The loss in weight of a body taken from earth's surface to a height h is 1%. The change in weight taken into a mine of depth h will be

The correct option is 0.5% loss.

Let us analyze the variation of acceleration due to gravity with height and depth. Let $g$ be the acceleration due to gravity on the surface of the Earth.

Case 1: At a height $h$ above the Earth's surface
When a body is taken to a height $h$ (where $h\ll R$, and $R$ is the radius of the Earth), the acceleration due to gravity $g_h$ is given by the approximation:
$g_h=g\left(1-\frac{2h}{R}\right)$

The weight of the body of mass $m$ at height $h$ is $W_h=m{g}_h$, and its weight on the surface is $W=mg$.
The fractional change in weight (loss in weight) at height $h$ is:
$\frac{W-W_h}{W}=\frac{g-g_h}{g}=\frac{2h}{R}$

Given that the loss in weight at height $h$ is 1%:
$\frac{2h}{R}=1\%=0.01$
This gives:
$\frac{h}{R}=0.5\%=0.005$

Case 2: At a depth $h$ in a mine
The acceleration due to gravity at depth $d=h$ is given by:
$g_d=g\left(1-\frac{h}{R}\right)$

The fractional change in weight (loss in weight) at depth $h$ is:
$\frac{W-W_d}{W}=\frac{g-g_d}{g}=\frac{h}{R}$

Substituting the value of $\frac{h}{R}$ from Case 1:
$\frac{W-W_d}{W}=0.5\%$

Since the value is positive, it represents a decrease in acceleration due to gravity, which means there is a 0.5% loss in weight.