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A bullet moving with a uniform velocity $v$ , stops suddenly after hitting the target and the whole mass melts. Let mass be m, specific heat S, initial temperature $25 ° C$ , melting point $475 ° C$ and the latent heat $l$ . Then $v$ is given by

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$m L = m S (475 - 25) + \frac{1}{2} \cdot \frac{m v^{2}}{J}$

$m S (475 - 25) + m L = \frac{m v^{2}}{2 J}$

$m S (475 - 25) + m L = \frac{m v^{2}}{J}$

$m S (475 - 25) - m L = \frac{m v^{2}}{2 J}$

answer is B.

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Detailed Solution

Firstly the temperature of bullet rises up to melting point, then it melts. Hence according to $W = J H$ .

$\Rightarrow \frac{1}{2} m v^{2} = J . [m . c . Δ θ + m L] = J [m S (475 - 25) + m L]$

$\Rightarrow m S (475 - 25) + m L = \frac{m v^{2}}{2 J}$

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