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If earth is supposed to be a sphere of radius $R$, if $g_{30}$ is value of acceleration due to gravity at latitude of $30^{\circ}$ and $g$ at the equator, the value of $g-g_{30^{\circ}}$ is
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The correct answer is:
$\frac{3}{4} \omega^2 R$
Acceleration due to gravity at latitude $\lambda$ is given by
$\begin{aligned} & g^{\prime}=g-R \omega^2 \cos ^2 \lambda \\ & \text { At } 30^{\circ}, g_{30^{\circ}}=g-R \omega^2 \cos ^2 30^{\circ}=g-\frac{3}{4} R \omega^2 \\ & \therefore \quad g-g_{30}=\frac{3}{4} \omega^2 R .\end{aligned}$
$\begin{aligned} & g^{\prime}=g-R \omega^2 \cos ^2 \lambda \\ & \text { At } 30^{\circ}, g_{30^{\circ}}=g-R \omega^2 \cos ^2 30^{\circ}=g-\frac{3}{4} R \omega^2 \\ & \therefore \quad g-g_{30}=\frac{3}{4} \omega^2 R .\end{aligned}$
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