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

The correct option is 5mgh/6.

To find the increase in gravitational potential energy when a body of mass m is raised to a height h from the Earth's surface, we use the general formula for gravitational potential energy.

The gravitational potential energy of a mass $m$ at a distance $r$ from the center of the Earth (of mass $M$ and radius $R$) is given by:

$U=-\frac{GMm}{r}$

Initially, the body is on the surface of the Earth, so the initial distance is $r_i=R$. The initial potential energy $U_i$ is:

$U_i=-\frac{GMm}{R}$

Finally, the body is raised to a height $h$ above the surface, so the final distance from the center is $r_f=R+h$. The final potential energy $U_f$ is:

$U_f=-\frac{GMm}{R+h}$

The increase in potential energy ($\Delta U$) is the difference between the final and initial potential energies:

$\Delta U=U_f-U_i$

$\Delta U=-\frac{GMm}{R+h}-\left(-\frac{GMm}{R}\right)$

$\Delta U=GMm\left(\frac{1}{R}-\frac{1}{R+h}\right)$

$\Delta U=GMm\left(\frac{R+h-R}{R(R+h)}\right)$

$\Delta U=\frac{GMmh}{R(R+h)}$

We know that the acceleration due to gravity at the Earth's surface $g$ is related to the gravitational constant $G$ by the relation:

$g=\frac{GM}{R^2}\Rightarrow GM=g{R}^2$

Substituting $GM=g{R}^2$ into our equation for $\Delta U$:

$\Delta U=\frac{\left(g{R}^2\right)mh}{R(R+h)}$

$\Delta U=\frac{mgh}{1+\frac{h}{R}}$

According to the problem, the height to which the body rises is $h=\frac{R}{5}$. Substituting this value into the equation:

$\Delta U=\frac{mgh}{1+\frac{R/5}{R}}$

$\Delta U=\frac{mgh}{1+\frac{1}{5}}$

$\Delta U=\frac{mgh}{\frac{6}{5}}$

$\Delta U=\frac{5}{6}mgh$

Thus, the increase in potential energy is 5mgh/6.