A) \[\left( \frac{n}{n+1} \right)mgR\]
B) \[\left( \frac{n}{n-1} \right)mgR\]
C) \[nmgR\]
D) \[\frac{mgR}{n}\]
Correct Answer: A
Solution :
The acceleration due to gravity varies with height as \[g=\frac{g}{\left( 1+\frac{h}{R} \right)}\] Hence, \[h=nR\Rightarrow g=\frac{g}{\left( 1+\frac{nR}{R} \right)}=\frac{g}{(1+n)}\] The change in potential energy is equal to \[\Delta U=mgh=\frac{mg(nR)}{(1+n)}=\left( \frac{n}{n+1} \right)mgR\]You need to login to perform this action.
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