MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Moment of inertia of a body about a tangent is found using the parallel axis theorem: $I_{\text{tangent}} = I_{\text{centre}} + MR^2$.Explanation:For the solid sphere ($M = 5\,\text{kg}$), moment of inertia about its diameter is:$$I_{\text{diameter}} = \frac{2}{5} MR^2$$A tangent to the sphere is parallel to a diameter, at distance $R$ from the centre.Using the parallel axis theorem:$$I_{\text{sphere}} = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2$$Substituting $M = 5\,\text{kg}$:$$I_{\text{sphere}} = \frac{7}{5} \times 5R^2 = 7R^2$$For the disc ($M = 4\,\text{kg}$), moment of inertia about a diameter in its plane is:$$I_{\text{diameter}} = \frac{1}{4} MR^2$$A tangent in the plane of the disc is parallel to a diameter, at distance $R$ from the centre.Using the parallel axis theorem:$$I_{\text{disc}} = \frac{1}{4} MR^2 + MR^2 = \frac{5}{4} MR^2$$Substituting $M = 4\,\text{kg}$:$$I_{\text{disc}} = \frac{5}{4} \times 4R^2 = 5R^2$$Therefore, the required ratio is:$$I_{\text{sphere}} : I_{\text{disc}} = 7R^2 : 5R^2 = 7 : 5$$Hence, $x = 7$ and $y = 5$.Answer:$x = 7$, $y = 5$Correct option: C (7, 5).