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