A) \[140{}^\circ C\]
B) \[106{}^\circ C\]
C) \[90{}^\circ C\]
D) \[100{}^\circ C\]
Correct Answer: C
Solution :
Since specific heat of lead is given in joule. Specific heat of lead \[=0.120\,J/g{{\,}^{o}}C=120\,J/kg\] The two-third of head produced goes into the bullet So, \[m\times s\times \Delta \theta =\frac{2}{3}\times \frac{1}{2}m{{v}^{2}}\] \[\Delta \theta =\frac{{{v}^{2}}}{3\times s}\] \[=\frac{180\times 180}{3\times 120}=90{{\,}^{o}}C\]You need to login to perform this action.
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