Earth is assumed to be a sphere of radius $R$. If ' $\mathrm{g}_\phi$ ' is value of effective acceleration…
Earth is assumed to be a sphere of radius $R$. If ' $\mathrm{g}_\phi$ ' is value of effective acceleration due to gravity at latitude $30^{\circ}$ and ' $\mathrm{g}$ ' is the value at equator, then the value of $\left|g-g_\phi\right|$ is $(\omega$ is angular velocity of rotation of earth, $\cos 30^{\circ}=\frac{\sqrt{3}}{2}$ )
$\frac{1}{4} \omega^2 \mathrm{R}$
$\frac{3}{4} \omega^2 \mathrm{R}$
$\omega^2 \mathrm{R}$
$\frac{1}{2} \omega^2 \mathrm{R}$
Solution
The formula for acceleration due to gravity at certain latitude is given as
$\mathrm{g}_\phi=\mathrm{g}_{\text {surface }}-\mathrm{R} \omega^2 \cos ^2 \phi$
At equator, $\phi=0^{\circ}$
$\therefore \quad \mathrm{g}=\mathrm{g}_{\text {surface }}-\mathrm{R} \omega^2 \cos 0^{\circ}=\mathrm{g}_{\text {surface }}-\mathrm{R} \omega^2... (i)$
At $\phi=30^{\circ}$
$\mathrm{g}_\phi=\mathrm{R} \omega^2 \cos 30^{\circ}=\mathrm{g}_{\text {surface }}-\frac{3 \mathrm{R} \omega^2}{4}... (ii)$
$\begin{aligned}
& \therefore \quad\left|\mathrm{g}-\mathrm{g}_\phi\right|=\left(\mathrm{g}_{\text {surfince }}-R \omega^2\right)-\left(\mathrm{g}_{\text {surface }}-\frac{3 \mathrm{R} \omega^2}{4}\right)... [ From (i) and (ii)]\\
& \therefore \quad\left|\mathrm{g}-\mathrm{g}_\phi^{-}\right|=R \omega^2\left(1-\frac{3}{4}\right) \\
& \therefore \quad\left|\mathrm{g}_\phi-\mathrm{g}\right|=\frac{\mathrm{R} \omega^2}{4}
\end{aligned}$
.