A cricket ball of mass $250g$ collides with a bat with velocity $10m/s$ and returns with the same velocity in 0.01 seconds. The force acted on the bat is: 

Therefore, the mass of the ball in question is 250 g, and the initial velocity of the ball before hitting the racket is $10m/s$. After hitting the racket, the ball should move at a speed of $10m/s$. 

Therefore, the initial momentum is given by the following equation

$P_i=m{v}_i$

 Where $v_i$ is the initial velocity. 

$$\begin{gathered} P_i=(0.25kg)\times (10m/s) \\ {P}_i=2.5kgm/s \_\_\_\_\_\_\left(1\right) \end{gathered}$$

 The final momentum of the ball is given by 

$P_f=m{v}_f \_\_\_\_\_\_\_\left(2\right)$

. Where $v_f$ is the final velocity.

 From equations   (1) and (2), we can see that the magnitudes of the first and last momentums are the same, but the directions are opposite to each other. Therefore, we can write: $P_f=-P_i$
Therefore, since the rate of change in momentum in a cycle of 0.01 seconds is the force applied to the bat, it can be written as follows. 

$$\begin{gathered} F=\frac{P_f-P_i}{t} \\ F=\frac{2P_f}{t} \\ F=\frac{2\times (2.5kgm/s)}{0.01s} \\ F=500N \end{gathered}$$
Therefore, the force applied to the bat is $F = 500 N$. Hence, the answer to the question is option 4.