JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Waves & Sound

In an open organ pipe

v_{3}

and

v_{6}

are

3^{rd}

and

6^{th}

harmonic frequencies, respectively. If

v_{6} - v_{3} = 2200

Hz then length of the pipe is

____

mm. (Take velocity of sound in air is

330

m/s.)

Answer: (B)

Solution

For an open pipe the

n

th harmonic is

f_{n} = \frac{n v}{2 L}

. Then

v_{6} - v_{3} = \frac{( 6 - 3 ) v}{2 L} = \frac{3 v}{2 L} = 2200 Hz .

With

v = 330 m/s

L = \frac{3 \times 330}{2 \times 2200} = 0.225 m .

The length of the pipe is

225 mm

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option