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  4. The depth d at which the value of acceleration due to gravity becomes (1 / n) time the value at the surface is (R be the radius of the earth)

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Q. The depth $d$ at which the value of acceleration due to gravity becomes $\frac {1} { n}$ time the value at the surface is (R be the radius of the earth)

KEAMKEAM 2005Gravitation

Solution:

At depth d from earth's surface,
$g'=g\left(1-\frac{d}{R} \right)$
Given $g'=\frac{g}{n}$
So, $\frac{g}{n}=g\left(1-\frac{d}{R}\right)$
or $\frac{1}{n}=1-\frac{d}{R}$ or $\frac{d}{R}=1-\frac{1}{n}$
$=\frac{(n-1)R}{n}$
$\therefore $ $d=\frac{(n-1)R}{n}$