Closed pipe resonance Steps: - The air column above water forms a pipe closed at one end (water surface), with resonances at L = (2n-1)λ/4. - As water level rises from 2.9 cm to 67 cm (noting likely transcription error in problem for second value), air column shortens from longer L_1 ≈ 3λ/4 to shorter L_2 ≈ λ/4. - Difference in water levels = 67 - 2.9 = 64.1 cm = 0.641 m = change in air column = λ/2. - λ = 2 × 0.641 = 1.282 m. - f = 330 / 1.282 ≈ 257 Hz ≈ 256 Hz. Why B is correct: - 256 Hz gives λ = 330/256 = 1.289 m, so λ/2 = 0.645 m (64.5 cm), matching the level difference via closed pipe formula L_{n+1} - L_n = λ/2. Why the others are wrong: - A. 128 Hz gives λ/2 = 1.29 m (129 cm), level difference too large. - C. 512 Hz gives λ/2 = 0.322 m (32.2 cm), level difference too small. - D. 1024 Hz gives λ/2 = 0.161 m (16.1 …