Q. For a body lying on the equator to appear weightless, what should be the angular speed of the earth?(Take $g = 10ms^{- 2};$ radius of earth = 6400 km)
A) $0.125\ rads^{- 1}$
B) $1.25\ rads^{- 1}$
C) $1.25 \times 10^{- 3}rads^{- 1}$
D) $1.25 \times 10^{- 2}rads^{- 1}$
Explanation
(c) At equator, $g’ = g - R\omega^{2} = 0$ or $\omega = \sqrt{g/R}$
or $\omega = \sqrt{10/(6.4 \times 10^{6})} = 1.25 \times 10^{- 3}\ rads^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.