Explanation

(d) Here, $m_{1} = 20$ kg,
$m_{2} = 0.1$ kg,
$v_{1} =$ velocity of recoil of gun,
$v_{2}\ =$ velocity of bullet
As $m_{1}v_{1} = m_{2}v_{2}$
$v_{1} = \frac{m_{2}}{m_{1}}v_{2} = \frac{0.1}{20}v_{2} = \frac{v_{2}}{200}$
Recoil energy of gun$= \frac{1}{2}m_{1}v_{1}^{2}$
$= \frac{1}{2} \times 20\left( \frac{v_{2}}{200} \right)^{2}$
$804 = \frac{10v_{2}^{2}}{4 \times 10^{4}} = \frac{v_{2}^{2}}{4 \times 10^{3}}$
$v_{2} = \sqrt{804 \times 4 \times 10^{3}}ms^{- 1}$