The rotation of the earth about its axis speeds up such that a man on the equator becomes weightless. In such a situation, what would be the duration of one day ?

Options

A. `sqrt(Rg)`
B. `sqrt((R )/(g))`
C. `2pi sqrt((R )/(g))`
D. `2pi sqrt(Rg)`

Answer

C

Explanation

`g_(eq)=g-omega^(2)R`
`g_(eq) =0`
`g-omega^(2) R=0 rArr omega =sqrt((g)/(R ))`