Calculate angular velocity of the earth so that acceleration due to gravity at `60^(@)` latitude becomes zero (radius of the earth = 6400 km, gravitational acceleration at poles = `10 m//s^(2) , cos60^(@) = 0.5`)

Options

A. `7.8xx10^(-2)` rad/s
B. `0.5xx10^(-3)` rad/s
C. `1xx10^(-3)` rad/s
D. `2.5xx10^(-3)` rad/s

Answer

D

Explanation

`g'=g-omega^(2)R cos^(2)lambda` and `lambda = 60^(@)`
But `g'=0 " " therefore g=omega^(2)R cos^(2)60^(@)=omega^(2)R xx (1)/(4)`
`4g=omega^(2)R " " therefore omega = 2sqrt((g)/(R ))`
`therefore omega = 2sqrt((10)/(64xx10^(5)))=(2)/(8)xx10^(-2)`
`=0.25xx10^(-2)=2.5xx10^(-3)` rad/s