A tube of diameter d and of length $l$ unit is open at both ends. Its fundamental frequency of resonance is found to be $ν_{1} .$ The velocity of sound in air is 330 m/sec. One end of tube is now closed. The lowest frequency of resonance of tube is $ν_{2} .$ Taking into consideration the end correction, $\frac{ν_{2}}{ν_{1}}$ is

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$\frac{(l + 0.6 d)}{(l + 0.3 d)}$

$\frac{1}{2} \frac{(l + 0.3 d)}{(l + 0.6 d)}$

$\frac{1}{2} \frac{(l + 0.6 d)}{(l + 0.3 d)}$

$\frac{1}{2} \frac{(d + 0.3 l)}{(d + 0.6 l)}$

answer is C.

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Detailed Solution

$\frac{λ_{1}}{2}$ = $l$ + 0.6d, $υ_{1} = \frac{V}{λ_{1}}$
Question Image
$\frac{λ_{2}}{4}$ = $l$ + 0.3d, $υ_{2} = \frac{V}{λ_{2}}$

$\frac{υ_{2}}{υ_{1}} = \frac{2 (l + .6 d)}{4 (l + .3 d)} = \frac{(l + .6 d)}{2 (l + .3 d)}$

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