In an experiment with a closed organ pipe, it is filled with water by $\left(\frac{1}{5}\right)$ th of its volume. The frequency of the fundamental note will change by

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Waves & Sound

In an experiment with a closed organ pipe, it is filled with water by

(\frac{1}{5})

th of its volume. The frequency of the fundamental note will change by

Answer: (A)

For a closed organ pipe the fundamental frequency is

f = v /4 L

, set by the length

L

of the air column. Pouring in water to fill one-fifth of the volume shortens the air column uniformly to

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

, so

f^{'} = \frac{v}{4 L ^{'}} = \frac{v}{4 \cdot \frac{4}{5} L} = \frac{5 v}{16 L} .

The fractional change is

\frac{f ^{'} - f}{f} = \frac{\frac{5 v}{16 L} - \frac{v}{4 L}}{\frac{v}{4 L}} = \frac{1}{4},

so the fundamental frequency rises by

25%

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option