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A hunter fired a metallic bullet of mass $m \mathrm{~kg}$ from a gun towards an obstacle and it just melts when it is stopped by the obstacle. The initial temperature of the bullet is $300 \mathrm{~K}$. If $\frac{1}{4}$ th of heat is absorbed by the obstacle, then the minimum velocity of the bullet is (Melting point of bullet $=600 \mathrm{~K}$, Specific heat of bullet $=0.03 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$, Latent heat of fusion of bullet $=6 \mathrm{cal} \mathrm{g}^{-1}$ )
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The correct answer is:
$410 \mathrm{~ms}^{-1}$
As per question $\frac{1}{4}$ th of heat is absorbed by the obstacle so, $\frac{3}{4}\left(\frac{1}{2} m v^2\right)=m s \Delta \theta+m L$
$\begin{aligned}
& \frac{3}{4} v^2=0.03 \times 4200 \times 300+6 \times 4200 \\
& \Rightarrow \quad v^2=\frac{8}{3}[0.01 \times 4200 \times 300+2 \times 4200] \\
& \Rightarrow v^2=40 \times 4200 \\
& \Rightarrow \quad v=\sqrt{168000}=410 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\begin{aligned}
& \frac{3}{4} v^2=0.03 \times 4200 \times 300+6 \times 4200 \\
& \Rightarrow \quad v^2=\frac{8}{3}[0.01 \times 4200 \times 300+2 \times 4200] \\
& \Rightarrow v^2=40 \times 4200 \\
& \Rightarrow \quad v=\sqrt{168000}=410 \mathrm{~m} / \mathrm{s}
\end{aligned}$
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