MHT CET · Physics · Waves and Sound
In resonance tube experiment, the first and second resonance are heard when water level is \(24.1 \mathrm{~cm}\) and \(74.1 \mathrm{~cm}\) respectively, below the open end of the tube. The inner diameter of the tube is
- A 3 cm
- B 2 cm
- C 5 cm
- D 4 cm
Answer & Solution
Correct Answer
(A) 3 cm
Step-by-step Solution
Detailed explanation
Let the end correction be \(x\), then the diameter \(D\) of the one end open tube will be given by, \(D=\frac{x}{0.3}\)
For first resonance: \(x+h=\frac{\lambda}{4}\)
\(\Rightarrow x+24.1=\frac{\lambda}{4} \quad---(1)\)
And, for second resonance: \(x+H=\frac{3 \lambda}{4}\)
\(\Rightarrow x+74.1=\frac{3 \lambda}{4} \quad---(2)\)
On taking the ratio of equation (1) and (2) and solving
\(\begin{aligned}
& \Rightarrow x=0.9 \mathrm{~cm} \\
& \Rightarrow 0.3 D=0.9 \mathrm{~cm} \\
& \therefore D=3 \mathrm{~cm}
\end{aligned}\)
For first resonance: \(x+h=\frac{\lambda}{4}\)
\(\Rightarrow x+24.1=\frac{\lambda}{4} \quad---(1)\)
And, for second resonance: \(x+H=\frac{3 \lambda}{4}\)
\(\Rightarrow x+74.1=\frac{3 \lambda}{4} \quad---(2)\)
On taking the ratio of equation (1) and (2) and solving
\(\begin{aligned}
& \Rightarrow x=0.9 \mathrm{~cm} \\
& \Rightarrow 0.3 D=0.9 \mathrm{~cm} \\
& \therefore D=3 \mathrm{~cm}
\end{aligned}\)
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