A) \[mg2R\]
B) \[\frac{2}{3}mgR\]
C) \[3mgR\]
D) \[\frac{1}{3}mgR\]
Correct Answer: B
Solution :
Change in potential energy \[\Delta U=-\frac{GMm}{R+2R}-\left( -\frac{GMm}{R} \right)\] \[=-\frac{GMm}{3R}+\frac{GMm}{R}\] \[=\frac{2GMm}{3R}=\frac{2}{3}mgR\] \[\left[ \because g=\frac{GM}{{{R}^{2}}} \right]\]You need to login to perform this action.
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