Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take `g = 10 m//s^(2)` for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and `phi = 60^(@)`)

Options

A. `1.25 xx 10^(-3) "rad s"^(-1)`
B. `2.50 xx 10^(-3) " rad s"^(-1)`
C. `3.75 xx 10^(-3) " rad s"^(-1)`
D. `5 xx 10^(-3) " rad s"^(-1)`

Answer

B

Explanation

Acceleration due to gravity
`g'=g-omega^(2)R cos^(2)lamdarArr 0=g-omega^(2)cos^(2)60^(@)`
`0=g-(omega^(2)R)/(4)rArr omega=2sqrt((g)/(R))=2sqrt((10)/(6400xx1000))`
Angular velocity, `omega'=(1)/(400)=2.5xx10^(-3)"rad s"^(-1)`.