NEET Physics Question - Gravitation - Acceleration Due to Gravity and its variation

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)

(2020-Covid)

$\frac{R(n-1)}{n}$

$\frac{Rn}{(n-1)}$

$\frac{R}{n}$

$\frac{R}{n^2}$

Solution:

The acceleration due to gravity at depth $d$ is $g_d = g(1 - \frac{d}{R})$.nGiven $g_d = \frac{g}{n}$, we have $\frac{g}{n} = g(1 - \frac{d}{R})$. Solving for $d$: $$1 - \frac{d}{R} = \frac{1}{n} \Rightarrow \frac{d}{R} = \frac{n-1}{n} \Rightarrow d = \frac{R(n-1)}{n}$$.

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