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

Let the original length of the air column in the closed organ pipe be $$L$$. The pipe has uniform cross-section, so filling it with water to $$\frac{1}{5}$$ of its volume means water occupies $$\frac{1}{5}$$ of its length as well.

Therefore, the new length of the air column is
$$L' = L - \frac{L}{5} = \frac{4L}{5}$$

For a closed organ pipe, the fundamental (first harmonic) frequency is given by the relation
$$f = \frac{v}{4L}$$
where $$v$$ is the speed of sound in air.

After adding water, the new fundamental frequency becomes
$$f' = \frac{v}{4L'} = \frac{v}{4\left(\frac{4L}{5}\right)} = \frac{5v}{16L}$$

To compare the two frequencies, form the ratio
$$\frac{f'}{f} = \frac{\frac{5v}{16L}}{\frac{v}{4L}} = \frac{5v}{16L} \times \frac{4L}{v} = \frac{5}{4} = 1.25$$

This shows the new frequency is $$1.25$$ times the original. The percentage change is
$$\left(1.25 - 1\right) \times 100\% = 0.25 \times 100\% = 25\%$$

Hence, the fundamental frequency increases by $$25\%$$.

Correct option: Option A