A body of mass m kg. starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant]

The correct option is GMm/6R.

To find the kinetic energy of the falling body, we can apply the law of conservation of mechanical energy. Since gravity is a conservative force, the total mechanical energy (the sum of kinetic energy and gravitational potential energy) remains constant throughout the motion of the body.

The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth (of mass $M$) is given by the formula:
$U=-\frac{GMm}{r}$
where $G$ is the universal gravitational constant.

Step 1: Determine the initial state of the body
The body starts falling from a height of $2R$ above the Earth's surface. The distance from the center of the Earth to the initial position is:
$r_1=R+2R=3R$
Since the body starts falling from rest, its initial kinetic energy is:
$K_1=0$
The initial potential energy is:
$U_1=-\frac{GMm}{3R}$

Step 2: Determine the final state of the body
The body falls to a point at a height of $R$ above the Earth's surface. The distance from the center of the Earth to this final position is:
$r_2=R+R=2R$
Let the kinetic energy at this point be $K_2$.
The final potential energy is:
$U_2=-\frac{GMm}{2R}$

Step 3: Apply the Law of Conservation of Energy
The total mechanical energy in the initial state equals the total mechanical energy in the final state:
$K_1+U_1=K_2+U_2$
Substituting the values we obtained:
$0-\frac{GMm}{3R}=K_2-\frac{GMm}{2R}$

Solving for the final kinetic energy $K_2$:
$K_2=\frac{GMm}{2R}-\frac{GMm}{3R}$
Taking the common denominator, we get:
$K_2=\frac{3GMm-2GMm}{6R}$
$K_2=\frac{GMm}{6R}$

Therefore, the kinetic energy of the body when it has fallen to a point $R$ above the earth's surface is $\frac{GMm}{6R}$.