In resonance tube experiment, the first and second resonance are heard when water level is $24.1…
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
3 cm
2 cm
5 cm
4 cm
Solution
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
$\begin{aligned}
& \Rightarrow x=0.9 \mathrm{~cm} \\
& \Rightarrow 0.3 D=0.9 \mathrm{~cm} \\
& \therefore D=3 \mathrm{~cm}
\end{aligned}$