A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to: 753 Hz, 500 Hz, 666 Hz, 333 Hz

Frequency of the sound produced by flute, f = 2(v/2L) = 2 × 330 / (2 × 0.5) = 660 Hz
Velocity of observer, v₀ = 10 × 5/18 = 25/9 m/s ≈ 2.78 m/s
∴ frequency detected by observer, f' = [(v + v₀)/v]f
∴ f' = [(25/9 + 330)/330]660 = (330 + 25/9)/330 × 660 ≈ 660(1 + 25/(9×330)) ≈ 660(1 + 1/118.8) ≈ 660(1.0084) ≈ 665.5 Hz
However, if we use v₀ = 10 × 1000/3600 = 25/9 m/s ≈ 2.77 m/s
Then f' = (330 + 25/9)/330 × 660 ≈ 665.5 Hz
Considering the approximation in the question, the closest answer is 666 Hz. Therefore, the closest answer is (4).