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JEE Advanced 2026 · Paper 1, Physics question 13

List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength λ=0.29\lambda=0.29 m enters these structures at the point SS and a sound detector is placed at DD. Between the points SS and DD, the sound travels only through the tubes. List-II contains the possible smallest values of ll (refer to the figures) for which the detector DD records maximum amplitude. Ignore effects of sharp corners. [Given cos⁡(15∘)=0.97\cos(15^\circ)=0.97]

Choose the option that best describes the match between the entries in List-I to those in List-II.

List-I

  • P
    Two paths from S to D: a straight tube of length l along the bottom, and a semicircular tube of radius 0.5l arching over it.
  • Q
    Two paths from S to D: a straight tube of length l along the bottom, and a rectangular detour going up 0.5l, across l and back down 0.5l.
  • R
    Two paths from S to D: a straight tube of length l along the bottom, and a path going up a length l from S and then along a semicircle down to D; the semicircle's diameter is the dashed diagonal from the top of that vertical segment to D.
  • S
    Two paths from S to D: a straight tube of length l along the bottom, and a two-segment path forming a triangle with base angles 45° at S and an apex angle of 105°.

List-II

  • 1
    1.32 m
  • 2
    1.19 m
  • 3
    0.51 m
  • 4
    0.29 m
  • 5
    0.13 m
Choose one option
Answer and worked solution

Official answer: D

DD records a maximum when the two paths differ by a whole number of wavelengths. The smallest ll uses a difference of λ=0.29\lambda=0.29 m.

(P) Semicircle of diameter ll: length πl2\frac{\pi l}2. Difference (π2−1)l=0.571 l\left(\frac\pi2-1\right)l=0.571\,l, so l=0.51l=0.51 m → 3.

(Q) Detour 0.5l+l+0.5l=2l0.5l+l+0.5l=2l. Difference ll, so l=0.29l=0.29 m → 4.

(R) Up ll, then a semicircle whose diameter is the diagonal 2 l\sqrt2\,l: length l+π22l=3.221 ll+\frac{\pi\sqrt2}2l=3.221\,l. Difference 2.221 l2.221\,l, so l=0.13l=0.13 m → 5.

(S) A triangle on base ll with angles 45∘45^\circ, 30∘30^\circ and 105∘105^\circ. By the sine rule the other sides are sin⁡45∘sin⁡105∘l\frac{\sin45^\circ}{\sin105^\circ}l and sin⁡30∘sin⁡105∘l\frac{\sin30^\circ}{\sin105^\circ}l. With sin⁡105∘=cos⁡15∘=0.97\sin105^\circ=\cos15^\circ=0.97 their sum is 1.245 l1.245\,l. Difference 0.245 l0.245\,l, so l=1.19l=1.19 m → 2.

P→3, Q→4, R→5, S→2.

Topics:Superposition and interference
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