In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:

For a closed pipe (closed at one end), the fundamental frequency is $$f = \dfrac{v}{4L}$$.

For the 100 cm pipe: $$f_1 = \dfrac{v}{4 \times 1.00} = \dfrac{v}{4}$$

For the 120 cm pipe: $$f_2 = \dfrac{v}{4 \times 1.20} = \dfrac{v}{4.8}$$

The beat frequency is $$|f_1 - f_2| = 15$$ Hz.

$$\dfrac{v}{4} - \dfrac{v}{4.8} = 15$$

$$v\left(\dfrac{1}{4} - \dfrac{1}{4.8}\right) = 15$$

$$v\left(\dfrac{4.8 - 4}{4 \times 4.8}\right) = 15$$

$$v \cdot \dfrac{0.8}{19.2} = 15$$

$$v = 15 \times \dfrac{19.2}{0.8} = 15 \times 24 = 360$$ m/s

Hence, the correct answer is Option D.