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>>Variation in Value of Acceleration due to Gravity
>> $The depth 'd' at which the value of acce$

Question

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

A

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

B

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

C

$d = R (\frac{n - 1}{2 n})$

D

$d = R^{2} (\frac{n - 1}{n})$

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Updated on : 2022-09-05

Solution

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Correct option is B)

Acceleration due to gravity at a depth $d$ under the earth surface $g^{'} = g_{s} (1 - \frac{d}{R})$
Given : $g^{'} = \frac{g _{s}}{n}$
$∴$ $\frac{g _{s}}{n} = g_{s} (1 - \frac{d}{R})$
Or $\frac{1}{n} = (1 - \frac{d}{R})$
$⟹$ $d = R (\frac{n - 1}{n})$

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