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(C) The gravitational potential energy of a body of mass $m$ at the surface of the Earth is $U_i = -\frac{GMm}{R}$.At a height $h$ above the surface,t

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$\frac{R}{4}$

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(C) The gravitational potential energy of a body of mass $m$ at the surface of the Earth is $U_i = -\frac{GMm}{R}$.At a height $h$ above the surface,the potential energy is $U_f = -\frac{GMm}{R+h}$.The increase in potential energy is $\Delta U = U_f - U_i = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = GMm \left( \frac{h}{R(R+h)} \right)$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this,$\Delta U = mgR^2 \left( \frac{h}{R(R+h)} \right) = mgR \left( \frac{h}{R+h} \right)$.Given $\Delta U = \frac{mgR}{5}$,we equate: $\frac{mgR}{5} = mgR \left( \frac{h}{R+h} \right)$.$\frac{1}{5} = \frac{h}{R+h} \implies R+h = 5h \implies 4h = R \implies h = \frac{R}{4}$.

$A$ body of mass $m$ is raised through a height $h$ above the Earth's surface such that the increase in potential energy is $\frac{mgR}{5}$. The height to which the body is raised is ($R=$ radius of Earth,$g=$ acceleration due to gravity).

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