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 .)

Physics · Wave Optics

JEE Main 2026 — 22 January, Evening Shift — Question 33

In an open organ pipe $v_{3}$ and $v_{6}$ are $3^{\text {rd }}$ and $6^{\text {th }}$ harmonic frequencies, respectively.

If $v_{6}-v_{3}=2200 \mathrm{~Hz}$ then length of the pipe is $\_\_\_\_$ mm. (Take velocity of sound in air is $330 \mathrm{~m} / \mathrm{s}$ .)

Option A:
275
Option B:
225
Correct
Option C:
200
Option D:
250

Answer: B

Step-by-step solution

$\mathrm{f}=\mathrm{n}\left(\frac{\mathrm{V}_{0}}{2 \mathrm{~L}}\right)$ $\frac{6 \mathrm{~V}_{0}}{2 \mathrm{~L}}-\frac{3 \mathrm{~V}_{0}}{2 \mathrm{~L}}=2200$ $\frac{3 \mathrm{~V}_{0}}{2 \mathrm{~L}}=2200$ $\frac{3 \times 330}{2 \times L}=2200$ $L=\frac{3 \times 330}{2 \times 2200}$ $\mathrm{L}=0.225 \mathrm{~m}$ $\mathrm{L}=225 \mathrm{~mm}$

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Paper: 22 January, Evening Shift
Subject: Physics
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