A closed organ and an open organ tube filled by two different gases having same bulk modulus but different densities ρ 1 and ρ 2 respectively. The frequency of 9 th harmonic of closed tube is identical with 4 th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ 1 : ρ 2 =1: 16 , then the length of the open tube is :

Physics · Sound Waves

JEE Main 2025 — 22 January, Morning Shift — Question 52

A closed organ and an open organ tube filled by two different gases having same bulk

modulus but different densities $\rho_{1}$ and $\rho_{2}$ respectively. The frequency

of $9^{\text {th }}$ harmonic of closed tube is identical with $4^{\text {th }}$ harmonic

of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases

is $\rho_{1}: \rho_{2}=1: 16$ , then the length of the open tube is :

Option A:
$\frac{20}{7} \mathrm{~cm}$
Option B:
$\frac{15}{7} \mathrm{~cm}$
Option C:
$\frac{20}{9} \mathrm{~cm}$
Correct
Option D:
$\frac{15}{9} \mathrm{~cm}$

Answer: C

Step-by-step solution

$9^{\text {th }}$ harmonic of closed pipe

$=\frac{9 \mathrm{~V}_{1}}{4 \ell_{1}}$ $4^{\text {th }}$

harmonic of open pipe $=\frac{2 \mathrm{~V}_{2}}{\ell_{2}}$

$\therefore \frac{9 \mathrm{~V}_{1}}{4 \ell_{1}}=\frac{2 \mathrm{~V}_{2}}{\ell_{2}}$

$\therefore \frac{9}{4 \ell_{1}} \sqrt{\frac{\mathrm{~B}}{\mathrm{P}_{1}}}=\frac{2}{\ell_{2}} \sqrt{\frac{\mathrm{~B}}{\mathrm{P}_{2}}}$ $\ell_{2}=\frac{20}{9} \mathrm{~cm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Sound Waves
Topic: Vibrations in rod and Air Columns - Organ pipes