Rotational Motion

Concept:The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through its end at angle $\theta$ is $I_{\text{end}} = \frac{1}{3} m l^2 \sin^2 \theta$; through its center is $I_{\text{center}} = \frac{1}{12} m l^2 \sin^2 \theta$; a rod parallel to the axis at distance $d$ contributes $m d^2$.Explanation:For structure (P): Two rods are perpendicular at junction C.The axis OCO' bisects the 90° angle, making $\theta = 45^\circ$ with each rod.Both rods have the axis through their end: $I_1 = I_2 = \frac{1}{3} m l^2 \sin^2 45^\circ = \frac{1}{3} m l^2 \cdot \frac{1}{2} = \frac{1}{6} m l^2$.Total: $I_P = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} m l^2$. Hence P → 5.For structure (Q): Two inclined rods CA and CB each have axis through their end at $\theta = 60^\circ$: $I_{CA} = I_{CB} = \frac{1}{3} m l^2 \sin^2 60^\circ = \frac{1}{3} m l^2 \cdot \frac{3}{4} = \frac{1}{4} m l^2$.Base rod AB is parallel to axis at distance $h = l \sin 60^\circ = \frac{\sqrt{3}}{2} l$.Its inertia: $I_{AB} = m h^2 = m \cdot \frac{3}{4} l^2 = \frac{3}{4} m l^2$.Total: $I_Q = \frac{1}{4} + \frac{1}{4} + \frac{3}{4} = \frac{5}{4} m l^2$. Hence Q → 1.For structure (R): Four rods of a square loop are symmetric about the diagonal axis AC.Each rod has axis through one end at $\theta = 45^\circ$: $I_{\text{each}} = \frac{1}{3} m l^2 \sin^2 45^\circ = \frac{1}{6} m l^2$.Total: $I_R = 4 \times \frac{1}{6} m l^2 = \frac{2}{3} m l^2$. Hence R → 4.For structure (S): Two rods CA and CB meet at vertex C on the axis, each at $\theta = 30^\circ$.$I_{CA} = I_{CB} = \frac{1}{3} m l^2 \sin^2 30^\circ = \frac{1}{3} m l^2 \cdot \frac{1}{4} = \frac{1}{12} m l^2$.Total: $I_S = \frac{1}{12} + \frac{1}{12} = \frac{1}{6} m l^2$. Hence S → 2.Answer:The correct matches are P → 5, Q → 1, R → 4, S → 2.Therefore, option (A) is correct.