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A lead bullet at $27°C$ just melts when stopped by an obstacle. Assuming that $25\%$ of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking........ $m/sec$ ($M.P.$ of lead $= 327°C,$ specific heat of lead $= 0.03\, cal/gm°C,$ latent heat of fusion of lead $= 6\, cal/gm$ and $J = 4.2\, joule/cal)$
$410$
$1230$
$307.5$
None of the above
Solution
(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. $\frac{3}{8}m{v^2} \times {10^{ – 3}} \ge 15m \times 4.2$==> ${v_{\min }} = 410\;m/s$
