The depth 'd' below the surface of the earth where the value of acceleration due to gravity becomes (1/n) times the value at the surface of the earth is (R = radius of the earth) (a) R(n-1/n) (b) R(n/n+1) (c) R/n (d) R/n²

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

The depth 'd' below the surface of the earth where the value of… | MHT CET 2021 (20 Sep, Shift 1)

Q89MHT CET 2021 · 20 SEP · SHIFT 1Easy

Acceleration Due to Gravity

The depth 'd' below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is $(R =$ radius of the earth)

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Answer: (a)

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

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