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A cylindrical tube open at both ends has fundamental frequency $f$ in air. When the tube is dipped vertically in water so that one-third part of the tube is in water, the fundamental frequency of air column becomes (neglect end correction)
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The correct answer is:
$\frac{3 f}{2}$
We know. Frequency is ratio of velocity of sound and wavelength:
$f=\frac{v}{\lambda}$
For tube of length , in fundamental mode wavelength is given by $\frac{\lambda}{2}=L$
$\frac{\lambda}{2}=L$.
$\therefore f=\frac{v}{2 L}$
Now, if tube is dipped one third, then for fundamental mode in air column has:
$\begin{aligned} & \frac{\lambda^{\prime}}{2}=\left(\frac{2}{3} L\right) \\ & \therefore f^{\prime}=\frac{3 v}{4 L}=\frac{3}{2} f\end{aligned}$
$f=\frac{v}{\lambda}$
For tube of length , in fundamental mode wavelength is given by $\frac{\lambda}{2}=L$
$\frac{\lambda}{2}=L$.
$\therefore f=\frac{v}{2 L}$
Now, if tube is dipped one third, then for fundamental mode in air column has:
$\begin{aligned} & \frac{\lambda^{\prime}}{2}=\left(\frac{2}{3} L\right) \\ & \therefore f^{\prime}=\frac{3 v}{4 L}=\frac{3}{2} f\end{aligned}$
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