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Imagine earth to be a solid sphece of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth d below earth's surface is same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of earth), the ratio of $\frac{h}{d}$ will be
$\frac{4}{3}$
$\;\frac{3}{2}$
$\;\frac{2}{3}$
$1$
Solution
$\Rightarrow g(h)=\frac{g}{\left(1+\frac{h}{R}\right)^2}$
$\therefore \frac{g}{4}=\frac{g}{\left(1+\frac{h}{R}\right)^2}$
$\therefore 4=\left(1+\frac{h}{R}\right)^2$
$\therefore 2=1+\frac{h}{R}$
$\therefore \frac{h}{R}=1 \quad \ldots(1)$
$\Rightarrow g(d)=g\left(1-\frac{d}{R}\right)^2$
$\therefore \frac{g}{4}=g\left(1-\frac{d}{R}\right)^2$
$\therefore \frac{1}{4}=1-\frac{d}{R}$
$\therefore \frac{d}{R}=1-\frac{1}{4}$
$\therefore \frac{d}{R}=\frac{3}{4} \quad \ldots(2)$
$\therefore$ Taking ratio of equ.$(1)$ and $(2)$
$\frac{h}{d}=\frac{4}{3}$
