A) \[\frac{F}{2\,l}\]
B) \[Fl\]
C) \[2F\,l\]
D) \[\frac{F\,l}{2}\]
Correct Answer: D
Solution :
Work done in stretching the wire = potential energy stored \[=\frac{1}{2}\,\times Stress\,\times Strain\times \,Volume\] \[=\frac{1}{2}\,\times \frac{F}{A}\,\times \frac{l}{L}\,\times AL\,=\frac{1}{2}\,Fl\]You need to login to perform this action.
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