A) \[\left( \frac{n}{n+1} \right)mgR\]
B) \[\left( \frac{n-1}{n} \right)mgR\]
C) \[\left( \frac{mgR}{n} \right)\]
D) \[\frac{mgR}{(n-1)}\]
Correct Answer: A
Solution :
Gravitational potential energy of mass m at earth's surface \[{{U}_{e}}=-\frac{GMm}{R}\] Gravitational potential energy of same mass at a height\[nR\]from the earth's surface \[{{U}_{h}}=-\frac{GMm}{(R+nR)}=-\frac{GMm}{R(n+1)}\] Thus, magnitude of the change in gravitational potential energy \[\Delta U={{U}_{h}}-{{U}_{e}}\] \[=\frac{GMm}{R}\left\{ 1-\frac{1}{(n+1)} \right\}\] \[=\left( \frac{n}{n+1} \right)\frac{GMm}{R}\] \[=\left( \frac{n}{n+1} \right)mgR\] \[[\because GM=g{{R}^{2}}]\]
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