MHTCET 11 May 2024 Shift 2 Paper

Concept:For a disc, the moment of inertia about its centre is $I = \frac{MR^2}{2}$. Use the parallel axis theorem to find moments about parallel axes.Explanation:Given, moment of inertia about the central axis perpendicular to the plane is $I = \frac{MR^2}{2}$.For an axis tangential to the rim, the distance from the centre is $h = R$.By the parallel axis theorem, $I_1 = I_{cm} + Mh^2 = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2$.For an axis through a point midway between the centre and the rim, $h = \frac{R}{2}$.Thus, $I_2 = \frac{MR^2}{2} + M\left(\frac{R}{2}\right)^2 = \frac{MR^2}{2} + \frac{MR^2}{4} = \frac{3}{4} MR^2$.Therefore, the ratio is $\frac{I_1}{I_2} = \frac{\frac{3}{2} MR^2}{\frac{3}{4} MR^2} = \frac{2}{1}$.Hence, $I_1 : I_2 = 2 : 1$.Answer:The required ratio is $2 : 1$, which corresponds to option A.