A) \[\left( \frac{n}{n+1} \right)mgR\]
B) \[\left( \frac{n}{n-1} \right)mgR\]
C) \[nmgR\]
D) \[\left( \frac{mgR}{n} \right)\]
Correct Answer: A
Solution :
\[{{U}_{e}}=-\frac{GMm}{R}\] \[{{U}_{h}}=-\frac{GMm}{(R+nR)}=-\frac{GMm}{R(n+1)}\] Thus, magnitude of the change in gravitational potential energy \[\Delta U={{U}_{h}}-{{U}_{e}}\] \[=\frac{GMm}{R}\left( 1-\frac{1}{(n+1)} \right)\] \[=\left( \frac{n}{n+1} \right)\frac{GMm}{R}=\left( \frac{n}{n+1} \right)mgR\]
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