(a) If mass of the bullet is $m \,gm,$

then total heat required for bullet to just melt down

$Q_1 = m c \Delta \theta + m L = m \times 0.03 (327 -27) + m \times 6$

     $= 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}(m \times {10^{ – 3}}){v^2}J$

(As $m\;gm = m \times {10^{ – 3}}kg$)

As $25\%$ of this energy is absorbed by the obstacle,

The energy absorbed by the bullet

${Q_2} = \frac{{75}}{{100}} \times \frac{1}{2}m{v^2} \times {10^{ – 3}} = \frac{3}{8}m{v^2} \times {10^{ – 3}}J$

Now the bullet will melt if ${Q_2} \ge {Q_1}$

i.e. $\frac{3}{8}m{v^2} \times {10^{ – 3}} \ge 15m \times 4.2$==> ${v_{\min }} = 410\;m/s$