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Let $\omega$ be the angular velocity of the earth’s rotation about its axis. Assume that the acceleration due to gravity on the earth’s surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth d below earth’s surface at a pole $(d < < R)$ The value of $d$ is
$\frac{\omega ^2 R^2}{ g}$
$\frac{\omega ^2 R^2}{ 2g}$
$\frac{2 \omega ^2 R^2}{ g}$
$\frac{\sqrt{ Rg}}{ g}$
Solution
Lets $\omega=$ the angular velocity of the earth's rotation about it's axis.
$g=$ acceleration due to gravity.
$R=$ radius of earth
$d=$ depth a below earth surface at which the acceleration due to gravity is same at the equator and the poles.
$g_{e f f}$ due to rotation $=g_{e f f}$ due to depth
$g-\omega^{2} R=g\left(1-\frac{d}{R}\right)$
$g-R \omega^{2}=g-\frac{g d}{R}$
$-R \omega^{2}=-\frac{g d}{R}$
$d=\frac{\omega^{2} R^{2}}{g}$
