Physics - Gravitation Question with Solution | TestHub

The formula for acceleration due to gravity at certain latitude is given as ${\mathrm{g}}_{\varphi }={\mathrm{g}}_{\text{surface }}-\mathrm{R}{\omega }^2{\cos }^2\varphi$ At equator, $\varphi =0^{\circ }$ $\therefore \mathrm{g}={\mathrm{g}}_{\text{surface }}-\mathrm{R}{\omega }^2\cos 0^{\circ }={\mathrm{g}}_{\text{surface }}-\mathrm{R}{\omega }^2...(i)$ At $\varphi =30^{\circ }$ ${\mathrm{g}}_{\varphi }=\mathrm{R}{\omega }^2\cos 30^{\circ }={\mathrm{g}}_{\text{surface }}-\frac{3\mathrm{R}{\omega }^2}{4}...(ii)$ $\begin{array}{rl}& \therefore \left|\mathrm{g}-{\mathrm{g}}_{\varphi }\right|=\left({\mathrm{g}}_{\text{surfince }}-R{\omega }^2\right)-\left({\mathrm{g}}_{\text{surface }}-\frac{3\mathrm{R}{\omega }^2}{4}\right)...[From(i)and(ii)]\\ & \therefore \left|\mathrm{g}-{\mathrm{g}}_{\varphi }^{-}\right|=R{\omega }^2\left(1-\frac{3}{4}\right)\\ & \therefore \left|{\mathrm{g}}_{\varphi }-\mathrm{g}\right|=\frac{\mathrm{R}{\omega }^2}{4}\end{array}$