MHTCET 5 May 2025 Shift 2 Paper

Concept:Use the parallel axis theorem to shift the moment of inertia from the centre of the rod to the required axis.Explanation:For a thin uniform rod of mass $M$ and length $L$, the moment of inertia about its central perpendicular axis is:$$I_0 = \frac{ML^2}{12}$$The required axis is at a distance $\frac{L}{4}$ from one end.The centre of the rod is at distance $\frac{L}{2}$ from that same end.So the distance between the centre and the given axis is:$$d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$$Using the parallel axis theorem:$$I = I_0 + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2$$$$I = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{7ML^2}{48}$$This matches option D.Answer:$$\frac{7ML^2}{48}$$ (Option D)