Thank you for reporting, we will resolve it shortly
Q.
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
Solution:
For height $\frac{\Delta g}{g} \times 100 \%=\frac{2 h}{R}=1 \%$;
For depth $\frac{\Delta g}{g} \times 100 \%$
$=\frac{d}{R}=\frac{h}{R}=\frac{1}{2} \%=0.5 \%$