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
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$Weight of the body at equator = \frac{3}{5} of initial weight \therefore g' = \frac{3}{5} g \text{ (because mass remains constant)} g' = g - \omega^2 R \cos^2 \lambda \Rightarrow \frac{3}{5} g = g - \omega^2 R \cos^2 (0^\circ) \Rightarrow \omega^2 = \frac{2g}{5R} \Rightarrow \omega = \sqrt{\frac{2g}{5R}} = \sqrt{\frac{2 \times 10}{5 \times 6400 \times 10^3}} = 7.8 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{sec}}$
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