The acceleration due to gravity at the poles is `10ms^(-2)` and equitorial radius is `6400 km` for the earth. Then the angular velocity of rotaiton of the earth about its axis so that the weight of a body at the equator reduces to `75%` is

Options

A. `(1)/(1600)"rad/s"`
B. `(1)/(800)"rad/s"`
C. `(1)/(400)"rad/s"`
D. `(1)/(200)"rad/s"`

Answer

a

Explanation

`g_(phi)=g-Romega^(2)cos^(2)phi`
`7.5=10-6.4xx10^(6)xx omega^(2)xx"1 (at equator)"`
`6.4xx10^(6)omega^(2)=2.5`
`omega^(2)=sqrt((25)/(64xx10^(6)))=(5)/(8)xx10^(-3)`
`=8.25xx10^(-4)=(1)/(1600)"rad/s."`