What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

Options

A. `7.8xx10^(-4)"rad/s"`
B. `7.8" rad/s"`
C. `0.8xx10^(-4)"rad/s"`
D. 1 rad/s

Answer

a

Explanation

`g_(e)=g-R omega^(2)`
`(3)/(5)g-g=-Romega^(2)`
`(5g-3g)/(5)=R omega^(2)`
`(2)/(5)g=Romega^(2)`
`omega^(2)=(2g)/(5R)=(2xx9.8)/(5xx6.4xx10^(6))`
`omega^(2)=(196xx10^(-6))/(5xx64)`
`=(14)/(8sqrt5)xx10^(-3)=0.78xx10^(-3)`
`omega=7.8xx10^(-4)"rad/s."`