MHTCET 14 May 2023 Shift 2 Solved Paper

The acceleration due to gravity on the surface of the Earth varies with latitude due to the centrifugal force resulting from the Earth's rotation. The effective acceleration due to gravity at a latitude $\phi$, denoted as $g_\phi$, takes into account the centrifugal force and is less than the acceleration due to gravity that would be experienced if the Earth were not rotating (denoted as $g_0$ ). At the equator ( $\phi = 0^{\circ}$ ), this effect is maximal because the velocity due to Earth's rotation is maximal. As we go to the poles, the effect of rotation becomes minimal.The effective acceleration due to gravity at latitude $\phi$ is given by:$g_\phi = g_0 - \omega^2 R \cos^2 \phi$ Where:$g_0$ is the acceleration due to gravity without the Earth's rotation, at the equator $\omega$ is the angular velocity of the Earth's rotation$R$ is the radius of the Earth $\phi$ is the latitudeAt the equator, the effective acceleration due to gravity, $g$, is:$g = g_0 - \omega^2 R$because $\cos 0^{\circ} = 1$.At latitude $30^{\circ}$, the effective acceleration due to gravity, $g_\phi$, using $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, is: $g_\phi = g_0 - \omega^2 R \left(\frac{\sqrt{3}}{2}\right)^2$$g_\phi = g_0 - \omega^2 R \times \frac{3}{4}$To find $\left|g - g_\phi\right|$, we subtract the above two equations:$\left|g - g_\phi\right| = \left| (g_0 - \omega^2 R) - \left(g_0 - \omega^2 R \times \frac{3}{4}\right) \right|$$\left|g - g_\phi\right| = \omega^2 R - \omega^2 R \times \frac{3}{4}$$\left|g - g_\phi\right| = \omega^2 R \left(1 - \frac{3}{4}\right)$$\left|g - g_\phi\right| = \omega^2 R \times \frac{1}{4}$So the value of $\left|g - g_\phi\right|$ is given by the option $A$, which is:$\frac{1}{4} \omega^2 R$