`R` is the radius of the earth and `omega` is its angular velocity and `g_(p)` is the value of `g` at the poles. The effective value of `g` at the latitude `lambda=60^(@)` will be equal to

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

B

Explanation

The effective acceleration due to gravity at a certain latitude `(lambda)` is given by
`g'=g_(P)-omega^(2)R cos^(2)lambda " "` …….(1)
In this problem, `g'=g_(P)-(3)/(4)omega^(2)R " "` ……(2)
Comparing (1) and (2), we get
`cos^(2)lambda = (3)/(4) " " therefore cos lambda =(sqrt(3))/(2) " " therefore lambda = 30^(@)`