Q. The speed of earth’s rotation about its axis is $\omega$. Its speed is increased to $x$ times to make the effective acceleration due to gravity equal to zero at the equator, then $x$ is around $(g = 10ms^{- 2},R = 6400\ km)$

A) 1
B) 8.5
C) 17
D) 34
Explanation

(c) At equator, $g^{'} = g - R\omega^{2}.$ When angular velocity be
$\omega^{'}( = x\omega),$ then, $0 = g - R\omega^{'2}$ or $\omega^{'} = \sqrt{g/R} = x\omega$
or $x = (\sqrt{g/R})/\ \omega$
or $x = \frac{\sqrt{10/(6.4 \times 10^{6})}}{2\pi} \times 24 \times 60 \times 60 = 17$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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