If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an abject of mass m raised from the surface of earth to a height equal to the radius of the earth R, is

The correct option is mgR/2.

Let us find the gain in gravitational potential energy step-by-step.

1. Potential Energy on the Earth's Surface:
The gravitational potential energy $U_i$ of an object of mass $m$ at the surface of the Earth (at a distance $R$ from the center of the Earth) is given by:
$U_i=-\frac{GMm}{R}$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth.

2. Potential Energy at Height R above the Earth's Surface:
When the object is raised to a height $h=R$ above the surface, its total distance from the center of the Earth becomes:
$r=R+h=R+R=2R$
The potential energy $U_f$ at this height is:
$U_f=-\frac{GMm}{2R}$

3. Gain in Potential Energy:
The gain in potential energy $\Delta U$ is the difference between the final potential energy and the initial potential energy:
$\Delta U=U_f-U_i$
$\Delta U=-\frac{GMm}{2R}-\left(-\frac{GMm}{R}\right)$
$\Delta U=\frac{GMm}{R}-\frac{GMm}{2R}$
$\Delta U=\frac{GMm}{2R}$

4. Expressing in terms of acceleration due to gravity (g):
We know that the acceleration due to gravity on the Earth's surface is:
$g=\frac{GM}{R^2}$
which gives:
$GM=g{R}^2$
Substituting $GM$ into the expression for $\Delta U$:
$\Delta U=\frac{\left(g{R}^2\right)m}{2R}$
$\Delta U=\frac{mgR}{2}$

Thus, the gain in potential energy is $\frac{mgR}{2}$ (or $\mathrm{mgR}/2$).