MHT CET · Physics · Waves and Sound
A resonance tube closed at one end is of height 1.5 m . A tuning fork of frequency 340 Hz is vibrating above the tube. Water is poured in the tube gradually. The minimum height of water column for which resonance is obtained is . (Neglect end correction, speed of sound in air \(=340 \mathrm{~m} / \mathrm{s}\) )
- A 75 cm
- B 50 cm
- C 30 cm
- D 25 cm
Answer & Solution
Correct Answer
(D) 25 cm
Step-by-step Solution
Detailed explanation
For a resonance tube, In the first mode, we have \(\mathrm{n}=\frac{\mathrm{V}}{4 \mathrm{~L}_{\mathrm{j}}}\) or \(\mathrm{L}_1=\frac{\mathrm{V}}{4 \mathrm{n}}=\frac{340}{4 \times 340}=0.25 \mathrm{~m}=25 \mathrm{~cm}\)
In this case \(\mathrm{L}_1=\frac{\lambda}{4}=25 \mathrm{~cm} \Rightarrow \lambda=100 \mathrm{~cm}\)
Resonance can also be obtained when the length of the air column is \(\frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \frac{7 \lambda}{4}\) etc.
At \(\frac{5 \lambda}{4}=\frac{5 \times 100}{4}=125 \mathrm{~cm}\) it stops as length of tube is \(1.5 \mathrm{~m}=150 \mathrm{~cm}\).
Hence the minimum height of water \(=150-125=25 \mathrm{~cm}\)
In this case \(\mathrm{L}_1=\frac{\lambda}{4}=25 \mathrm{~cm} \Rightarrow \lambda=100 \mathrm{~cm}\)
Resonance can also be obtained when the length of the air column is \(\frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \frac{7 \lambda}{4}\) etc.
At \(\frac{5 \lambda}{4}=\frac{5 \times 100}{4}=125 \mathrm{~cm}\) it stops as length of tube is \(1.5 \mathrm{~m}=150 \mathrm{~cm}\).
Hence the minimum height of water \(=150-125=25 \mathrm{~cm}\)
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