Find the percentage decrease in weight of a body, when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.

Here, R = 6400km; x = 16km
The value of acceleration due to gravity at a depth A

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{g}}^{\mathrm{\prime }}=\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{16}{6,400}\right)=\frac{399}{400}\mathrm{g}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{g}-{\mathrm{g}}^{\mathrm{\prime }}=\mathrm{g}-\frac{399}{400}\mathrm{g}=\frac{1}{400}\mathrm{g}=2.5\times {10}^{-3}\mathrm{g}\end{array}$

If m is mass of the body, then mg and mg' will be respectively weight of the body on the surface of earth and at a depth of m below the surface of the earth. Then, % decrease in the weight of the body

$\begin{array}{l}=\frac{\mathrm{mg}-{\mathrm{mg}}^{\mathrm{\prime }}}{\mathrm{mg}}\times 100=\frac{\mathrm{g}-{\mathrm{g}}^{\mathrm{\prime }}}{\mathrm{g}}\times 100\\ =\frac{2\times 5\times {10}^{-3}\mathrm{g}}{\mathrm{g}}\times 100=0.25\mathrm{\%}\end{array}$