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 at the equator gives the same reading as a reading taken at a depth `d` below earth's surface at a pole `(dlt ltR)`. the value of `d` is-

Options

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

Answer

A

Explanation

`g_(ef f)` due to rotation `= g_(ef f)` due to depth d
`g-R omega^(2)=g(1-(d)/(R ))rArr R omega^(2)=(gd)/(R )rArr d=(R^(2)omega^(2))/(g)`