A) \[\frac{F}{2\Delta l}\]
B) \[F\Delta l\]
C) \[2F\Delta l\]
D) \[\frac{F\Delta l}{2}\]
Correct Answer: D
Solution :
Work done in stretching the wire= potential energy stored \[=-5\times stress\times strain\times volume\] \[=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta l}{l}\times Al=\frac{1}{2}F\Delta l\]
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