A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand. The average force exerted by sand on the ball is ______ N.
(Take g=10 m/$$s^{2}$$)

The ball is released from rest, so its entire loss of gravitational potential energy is converted into the work done against the resisting force of the sand during the 10 cm penetration.

Total vertical drop before coming to rest
= free-fall height + penetration depth
$$h = 10 \text{ m} + 0.1 \text{ m} = 10.1 \text{ m}$$

Loss of gravitational potential energy
$$\Delta U = m g h$$
Here, $$m = 2 \text{ kg}$$ and (as usual for JEE) we take $$g = 10 \text{ m s}^{-2}$$.
$$\Delta U = 2 \times 10 \times 10.1 = 202 \text{ J}$$

This entire 202 J is dissipated only while the ball moves the last $$0.1 \text{ m}$$ inside the sand. If $$F_{\text{avg}}$$ is the average upward (resisting) force exerted by the sand, the work it does is

$$W_{\text{sand}} = F_{\text{avg}} \times s$$ where $$s = 0.1 \text{ m}$$ is the penetration depth.

Because the ball finally stops, the work done by the sand equals the loss in potential energy (energy conservation):

$$F_{\text{avg}} \times 0.1 = 202$$

$$\Rightarrow \; F_{\text{avg}} = \frac{202}{0.1} = 2020 \text{ N}$$

Thus the average force exerted by the sand on the ball is

Option B which is: $$2020 \text{ N}$$