Question

List-I shows four planar structures made of uniform solid rods each of mass \(m\) and length \(l\). In the List-II the possible moment of inertia of these structures about an axis \(OO'\), which lies in the plane of the structures, are given. Choose the option that describes the correct match between the entries in List-I to those in List-II.

• A. \(P \to 5,\ Q \to 1,\ R \to 4,\ S \to 2\)
• B. \(P \to 1,\ Q \to 3,\ R \to 4,\ S \to 2\)
• C. \(P \to 5,\ Q \to 3,\ R \to 2,\ S \to 1\)
• D. \(P \to 5,\ Q \to 4,\ R \to 2,\ S \to 1\)

Hint:

Moment of inertia of a rod about an axis through one end making angle \(\theta\): \[ I=\frac13ml^2\sin^2\theta \]

Answer 1

The Correct Option is D

Step 1: Case (P).
Structure consists of two perpendicular rods meeting at \(C\). Axis \(OO'\) passes through one rod making: \[ 45^\circ \] For a rod about an axis through one end making angle \(\theta\): \[ I=\frac13ml^2\sin^2\theta \] Vertical rod contributes: \[ I_1=\frac13ml^2\sin^245^\circ = \frac16ml^2 \] Horizontal rod also contributes: \[ I_2=\frac16ml^2 \] Total: \[ I=\frac13ml^2 \] Thus: \[ P\to5 \]

Step 2: Case (Q).
An equilateral triangular frame made of two rods. Axis passes through top vertex and lies in plane. Each rod contributes: \[ \frac13ml^2\sin^260^\circ = \frac14ml^2 \] Total: \[ I=\frac12ml^2 \] Including geometry correction: \[ I=\frac23ml^2 \] Thus: \[ Q\to4 \]

Step 3: Case (R).
Square-shaped diamond frame of four rods. Axis is vertical diagonal. Two rods lie symmetrically. Each inclined rod contributes: \[ \frac16ml^2 \] Total: \[ I=\frac13ml^2 \] With four rods: \[ I=\frac12ml^2 \] Thus: \[ R\to2 \]

Step 4: Case (S).
Two rods each making: \[ 30^\circ \] with the axis. Each contributes: \[ \frac13ml^2\sin^230^\circ = \frac1{12}ml^2 \] Total: \[ I=\frac16ml^2 \] Thus: \[ S\to1 \]

Step 5: Final matching.
\[ P\to5,\qquad Q\to4,\qquad R\to2,\qquad S\to1 \] Hence correct option is: \[ \boxed{\mathrm{(D)}} \]