A) 0.2 J
B) 10 J
C) 20 J
D) 0.1 J
Correct Answer: D
Solution :
Elastic energy stored in the wire is \[\operatorname{U}= \frac{1}{2} \times stress\,\,\times strain \times volume\] \[=\,\,\,\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta \,l}{L}\times AL=\frac{1}{2}F\Delta l\] \[=\,\,\frac{1}{2}\times 200\times 1\times {{10}^{-\,3}}=0.1\,\,J\]You need to login to perform this action.
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