Moment of Inertia of Planar Structures About OCO Axis

To find the moment of inertia of each planar structure about the given axis $OCO'$, we use the fundamental results for a thin uniform rod of mass $m$ and length $l$:

1. Moment of inertia about an axis passing through one end at an angle $\theta$ to the rod: $I = \frac{1}{3} m l^2 \sin^2\theta$

2. Moment of inertia about an axis parallel to the rod at a perpendicular distance $h$: $I = m h^2$

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Step-by-Step Analysis of Each Structure:

(P) Two perpendicular rods ($AC \perp CB$) connected at $C$:

• The axis $OCO'$ passes through the vertex $C$ and makes an angle of $45^\circ$ with rod $AC$.
• Since $AC \perp CB$, the angle between axis $OCO'$ and rod $CB$ is also $90^\circ - 45^\circ = 45^\circ$.
• Both rods have one end on the axis $OCO'$.

The moment of inertia of each rod is: $I_{AC} = \frac{1}{3} m l^2 \sin^2(45^\circ) = \frac{1}{3} m l^2 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6} m l^2$ $I_{CB} = \frac{1}{3} m l^2 \sin^2(45^\circ) = \frac{1}{6} m l^2$

Total moment of inertia for structure (P): $I_P = I_{AC} + I_{CB} = \frac{1}{6} m l^2 + \frac{1}{6} m l^2 = \frac{1}{3} m l^2$

$\implies \mathbf{P \rightarrow 5}$

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(Q) Equilateral triangle $ABC$:

• The axis $OCO'$ passes through vertex $C$ and is parallel to side $AB$.
• The angle made by the axis $OCO'$ with sides $AC$ and $BC$ is $60^\circ$.
• For sides $AC$ and $BC$: $I_{AC} = I_{BC} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} m l^2$
• Side $AB$ is parallel to $OCO'$ at a perpendicular distance equal to the height of the triangle: $h = l \sin(60^\circ) = \frac{\sqrt{3}}{2} l$ $I_{AB} = m h^2 = m \left(\frac{\sqrt{3}}{2} l\right)^2 = \frac{3}{4} m l^2$

Total moment of inertia for structure (Q): $I_Q = I_{AC} + I_{BC} + I_{AB} = \frac{1}{4} m l^2 + \frac{1}{4} m l^2 + \frac{3}{4} m l^2 = \frac{5}{4} m l^2$

$\implies \mathbf{Q \rightarrow 1}$

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(R) Square frame made of four rods:

• The axis $OCO'$ lies along the diagonal $AC$ of the square.
• The diagonal makes an angle of $45^\circ$ with each of the four rods ($AB, BC, CD, DA$).
• The axis passes through one vertex (end point) of each rod.

For each of the 4 rods: $I_{\text{rod}} = \frac{1}{3} m l^2 \sin^2(45^\circ) = \frac{1}{6} m l^2$

Total moment of inertia for structure (R): $I_R = 4 \times \left(\frac{1}{6} m l^2\right) = \frac{2}{3} m l^2$

$\implies \mathbf{R \rightarrow 4}$

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(S) V-shaped structure of two rods $CA$ and $CB$:

• The axis $OCO'$ passes through vertex $C$ and bisects the angle between the two rods, making an angle of $30^\circ$ with each rod.

For each rod: $I_{CA} = I_{CB} = \frac{1}{3} m l^2 \sin^2(30^\circ) = \frac{1}{3} m l^2 \left(\frac{1}{2}\right)^2 = \frac{1}{12} m l^2$

Total moment of inertia for structure (S): $I_S = I_{CA} + I_{CB} = \frac{1}{12} m l^2 + \frac{1}{12} m l^2 = \frac{1}{6} m l^2$

$\implies \mathbf{S \rightarrow 2}$

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Conclusion:

The correct matching is: $\text{P} \rightarrow 5, \quad \text{Q} \rightarrow 1, \quad \text{R} \rightarrow 4, \quad \text{S} \rightarrow 2$

This corresponds to Option (A).