At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of earth is (R = radius of earth)

The correct option/answer is 32N.

Let's understand the step-by-step variation of gravity and weight with height above the Earth's surface.

The weight of a person of mass $m$ at the surface of the Earth is given by:
$W=mg=72\text{N}$
where $g$ is the acceleration due to gravity at the surface of the Earth, which is defined as:
$g=\frac{GM}{R^2}$
Here, $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.

At a height $h$ above the surface of the Earth, the acceleration due to gravity $g^{\prime }$ is given by:
$g^{\prime }=\frac{GM}{{\left(R+h\right)}^2}$

Given that the height is $h=\frac{R}{2}$, we substitute this value into the equation:
$g^{\prime }=\frac{GM}{{\left(R+\frac{R}{2}\right)}^2}$
$g^{\prime }=\frac{GM}{{\left(\frac{3R}{2}\right)}^2}$
$g^{\prime }=\frac{GM}{\frac{9}{4}{R}^2}$
$g^{\prime }=\frac{4}{9}\left(\frac{GM}{R^2}\right)$
Since $g=\frac{GM}{R^2}$, we can write:
$g^{\prime }=\frac{4}{9}g$

Now, the weight of the person at height $h=\frac{R}{2}$ is $W^{\prime }$:
$W^{\prime }=m{g}^{\prime }$
$W^{\prime }=m\left(\frac{4}{9}g\right)$
$W^{\prime }=\frac{4}{9}\left(mg\right)$
Substituting the initial weight $W=mg=72\text{N}$:
$W^{\prime }=\frac{4}{9}\times 72$
$W^{\prime }=4\times 8$
$W^{\prime }=32\text{N}$

Thus, the weight of the person at a height of R/2 from the surface of the Earth is 32N.