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A particle starts from rest at a distance $R$ from the centre and along the axis of a fixed ring of radius $R \&$ mass $M.$ Its velocity at the centre of the ring is :
$\sqrt {\frac{{\sqrt 2 GM}}{R}} $
$\sqrt {\frac{{2GM}}{R}} $
$\sqrt {\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\frac{{GM}}{R}} $
$\sqrt {\left( {2 - \sqrt 2 } \right)\frac{{GM}}{R}} $
Solution
Applying conservation of energy
$v=\frac{-G \times M}{\sqrt{x^{2}+R^{2}}}$
Potential energy $=v \times m=\frac{-G m M}{\sqrt{x^{2}+R^{2}}}$
Initial kinetic energyr $_{i}=0$
Potential energy at center $=\frac{-G M m}{R}\{x=0\}$
$\frac{-G M m}{\sqrt{x^{2}+R^{2}}}+0=\frac{G M m}{R}+\frac{1}{2} m v^{2}$
$\frac{1}{2} m v^{2}=\frac{G M m}{R}-\frac{G M m}{\sqrt{x^{2}+R^{2}}}=G M m\left\{\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right\}$
$v^{2}=\frac{2 G M m}{m}\left\{\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right\}$
$v^{2}=\sqrt{\left(2-\frac{2}{\sqrt{2}}\right)\left(\frac{G M}{R}\right)}=\sqrt{(2 \sqrt{2}) \frac{G M}{R}}$
