If the earth has no rotational motion, the weight of a person on the equation is W. Detrmine the speed with which the earth would have to rotate about its axis so that the person at the equator will weight `(3)/(4)` W. Radius of the earth is 6400 km and g = 10 `m//s^(2)`.

Options

A. `1.1xx 10^(-3) ` rad/s
B. `0.83 xx 10^(-3)` rad/s
C. `0.63 xx 10^(-3)` rad/s
D. `0.28 xx 10^(-3)` rad/s

Answer

C

Explanation

We know , g. = g - `omega^(2) R cos^(2) theta`
`(3g)/(4) = g - omega^(2)` R
Given, g. = `(3)/(4) ` g `" " omega^(2) R = (g)/(4)`
`omega = sqrt((g)/(4R)) = sqrt((1)/(4 xx 6400 xx 10^(3))) = (1)/(2 xx 8 xx 100)`
`= 0.6 xx 10^(-3) ` rad/s