Q. What should be the angular speed of earth in $rad^{- 1}$ so that a body 5kg weighs zero at the equator? (Take g = $10\ ms^{- 2}$ and radius of earth = 6400 km)
A) 1/1600
B) 1/800
C) 1/400
D) 1/80
Explanation
(b) Hence, $g^{'} = g - R\omega^{2} = 0;$
$\omega = \sqrt{g/R} = \sqrt{10/(6400 \times 10^{3})} = 1/800$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.