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 :
(Take : $\mathrm{g}=10 \mathrm{~ms}^{-2}$, the radius of earth, $\mathrm{R}=6400 \times 10^{3} \mathrm{~m}$, Take $\left.\pi=3.14\right)$
Correct Option: , 4
For objects to float
$m g=m \omega^{2} R$
$\omega=$ angular velocity of earth.
$\mathrm{R}=$ Radius of earth
$\omega=\sqrt{\frac{g}{R}}$ ............(1)
Duration of day $=\mathrm{T}$
$\mathrm{T}=\frac{2 \pi}{\omega}$ ............(2)
$\Rightarrow T=2 \pi \sqrt{\frac{R}{g}}$
$=2 \pi \sqrt{\frac{6400 \times 10^{3}}{10}}$
$\Rightarrow \frac{\mathrm{T}}{60}=83.775$ minutes
$\simeq 84$ minutes
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