Explanation

(a) If mass of the bullet is $m\ gm$,
Then total heat required for bullet to just melt down
$Q_{1} = m\ c\ \mathrm{\Delta}\theta + m\ L = m \times 0.03(327 - 27) + m \times 6$
$= 15\ m\ cal = (15m \times 4.2)J$
Now when bullet is stopped by the obstacle, the loss in its mechanical energy $= \frac{1}{2}\left( m \times 10^{- 3} \right)v^{2}J$
(As $m\ g = m \times 10^{- 3}kg$)
As 25% of this energy is absorbed by the obstacle,
The energy absorbed by the bullet
$Q_{2} = \frac{75}{100} \times \frac{1}{2}mv^{2} \times 10^{- 3} = \frac{3}{8} \times 10^{- 3}J$
Now the bullet will melt if $Q_{2} \geq Q_{1}$
$i.e.,\frac{3}{8}mv^{2} \times 10^{- 3} \geq 15m \times 4.2 \Rightarrow v_{\min} = 410m/s$