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}] $

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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}] $



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