A) \[{{140}^{o}}C\]
B) \[{{106}^{o}}C\]
C) \[{{90}^{o}}C\]
D) \[{{100}^{o}}C\]
Correct Answer: C
Solution :
Specific heat of lead\[=0.120\,\,J/{{g}^{o}}C\] \[=120\,\,J/kg\] The two-third of heat 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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