A) \[mgR=\frac{n}{(n-1)}\]
B) \[mgR\]
C) \[mgR=\frac{n}{(n+1)}\]
D) \[mgR=\frac{{{n}^{2}}}{({{n}^{2}}+1)}\]
E) \[\frac{mgR}{n}\]
Correct Answer: C
Solution :
Change in potential energy \[\Delta U={{U}_{2}}-{{U}_{1}}\] \[\therefore \] \[\Delta U=-\frac{GMm}{(R+nR)}+\frac{GMm}{R}\] Or \[\Delta U=-\frac{GMm}{R(1+n)}+\frac{GMm}{R}\] Or \[\Delta U=\frac{GMm}{R}\left[ -\frac{1}{1+n}+1 \right]\] Or \[\Delta U=\frac{({{R}^{2}}g)m}{R}\times \frac{n}{(1+n)}\]\[\left[ \because g=\frac{GM}{{{R}^{2}}} \right]\] Or \[\Delta U=mgR\left( \frac{n}{n+1} \right)\]
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