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(D) The gravitational potential $V$ at a distance $x$ from the centre of a ring of radius $R$ and mass $M$ is given by $V(x) = -\frac{GM}{\sqrt{x^2 +

Vedclass · Accepted Answer

$\sqrt {2\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\frac{{GM}}{R}} $

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(D) The gravitational potential $V$ at a distance $x$ from the centre of a ring of radius $R$ and mass $M$ is given by $V(x) = -\frac{GM}{\sqrt{x^2 + R^2}}$.Applying the principle of conservation of mechanical energy between the initial position $(x = R)$ and the centre $(x = 0)$:$K_i + U_i = K_f + U_f$Since the particle starts from rest,$K_i = 0$.$0 + m V(R) = \frac{1}{2}mv^2 + m V(0)$$0 + m \left( -\frac{GM}{\sqrt{R^2 + R^2}} \right) = \frac{1}{2}mv^2 + m \left( -\frac{GM}{R} \right)$$-\frac{GM}{\sqrt{2}R} = \frac{1}{2}v^2 - \frac{GM}{R}$$\frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{\sqrt{2}R} = \frac{GM}{R} \left( 1 - \frac{1}{\sqrt{2}} \right)$$v^2 = \frac{2GM}{R} \left( 1 - \frac{1}{\sqrt{2}} \right)$$v = \sqrt{2\left( 1 - \frac{1}{\sqrt{2}} \right) \frac{GM}{R}}$

$A$ particle of mass $m$ starts from rest at a distance $R$ from the centre and along the axis of a fixed ring of radius $R$ and mass $M$. Its velocity at the centre of the ring is:

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