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) ((n-1))/(nR) (c) (Rn)/((n-1)) (d) (n)/(R(n-1))

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

The depth at which acceleration due to gravity becomes g/n is [ R =… | MHT CET 2021 (20 Sep, Shift 2)

Q88MHT CET 2021 · 20 SEP · SHIFT 2Easy

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

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