Q. Assuming earth to be a sphere of radius R, if $g_{30{^\circ}}$ is value of acceleration due to gravity at latitude of $30{^\circ}$ and $g$ at the equator, the value of $g - g_{30{^\circ}}$is
A) $\frac{1}{4}\omega^{2}R$
B) $\frac{3}{4}\omega^{2}R$
C) $\omega^{2}R$
D) $\frac{1}{2}\omega^{2}R$
Explanation
(b) The value of acceleration due to gravity at latitude λ is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g_{\lambda} = g - R\omega^{2}\cos^{2}\lambda$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ g - g_{\lambda} = R\omega^{2}\cos^{2}\lambda$
$\ \ \ \ \ \ \ \ At\ \lambda = 30{^\circ},$
$\ \ \ \ \ \ \ \ \ \ \ \ \ g - g_{30{^\circ}} = \ R\omega^{2}\cos^{2}30{^\circ}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = R\omega^{2}\left( \frac{\sqrt{3}}{2} \right)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3}{4}R\omega^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.