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

# 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

1. A

$mL=mS\text{\hspace{0.17em}}\left(475-25\right)+\frac{1}{2}\cdot \frac{m{v}^{2}}{J}$

2. B

$mS\left(475-25\right)+mL=\frac{m{v}^{2}}{2J}$

3. C

$mS\text{\hspace{0.17em}}\left(475-25\right)+mL=\frac{m{v}^{2}}{J}$

4. D

$mS\text{\hspace{0.17em}}\left(475-25\right)-mL=\frac{m{v}^{2}}{2J}$

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### Solution:

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

$⇒\frac{1}{2}m{v}^{2}=J.\left[m.c.\Delta \theta +mL\right]=J\left[m\text{\hspace{0.17em}}S\text{\hspace{0.17em}}\left(475-25\right)+mL\right]$

$⇒mS\left(475-25\right)+mL=\frac{m{v}^{2}}{2J}$

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