A) Infinite
B) \[{{(gR)}^{1/2}}\]
C) \[{{(2gR)}^{1/2}}\]
D) gR
Correct Answer: C
Solution :
Apply C.O.M.E. T.E. at surface = T.E. at infinity \[\frac{-GMm}{R}+\frac{1}{2}m.{{V}^{2}}=0\] \[\Rightarrow \,V=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\]
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