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Standard XII

Physics

Latent Heat

A bullet movi...

Question

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

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

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

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

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

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Solution

The correct option is A $m L = m S (475 - 25) + \frac{1}{2} \frac{m v^{2}}{J}$
Firstly the temperature of bullet rises up to melting point, then it melts. Hence according to $W = J Q$
$\Rightarrow \frac{1}{2} m v^{2} = J . [m . c . Δ θ + m L] = J [m S (475 - 25) + m L]$
$\Rightarrow m S (475 - 25) + m L = \frac{m v^{2}}{2 J}$

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

The correct option is A mL=mS(475−25)+12mv2JFirstly the temperature of bullet rises up to melting point, then it melts. Hence according to W=JQ⇒12mv2=J.[m.c.Δθ+mL]=J[mS(475−25)+mL]⇒mS(475−25)+mL=mv22J

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