Q: Find the percentage decrease in the weight of the body when taken to a depth of 32 Km below the surface of earth.

Sol: Weight of the body at depth d is

$\large m g’ = m g(1-\frac{d}{R})$

% decrease in weight $\large = \frac{mg-mg’}{mg}\times 100$

$\large = \frac{d}{R}\times 100$

$\large = \frac{32}{6400}\times 100$

= 0.5 %