A cylindrical tube open at both ends has fundamental frequency ' \(n\) ' in air. The tube is dipped vertically in water so that one-fourth of it is in water. The fundamental frequency of the air column becomes

The fundamental frequency of open tube is
\(\mathrm{n}_1=\frac{\mathrm{v}}{2 \ell_1}\)
When tube is dipped in water, one-fourth of it is in water and three-fourth is in air.

Hence, it becomes a tube closed at one end with length \(\ell_2=\frac{3}{4} \ell_1\) The fundamental frequency of closed tube is
\(\begin{aligned}
& \mathrm{n}_2=\frac{\mathrm{v}}{4 \ell_2} \\
& \therefore \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{1}{4 \ell_2} \times 2 \ell_1=\frac{\ell_1}{2 \ell_2}=\frac{4}{2 \times 3} \quad\left[\because \frac{\ell_1}{\ell_2}=\frac{4}{3}\right] \\
& \therefore \mathrm{n}_2=\frac{2}{3} \mathrm{n}_1=\frac{2}{3} \mathrm{n}
\end{aligned}\)