The acceleration due to gravity on the earth's surface at the poles is g and angular velocity of the…

Acceleration due to gravity at height $h$ above the pole

is given by

$g_h=g\left(1-\frac{2h}{R}\right) ...\left(1\right)$

Acceleration due to gravity at the equator due to rotation of the earth is given by

$g\text{'}=g-{\omega }^{2}R {\cos }^2\lambda ...\left(2\right)$

At the equator, $\lambda =0$

$\Rightarrow g\text{'}=g-{\omega }^{2}R ...\left(3\right)$

As both weights are equal, i.e.; ${\mathrm{g}}_h={\mathrm{g}}^{\text{'}}$

From equation $\left(1\right)$ and $\left(2\right)$, we get,
$\Rightarrow {\omega }^{2}R=\frac{2g}{R}h$
$\therefore h=\frac{R^2{\omega }^2}{2g}$