Question

A pipe open at both ends has a fundamental frequency 'f' in air. The pipe is dipped in water, so that \(\frac{2}{3}^{rd}\) length of pipe is in water, Now the fundamental frequency of the air column is

• A. \(\frac{1}{2}f\)
• B. \(\frac{3}{2}f\)
• C. \(\frac{5}{2}f\)
• D. \(\frac{7}{2}f\)

Hint:

The part of the pipe above water behaves as a closed pipe of length L/3.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
An open pipe of length \(L\) has fundamental frequency \(f = \dfrac{v}{2L}\). A pipe dipped in water becomes a closed pipe whose length is the air column above the water, with fundamental \(f' = \dfrac{v}{4\ell}\).

Step 2: Compute
Two thirds of the pipe is in water, so the air column is \(\ell = L/3\).
\[ f' = \frac{v}{4(L/3)} = \frac{3v}{4L} = \frac32\cdot\frac{v}{2L} = \frac32 f \]
So the new fundamental frequency is \(\frac32 f\).

Final Answer:
The new fundamental frequency is \(\frac32 f\), option (B). \[ \boxed{\frac{3}{2}f} \]