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A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre $C$. When released, it reaches $C$ under the gravitational attraction of the ring. Its speed at $C$ will be
$\sqrt {\frac{{2Gm}}{r}\left( {\sqrt 2 - 1} \right)} $
$\sqrt {\frac{{Gm}}{r}} $
$\sqrt {\frac{{2Gm}}{r}\left( {1 - \frac{1}{{\sqrt 2 }}} \right)} $
$\sqrt {\frac{{2Gm}}{r}} $
Solution
Let $'M'$ be the mass of the particle
Now, ${E_{initial}} = {E_{final}}$
$i.e.,\frac{{GMm}}{{\sqrt 2 r}} + 0 = \frac{{GMm}}{r} + \frac{1}{2}M{V^2}$
$or,\frac{1}{2}M{V^2} = \frac{{GMm}}{r}\left[ {1 – \frac{1}{{\sqrt 2 }}} \right]$
$ \Rightarrow \frac{1}{2}{V^2} = \frac{{Gm}}{r}\left[ {1 – \frac{1}{{\sqrt 2 }}} \right]$
$or,\,\,V = \sqrt {\frac{{2Gm}}{r}\left( {1 – \frac{1}{{\sqrt 2 }}} \right)} $
