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A lead bullet of mass $10 \mathrm{~g}$ travelling at $300 \mathrm{~m} / \mathrm{s}$ strikes against a block of wood and comes to rest. Assuming $50 \%$ of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead $=150 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$ )
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The correct answer is:
$100^{\circ} \mathrm{C}$
$m=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}, v=300 \mathrm{~m} / \mathrm{s}$
$\begin{aligned} \text { Heat produced }= & \frac{1}{2} \times \mathrm{KE} \text { of the bullet } \\ & =\frac{1}{2}\left(\frac{1}{2} m v^2\right) \\ & =\frac{1}{4} \times 10 \times 10^{-3} \times(300)^2 \\ & =2.5 \times 10^{-3} \times 9 \times 10^4 \\ & =225 \mathrm{~J}\end{aligned}
From
$\begin{aligned} Q & =m s \Delta t \\ 225 & =10 \times 10^{-3} \times 150 \times \Delta t \\ \Delta t & =\frac{225}{10 \times 10^{-3} \times 150} \\ & =\frac{225}{1.5} \\ & =150^{\circ} \mathrm{C}\end{aligned}$
$\begin{aligned} \text { Heat produced }= & \frac{1}{2} \times \mathrm{KE} \text { of the bullet } \\ & =\frac{1}{2}\left(\frac{1}{2} m v^2\right) \\ & =\frac{1}{4} \times 10 \times 10^{-3} \times(300)^2 \\ & =2.5 \times 10^{-3} \times 9 \times 10^4 \\ & =225 \mathrm{~J}\end{aligned}
From
$\begin{aligned} Q & =m s \Delta t \\ 225 & =10 \times 10^{-3} \times 150 \times \Delta t \\ \Delta t & =\frac{225}{10 \times 10^{-3} \times 150} \\ & =\frac{225}{1.5} \\ & =150^{\circ} \mathrm{C}\end{aligned}$
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