A) 2 mgl
B) mgl
C) \[\frac{mgl}{2}\]
D) \[\frac{mgl}{4}\]
Correct Answer: C
Solution :
Centre of mass of rod will lie at length - \[\therefore Work done = change in potential energy\] \[=\,\, Final PE - Initial PE\] \[=mg\left( \frac{l}{2} \right)-mg\,(0)=mg\frac{l}{2}\]You need to login to perform this action.
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