Two-Path Sound Tube Lab — interference of sound in tube networks — JEE Advanced 2026 Paper 1 Q.13

01The question

Q.13 List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength λ = 0.29 m enters these structures at the point S and a sound detector is placed at D. Between the points S and D, the sound travels only through the tubes. List-II contains the possible smallest values of l (refer to the figures) for which the detector D records maximum amplitude. Ignore effects of sharp corners. [Given cos(15°) = 0.97]
Choose the option that best describes the match between the entries in List-I to those in List-II.

List-I

(P)

(Q)

(R)

(S)

The four figures are redrawn from the paper.

List-II

(1) 1.32 m
(2) 1.19 m
(3) 0.51 m
(4) 0.29 m
(5) 0.13 m

(A)P→4, Q→3, R→5, S→1
(B)P→4, Q→3, R→1, S→5
(C)P→3, Q→4, R→1, S→2
(D)P→3, Q→4, R→5, S→2

02The experiment

Answer— Press START: each tube is lengthened until the detector first hears a maximum.

Ready.

Tube

Length l0.300 m

Speed

03Detailed solution

Answer: — P→3, Q→4, R→5, S→2

The condition. The sound splits at S and meets itself at D. The detector hears a maximum when the path difference Δ = (detour length) − l is a whole number of wavelengths. The smallest l > 0 has Δ = λ = 0.29 m. Each detour is a fixed multiple k of l, so l = λ/(k − 1).
(P) A semicircle of diameter l: its length is πl/2, so Δ = (π/2 − 1)l = 0.571l and l = 0.508 m → (3) 0.51 m.
(Q) Up 0.5l, across l, down 0.5l: the detour is 2l, so Δ = l and l = 0.29 m → (4).
(R) Up l, then a semicircle whose diameter is the diagonal l√2. The arc is πl√2/2 = 2.221l, so the detour is 3.221l, Δ = 2.221l and l = 0.131 m → (5) 0.13 m.
(S) The triangle has angles 45° at S, 105° at the top, and so 30° at D. By the sine rule its two sides are l sin 30°/sin 105° and l sin 45°/sin 105°, with sin 105° = cos 15° = 0.97. The detour is (0.5 + 0.707)/0.97 · l = 1.244l, so Δ = 0.244l and l = 1.19 m → (2).
That is P→3, Q→4, R→5, S→2: option (D). Using the exact cos 15° = 0.9659 would give 1.16 m for S, which is still nearest to (2).

Understand the concept

This is interference with a fixed geometry. Both paths start and end at the same points, so only their difference in length matters. Constructive interference needs Δ = mλ. Because every length in each figure scales with l, the difference is a fixed fraction of l. The first maximum therefore needs the shortest l that makes that fraction equal one wavelength. A long detour, as in R, needs only a small l; a detour barely longer than the straight tube, as in S, needs a large one.

Key takeaways

Maximum at the detector: Δ = mλ. The smallest non-zero case is Δ = λ.
Detour = k × l, so l_min = λ/(k − 1).
The detour factors are: semicircle on l, π/2; rectangle, 2; side plus a semicircle on the diagonal, 1 + π/√2; triangle, by the sine rule.

Verification: an independent derivation, a sympy route, a numerical route (the two waves added as phasors and the first maximum of the detector amplitude found by sweeping l, with the path lengths measured from the drawn geometry), and the page's own engine, run by tests/verify_p1phyq13.py. The official key (D) and the MathonGo solution (D) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. The tubes are drawn at a fixed size, with l labelled, and the wave is shown in slow motion.

How to use this simulation

Press ▶ START EXPERIMENT. For each tube P, Q, R, S the run lengthens l from zero.
Watch the crests travel along both paths and meet at D. The detector's amplitude is plotted against l.
At the first maximum the path difference is exactly one wavelength. The run records that l and matches it to List-II.
Tap a Tube button and drag Length l to hear the maximum and the silence (Δ = λ/2) yourself.
Replay runs it again. Reset clears it. Classroom fills the screen for a projector.