question_answer

A rod of mass m and length (is made to stand at an angle of\[{{60}^{o}}\]with the vertical. Potential energy of the rod in this position is

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 \[U=mgL\] \[U=mg\frac{l}{4}\]