In order to make the effective acceleration due to gravity at the equator to zero, the numerical value of the angular velocity of rotation of the earth should be
[Radius of the earth=6400 km, `g=10m/s^(2)`]

Options

A. `(1)/(400)` rad/s
B. `(1)/(200)` rad/s
C. `(1)/(800)` rad/s
D. `(1)/(500)` rad/s

Answer

C

Explanation

At the equator `g'=g-omega^(2)R`
If `g'=0`, then `0=g-omega^(2)R` or `omega^(2)=(g)/(R )`
`therefore omega = sqrt((g)/(R ))=sqrt((10)/(6400xx10^(3)))=(1)/(800)` rad/s