(2003)
Solution:
Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. Given $h = 3R$, $\Delta U = \frac{mg(3R)}{1+3} = \frac{3}{4}mgR$.
A body of mass $m$ is placed on earth surface which is taken from earth surface to a height of $h = 3R$ then change in gravitational potential energy is:
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