Q. A rod of mass $m$ and length $l$ is made to stand at an angle of $60{^\circ}$ with the vertical potential energy of the rod in this position is
A) $mgl$
B) $\frac{mgl}{2}$
C) $\frac{mgl}{3}$
D) $\frac{mgl}{4}$
Explanation
(d) For any uniform rod, the mass is concentrated at its centre
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Height of the mass from ground is, $h = (l/2)\sin{30{^\circ}}$
Potential energy of the rod $= mgh$
$= m \times g \times \frac{l}{2}\sin{30{^\circ} = m \times g \times \frac{l}{2} \times \frac{1}{2} = \frac{mgl}{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.