Question

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

• A. $410m{s}^{-1}$
• B. $260m{s}^{-1}$
• C. $460m{s}^{-1}$
• D. $310m{s}^{-1}$

Answer 1

Answer: (A)

Given, mass of metallic bullet $=mkg$
Initial temperature of bullet $=300K$
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 $=600K$
Specific heat of bullet $=0.03cal{g}^{-1}{}^{\circ }{C}^{-1}$
Latent heat of fusion of bullet $=6cal{g}^{-1}$
According to the question,
$\frac{3}{4}\left(\frac{1}{2}m{v}^2\right)=ms\Delta \theta +mL$
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, $v=410m/s$