Question:
A closed organ pipe of length $L_{1}$ and an open organ pipe contain diatomic gases of densities $\rho_{1}$ and $\rho_{2}$ respectively. The compressibilities of the gases are same in both pipes, which are vibrating in their first overtone with same frequency. The length of the open organ pipe is (Neglect end correction)
A closed organ pipe of length $L_{1}$ and an open organ pipe contain diatomic gases of densities $\rho_{1}$ and $\rho_{2}$ respectively. The compressibilities of the gases are same in both pipes, which are vibrating in their first overtone with same frequency. The length of the open organ pipe is (Neglect end correction)
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Logic Tip: The first overtone of a closed pipe is the 3rd harmonic, while for an open pipe, it is the 2nd harmonic (which has frequency $v/L$).
Updated On: Apr 28, 2026
- $\frac{4L_{1{3}$
- $\frac{4L_{1{3}\sqrt{\frac{\rho_{1{\rho_{2}$
- $\frac{4L_{1{3}\sqrt{\frac{\rho_{2{\rho_{1}$
- $\frac{3}{4L_{1\sqrt{\frac{\rho_{1{\rho_{2}$
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The Correct Option is B
Solution and Explanation
Concept:
The frequency of the first overtone for a closed pipe is $f_c = \frac{3v_1}{4L_1}$, and for an open pipe, it is $f_o = \frac{v_2}{L_2}$.
Step 1: Set the frequencies equal.
Given $f_{closed} = f_{open}$: $$\frac{3v_1}{4L_1} = \frac{v_2}{L_2} \implies L_2 = \frac{4L_1v_2}{3v_1} \text{ }$$
Step 2: Relate velocity to density.
According to Laplace's correction, $v = \sqrt{\frac{\gamma P}{\rho$. Since both are diatomic gases ($\gamma$ same) and compressibilities are same, we use the density relation: $$\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2 \text{ }$$
Step 3: Substitute to find $L_2$.
$$L_2 = \frac{4L_1}{3}\sqrt{\frac{\rho_1}{\rho_2 \text{ }$$
The frequency of the first overtone for a closed pipe is $f_c = \frac{3v_1}{4L_1}$, and for an open pipe, it is $f_o = \frac{v_2}{L_2}$.
Step 1: Set the frequencies equal.
Given $f_{closed} = f_{open}$: $$\frac{3v_1}{4L_1} = \frac{v_2}{L_2} \implies L_2 = \frac{4L_1v_2}{3v_1} \text{ }$$
Step 2: Relate velocity to density.
According to Laplace's correction, $v = \sqrt{\frac{\gamma P}{\rho$. Since both are diatomic gases ($\gamma$ same) and compressibilities are same, we use the density relation: $$\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2 \text{ }$$
Step 3: Substitute to find $L_2$.
$$L_2 = \frac{4L_1}{3}\sqrt{\frac{\rho_1}{\rho_2 \text{ }$$
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