Q. What would 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 $6400\ km$
A) $7.4 \times 10^{- 4}rad/sec$
B) $6.7 \times 10^{- 4}rad/sec$
C) $7.8 \times 10^{- 4}rad/sec$
D) $8.7 \times 10^{- 4}rad/sec$
Explanation
(c) Weight of the body at equator $= \frac{3}{5}$ of initial weight
$\therefore g^{'} = \frac{3}{5}g$ (because mass remains constant)
$g^{'} = g - \omega^{2}R\cos^{2}{\lambda \Rightarrow \frac{3}{5}g = g - \omega^{2}R\cos^{2}{(0{^\circ})}}$
$\Rightarrow \omega^{2} = \frac{2g}{5R} \Rightarrow \omega = \sqrt{\frac{2g}{5R}} = \sqrt{\frac{2 \times 10}{5 \times 6400 \times 10^{3}}}$
$= 7.8 \times 10^{- 4}\ \frac{rad}{\sec}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.