MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Moment of inertia of solid spheres about a given axis, using the parallel axis theorem when the axis does not pass through the centre.Explanation:The axis AA' passes through the centres of three spheres.For a solid sphere about its own centre, $I_{cm} = \frac{2}{5}mr^2$.So, moment of inertia of these three spheres:$$I_1 = 3 \times \frac{2}{5}mr^2 = \frac{6}{5}mr^2$$The other two spheres touch the axis at their point of contact.Their centres are at a distance $d = r$ from the axis.Using the parallel axis theorem:$$I = I_{cm} + md^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2$$For two such spheres:$$I_2 = 2 \times \frac{7}{5}mr^2 = \frac{14}{5}mr^2$$Total moment of inertia of the system:$$I = I_1 + I_2 = \frac{6}{5}mr^2 + \frac{14}{5}mr^2 = \frac{20}{5}mr^2 = 4mr^2$$Answer:$I = 4mr^2$(Correct option: C)