A) \[\sqrt{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
At a platform at a height h, escape energy = binding energy of sphere or \[\frac{1}{2}mv_{e}^{2}=\frac{GMm}{R+h}\] or \[v{{}_{e}}=\sqrt{\frac{2G\,\,M}{R+h}}=\sqrt{\frac{2G\,M}{2R}}\] (\[\because \] \[h=R\]) But at surface of earth, \[{{v}_{e}}=\sqrt{\frac{2G\,\,M}{R}}\] As given, \[v{{}_{e}}=f\,\,{{v}_{e}}\] Hence, \[\sqrt{\frac{2GM}{2R}}=f\sqrt{\frac{2G\,M}{R}}\] or \[\frac{1}{2R}=\frac{{{f}^{2}}}{R}\] \[\therefore \] \[f=\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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