The angular speed of rotation of earth about its axis at which the weight of man standing on the equator becomes half of his weight at the poles is given by

Options

A. `0.034 rad s^(-1)`
B. `8.75 xx 10^(-4) rad s^(-1)`
C. `1.23 xx 10^(-2) rad s^(-1)`
D. `7.65 xx 10^(-7) rad s^(-1)`

Answer

B

Explanation

`g_("equator") = g - R omega^(2)` and `g_("pole") - R omega^(2)`
`:. (g_("pole"))/(2)` or `(g)/(2) = g - R omega^(2)`
or `omega sqrt((g)/(2R))`
`= sqrt((9.8)/(2 xx 6.4 xx 10^(6))`
`= 8.75 xx 10^(-4) rad//s`