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प्रश्न
A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. Find the gravitational potential energy of the chain with reference level at the centre of the sphere.
उत्तर
Let us consider a small element, which makes angle 'dθ' at the centre.
\[\therefore dm = \rho \left( \frac{m}{L} \right) Rd\theta\]
Gravitational potential energy of 'dm' with respect to centre of the sphere
\[= \left( \text{ dm }\right) \text{ g R }\cos \theta\]
\[ = \left( \frac{\text{ mg }}{\text{ L}} \right) R^2 \cos \theta d\theta\]
\[\therefore \text{ Total gravitational potential energy, }E_P = \int\limits_0^{L/R} \text{ mg }\frac{R^2}{L} \cos \theta d\theta\]
\[ E_P = \frac{m R^2 g}{L}\left[ \sin \theta \right] \left[ \text{ As }, \theta = \frac{L}{R} \right]\]
\[ E_P = \frac{m R^2 g}{L}\sin \left( \frac{L}{R} \right)\]
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