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The depth from the surface of the earth at which the acceleration due to gravity will be 75% of the value on the surface of the earth. (radius of the earth = R)

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$\frac{3 R}{4}$

$\frac{R}{2}$

$\frac{R}{4}$

$\frac{R}{8}$

answer is A.

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Detailed Solution

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

$g^{1} = \frac{75}{100} g$

$\frac{3 g}{4} = g (1 - \frac{d}{R})$

$d = \frac{R}{4}$

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