List-I shows four planar structures made of uniform solid rods each...

The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through one end and making an angle $\theta$ with the rod is given by $I = \dfrac{m l^2}{3} \sin^2\theta$.

For structure (P):
The axis $OO'$ makes an angle of $45^\circ$ with both rods $CA$ and $CB$.
$I_P = I_{CA} + I_{CB} = \dfrac{m l^2}{3} \sin^2(45^\circ) + \dfrac{m l^2}{3} \sin^2(45^\circ) = 2 \times \dfrac{m l^2}{3} \times \dfrac{1}{2} = \dfrac{m l^2}{3}$.
This matches with (5).

For structure (Q):
The axis $OO'$ makes an angle of $60^\circ$ with rods $CA$ and $CB$. Rod $AB$ is parallel to the axis at a distance $d = l \sin(60^\circ) = \dfrac{\sqrt{3}}{2} l$.
$I_Q = I_{CA} + I_{CB} + I_{AB} = 2 \times \left( \dfrac{m l^2}{3} \sin^2(60^\circ) \right) + m d^2$
$I_Q = 2 \times \dfrac{m l^2}{3} \times \dfrac{3}{4} + m \left( \dfrac{\sqrt{3}}{2} l \right)^2 = \dfrac{m l^2}{2} + \dfrac{3 m l^2}{4} = \dfrac{5 m l^2}{4}$.
This matches with (1).

For structure (R):
The axis $OO'$ is the diagonal of the square, making an angle of $45^\circ$ with all four rods.
$I_R = 4 \times \left( \dfrac{m l^2}{3} \sin^2(45^\circ) \right) = 4 \times \dfrac{m l^2}{3} \times \dfrac{1}{2} = \dfrac{2 m l^2}{3}$.
This matches with (4).

For structure (S):
The axis $OO'$ makes an angle of $30^\circ$ with both rods $CA$ and $CB$.
$I_S = 2 \times \left( \dfrac{m l^2}{3} \sin^2(30^\circ) \right) = 2 \times \dfrac{m l^2}{3} \times \dfrac{1}{4} = \dfrac{m l^2}{6}$.
This matches with (2).

Therefore, the correct matching is P$\rightarrow$ 5, Q$\rightarrow$ 1, R$\rightarrow$ 4, S$\rightarrow$ 2.

Answer: P$\rightarrow$ 5, Q$\rightarrow$ 1, R$\rightarrow$ 4, S$\rightarrow$ 2