Solution:

When the lift is stationary, the time taken by the ball to return to the boy's hand is 10 s.

Now, when the lift starts moving up at a speed of 5 m/s, we need to find the time taken by the ball to return to the boy's hand.

Let's consider the motion of the ball in two parts:

1. Motion of the ball when thrown up by the boy in the stationary lift
2. Motion of the ball when thrown up by the boy in the moving lift

1. Motion of the ball when thrown up by the boy in the stationary lift:
When the lift is stationary, the motion of the ball is a simple case of vertical motion under gravity.

The time taken by the ball to reach the topmost point (highest point) can be calculated using the formula:
t = u/g, where u is the initial velocity of the ball and g is the acceleration due to gravity.

Here, the initial velocity of the ball (u) is zero, as the boy throws the ball vertically upward from rest.

So, t = 0/g = 0 s.

The time taken by the ball to reach the topmost point is zero seconds.

The total time taken by the ball to return to the boy's hand is 10 s. Therefore, the time taken by the ball to come back to the boy's hand from the topmost point is 10 - 0 = 10 s.

2. Motion of the ball when thrown up by the boy in the moving lift:
When the lift is moving up at a speed of 5 m/s, the motion of the ball is a case of projectile motion.

The initial vertical velocity (u) of the ball is zero, as the boy throws the ball vertically upward from rest.

The horizontal velocity (v) of the lift is 5 m/s. This horizontal velocity does not affect the vertical motion of the ball. Hence, we can consider the vertical motion of the ball independently of the horizontal motion of the lift.

The time taken by the ball to reach the topmost point (highest point) can be calculated using the formula:
t = u/g, where u is the initial velocity of the ball and g is the acceleration due to gravity.

Here, the initial velocity of the ball (u) is zero, as the boy throws the ball vertically upward from rest.

So, t = 0/g = 0 s.

The time taken by the ball to reach the topmost point is zero seconds.

The total time taken by the ball to return to the boy's hand is 10 s. Therefore, the time taken by the ball to come back to the boy's hand from the topmost point is 10 - 0 = 10 s.

Therefore, the time taken by the ball to return to the boy's hand is the same in both cases, i.e., 10 s.

Hence, the correct option is D, i.e., equal to 10 s.

Solution:

Given:

Time taken by the ball to return to his hands in a stationary lift = 10 s

Speed of the lift = 5 m/s

To find:

The time taken for a ball thrown straight up to return to his hands

Explanation:

Let's assume that the height to which the ball is thrown is 'h'.

When the lift is stationary, the time taken by the ball to reach the maximum height and come back to the boy's hands is given by:

t = 2u/g

where,

u = initial velocity of the ball = √(2gh)

g = acceleration due to gravity = 9.8 m/s²

∴ t = 2√(2h/g)

Given, t = 10 s

∴ 10 = 2√(2h/g)

∴ h = g/8

Now, when the lift starts moving up, the initial velocity of the ball will be the sum of the velocity of the lift and the velocity with which the ball is thrown.

Let the velocity with which the ball is thrown be 'v'.

∴ Initial velocity of the ball = 5 + v

Using the formula for time of flight, the time taken by the ball to reach the maximum height and come back to the boy's hands is given by:

t' = 2u/g

where,

u = initial velocity of the ball = 5 + v

∴ t' = 2(5 + v)sinθ/g

where,

θ = angle of projection = 90° (as the ball is thrown straight up)

∴ t' = 2(5 + v)/g

Given, t = 10 s

∴ t' = t = 10 s

∴ 2(5 + v)/g = 10

∴ 5 + v = 49

∴ v = 44 m/s

Now, using the formula for time of flight, the time taken by the ball to reach the maximum height and come back to the boy's hands is given by:

t'' = 2u/g

where,

u = initial velocity of the ball = 5 + 44 = 49 m/s

∴ t'' = 2(49)sinθ/g

where,

θ = angle of projection = 90° (as the ball is thrown straight up)

∴ t'' = 2(49)/g

∴ t'' = 10 s

Therefore, the time taken for a ball thrown straight up to return to his hands is equal to 10 s. Hence, the correct option is (D).