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A cylindrical tube partially filled with water resonates with a tuning fork when the air column height is 0.1 m. When the water level is lowered further, resonance occurs again at an air column height of 0.35 m. The end correction for the tube is:

1. 0.025 m
2. 0.015 m
3. 0.001 m
4. 0.002 m

Correct answer: 0.025 m

Solution

For a closed pipe: Lₙ + e = (2n-1)*lambda/4. First resonance: L1 + e = lambda/4. Second resonance: L2 + e = 3*lambda/4. Subtracting: L2 - L1 = lambda/2, so lambda = 2*(0.35-0.1) = 0.5 m. Then e = lambda/4 - L1 = 0.125 - 0.1 = 0.025 m.

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