A body of mass m rises to height h =R/ 5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is

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(4/5) mgh

(5/6) mgh

mgh

(6/7) mgh

answer is C.

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Detailed Solution

$\begin{array}{l} ΔV = \frac{GMm}{R} - \frac{GMm}{R + (R / 5)} = \frac{GMm}{6 R} \\ = \frac{GM}{R^{2}} \times (\frac{mR}{6}) = \frac{mgR}{6} (∵ \frac{GM}{R^{2}} = g) \\ = \frac{mg (5 h)}{6} = \frac{5}{6} mgh \end{array}$

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