QJEEM23Ph756f: JEE Main 2023 Physics - Class 11 JEEM23Ph756f Solution: A body is dropped on ground from a height ' ' and after hitting …

The velocity of the body just before hitting the ground, due to gravitational acceleration, is given by $v_{1} = \sqrt{2gh_{1}}$ , and the velocity just after hitting the ground, when it rebounds to a height $h_{2}$ , is given by $v_{2} = \sqrt{2gh_{2}}$ .

According to the problem, the ratio $\frac{v_{1}}{v{2}} = 4$ . Therefore, we can write $\frac{\sqrt{2gh_{1}}}{\sqrt{2gh_{2}}} = 4$ or equivalently $\frac{h_{1}}{h_{2}} = 4^2 = 16$ .

The loss in kinetic energy due to the collision with the ground is given by the difference between the initial kinetic energy $K_{1} = \frac{1}{2} m v_{1}^2$ and the final kinetic energy $K_{2} = \frac{1}{2} m v_{2}^2$ , where m is the mass of the body.

Substituting $v_{1} = \sqrt{2gh_{1}}$ and $v_{2} = \sqrt{2gh_{2}}$ into these expressions, we get $K_{1} = mgh_{1}$ and $K_{2} = mgh_{2}$ .

The loss in kinetic energy is then $\Delta K = K_{1} - K_{2} = mgh_{1} - mgh_{2}$ .

The percentage loss in kinetic energy is given by

$\frac{\Delta K}{K_{1}} \times 100 = \frac{mgh_{1} - mgh_{2}}{mgh_{1}} \times 100 = \frac{h_{1} - h_{2}}{h_{1}} \times 100$ .

Since $h_{1}/h_{2} = 16$ , we can write $h_{2} = h_{1}/16$ , so the percentage loss in kinetic energy is

$\frac{h_{1} - h_{1}/16}{h_{1}} \times 100 = 100(1 - \frac{1}{16}) = 100 \times \frac{15}{16} = \frac{375}{4}$ .

So, the value of $x$ is 375.