Physics - Sound Wave Question with Solution | TestHub

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: $x+h=\frac{\lambda }{4}$ $\Rightarrow x+24.1=\frac{\lambda }{4}---(1)$ And, for second resonance: $x+H=\frac{3\lambda }{4}$ $\Rightarrow x+74.1=\frac{3\lambda }{4}---(2)$ On taking the ratio of equation (1) and (2) and solving $\begin{array}{rl}& \Rightarrow x=0.9\mathrm{cm}\\ & \Rightarrow 0.3D=0.9\mathrm{cm}\\ & \therefore D=3\mathrm{cm}\end{array}$