PRAYOGX
ADV-2026-P1-PHY-Q16

Spin-Up Lab — which moment of inertia does each rod frame have about OCO′?

JEE Advanced 2026 · Paper 1 · Physics · Q.16 · matching list, single option (+4 / 0 / −1) · System of Particles and Rotational Motion

01The question

Q.16 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 OCO′, 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.

List-I

(P)CABOO′45°90°
(Q)CABOO′60°60°
(R)CBADOO′
(S)CABOO′30°30°

The figures are redrawn from the paper.

List-II

  1. (1)  5⁄4 ml²
  2. (2)  1⁄6 ml²
  3. (3)  1⁄12 ml²
  4. (4)  2⁄3 ml²
  5. (5)  1⁄3 ml²
  1. (A)P→5, Q→1, R→4, S→2
  2. (B)P→1, Q→3, R→4, S→2
  3. (C)P→5, Q→3, R→2, S→1
  4. (D)P→5, Q→4, R→2, S→1

02The experiment

Target:the match P, Q, R, S → List-II

Ready.

Frame
30°
Speed

03Detailed solution

Answer:  P→5, Q→1, R→4, S→2

  1. The idea. A uniform rod of mass m and length l with one end on the axis, making an angle θ with it: a piece at distance s along the rod is r⊥ = s sin θ from the axis, so I = ∫0l (m/l)(s sin θ)² ds = (ml²/3) sin² θ. A rod parallel to the axis at distance d has every piece at d: I = md². The moments of the rods of a frame add.
  2. (P) Two perpendicular rods, the axis at 45° to each → (5). I = 2 × (ml²/3)(sin² 45°) = 2 × (ml²/3)(½) = ml²/3.
  3. (Q) Equilateral triangle, the axis through C parallel to AB → (1). CA and CB are at 60°: 2 × (ml²/3)(¾) = ml²/2. AB is parallel to the axis at d = l sin 60° = (√3/2)l: m(3l²/4). Total ml²/2 + 3ml²/4 = 5ml²/4.
  4. (R) Square, the axis along the diagonal CA → (4). All four sides make 45° with the diagonal and each has one end on it: 4 × (ml²/3)(½) = 2ml²/3.
  5. (S) Two rods at 30° to the axis → (2). I = 2 × (ml²/3)(sin² 30°) = 2 × (ml²/3)(¼) = ml²/6.
  6. That is P→5, Q→1, R→4, S→2: option (A). Value (3), ml²/12, belongs to a single rod about a perpendicular axis through its centre; no frame here has it.

Understand the concept

The moment of inertia measures how much mass sits far from the axis: each piece counts with the square of its perpendicular distance, r⊥². Tilt a rod towards the axis and its far end comes closer, so its share falls as sin² θ; a rod held parallel to the axis keeps all its mass at one distance. With the same torque, the frame with the larger I spins up more slowly: α = τ/I.

Key takeaways

  • Rod through the axis at angle θ (one end on it): I = (ml²/3) sin² θ; sin 90° gives the familiar ml²/3.
  • Rod parallel to the axis at distance d: I = md².
  • Only the distance perpendicular to the axis counts; moments of the parts of a body add.
  • Same torque, bigger I, slower spin-up: I = τ/α.

Verification: an independent derivation per frame; the rod integral done exactly in sympy; a second route through each rod's centre (ml² sin² θ/12 + mdcm², the parallel-axis theorem); a third from the full inertia tensor of the frames placed in space; a rigid-body spin-up under a constant torque that recovers I = τ/α; every option audited; and the page's own engine, run by tests/verify_p1phyq16.py. The MathonGo solution was read first only for the approach; it, the independent result and the official key, read last, all give (A). Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.

How to use this simulation

  1. Press ▶ START EXPERIMENT. Frames P, Q, R and S are each spun about OCO′ by the same constant torque.
  2. Watch the dashed lines: each rod's distance from the axis. Mass far from the axis makes a frame spin up slowly.
  3. The angular speed is plotted against time; its slope is α, and I = τ/α is matched to List-II.
  4. Tap a Frame button and drag Turn it to look to see it from any angle, with each rod's share of I.
  5. Replay runs it again. Reset clears it. Classroom fills the screen for a projector.