A pipe $\mathrm{P}_{\mathrm{C}}$ closed at one end and pipe $\mathrm{P}_{\mathrm{O}}$ open at both ends are…
- $\frac{4}{5}$
- $\frac{5}{6}$
- $\frac{2}{3}$
- $\frac{3}{5}$
Solution
If they are in the resonance with the same tuning fork, using:
$\lambda \mathrm{f}=\mathrm{v}$
$f$ and $v$ are the same, so
$\begin{aligned} & \lambda_C=\lambda_0 \\ & \therefore\left(\frac{4}{5} L_C\right)=\frac{2}{3} L_0 \\ & \Rightarrow \frac{L_C}{L_0}=\frac{5}{6}\end{aligned}$Asked in: MHT CET 2022 (05 Aug Shift 2)