Question

What is the depth at which the value of acceleration due to gravity becomes 1/ n times the value that the
surface of earth? (Radius of earth =R) [NEET 2020]

Moderate

Solution

Radius of earth = R
Let at depth d, acceleration due to gravity be g_d which becomes $\frac{g}{n}$ .
i.e., $g_{d} = \frac{g}{n} \Rightarrow g (1 - \frac{d}{R}) = \frac{g}{n}$
$\Rightarrow 1 - \frac{d}{R} = \frac{1}{n} \Rightarrow 1 - \frac{1}{n} = \frac{d}{R}$
$\Rightarrow \frac{n - 1}{n} = \frac{d}{R} \Rightarrow d = (\frac{n - 1}{n}) R$

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