Assuming that the earth is a sphere of uniform mass density, what is the percentage decrease in the weight of a body when taken to the end of a tunnel $32 \mathrm{km}$ below the surface of the earth? Radius of earth $=6400 \mathrm{km}$.

Acceleration due to gravity at depth d is given by
$g^{\text{'}}=g\left(1-\frac{d}{R}\right)=g\left(1-\frac{32}{6400}\right)=\frac{199g}{200}$
If m is the mass of the body, its weight on the surface of the earth is mg and at the end of the tunnel it is $m{g}^{\text{'}}$. Therefore, decrease in weight is
$$\begin{gathered} mg-m{g}^{\text{'}}=m\left(g-\frac{199g}{200}\right)=\frac{mg}{200} \\ \therefore \text{ Percentage decrease }=\frac{mg/200}{mg}\times 100=0.5\% \end{gathered}$$
Hence the correct choice is (b).