Question

A hunter fired a metallic bullet of mass $m\, kg$ from a gun towards an obstacle and it just melts when it is stopped by the obstacle. The initial temperature of the bullet is $300\, K$. If $\frac{1}{4} \,th$ of heat is absorbed by the obstacle, then the minimum velocity of the bullet is
(Melting point of bullet $=600 \,K$,
Specific heat of bullet $=0.03$ cal $g^{-1}{ }^{\circ} C ^{-1}$,
Latent heat of fusion of bullet $=6$ cal $g ^{-1}$ )

• A. $410 \,ms ^{-1}$
• B. $260 \,ms ^{-1}$
• C. $460 \,ms ^{-1}$
• D. $310 \,ms ^{-1}$

Answer 1

Answer: (A)

Given, mass of metallic bullet $=m\, kg$
Initial temperature of bullet $=300 \,K$
Heat absorbed by the obstacle $=\frac{1}{4}$ th of total heat
Let the total heat be one, then remaining heat $=1-\frac{1}{4}=\frac{3}{4}$ th part
Melting point of bullet $=600\, K$
Specific heat of bullet $=0.03 \,calg ^{-1}{ }^{\circ} C ^{-1}$
Latent heat of fusion of bullet $=6 \,calg ^{-1}$
According to the question,
$\frac{3}{4}\left(\frac{1}{2} m v^{2}\right)=m s \Delta \theta+m L$
or,$\frac{3}{4} v^{2}=0.03 \times 4200 \times 300+6 \times 4200$
or, $v^{2}=\frac{8}{3}[0.01 \times 4200 \times 300+2 \times 4200]$
or, $v^{2}=40 \times 4200$
or, $v=\sqrt{168000}$
or, $v=410 \,m / s$