The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{2 \mathrm{n}}$ is $(\mathrm{R}=$…
The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{2 \mathrm{n}}$ is $(\mathrm{R}=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity on earth's surface, $\mathrm{n}$ is integer)
$\frac{\mathrm{R}(1-2 \mathrm{n})}{\mathrm{n}}$
$\frac{\mathrm{R}(1-\mathrm{n})}{2 \mathrm{n}}$
$\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}$
$\frac{\mathrm{R}(2 \mathrm{n}-1)}{2 \mathrm{n}}$
Solution
The gravitational acceleration at depth is given as $g_d=g\left[1-\frac{d}{R}\right]$ Given $\mathrm{g}_{\mathrm{d}}=\frac{\mathrm{g}}{2 \mathrm{n}}$
$\begin{aligned} \therefore \quad \frac{g}{2 n} & =g\left[1-\frac{d}{R}\right] \\ \frac{d}{R} & =1-\frac{1}{2 n} \\ d & =\left[\frac{2 n-1}{2 n}\right] R\end{aligned}$