Let `omega` be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed by a spring balance gives the same reading at the equator as at a height `h` above the poles `(h lt lt R)`. The value of `h` is

Options

A. `omega^(2)R^(2)//g`
B. `omega^(2)R^(2)//2g`
C. `2omega^(2)R^(2)//g`
D. `sqrt(Rg)//omega`

Answer

B

Explanation

At equator, apparent weight `W_(1)=mg-momega^(2)R`
At poles, apparent weight `W_(2)=mg(1-(2h)/R)`
`W_(1)=W_(2)`
`mg-momega^(2)R=mg(1-(2h)/R)`
`-momega^(2)R=(-2mgh)/R`
`h=(omega^(2)R^(2))/g`