In an experiment with a closed organ pipe, it is filled with water by (\frac15) th of its volume. The frequency of the fundamental note will change by

Waves PYQ — JEE Main 2025 · 4 Apr · Shift 1 · Q62 with Solution

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Correct answer: B

For a closed organ pipe of air-column length $L$ , the fundamental frequency is

f = \frac{v}{4L}

where $v$ is the speed of sound.

Let the total length of the pipe be $L$ and cross-sectional area be $A$ . So total volume is

V = AL

Initially, the pipe is empty of water, so the air-column length is $L$ . Hence initial fundamental frequency is

f_1 = \frac{v}{4L}

Since the pipe has uniform cross-section, filling $\frac{1}{5}$ of the volume means filling $\frac{1}{5}$ of the length.

So water occupies length

\frac{L}{5}

Thus remaining air-column length is

L' = L - \frac{L}{5} = \frac{4L}{5}

Now the fundamental frequency becomes

f_2 = \frac{v}{4L'} = \frac{v}{4(4L/5)} = \frac{5v}{16L}

Compare with $f_1 = \frac{v}{4L}$ :

\frac{f_2}{f_1} = \frac{5v/(16L)}{v/(4L)} = \frac{5}{4}

So,

f_2 = \frac{5}{4} f_1

This means frequency increases by

\left(\frac{5}{4}-1\right)\times 100 = \frac{1}{4}\times 100 = 25\%

The frequency of the fundamental note increases by $25\%$ .

Therefore, the correct option is:

\boxed{\text{B: }25\%}