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. \[m\times s\times \Delta \theta =\frac{2}{3}\times \frac{1}{2}m{{v}^{2}}\] \[\Delta \theta =\frac{{{v}^{2}}}{3\times s}=\frac{180\times 180}{3\times 120}=90{{\,}^{o}}C\]
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