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A closed organ pipe with cross-sectional area 100 cm² resonates with a 1000 Hz tuning fork at its fundamental mode. What is the minimum volume of water that must be removed so the pipe resonates again with the same fork? (Speed of sound = 320 m/s)

1. 800 cm³
2. 1200 cm³
3. 1600 cm³
4. 2000 cm³

Correct answer: 1600 cm³

Solution

A closed pipe resonates when its length equals odd multiples of lambda/4. Fundamental: L = lambda/4 = v/(4f) = 8 cm. Next resonance (3rd harmonic): L = 3*lambda/4 = 24 cm. The water level must drop by 16 cm, so volume drained = 16 * 100 = 1600 cm³.

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