Let $ω$ be the angular velocity of the earth’s rotation about its axis. An object weighed by a spring balance gives the same reading at the equator as at a height h above the poles (h<<R). The value of h is

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$\frac{2 ω^{2} R^{2}}{g}$

$\frac{ω^{2} R^{2}}{g}$

$\frac{\sqrt{Rg}}{ω}$

$\frac{ω^{2} R^{2}}{2 g}$

answer is B.

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Detailed Solution

$g_{equator} = g_{h}$
$\begin{array}{l} g - {Rω}^{2} = g (1 - \frac{2 h}{R}) (∵ h << R) \\ {Rω}^{2} = \frac{2 hg}{R} \\ h = \frac{R^{2} ω^{2}}{2 g} \end{array}$

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