Weight of a body of mass m decreases by 1% when it is raised to height h above the earth's surface. If the body is taken to a depth h in a mine, change in its weight is

The correct option is 0.5% decrease.

Let the mass of the body be $m$ and the radius of the Earth be $R$. The weight of the body at the Earth's surface is given by:
$W=mg$
where $g$ is the acceleration due to gravity at the surface of the Earth.

When the body is raised to a height $h$ (where $h\ll R$) above the Earth's surface, the acceleration due to gravity becomes:
$g_h=g\left(1-\frac{2h}{R}\right)$
Consequently, the weight of the body at height $h$ is:
$W_h=m{g}_h=mg\left(1-\frac{2h}{R}\right)=W\left(1-\frac{2h}{R}\right)$

The fractional change (decrease) in weight at height $h$ is given by:
$\frac{W-W_h}{W}=\frac{2h}{R}$
Given that the weight decreases by 1% at height $h$:
$\frac{2h}{R}=1\%$
Which gives:
$\frac{h}{R}=0.5\%$

Now, if the body is taken to a depth $d=h$ in a mine, the acceleration due to gravity at depth $h$ is:
$g_d=g\left(1-\frac{h}{R}\right)$
The weight of the body at depth $h$ is:
$W_d=m{g}_d=mg\left(1-\frac{h}{R}\right)=W\left(1-\frac{h}{R}\right)$

The fractional change (decrease) in weight at depth $h$ is:
$\frac{W-W_d}{W}=\frac{h}{R}$
Since we established that $\frac{h}{R}=0.5\%$, the decrease in weight at depth $h$ is:
$0.5\%$

Therefore, the change in its weight is a 0.5% decrease.