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 $=0.03 \,cal / g^{\circ} C$, latent heat of fusion of lead $=6\, cal / g )$

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

Answer 1

Answer: (B)

Let mass of the bullet be $m$ gram, then total heat required for bullet to just melt down
$Q_1 =m c \Delta T+m L $
$= m \times(0.03)(327-27)+m \times 6$
$= 15 m-c a l$
$= (16 m \times 4 \cdot 2) J$
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 $Q_2 \geq Q_1$
$\text { i.e., } \quad \frac{3}{8} m v^2 \times 10^{-3} \geq 15 m \times 4.2$
$ \Rightarrow v_{\min }=410\, m / s$