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JEE Advanced2026PhysicsWaves and SoundActual

List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength = 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 effect

Options

  1. AP 4, Q 3, R 5, S 1
  2. BP 4, Q 3, R 1, S 5
  3. CP 3, Q 4, R 1, S 2
  4. DP 3, Q 4, R 5, S 2

Correct answer

D. P 3, Q 4, R 5, S 2

Step-by-step solution

For maximum amplitude at the detector D , the path difference x between the two paths must be an integer multiple of the wavelength . For the smallest value of l , we take x = = 0.29 m. For configuration (P): The sound travels through a straight tube of length l and a semi-circular tube of diameter l . Path difference x = l 2 - l = l ( 2 - 1 ) . Equating to : l ( 3.1416 2 - 1 ) = 0.29 l(0.5708) = 0.29 l 0.51 m. Thus, P 3. For configuration (Q): The sound travels through a straight tube of length l and a rectangular

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