A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities _1 and _ 2^ , respectively. The frequency of 9^ th harmonic of closed tube is identical with 4^ th

Correct answer: D. (20)/(9) ~cm aligned & 9^ th harmonic of closed pipe = 9 V_1 4 _1 & 4^ th harmonic of open pipe = 2 V_2 _2 & 9 V_1 4 _1 = 2 V_2 _2 & (9)/(4 _1) B _1 =(2)/( _2) B _2 ( _2)/( _1)=(8)/(9) ( _1)/( _2) & _2= _1 × (8)/(9) × (1)/(4)=(20)/(9) cm aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2025 · 22 January · Morning Shift · Physics · Waves and Sound

A closed organ and an open organ tube are filled by two different...

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities

\rho_1

and

\rho_{2^{\prime}}

, 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 :

Options

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

Correct answer

D. $\frac{20}{9} \mathrm{~cm}$

Step-by-step solution

\begin{aligned} & 9^{\text{th }} \text{ harmonic of closed pipe }=\frac{9 \mathrm{ V}_1}{4 \ell_1} \\ & 4^{\text{th }} \text{ 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}}{\rho_1}}=\frac{2}{\ell_2} \sqrt{\frac{\mathrm{ B}}{\rho_2}} \Rightarrow \frac{\ell_2}{\ell_1}=\frac{8}{9} \sqrt{\frac{\rho_1}{\rho_2}} \\ & \ell_2=\ell_1 \times \frac{8}{9} \times \frac{1}{4}=\frac{20}{9} \mathrm{ cm}\end{aligned}

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