The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{\mathrm{n}}$ is [ $\mathrm{R}$ =…

The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{\mathrm{n}}$ is [ $\mathrm{R}$ = radius of earth, $\mathrm{g}$ = acceleration due to gravity, $\mathrm{n}=$ integer]
  1. $\frac{R(n-1)}{n}$
  2. $\frac{(\mathrm{n}-1)}{\mathrm{nR}}$
  3. $\frac{\mathrm{Rn}}{(\mathrm{n}-1)}$
  4. $\frac{\mathrm{n}}{\mathrm{R}(\mathrm{n}-1)}$

Solution

$\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \\ & \frac{\mathrm{g}}{\mathrm{n}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \\ & \mathrm{d}=\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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