Question

$R$ is the radius of the Earth and $\omega$ is its angular velocity and $g_{p}$ is the value of $g$ at 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. $g_{p}+\frac{1}{4} R \omega^{2}$

Answer 1

Answer: (A)

$g=g_{p}-\omega^{2} R\, \cos ^{2}\, \lambda$
Here $\lambda=60^{\circ}$
$\Rightarrow g=g_{p}-\omega^{2} R \cos ^{2} 60^{\circ}=g_{p}-\frac{\omega^{2} R}{4}$