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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
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B
54mgR
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C
34mgR
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D
2mgR
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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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