A) \[2mgR\]
B) \[mgR\]
C) \[\frac{1}{2}mgR\]
D) \[\frac{1}{4}mgR\]
Correct Answer: C
Solution :
Increase in potential energy \[\Delta U=-\frac{GMm}{(R+R)}-\left( -\frac{GMm}{R} \right)\] \[\Delta U=\frac{1}{2}\frac{GMm}{R}=\frac{1}{2}\left( \frac{GM}{{{R}^{2}}} \right)mR\] \[\Delta U=\frac{1}{2}mgR\] \[\left( \because g=\frac{GM}{{{R}^{2}}} \right)\]
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