerted by water) is greater than the weight of the plastic block (downward force). This net upward force causes the plastic block to rise and float on the water surface.

Q21. 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?
Answer:
Given:
Mass of substance,

$m = 50$

g
Volume of substance,

$V = 20$

cm³
Density of water,

$\rho_{\text{water}} = 1$

g/cm³
Density of substance,

$\rho_{\text{substance}} = ?$

Formula:
Density,

$$\rho = \frac{m}{V}$$

Substitution:

$$\rho_{\text{substance}} = \frac{50}{20} = 2.5 \text{ g/cm}^3$$

Since the density of the substance (2.5 g/cm³) is greater than the density of water (1 g/cm³),

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/cm³? What will be the mass of the water displaced by this packet?
Answer:
Given:
Mass of packet,

$m = 500$

g
Volume of packet,

$V = 350$

cm³
Density of water,

$\rho_{\text{water}} = 1$

g/cm³
Density of packet,

$\rho_{\text{packet}} = ?$

Mass of water displaced,

$m_{\text{water}} = ?$

Formula:
Density

$$\rho = \frac{m}{V}$$

Mass of water displaced (equal to volume × density of water),

$$m_{\text{water}} = V \times \rho_{\text{water}}$$

Substitution:

$$\rho_{\text{packet}} = \frac{500}{350} \approx 1.43 \text{ g/cm}^3$$

$$m_{\text{water}} = 350 \times 1 = 350 \text{ g}$$

Since the density of the packet (1.43 g/cm³) is greater than the density of water (1 g/cm³),

• The packet will sink in water.
• Mass of water displaced by the packet is 350 g.