Question

A lead bullet at $27^{\circ }C$ 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 is nearly (Melting point of lead $=327^{\circ }C$, specific heat of lead $=0.03cal/g^{\circ }C$, latent heat of fusion of lead $=6cal/g$ and $J=4.2Jca{l}^{-1})$

• A. $410m{s}^{-1}$
• B. $1230m{s}^{-1}$
• C. $307.5m{s}^{-1}$
• D. none of these

Answer 1

Answer: (A)

$(75\%)\frac{1}{2}m{v}^2=(ms\Delta T+mL)$
$\left(\frac{3}{4}\right)\frac{m{v}^2}{2}=m\left(0.03\times 4.2J\times 10^{3}k{g}^{-1}\times {}^{\circ }{C}^{-1}\right)$
$(327-27{)}^{\circ }C+m\left(6\times 4.2J\times 10^{3}k{g}^{-1}\right)$
$\frac{3}{8}\left(v^2\right)=(37.8+25.2)\times 10^3$
$v=\sqrt{\frac{8\times 63000}{3}}\approx 410m{s}^{-1}$