Question

R is the radius of the earth and to is its angular velocity and gp is the value of g at the poles. The effective value of g at the latitude A = ${60}^{\circ }$ 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

The correct option is A $g_p-\frac{1}{4}R{\omega }^2$

$\begin{array}{cc}{g}_{\text{eff}}& =g_p-R{\omega }^2{\cos }^2\varphi \\ & =g-R{\omega }^2\times \frac{1}{4}\\ & =g-\frac{R{\omega }^2}{4}\\ \text{Hence option}A\text{is correct}& \end{array}$