If the earth is supposed to be a sphere of radius R, $g_{30^{o}}$ is the value of acceleration due to gravity at latitude of 30° and g at the pole, the value of $g - g_{30^{o}}$ is (where $ω$ is the angular velocity of earth)

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$1 / 4 ω^{2} R$

$ω^{2} R$

$3 / 4 ω^{2} R$

$1 / 2 ω^{2} R$

answer is B.

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Detailed Solution

Acceleration due to gravity at latitude $λ$ is given by:
$g^{'} = g - {Rω}^{2} \cos^{2} λ$

$At 30^{\circ} g_{30^{\circ}} = g - {Rω}^{2} \cos^{2} 30^{\circ} = g - \frac{3}{4} {Rω}^{2}$

$\Rightarrow g - g_{30^{\circ}} = \frac{3}{4} ω^{2} R$

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