What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

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

$\therefore {\mathrm{g}}^{\mathrm{\prime }}=\frac{3}{5}\mathrm{g}$ (because mass remains constant)

$\begin{array}{l}{\mathrm{g}}^{\mathrm{\prime }}=\mathrm{g}-{\mathrm{\omega }}^2{\mathrm{Rcos}}^2\mathrm{\lambda }\Rightarrow \frac{3}{5}\mathrm{g}=\mathrm{g}-{\mathrm{\omega }}^2{\mathrm{Rcos}}^2\left(0^{\circ }\right)\\ \Rightarrow {\mathrm{\omega }}^2=\frac{2\mathrm{g}}{5\mathrm{R}}\Rightarrow \mathrm{\omega }=\sqrt{\frac{2\mathrm{g}}{5\mathrm{R}}}=\sqrt{\frac{2\times 10}{5\times 6400\times {10}^3}}\\ =7.8\times {10}^{-4}\frac{\mathrm{rad}}{\sec }\end{array}$