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A body stretches a spring by a certain amount at the equator. Assuming earth to be a perfect sphere of radius R; g, the acceleration due to gravity on the surface of the earth and $ω$ , the angular speed of the earth, the height at which at poles it will stretch the spring by the same amount at the equator is

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$\frac{R}{\sqrt{1 - (ω^{2} R / g)}}$

$R [1 - \sqrt{1 - (ω^{2} R / g)}]$

$R [\frac{1}{\sqrt{1 - (ω^{2} R / g)}} - 1]$

answer is D.

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Detailed Solution

At Equator $F_{s} = kx = {mg}_{e} . . . . . . . . . . . . (1)$

At poles at a height ‘h’

$kx = m {(g_{p})}_{h} = m g_{p} {[\frac{R}{R + h}]}^{2} . . . . . . . . . . . . . (2)$

$∴ From data (1) = (2)$

$\Rightarrow g_{e} = g_{p} {[\frac{R}{R + h}]}^{2} \Rightarrow g - R ω^{2} = g {[\frac{R}{R + h}]}^{2}$

$\Rightarrow 1 - \frac{R ω^{2}}{g} = {(\frac{R}{R + h})}^{2} \Rightarrow \sqrt{1 - \frac{R ω^{2}}{g}} = \frac{R}{R + h}$

$∴ h = R [\frac{1}{\sqrt{1 - \frac{R ω^{2}}{g}}} - 1]$

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