The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{\mathrm{n}-1}$ times the…

The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{\mathrm{n}-1}$ times the value at the earth's surface is ( $R=$ radius of the earth)
  1. $\mathrm{R}\left(\frac{\mathrm{n}}{\mathrm{n}-1}\right)$
  2. $\mathrm{R}\left(\frac{\mathrm{n}-2}{\mathrm{n}-1}\right)$
  3. ${ }^{\wedge} \mathrm{R}\left(\frac{2 \mathrm{n}-1}{\mathrm{n}}\right)$
  4. $\mathrm{R}\left(\frac{\mathrm{n}-1}{2 \mathrm{n}-1}\right)$

Solution

$\begin{aligned} & g_d=g\left(1-\frac{d}{R}\right) \\ & g\left(\frac{1}{n-1}\right)=g\left(1-\frac{d}{R}\right) \quad \ldots\left(\text { Given: } g_d=g\left(\frac{1}{n-1}\right)\right. \\ \therefore \quad & \frac{d}{R}=1-\left(\frac{1}{n-1}\right) \\ \therefore \quad & d=R\left(\frac{n-2}{n-1}\right)\end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 1)

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