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