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A point P lies on the axis of a fixed ring of mass M and radius R, at a distance 2R from its centre O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be

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$\sqrt{\frac{2 G M}{R}}$

$\sqrt{\frac{2 G M}{R} (\sqrt{5} - 1)}$

$\sqrt{\frac{2 G M}{R} (1 - \frac{1}{\sqrt{5}})}$

answer is D.

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Detailed Solution

$Potential energy at 2 R = - \frac{GMm}{\sqrt{5} R} and at the center - \frac{GMm}{R}$

$Difference in potential energy = kinetic enegry$

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