Resonance in a closed air column Steps:

• The tube forms a closed pipe with water at one end; resonances occur at lengths h = (2n-1)λ/4, ignoring end correction.
• First resonance (n=1) at h₁ = 18.6 cm = 0.186 m corresponds to λ/4.
• Second resonance (n=3) at h₂ = 56.4 cm = 0.564 m corresponds to 3λ/4.
• Difference h₂ - h₁ = 0.378 m = λ/2, so λ = 0.756 m.
• Frequency f = v/λ = 350 m/s / 0.756 m ≈ 463 Hz, closest to 440 Hz. Why B is correct:
• Matches the calculated f ≈ 463 Hz using the resonance condition Δh = λ/2 and v = fλ. Why the others are wrong:
• A: 220 Hz yields λ ≈ 1.59 m, Δh ≈ 0.80 m (too large).
• C: 660 Hz yields λ ≈ 0.53 m, Δh ≈ 0.27 m (too small).
• D: 880 Hz yields λ ≈ 0.40 m, Δh ≈ 0.20 m (too small).

Final answer: B