The depth below the earth's surface at which the acceleration due to gravity ' \(g\) ' becomes \(\frac{\mathrm{g}}{\mathrm{n}}\) is \((\mathrm{R}=\) radius of the earth, \(\mathrm{n}\) is an integer, \(\mathrm{n}>1)\)

Acceleration due to gravity at surface,
\(g=\left(\frac{G M}{R^2}\right)\)
where, \(\mathrm{G}\) is the universal gravitational constant, \(\mathrm{R}\) is the radius of earth.
At a depth \(\mathrm{h}\) below the surface, considering force balance:
\(\begin{aligned}
& \frac{G\left\{M\left(\frac{R-h}{R}\right)^3\right\} m}{(R-h)^2}=m g^{\prime} \\
& \Rightarrow g^{\prime}=\left(\frac{G M}{R^2}\right) \frac{(R-h)}{R}=g\left(\frac{R-h}{R}\right)=\frac{g}{n} \\
& \Rightarrow \frac{R-h}{R}=\frac{1}{n} \\
& \Rightarrow \frac{-h}{R}=\frac{1-n}{n}
\end{aligned}\)
\(\Rightarrow \mathrm{h}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}\)