A) \[mL=mS\,(475-25)+\frac{1}{2}\cdot \frac{m{{v}^{2}}}{J}\]
B) \[mS(475-25)+mL=\frac{m{{v}^{2}}}{2J}\]
C) \[mS\,(475-25)+mL=\frac{m{{v}^{2}}}{J}\]
D) \[mS\,(475-25)-mL=\frac{m{{v}^{2}}}{2J}\]
Correct Answer: B
Solution :
Firstly the temperature of bullet rises up to melting point, then it melts. Hence according to \[W=JH\]. Þ \[\frac{1}{2}m{{v}^{2}}=J.[m.c.\Delta \theta +mL]=J[m\,S\,(475-25)+mL]\] Þ \[mS(475-25)+mL=\frac{m{{v}^{2}}}{2J}\]
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