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A lead bullet penetrates into a solid object and melts. Assuming that $50 \%$ of its kinetic energy was used to heat it, the initial speed of the bullet is (the initial temperature of the bullet is $25^{\circ} \mathrm{C}$ and its melting point is $300^{\circ} \mathrm{C}$. Latent heat of fusion of lead $=2.5 \times 10^4 \mathrm{~J} / \mathrm{kg}$ and specific heat capacity of lead $=125 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$
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490 m/s
Given, $S=125 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$
$\theta=25^{\circ} \mathrm{C}, \theta=300^{\circ} \mathrm{C}, m=2.5 \times 10^4 \mathrm{~J} / \mathrm{kg}$, then
$\frac{1}{2} m v^2 \times \frac{1}{2}=m S \Delta \theta+m L$
$\begin{array}{rlrl}\Rightarrow & & \frac{v^2}{4} & =125(300-25)+2.5 \times 10^4 \\ \Rightarrow & & \frac{v^2}{4} & =59375 \\ \Rightarrow & & v^2 & =59375 \times 4 \\ \Rightarrow & & v & =\sqrt{237500} \Rightarrow v=487.3 \mathrm{~m} / \mathrm{s} \\ \Rightarrow & & \approx 490 \mathrm{~m} / \mathrm{s}\end{array}$
$\theta=25^{\circ} \mathrm{C}, \theta=300^{\circ} \mathrm{C}, m=2.5 \times 10^4 \mathrm{~J} / \mathrm{kg}$, then
$\frac{1}{2} m v^2 \times \frac{1}{2}=m S \Delta \theta+m L$
$\begin{array}{rlrl}\Rightarrow & & \frac{v^2}{4} & =125(300-25)+2.5 \times 10^4 \\ \Rightarrow & & \frac{v^2}{4} & =59375 \\ \Rightarrow & & v^2 & =59375 \times 4 \\ \Rightarrow & & v & =\sqrt{237500} \Rightarrow v=487.3 \mathrm{~m} / \mathrm{s} \\ \Rightarrow & & \approx 490 \mathrm{~m} / \mathrm{s}\end{array}$
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