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 o and g at the equator, the value of $\mathbf{g} - \mathbf{g}_{30^\circ}$ is
$(a) \frac{1}{4} \omega^{2} R \quad (b) \frac{3}{4} \omega^{2} R (c) \omega^{2} R \quad (d) \frac{1}{2} \omega^{2} R$
Text Solution
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Acceleration due to gravity at latitude λ λ is given by
$g' = g - R \omega^{2} \cos^{2} \lambda$
At $30^{0}, g_{30^{0}} = g - R \omega^{2} \cos^{2} 30^{0} = g - \frac{3}{4} R \omega^{2}$
∴ ∴ $g - g_{30} = \frac{3}{4} \omega^{2} R^{2}.$
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