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If the earth is supposed to be a sphere of radius $R, and g^{'} is$ the value of acceleration due to gravity at latitude of 30ºand g at the pole, then the value of $g - g^{'} will be$ ( angular velocity of spin of earth = $ω$ )

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$\frac{ω^{2} R}{4}$

$\frac{3 ω^{2} R}{4}$

$ω^{2} R$

$\frac{ω^{2} R}{3}$

answer is B.

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Detailed Solution

$g - g^{'} = ω^{2} R \cos^{2} 30^{\circ} = ω^{2} R {[\frac{\sqrt{3}}{2}]}^{2} = \frac{3}{4} ω^{2} R$

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