A uniform chain AB of mass m and length l is placed with one end A at the highest point…
Physics · JEE Advanced · NTA Exams — Work, Energy and Power
A uniform chain AB of mass m and length l is placed with one end A at the highest point of a hemisphere of radius R. Referring to the top of the hemisphere as the datum level, the potential energy of the chain is (given that \(\ell<\frac{\pi R}{2}\))


- \(\frac{m R^{2} g}{\ell}\left(\frac{\ell}{R}-\sin \frac{\ell}{R}\right)\)
- \(\frac{m R^{2} g}{2 \ell}\left(\frac{\ell}{R}-\sin \frac{\ell}{R}\right)\)
- \(\frac{m R^{2} g}{2 \ell}\left(\sin \frac{\ell}{R}-\frac{\ell}{R}\right)\)
- \(-\frac{m R^{2} g}{\ell}\left(\frac{\ell}{R}-\sin \frac{\ell}{R}\right)\)
Answer
(D) - m R^ 2 g ( R - R )
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