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Q.

What is the change in gravitational potential energy when a body of mass m is raised to a height nR above the surface of earth of radius R

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a

(\frac{n + 1}{n}) mgR

b

(\frac{n}{n + 1}) mgR

c

\frac{mgR}{n}

d

mgR

answer is C.

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Detailed Solution

we know change in P.E

$Δ P .E = \frac{mgh}{(1 + \frac{h}{R})}$

h = nR

$P .E = \frac{mgnR}{1+ \frac{nR}{R}} = \frac{n}{1 + n} (mgR)$

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