A) 100°C
B) 125°C
C) 150°C
D) 200°C
Correct Answer: C
Solution :
Since specific heat of lead is given in Joules, hence use \[W=Q\] instead of \[W=JQ\]. Þ\[\frac{1}{2}\times \left( \frac{1}{2}m{{v}^{2}} \right)=m.c.\Delta \theta \]Þ\[\Delta \theta =\frac{{{v}^{2}}}{4c}=\frac{{{(300)}^{2}}}{4\times 150}=150{}^\circ C\].You need to login to perform this action.
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