JEE Main 2026 · 22 January · evening · Single correct · +4 / −1

In an open organ pipe ν₃ and ν₆ are 3ʳᵈ and 6ᵗʰ harmonic frequencies, respectively. If ν₆-ν₃=2200 Hz then length of the pipe is ____ mm .
(Take velocity of sound in air is 330 m/s.)

  1. (a) 200
  2. (b) 225
  3. (c) 275
  4. (d) 250
Answer and solution

Answer (b) 225

Solution

Physics sir · 3 steps

Open pipe harmonics are spaced by v/2L, so going from 3rd to 6th adds 3v/2L.

  1. Open pipe harmonics. Both ends of an open pipe are antinodes, so every harmonic is present. The n-th harmonic is νₙ = nv/2L.

    ν₆ - ν₃ = 6v/2L - 3v/2L = 3v/2L

  2. Solve for L. Set 3v/2L = 2200 Hz. Then L = (3 × 330)/(2 × 2200) = 990/4400 = 0.225 m.
  3. So. L = 0.225 m = 225 mm, which is option (b).

The idea to keep. In an open pipe the harmonics are equally spaced by v/2L. The difference between any two harmonic frequencies is (gap in n) × v/2L.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 60 times in JEE Main since 2020. This question is from 2026.