"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 g = 10 ms$-$2, the radius of earth, R = 6400 $\times$ 103 m, Take $\pi$ = 3.14]"

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"For objects to float

mg = 2$\omega$²R

$\omega$ = angular velocity of earth.

R = Radius of earth

$\omega = \sqrt {{g \over R}}$ ..... (1)

Duration of day = T

$T = {{2\pi } \over \omega }$ ..... (2)

$\Rightarrow T = 2\pi \sqrt {{R \over g}}$

$= 2\pi \sqrt {{{6400 \times {{10}^3}} \over {10}}}$

$\Rightarrow {T \over {60}}$ = 83.775 minutes

$\simeq$ 84 minuites"

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 g = 10 ms^$-$2, the radius of earth, R = 6400 $\times$ 10³ 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 : [Take g = 10 ms$-$2, the radius of earth, R = 6400 $\times$ 103 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 : [Take g = 10 ms^$-$2, the radius of earth, R = 6400 $\times$ 10³ m, Take $\pi$ = 3.14]