An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. The percentage change in the apparent frequency is

Using the Doppler effect formula when the observer moves towards a stationary source:

$$f' = f\left(\frac{v + v_o}{v}\right)$$

where $$v$$ is the velocity of sound, $$v_o$$ is the velocity of the observer, and $$f$$ is the actual frequency.

Given that $$v_o = \frac{v}{5}$$:

$$f' = f\left(\frac{v + v/5}{v}\right) = f\left(\frac{6v/5}{v}\right) = \frac{6f}{5}$$

The percentage change in apparent frequency is:

$$\frac{f' - f}{f} \times 100 = \frac{6f/5 - f}{f} \times 100$$

$$= \frac{f/5}{f} \times 100 = \frac{1}{5} \times 100 = 20\%$$

The correct answer is Option D.