MHT CET · Physics · Waves and Sound
The fundamental frequency of air column in pipe 'A' closed at one end is in unison with second overtone of an air column in pipe 'B' open at both ends. The ratio of length of air column in pipe 'A' to that of air column in pipe ' \(\mathrm{B}\) ' is
- A \(1: 6\)
- B \(3: 8\)
- C \(2: 3\)
- D \(3: 4\)
Answer & Solution
Correct Answer
(A) \(1: 6\)
Step-by-step Solution
Detailed explanation
Fundamental frequency of ta closed pipè, \(\mathrm{n}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1}\)
Frequency of the second overtone, \(\mathrm{n}_2=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}\)
Given \(\mathrm{n}_1=\mathrm{n}_2\)
\(\Rightarrow \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}\)
\(\therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{1}{6}\)
Frequency of the second overtone, \(\mathrm{n}_2=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}\)
Given \(\mathrm{n}_1=\mathrm{n}_2\)
\(\Rightarrow \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}\)
\(\therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{1}{6}\)
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