(c) Weight of the body at equator = $\frac{3}{5}$ of initial weight  

$g' = \frac{3}{5}g$ (because mass remains constant)

$g' = g – {\omega ^2}R{\cos ^2}\lambda $  $⇒$  $\frac{3}{5}g = g – {\omega ^2}R{\cos ^2}(0^\circ )$

$⇒$ ${\omega ^2} = \frac{{2g}}{{5R}}$  $⇒$  $\omega = \sqrt {\frac{{2g}}{{5R}}} = \sqrt {\frac{{2 \times 10}}{{5 \times 6400 \times {{10}^3}}}} $ = $7.8 \times {10^{ – 4}}\frac{{rad}}{{\sec }}$