Moment of Inertia & Radius of Gyration question from JEE Main 2026 (23 Jan, Shift 1)

Answer: (d)

1. Each cylinder has mass $m = \frac{M}{4}$.
2. Two sides cut by the axis at their centres (axis $\perp$ to their length): $m\left(\frac{R^2}{4} + \frac{L^2}{12}\right)$ each.
3. Two sides parallel to the axis at distance $\frac{L}{2}$: $\frac{mR^2}{2} + \frac{mL^2}{4}$ each.
4. $I = 2m\left(\frac{3R^2}{4} + \frac{L^2}{3}\right) = \frac{3}{8}MR^2 + \frac{1}{6}ML^2$
5. So the answer is (d).