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 B

Percentage change in g when the body is raised to height h, $\frac{\Delta g}{g}\times 100\%=\frac{2h\times 100}{R}=1\%$

Percentage change in g when the body is taken to a depth d,

$\frac{\Delta g}{g}\times 100\%=\frac{d}{R}\times 100\%=\frac{h}{R}\times 100\%[Asd=h]$

$\therefore$ Percentage decreases in weight $=\frac{1}{2}(\frac{2h}{R}\times 100)=\frac{1}{2}\left(1\%\right)=0.5\%$