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A cylindrical tube open at both the ends has a fundamental frequency of $390 \mathrm{~Hz}$ in air. If $\frac{1}{4}$ th of the tube is immersed vertically in water the fundamental frequency of air column is
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The correct answer is:
$260 \mathrm{~Hz}$
Fundamental frequency of cylindrical open tube
$$
\mathrm{n}=\frac{\mathrm{v}}{2 \mathrm{~L}}=390 \mathrm{~Hz}
$$
When it is immersed in water it becomes a closed tube of length $\frac{3}{4}$ th of the initial length.
Therefore, its fundamental frequency is
$$
\begin{aligned}
\mathrm{n}^{\prime} &=\frac{\mathrm{V}}{4\left(\frac{3}{4} \mathrm{~L}\right)}=\frac{\mathrm{v}}{3 \mathrm{~L}}=\frac{2}{3}\left(\frac{\mathrm{V}}{2 \mathrm{~L}}\right) \\
&=\frac{2}{3} \times 390 \mathrm{~Hz}=260 \mathrm{~Hz}
\end{aligned}
$$
$$
\mathrm{n}=\frac{\mathrm{v}}{2 \mathrm{~L}}=390 \mathrm{~Hz}
$$
When it is immersed in water it becomes a closed tube of length $\frac{3}{4}$ th of the initial length.
Therefore, its fundamental frequency is
$$
\begin{aligned}
\mathrm{n}^{\prime} &=\frac{\mathrm{V}}{4\left(\frac{3}{4} \mathrm{~L}\right)}=\frac{\mathrm{v}}{3 \mathrm{~L}}=\frac{2}{3}\left(\frac{\mathrm{V}}{2 \mathrm{~L}}\right) \\
&=\frac{2}{3} \times 390 \mathrm{~Hz}=260 \mathrm{~Hz}
\end{aligned}
$$
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