A) zero
B) \[\sqrt{\frac{2GM}{R}}\]
C) \[\sqrt{\frac{2GM}{R}(\sqrt{5}-1)}\]
D) \[\sqrt{\frac{2GM}{R}\left( 1-\frac{1}{\sqrt{5}} \right)}\]
Correct Answer: D
Solution :
Gravitational potential at a point P distance r from centre of ring lying on the axis is \[{{V}_{p}}=\frac{-GM}{\sqrt{{{R}^{2}}+{{r}^{2}}}}\] When, \[r=2R,{{V}_{p}}=-\frac{GM}{\sqrt{{{R}^{2}}-4{{R}^{2}}}}=-\frac{GM}{\sqrt{5}R}\] Gravitational potential at centre\[O\]of the ring is \[{{V}_{O}}=-\frac{GM}{R}\] \[\frac{1}{2}m{{v}^{2}}=m[{{V}_{p}}-{{V}_{O}}]\] or \[v=\sqrt{2({{V}_{p}}-{{V}_{Q}})}\] \[=\sqrt{2\left( -\frac{GM}{\sqrt{5}R}+\frac{GM}{R} \right)}\] \[=\sqrt{\frac{2GM}{R}\left( 1-\frac{1}{\sqrt{5}} \right)}\]
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