The moment of inertia of single rod about the axis passing through its center and making angle $\theta$ along the length of rod is :

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English Explanation 1

$$ \mathrm{I}_{\text {center }}=\frac{1}{12} \mathrm{ml}^2 \sin ^2 \theta $$

If the axis instead passes through one end of the rod at an angle $\theta$,

$$ \mathrm{I}_{\mathrm{end}}=\frac{1}{3} \mathrm{ml}^2 \sin ^2 \theta $$

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English Explanation 2

Structure (P) consists of two uniform rods, each of mass m and length l, joined at a right angle ($90^\circ$) at point C.

The axis OCO' passes through the junction C and bisects the angle, making an angle $\theta = 45^\circ$ with both rods.

Since the axis passes through the end of both rods,

$$I_1 = \frac{1}{3}ml^2\sin^2(45^\circ) = \frac{1}{3}ml^2\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6}ml^2$$

$$I_2 = \frac{1}{3}ml^2\sin^2(45^\circ) = \frac{1}{3}ml^2\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6}ml^2$$

The total moment of inertia of structure (P) is the sum of the two parts:

$$I_P = I_1 + I_2 = \frac{1}{6}ml^2 + \frac{1}{6}ml^2 = \frac{1}{3}ml^2$$

$$P \rightarrow 5$$

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English Explanation 3

Rods CA and CB have one end attached at vertex C, which lies directly on the axis of rotation. Each rod is inclined at an angle $\theta = 60^\circ$ to the axis.

$$I_{CA} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m l^2 \left(\frac{3}{4}\right) = \frac{1}{4} m l^2$$

$$I_{CB} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m l^2 \left(\frac{3}{4}\right) = \frac{1}{4} m l^2$$

Base rod AB is horizontal and completely parallel to the axis OCO'.

This distance is equal to the altitude (height h) of the equilateral triangle.

$$h = l \sin(60^\circ) = \frac{\sqrt{3}}{2} l$$

Since the entire mass m of rod AB is distributed uniformly at a constant distance h from the axis, its moment of inertia is:

$$I_{AB} = m h^2 = m \left(\frac{\sqrt{3}}{2} l\right)^2 = \frac{3}{4} m l^2$$

So, the total moment of inertia for structure Q

$$I_Q = I_{CA} + I_{CB} + I_{AB}$$

$$I_Q = \frac{1}{4} m l^2 + \frac{1}{4} m l^2 + \frac{3}{4} m l^2 = \frac{5}{4} m l^2$$

$Q \rightarrow 1$

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English Explanation 4

The rods $A B$ and $B C$ meet at vertex $B$.

The axis passing through A and C , it forms a right-angled triangle ABC where the diagonal AC is the axis. The angle made by axis with every rod is $45^{\circ}$. As all four sides are symmetric about axis of rotation therefore, the total moment of inertia of the square loop about its diagonal axis is:

$$ \mathrm{I}=\mathrm{I}_{\mathrm{AB}}+\mathrm{I}_{\mathrm{BC}}+\mathrm{I}_{\mathrm{CD}}+\mathrm{I}_{\mathrm{DA}}=4 \times \frac{1}{3} \mathrm{ml}^2 \sin ^2 45^{\circ}=4 \times \frac{1}{3} \mathrm{ml}^2 \times \frac{1}{2}=\frac{2}{3} \mathrm{ml}^2 $$

R→ 4

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English Explanation 5

Rods CA and CB have one end attached at vertex C, which lies directly on the axis of rotation. Each rod is inclined at an angle $\theta = 30^{\circ}$ to the axis.

$$I_{CA} = \frac{1}{3}ml^2 \sin^2(30^{\circ}) = \frac{1}{3}ml^2\left(\frac{1}{2}\right)^2 = \frac{1}{3}ml^2\left(\frac{1}{4}\right) = \frac{1}{12}ml^2$$

$$I_{CB} = \frac{1}{3}ml^2 \sin^2(30^{\circ}) = \frac{1}{3}ml^2\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3}ml^2\left(\frac{1}{4}\right) = \frac{1}{12}ml^2$$

So, the total moment of inertia for structure S

$$I_S = I_{CA} + I_{CB}$$

$$I_S = \frac{1}{12}ml^2 + \frac{1}{12}ml^2 = \frac{2}{12}ml^2 = \frac{1}{6}ml^2$$

$S \rightarrow 2$

So, the correct matches are $\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 2$.

Therefore, the correct option is (A).

List-I	List-II
(P)	(1) $$\frac{5}{4}ml^2$$
(Q)	(2) $$\frac{1}{6}ml^2$$
(R)	(3) $$\frac{1}{12}ml^2$$
(S)	(4) $$\frac{2}{3}ml^2$$
	(5) $$\frac{1}{3}ml^2$$

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