A) \[\frac{n}{n+1}(mgR)\]
B) \[\frac{n}{n-1}(mgR)\]
C) \[nmgR\]
D) \[\frac{(mgR)}{n}\]
Correct Answer: A
Solution :
general expression for gravitation potential energy (GPE) \[=-\frac{GM}{x}\operatorname{m}\] Where x= 2(distance/displacement from earth,s centre at surface)\[{{\operatorname{GPE}}_{1}}=\frac{-GMm}{R}\] At a height=nr \[{{\operatorname{GPE}}_{2}}=\frac{-GMm}{R+nR}\] \[\Delta \operatorname{GPE}={{\operatorname{GPE}}_{2}}-{{\operatorname{GPE}}_{1}}\] \[=\frac{-GMm}{nR+R}\left[ -\frac{GMm}{R} \right]\] \[=GMm\left[ \frac{1}{R}-\frac{1}{\left( n+1 \right)R} \right]\Rightarrow \frac{GMm}{R}\left[ \frac{1}{1}-\frac{1}{n+1} \right]\] \[=\frac{GMm}{R}\left( \frac{\operatorname{n}}{n+1} \right)\Rightarrow \frac{GMm}{{{R}^{2}}}=\operatorname{mgR}\Rightarrow g\frac{GM}{{{R}^{2}}}\] \[=mgR\left( \frac{n}{n+1} \right)\]You need to login to perform this action.
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