A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

To find the average force exerted by the bat on the ball, we can apply Newton's second law of motion in terms of momentum, which states that the average force is equal to the rate of change of momentum.

First, let's list the given values:
Mass of the ball, $m=150\text{gm}=0.15\text{kg}$
Initial velocity of the ball, $v_i=12\text{m/s}$
Since the ball is hit and turned back, its final velocity will be in the opposite direction. Taking the final direction as negative (or vice versa):
Final velocity of the ball, $v_f=-20\text{m/s}$
Time interval for which the force acts, $\Delta t=0.01\text{s}$

The change in momentum ($\Delta p$) of the ball is given by:
$\Delta p=m(v_f-v_i)$

Substituting the values:
$\Delta p=0.15\times (-20-12)$
$\Delta p=0.15\times (-32)$
$\Delta p=-4.8\text{kg m/s}$

The magnitude of the change in momentum is:
$\left|\Delta p\right|=4.8\text{N s}$

Now, the average force ($F$) exerted by the bat is:
$F=\frac{\left|\Delta p\right|}{\Delta t}$

Substituting the values:
$F=\frac{4.8}{0.01}=480\text{N}$

Therefore, the average force exerted by the bat on the ball is 480 N.