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(D) For bodies at the equator to start floating,the gravitational force must be equal to the required centripetal force.$mg = m \omega^2 R$where $\ome

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$84$

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(D) For bodies at the equator to start floating,the gravitational force must be equal to the required centripetal force.$mg = m \omega^2 R$where $\omega$ is the angular velocity of the Earth and $R$ is the radius of the Earth.Solving for $\omega$:$\omega = \sqrt{\frac{g}{R}}$The duration of the day $T$ is given by $T = \frac{2\pi}{\omega}$.Substituting $\omega$:$T = 2\pi \sqrt{\frac{R}{g}}$Given $g = 10 \, m/s^2$ and $R = 6400 	imes 10^3 \, m$:$T = 2 	imes 3.14 	imes \sqrt{\frac{6400 	imes 10^3}{10}}$$T = 6.28 	imes \sqrt{640000}$$T = 6.28 	imes 800 = 5024 \, s$To convert the duration into minutes:$T_{	ext{min}} = \frac{5024}{60} \approx 83.73 \, 	ext{minutes}$Rounding to the nearest integer,the duration is approximately $84 \, 	ext{minutes}$.

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating,the duration of the day would be approximately ........ minutes
(Take: $g = 10 \, m/s^2$,the radius of earth,$R = 6400 \times 10^3 \, m$,Take $\pi = 3.14$)

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If the angular velocity of earth's spin is increased such that the bodies at the equator start floating,the duration of the day would be approximately ........ minutes(Take: $g = 10 \, m/s^2$,the radius of earth,$R = 6400 \times 10^3 \, m$,Take $\pi = 3.14$)