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(D) The fundamental frequency of an open tube of length $\ell_1$ is given by $n_1 = \frac{v}{2\ell_1} = n$.When the tube is dipped vertically in water

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$\frac{2n}{3}$

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(D) The fundamental frequency of an open tube of length $\ell_1$ is given by $n_1 = \frac{v}{2\ell_1} = n$.When the tube is dipped vertically in water such that one-fourth of its length is submerged, the length of the air column becomes $\ell_2 = \ell_1 - \frac{1}{4}\ell_1 = \frac{3}{4}\ell_1$.Since the tube is now closed at one end by the water surface, it acts as a closed organ pipe.The fundamental frequency of a closed organ pipe is $n_2 = \frac{v}{4\ell_2}$.Substituting $\ell_2 = \frac{3}{4}\ell_1$ into the equation:$n_2 = \frac{v}{4(\frac{3}{4}\ell_1)} = \frac{v}{3\ell_1}$.Comparing $n_2$ with $n_1$:$n_2 = \frac{2}{3} 	imes (\frac{v}{2\ell_1}) = \frac{2}{3}n$.

$A$ cylindrical tube open at both ends has a 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

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