Gravitation Question 124
Question: The change in potential energy, when a body of mass m is raised to a height nR from the earth’s surface is (R = Radius of earth)[MP PMT 1996]
Options:
A) $ [mgR\frac{n}{n-1}] $
B) nmgR
C) $ [mgR\frac{n^{2}}{n^{2}+1}] $
D) $ [mgR\frac{n}{n+1}] $
Show Answer
Answer:
Correct Answer: D
Solution:
$ [\Delta U=\frac{mgh}{1+\frac{h}{R}}=\frac{mg,nR}{1+\frac{nR}{R}}=\frac{nm,gR}{n+1}] $
