Question:

A resonance tube closed at one end is of height $1.5\ \text{m}$. A tuning fork of frequency $340\ \text{Hz}$ is vibrating above the tube. Water is poured in the tube gradually. The minimum height of water for which resonance is obtained is (Neglect end correction. Speed of sound in air = $340\ \text{m/s}$)

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For resonance tubes closed at one end, the first resonance always occurs at one-fourth wavelength.
Updated On: Feb 4, 2026
  • $5\ \text{cm}$
  • $125\ \text{cm}$
  • $150\ \text{cm}$
  • $25\ \text{cm}$
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The Correct Option is D

Solution and Explanation

Step 1: Find wavelength of sound.
\[ \lambda = \dfrac{v}{f} = \dfrac{340}{340} = 1\ \text{m} \] Step 2: Condition for first resonance in closed tube.
For a tube closed at one end, first resonance occurs when: \[ L = \dfrac{\lambda}{4} \] Step 3: Calculate resonant air column length.
\[ L = \dfrac{1}{4} = 0.25\ \text{m} = 25\ \text{cm} \] Step 4: Find minimum height of water.
Total tube height = $1.5\ \text{m}$
\[ \text{Height of water} = 1.5 - 0.25 = 1.25\ \text{m} \] But minimum air column corresponds to first resonance, so minimum water height required is $25\ \text{cm}$.
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