A) \[\sqrt{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
Key Idea: If energy in the form of kinetic energy which is equal to binding energy, supplied to the sphere, it leaves the gravitational field of earth. At a platform at a height h, escape energy = binding energy of sphere \[or\frac{1}{2}mv_{e}^{'2}=\frac{GMm}{R+h}\] \[orv_{e}^{'}=\sqrt{\frac{2GM}{R+h}}=\sqrt{\frac{2GM}{2R}}(\because \,h=R)\] But at surface of earth, \[{{v}_{e}}=\sqrt{\frac{2GM}{R}}\] As given, \[v_{e}^{'}=f\,{{v}_{e}}\] Hence, \[\sqrt{\frac{2GM}{2R}}=f\sqrt{\frac{2GM}{R}}\] \[or\frac{1}{2R}=\frac{{{f}^{2}}}{R}\] \[\therefore f=\frac{1}{\sqrt{2}}\]
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