A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

To find the force the batsman must apply to hold the bat firmly, we can analyze the collision using Newton's second law of motion and the concept of impulse.

First, let's identify and convert the given values into standard SI units:
- Mass of the cricket ball, $m=150\text{g}=0.15\text{kg}$
- Initial speed of the ball, $u=126\text{km/h}=126\times \frac{5}{18}\text{m/s}=35\text{m/s}$
- Since the collision is completely elastic, the ball bounces straight back to the bowler with the same speed but in the opposite direction. Therefore, the final velocity of the ball is $v=-35\text{m/s}$.
- Time of contact between the ball and the bat, $\Delta t=0.001\text{s}$.

Next, we calculate the change in momentum ($\Delta p$) of the cricket ball:
$\Delta p=m(v-u)$
$\Delta p=0.15\text{kg}\times (-35\text{m/s}-35\text{m/s})$
$\Delta p=0.15\times (-70)\text{kg m/s}=-10.5\text{kg m/s}$

The magnitude of the change in momentum is $10.5\text{kg m/s}$. By the impulse-momentum theorem, the average force exerted by the bat on the ball is:
$F_{\text{average}}=\frac{\left|\Delta p\right|}{\Delta t}$
$F_{\text{average}}=\frac{10.5}{0.001}\text{N}=10500\text{N}=1.05\times {10}^4\text{N}$

According to Newton's third law of motion, the ball exerts an equal and opposite force on the bat. To hold the bat firmly at its position, the batsman must apply an equal and opposite force. Therefore, the force that the batsman had to apply is $1.05\times {10}^4\text{N}$.