The depth from the surface of the earth of radius $\mathrm{R}$, at which acceleration due to gravity will be…

The depth from the surface of the earth of radius $\mathrm{R}$, at which acceleration due to gravity will be $60 \%$ of the value of the value on the earth surface is
  1. $\frac{2 \mathrm{R}}{3}$
  2. $\frac{2 R}{5}$
  3. $\frac{3 R}{5}$
  4. $\frac{5 \mathrm{R}}{3}$

Solution

$\begin{aligned} & g^{\prime}=g\left(1-\frac{d}{R}\right) \\ & g^{\prime}=0.6 g \\ & \therefore 0.6=1-\frac{d}{R} \\ & \therefore \frac{d}{R}=1-0.6=0.4=\frac{2}{5} \\ & \therefore d=\frac{2}{5} R\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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