A) \[0.2\] J
B) 10 J
C) 20 J
D) \[0.1\]J
Correct Answer: D
Solution :
| [d] Elastic energy stored in the wire is |
| \[U=\frac{1}{2}\times \]Stress \[\times \] Strain \[\times \] Volume |
| \[=\frac{1}{2}\frac{F}{A}\times \frac{\Delta l}{L}\times AL=\frac{1}{2}F\Delta l\] |
| \[=\frac{1}{2}\times 200\times 1\times {{10}^{-3}}=0.1\,\,J\] |
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