Question:
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 $\pi=3.14)$
Correct Option: , 4
Solution:
(4)
For objects to float
$\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{R}$
$\omega=$ angular velocity of earth.
$\mathrm{R}=$ Radius of earth
$\omega=\sqrt{\frac{g}{R}} \ldots(1)$
Duration of day $=\mathrm{T}$
$\mathrm{T}=\frac{2 \pi}{\omega} \ldots(2)$
$\Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
$=2 \pi \sqrt{\frac{6400 \times 10^{3}}{10}}$
$\Rightarrow \frac{\mathrm{T}}{60}=83.775$ minutes
$\simeq 84$ minutes