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 × 10³ m, Take π = 3.14]

Solution

Physics sir · 5 steps

Floating means the whole of gravity is used up as centripetal force, so set ω² R = g.

1. Weight at equator. A body at the equator moves in a circle of radius R. Gravity mg pulls it in, and the normal force N pushes it out. The difference provides the centripetal force: mg - N = mω² R.
2. Condition to float. The body floats when the ground no longer pushes on it, that is N = 0. Then gravity alone supplies the centripetal force.

   mω² R = mg ⇒ ω = √(g/R)

3. Put in numbers. ω = √(10/(6.4 × 10⁶)) = √(1.5625 × 10⁻⁶) = 1.25 × 10⁻³ rad s⁻¹.
4. Length of day. T = 2π/ω = 6.28/(1.25 × 10⁻³) = 5024 s. Dividing by 60 gives about 83.7 minutes.
5. So. The day would last about 84 minutes, option (a).

The idea to keep. Apparent weight at the equator is mg - mω² R. Weightlessness means N = 0, so ω = √(g/R) and T = 2π√(R/g), about 84 minutes for Earth.

Answer from the NTA final answer key. Solution by PrepWiser.