If R is the radius of earth, $\omega$ is its angular velocjty and g_p is the value of g at the poles. The effective value of g at a latitude $\lambda$ = 60⁰ is    [JIPMER 2018]

The acceleration due to gravity at a latitude $\lambda$ is

$g^{\text{'}}=g-R{\omega }^2{\cos }^2\lambda$

At the poles,   $\lambda =90°$

  $\therefore g=g=g_p$              $\left(\because \cos 90°=0\right)$

For   $\lambda =60°, g^{\text{'}}=g_p-R{\omega }^2{\cos }^2\lambda$

or     $g^{\text{'}}=g_p-R{\omega }^2{\cos }^{2}60°$

Hence,   $g^{\text{'}}=g_p-R{\omega }^2{\left(\frac{1}{2}\right)}^2=g_p-\frac{1}{4}R{\omega }^2$