The depth below the earth's surface at which the acceleration due to gravity ' $g$ ' becomes…
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, $\mathrm{n}>1)$
$\frac{\mathrm{R}(\mathrm{n}+1)}{\mathrm{n}}$
$\frac{R(n-1)}{n}$
$\frac{\mathrm{Rn}}{(\mathrm{n}+1)}$
$\frac{\mathrm{R}}{\mathrm{n}}$
Solution
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:
$\begin{aligned}
& \frac{G\left\{M\left(\frac{R-h}{R}\right)^3\right\} m}{(R-h)^2}=m g^{\prime} \\
& \Rightarrow g^{\prime}=\left(\frac{G M}{R^2}\right) \frac{(R-h)}{R}=g\left(\frac{R-h}{R}\right)=\frac{g}{n} \\
& \Rightarrow \frac{R-h}{R}=\frac{1}{n} \\
& \Rightarrow \frac{-h}{R}=\frac{1-n}{n}
\end{aligned}$
$\Rightarrow \mathrm{h}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}$