A) 0.02J
B) 0.05J
C) 0.1 J
D) 0.2J
Correct Answer: B
Solution :
The elastic potential energy is given by \[U=\frac{1}{2}\times stress\times strain\times volume\] \[=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta l}{l}\times Al\] \[=\frac{1}{2}\times 100\times {{10}^{-3}}=\frac{0.1}{2}=0.05J\]You need to login to perform this action.
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