Q. $R$ is the radius of the earth and $\omega$ is its angular velocity and $g_{p}$ is the value of $g$ at the poles. The effective value of $g$ at the latitude $\lambda = 60{^\circ}$ will be equal to
A) $g_{p} - \frac{1}{4}R\omega^{2}$
B) $g_{p} - \frac{3}{4}R\omega^{2}$
C) $g_{p} - R\omega^{2}$
D) $g_{p} + \frac{1}{4}R\omega^{2}$
Explanation
(a) $g = g_{p} - R\omega^{2}\cos^{2}{\lambda = g_{p} - \omega^{2}R\cos^{2}{60{^\circ} = g_{p} - \frac{1}{4}R\omega^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.