QJEEM24Ph4172: JEE Main 2024 Physics - Class 11 JEEM24Ph4172 Solution: A player caught a cricket ball of mass moving at a speed of .…

The force exerted by the ball on the hand can be calculated using the formula derived from Newton's second law of motion, which is $F = \frac{\Delta p}{\Delta t}$ , where $F$ is the force, $\Delta p$ represents the change in momentum, and $\Delta t$ is the time over which this change occurs.

The change in momentum, $\Delta p$ , can be calculated as the difference between the final momentum, $p_f$ , and the initial momentum, $p_i$ . In this scenario, because the ball comes to a stop in the player's hand, its final velocity (and hence, its final momentum) is 0. Therefore, the change in momentum is equal to the initial momentum of the ball (since final momentum is zero).

The initial momentum, $p_i$ , of the ball can be calculated using the formula $p = mv$ , where $m$ is the mass of the ball and $v$ is its velocity. Given that the mass of the ball is $150 \, \mathrm{g} = 0.15 \, \mathrm{kg}$ (converting grams to kilograms) and its velocity is $20 \, \mathrm{m/s}$ , we have:

$p_i = (0.15 \, \mathrm{kg}) \times (20 \, \mathrm{m/s}) = 3 \, \mathrm{kg \cdot m/s}$

Since the change in momentum, $\Delta p$ , equals the initial momentum ( $p_i$ ) because the final momentum is 0, the force exerted can be found by substituting $\Delta p$ and $\Delta t$ into the first formula:

$F = \frac{3 \, \mathrm{kg \cdot m/s}}{0.1 \, \mathrm{s}} = 30 \, \mathrm{N}$

Therefore, the magnitude of force exerted by the ball on the hand of the player is 30 N, which corresponds to Option C.