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Question: A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mas...

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°C, melting point 475°C and the latent heat L. Then v is given by

A

mL=mS(475−25)+12⋅mv2JmL = mS(475 - 25) + \frac{1}{2} \cdot \frac{mv^{2}}{J}

B

mS(475−25)+mL=mv22JmS(475 - 25) + mL = \frac{mv^{2}}{2J}

C

mS(475−25)+mL=mv2JmS(475 - 25) + mL = \frac{mv^{2}}{J}

D

mS(475−25)−mL=mv22JmS(475 - 25) - mL = \frac{mv^{2}}{2J}

Answer

mS(475−25)+mL=mv22JmS(475 - 25) + mL = \frac{mv^{2}}{2J}

Explanation

Solution

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

⇒ 12mv2=J.[m.c.Δθ+mL]=J[mS(475−25)+mL]\frac{1}{2}mv^{2} = J.\lbrack m.c.\Delta\theta + mL\rbrack = J\lbrack mS(475 - 25) + mL\rbrack

⇒ mS(475−25)+mL=mv22JmS(475 - 25) + mL = \frac{mv^{2}}{2J}