MHT CET · Physics · Waves and Sound

A cylindrical tube open at both ends has fundamental frequency \(f\) in air. When the tube is dipped vertically in water so that one-third part of the tube is in water, the fundamental frequency of air column becomes (neglect end correction)

A \(\frac{f}{2}\)
B \(\frac{3 f}{2}\)
C \(\frac{f}{4}\)
D \(\frac{3 f}{4}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{3 f}{2}\)

Step-by-step Solution

Detailed explanation

We know. Frequency is ratio of velocity of sound and wavelength:
\(f=\frac{v}{\lambda}\)
For tube of length , in fundamental mode wavelength is given by \(\frac{\lambda}{2}=L\)
\(\frac{\lambda}{2}=L\).
\(\therefore f=\frac{v}{2 L}\)
Now, if tube is dipped one third, then for fundamental mode in air column has:
\(\begin{aligned} & \frac{\lambda^{\prime}}{2}=\left(\frac{2}{3} L\right) \\ & \therefore f^{\prime}=\frac{3 v}{4 L}=\frac{3}{2} f\end{aligned}\)

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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