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, $$R = 6400 \times 10^3$$ m, Take $$\pi = 3.14$$)

At the equator, bodies start floating when the centripetal acceleration equals the gravitational acceleration, i.e., $$\omega^2 R = g$$. This gives $$\omega = \sqrt{\frac{g}{R}}$$.

The duration of the day is $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}}$$.

Substituting the given values: $$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} = 6.28 \times \sqrt{6.4 \times 10^5} = 6.28 \times 800 = 5024$$ seconds.

Converting to minutes: $$T = \frac{5024}{60} \approx 84$$ minutes.