A) \[\frac{F}{2l}\]
B) \[Fl\]
C) \[2Fl\]
D) \[\frac{Fl}{2}\]
Correct Answer: D
Solution :
| [d] The work done in stretching the wire |
| = Potential energy store |
| \[=\frac{1}{2}\times \] Stress \[\times \] Strain \[\times \] Volume |
| \[=\frac{1}{2}\times \frac{F}{A}\times \frac{l}{L}\times AL=\frac{1}{2}Fl\] |
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