A cricket player catches a ball of mass 120 g moving with $$25 \text{ m s}^{-1}$$ speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be (in SI unit):

A cricket player catches a ball of mass $$m = 120 \text{ g} = 0.12 \text{ kg}$$ moving with velocity $$v = 25 \text{ m/s}$$. The catching process takes $$\Delta t = 0.1 \text{ s}$$.

The initial momentum of the ball is calculated as $$p_i = mv = 0.12 \times 25 = 3 \text{ kg m/s}$$, and since the ball comes to rest, the final momentum is $$p_f = 0$$. Therefore, the change in momentum is $$\Delta p = |p_f - p_i| = 3 \text{ kg m/s}$$.

The average force exerted during the catch is then $$F = \frac{\Delta p}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$$.

Hence, the answer is Option D: $$30$$.