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(D) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R \omega^2 \cos^2 \lambda$.For a man to feel weightle

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$4 \pi \sqrt{\frac{R}{g}}$

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(D) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R \omega^2 \cos^2 \lambda$.For a man to feel weightlessness,the effective gravity must be zero,so $g' = 0$.Substituting the given values,$0 = g - R \omega^2 \cos^2 60^{\circ}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $0 = g - R \omega^2 (\frac{1}{2})^2$.$g = R \omega^2 (\frac{1}{4}) \Rightarrow \omega^2 = \frac{4g}{R} \Rightarrow \omega = 2 \sqrt{\frac{g}{R}}$.The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.Substituting $\omega$,$T = \frac{2 \pi}{2 \sqrt{g/R}} = \pi \sqrt{\frac{R}{g}} 	imes 2 = 2 \pi \sqrt{\frac{R}{g}} 	imes 2 = 4 \pi \sqrt{\frac{R}{g}}$.

The rotation of the Earth,having radius $R$,about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case is:

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