The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is $ω$ . An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is : (h <<R, where R is the radius of the earth)

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$\frac{R^{2} ω^{2}}{8 g}$

$\frac{R^{2} ω^{2}}{4 g}$

$\frac{R^{2} ω^{2}}{2 g}$

$\frac{R^{2} ω^{2}}{g}$

answer is D.

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Detailed Solution

Weight of the object near the equator = $m (g - R ω^{2})$
And at height h near the poles = $m g \frac{R^{2}}{{(R + h)}^{2}}$
Given

$\begin{array}{l} m (g - R ω^{2}) = \frac{m g R^{2}}{{(R + h)}^{2}} = m g (1 - \frac{2 h}{R}), \\ \Rightarrow m (g - R ω^{2}) = m g (1 - \frac{2 h}{R}) [∵ h < < R] \end{array}$
$\begin{array}{l} \Rightarrow R ω^{2} = \frac{2 g h}{R} \\ \Rightarrow h = \frac{R^{2} ω^{2}}{2 g} \end{array}$