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

Options

A. `2pi sqrt((R)/(g))`
B. `(1)/(2pi)sqrt((R)/(g))`
C. `2pi sqrt(Rg)`
D. `(1)/(2pi)sqrt(Rg)`

Answer

A

Explanation

At equator, `g'=g-R omega^(2)=0`
`:. omega=sqrt((g)/(R))rArr(2pi)/(T)=sqrt((g)/(R))rArrT=2pi sqrt((R)/(g))`