A) \[mgl\]
B) \[\frac{mgl}{2}\]
C) \[\frac{mgl}{3}\]
D) \[\frac{mgl}{4}\]
E) \[\frac{mgl}{\sqrt{2}}\]
Correct Answer: D
Solution :
Centre of the mass of the rod lies at the midpoint and when the rod is displaced through an angle\[{{60}^{o}}\]it rises to point B. From the figure \[\sin {{30}^{o}}=\frac{BC}{AB}\] Or \[\sin {{30}^{o}}=\frac{L}{l/2}\] Or \[\frac{1}{2}=\frac{L}{l/2}\] Or \[L=\frac{l}{4}\] Then potential energy of the rod in this position is \[U=mgL\] \[U=mg\frac{l}{4}\]You need to login to perform this action.
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