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

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 :
  1. $\frac{15}{7} \mathrm{~cm}$
  2. $\frac{20}{7} \mathrm{~cm}$
  3. $\frac{15}{9} \mathrm{~cm}$
  4. $\frac{20}{9} \mathrm{~cm}$

Solution

$\begin{aligned} & 9^{\text {th }} \text { harmonic of closed pipe }=\frac{9 \mathrm{~V}_1}{4 \ell_1} \\ & 4^{\text {th }} \text { harmonic of open pipe }=\frac{2 \mathrm{~V}_2}{\ell_2} \\ & \therefore \frac{9 \mathrm{~V}_1}{4 \ell_1}=\frac{2 \mathrm{~V}_2}{\ell_2} \\ & \therefore \frac{9}{4 \ell_1} \sqrt{\frac{\mathrm{~B}}{\rho_1}}=\frac{2}{\ell_2} \sqrt{\frac{\mathrm{~B}}{\rho_2}} \Rightarrow \frac{\ell_2}{\ell_1}=\frac{8}{9} \sqrt{\frac{\rho_1}{\rho_2}} \\ & \ell_2=\ell_1 \times \frac{8}{9} \times \frac{1}{4}=\frac{20}{9} \mathrm{~cm}\end{aligned}$

Asked in: JEE Main 2025 (22 Jan Shift 1)

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