The depth 'd' at which the value of acceleration due to gravity becomes 1/n-1 times the value at the earth's surface is ( R = radius of the earth) (a) R(n/n-1) (b) R((n-2)/(n-1)) (c) R(2n-1/n) (d) R((n-1)/(2n-1))

Answer: (b). gd = g(1 - d/R) = g/n-1 d/R = 1 - 1/n-1 = (n-2)/(n-1) d = R((n-2)/(n-1)) So the answer is (b) R((n-2)/(n-1)).

The depth 'd' at which the value of acceleration due to gravity… | MHT CET 2024 (11 May, Shift 1)

Q34MHT CET 2024 · 11 MAY · SHIFT 1Easy

Acceleration Due to Gravity

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

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Answer: (b)

$g_d = g\left(1 - \frac{d}{R}\right) = \frac{g}{n-1}$
$\frac{d}{R} = 1 - \frac{1}{n-1} = \frac{n-2}{n-1}$
$d = R\left(\frac{n-2}{n-1}\right)$
So the answer is (b) $R\left(\frac{n-2}{n-1}\right)$ .

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