The depth at which acceleration due to gravity becomes g/n is ( R = radius of earth, g = acceleration due to gravity) ( n = integer) (a) (R(n-1))/(n) (b) (R(n+1))/(n) (c) (R(n-1)²)/(n) (d) (R(n+1)²)/(n)

Answer: (a). At depth d: gd = g(1 - d/R) g(1 - d/R) = g/n ⇒ d/R = 1 - 1/n d = (R(n-1))/(n) So the answer is (a).

The depth at which acceleration due to gravity becomes g/n is ( R =… | MHT CET 2025 (26 Apr, Shift 2)

Q21MHT CET 2025 · 26 APR · SHIFT 2Medium

Acceleration Due to Gravity

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is ( $R =$ radius of earth, $g =$ acceleration due to gravity) ( $n =$ integer)

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Answer: (a)

At depth $d$ : $g_d = g\left(1 - \frac{d}{R}\right)$
$g\left(1 - \frac{d}{R}\right) = \frac{g}{n} \Rightarrow \frac{d}{R} = 1 - \frac{1}{n}$
$d = \frac{R(n-1)}{n}$
So the answer is (a).

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