MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Use the parallel axis theorem to find the moment of inertia of both bodies about a tangent.Explanation:For the disc of mass $\frac{M}{2}$ and radius $R$, a tangent in its plane is parallel to a diameter and lies at a distance $R$ from the centre.The moment of inertia of the disc about a diameter is:$$I_{\text{diameter}} = \frac{1}{4} m R^2 = \frac{1}{4} \cdot \frac{M}{2} R^2 = \frac{M R^2}{8}$$By the parallel axis theorem:$$I_{\text{disc tangent}} = I_{\text{diameter}} + m R^2 = \frac{M R^2}{8} + \frac{M}{2} R^2 = \frac{5 M R^2}{8}$$For the solid sphere of mass $M$, the moment of inertia about a diameter is:$$I_{\text{diameter}} = \frac{2}{5} M R^2$$A tangent to the sphere is parallel to a diameter and is at a distance $R$ from the centre, so:$$I_{\text{sphere tangent}} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$$Therefore, the required ratio is:$$\frac{I_{\text{disc tangent}}}{I_{\text{sphere tangent}}} = \frac{\frac{5}{8} M R^2}{\frac{7}{5} M R^2} = \frac{5}{8} \times \frac{5}{7} = \frac{25}{56}$$Thus, the ratio is $25 : 56$.Answer:The ratio is $25 : 56$, which corresponds to option **B**.