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(B) The acceleration due to gravity at a latitude $\lambda$ is given by the formula:$g' = g - R{\omega ^2}{\cos ^2}\lambda$where $g$ is the accelerati

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$\frac{3}{4}{\omega ^2}R$

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(B) The acceleration due to gravity at a latitude $\lambda$ is given by the formula:$g' = g - R{\omega ^2}{\cos ^2}\lambda$where $g$ is the acceleration due to gravity at the equator (ignoring the centrifugal effect) or the effective gravity at the equator where $\lambda = 0^\circ$.At latitude $\lambda = 30^\circ$:$g_{30} = g - R{\omega ^2}{\cos ^2}30^\circ$Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$,we have $\cos^2 30^\circ = \frac{3}{4}$.Substituting this into the equation:$g_{30} = g - R{\omega ^2} \left( \frac{3}{4} \right)$Therefore,the difference is:$g - g_{30} = \frac{3}{4}{\omega ^2}R$.

If the Earth is assumed to be a sphere of radius $R$,and $g_{30}$ is the value of acceleration due to gravity at a latitude of $30^\circ$ and $g$ is the value at the equator,the value of $g - g_{30}$ is:

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