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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 $g = 10 m s^{- 2},$ the radius of earth, $R = 6400 \times 10^{3} m,$ Take $π = 3.14]$

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Correct option is B

Floating means weightlessness g1=g−Rω2cos2θ (θ=00 at equator) Given : g1=0 ⇒g=Rω2 ⇒ω=gR ⇒T=2πRg=23.14640000010=5024 s=84 min

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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=6400×103m, Take π=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 $g = 10 m s^{- 2},$ the radius of earth, $R = 6400 \times 10^{3} m,$ Take $π = 3.14]$