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JEE Main 2025 (22 Jan, Shift 1) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from JEE Main 2025 (22 Jan, Shift 1)

Q46JEE Main 2025 · 22 JAN · SHIFT 1Difficult

Organ Pipes & Resonance Tube

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities ρ1\rho_1 and ρ2\rho_2, respectively. The frequency of 99th harmonic of closed tube is identical with 44th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1:ρ2=1:16\rho_1 : \rho_2 = 1 : 16, then the length of the open tube is:

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Answer: (b)

  1. Closed tube 9th harmonic: 9v14Lc\frac{9v_1}{4L_c}; open tube 4th harmonic: 4v22Lo=2v2Lo\frac{4v_2}{2L_o} = \frac{2v_2}{L_o}
  2. Equate: Lo=8Lc9v2v1L_o = \frac{8L_c}{9}\frac{v_2}{v_1}
  3. Same BB: v2v1=ρ1ρ2=116=14\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}} = \sqrt{\frac{1}{16}} = \frac{1}{4}
  4. Lo=8×109×4=209 cmL_o = \frac{8 \times 10}{9 \times 4} = \frac{20}{9}\text{ cm}
  5. So the answer is (b) 209 cm\frac{20}{9}\text{ cm}.
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