The radius of earth is 6400 km and g = 10 m/s². In order that a body of 5 kg weight is zero at the equator, the angular speed is

$\frac{1}{800} rad / \sec$

$\frac{1}{1600} rad / \sec$

$\frac{1}{400} rad / \sec$

$\frac{1}{80} rad / \sec$

Solution

$\frac{1}{800} rad / \sec$

Given:- radius of earth=6400km=6400 $\times$ 10³, g=10m/s²

g^'=0, m=0(weight is Zero at equator)

g=R $ω$ ² $\Rightarrow$ $ω$ ²= $\frac{g}{R}$

Angular speed $ω$ = $\sqrt{\frac{g}{R}}$ = $\sqrt{\frac{10}{6400 \times 10^{3}}}$ = $\frac{1}{800} rad / \sec$

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