Question

A lead bullet at $27^{\circ }C$ melts when hit on a target. Assuming only $75\%$ of heat is used to melt the bullet, the velocity of bullet at time of striking is
(Take, melting point of lead $=327^{\circ }C$, specific heat of lead $=0.03cal/g^{\circ }C$, latent heat of fusion of lead $=6cal/g)$

• A. $330m/s$
• B. $410m/s$
• C. $470m/s$
• D. $510m/s$

Answer 1

Answer: (B)

Let mass of the bullet be $m$ gram, then total heat required for bullet to just melt down
$Q_1=mc\Delta T+mL$
$=m\times (0.03)(327-27)+m\times 6$
$=15m-cal$
$=(16m\times 4\cdot 2)J$
Now, when bullet is struck by obstacles, the loss in its mechanical energy
$=\frac{1}{2}\left(m\times 10^{-3}\right)v^2$
The energy absorbed by bullet,
$Q_2=\frac{72}{100}\times \frac{1}{2}m{v}^2\times 10^{-3}$
$=\frac{3}{8}m{v}^2\times 10^{-3}J$
Now, the bullet will melt if $Q_2\ge Q_1$
$\text{ i.e., }\frac{3}{8}m{v}^2\times 10^{-3}\ge 15m\times 4.2$
$\Rightarrow v_{\min }=410m/s$