Resonance in a closed-end tube

Steps:

• The setup is a closed pipe where water level determines the effective length L.
• Resonance occurs when L = (2n-1)λ/4 for odd harmonics (n=1,2,...).
• First resonance: L1 = λ/4; second: L2 = 3λ/4.
• Change in water level ΔL = L2 - L1 = λ/2 = 32 cm, so λ = 64 cm.

Why C is correct:

• In a closed pipe, the distance between consecutive resonances equals λ/2, per the resonance condition for standing waves.

Why the others are wrong:

• A: Far too small; ignores the λ/2 relationship.
• B: Matches ΔL directly but confuses it with λ instead of λ/2.
• D: Overestimates by assuming ΔL = λ/4 or open-pipe behavior.

Final answer: C