The rotation of the earth having radius R abo | vedclass

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(C) 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 weightles

Vedclass · Accepted Answer

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

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(C) 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 weightless,the effective gravity must be zero,so $g' = 0$.Substituting the given values: $0 = g - R\omega^2 \cos^2 60^{\circ}$.Since $\cos 60^{\circ} = 1/2$,we have $\cos^2 60^{\circ} = 1/4$.Thus,$0 = g - R\omega^2 (1/4)$,which implies $R\omega^2 / 4 = g$.Solving for $\omega$,we get $\omega^2 = 4g/R$,so $\omega = 2\sqrt{g/R}$.The duration of the day $T$ is related to angular velocity $\omega$ by $T = 2\pi / \omega$.Substituting $\omega$,we get $T = 2\pi / (2\sqrt{g/R}) = \pi \sqrt{R/g}$.

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 weightless. The duration of the day in such case will be:

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