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A lead bullet at $27\,^oC$ 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 is ........ $m/s$ ($M.P.$ of lead $= 327\,^oC,$ specific heat of lead $= 0.03\,cal/g\,^oC,$ latent heat of fusion of lead $= 6\,cal/g$ and $J = 4.2\,joule/cal$ )
$410$
$1230$
$307.5$
None of the above
Solution
If mass of the bullet is $\mathrm{m}\, \mathrm{gm}$.
then total heat iequired for bullet to just melt down
$\mathrm{Q}_{1}=\mathrm{m} \mathrm{c} \Delta \theta+\mathrm{mL}=\mathrm{m} \times 0.03(327-27)+\mathrm{m} \times 6$
$=15 \mathrm{m} \mathrm{cal}=(15 \mathrm{m} \times 4.2) \mathrm{J}$
Now when bullet is stopped by the obstacle, the loss in its mechanical energy $=\frac{1}{2}\left(\mathrm{m} \times 10^{-3}\right) \mathrm{v}^{2} \mathrm{J}$
(As $\left.\mathrm{mg}=\mathrm{m} \times 10^{-3} \mathrm{kg}\right)$
As $25 \%$ of this energy is absorbed by the obstacle, The energy absorbed by the bullet
$\mathrm{Q}_{2}=\frac{75}{100} \times \frac{1}{2} \mathrm{mv}^{2} \times 10^{-3}=\frac{3}{8} \mathrm{mv}^{2} \times 10^{-3} \mathrm{J}$
Now the bullet will melt if $\mathrm{Q}_{2} \geq \mathrm{Q}_{1}$
i.e., $\frac{3}{8} \mathrm{mv}^{2} \times 10^{-3} \geq 15 \mathrm{m} \times 4.2 \Rightarrow \mathrm{v}_{\min }=410 \mathrm{m} / \mathrm{s}$
