A) \[mgR\frac{\left( n-1 \right)}{n}\]
B) \[nmgR\]
C) \[mgR\left( \frac{{{n}^{2}}}{{{n}^{2}}+1} \right)\]
D) \[mgR\left( \frac{n}{n+1} \right)\]
E) mgR
Correct Answer: A
Solution :
The change in potential energy gravitational field is given by \[\Delta E=GMm\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\] In this problem, \[{{r}_{1}}=R\] and \[{{r}_{2}}=nR\] \[\Delta E=GMm\left( \frac{1}{R}-\frac{1}{nR} \right)\] \[=\frac{GMm}{R}\left( \frac{n-1}{n} \right)\] \[=mgR\left( \frac{n-1}{n} \right)\] \[\left[ \because g=\frac{GM}{{{R}^{2}}} \right]\]
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