MHT CET · Physics · Waves and Sound

A closed organ pipe of length ' \(\mathrm{L}_1\) ' 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 \(\frac{4 L_1}{3}\)
B \(\frac{4 \mathrm{~L}_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}\)
C \(\frac{4 \mathrm{~L}_1}{3} \sqrt{\frac{\rho_2}{\rho_1}}\)
D \(\frac{3}{4 L_1} \sqrt{\frac{\rho_1}{\rho_2}}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{4 \mathrm{~L}_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}\)

Step-by-step Solution

Detailed explanation

Given both gases are vibrating in the first overtone with same frequency, we get
\(\begin{aligned}
& f_{\text {closed }}=f_{\text {open }} \\
& \Rightarrow \frac{3 \mathrm{v}}{4 L_1}=\frac{\mathrm{V}}{L_2}
\end{aligned}\)
According to Laplace's correction
\(\begin{aligned}
& \mathrm{v}=\sqrt{\frac{\gamma \mathrm{P}}{\rho}} \\
& \frac{3}{4 \mathrm{~L}_1} \times \sqrt{\frac{\gamma \mathrm{P}}{\rho_1}}=\frac{1}{\mathrm{~L}_2} \times \sqrt{\frac{\gamma \mathrm{P}}{\rho_2}} \\
\therefore \quad \overline{\mathrm{L}}_2 & =\frac{4 \mathrm{~L}_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}
\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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