Question

# A body of mass $$m$$ is lifted up from the surface of earth to a height three times the radius of the earth $$R$$.The change in potential energy of the body is

A
3mgR
B
54mgR
C
34mgR
D
2mgR

Solution

## The correct option is B $$\displaystyle\frac{3}{4}mgR$$$$\Delta U=\displaystyle\dfrac{mgh}{1+\displaystyle\dfrac{h}{R}}$$$$h=3R$$hence ,$$\Delta U=\displaystyle\dfrac{mg(3R) }{1+\displaystyle\dfrac{3R}{R}} = \dfrac{3mgR}{4}$$Physics

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