A cylindrical tube (L = 120 , cm) is resonant | vedclass

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(A) The frequency of the tuning fork is $f = 330 \, Hz$ and the speed of sound is $v = 330 \, m/s$.The wavelength is $\lambda = \frac{v}{f} = \frac{33

Vedclass · Accepted Answer

$45$

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(A) The frequency of the tuning fork is $f = 330 \, Hz$ and the speed of sound is $v = 330 \, m/s$.The wavelength is $\lambda = \frac{v}{f} = \frac{330}{330} = 1 \, m = 100 \, cm$.For a tube closed at one end,the resonance condition is $l_n = \frac{(2n-1)\lambda}{4}$,where $n = 1, 2, 3, ...$.For $n=1$,$l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25 \, cm$.For $n=2$,$l_2 = \frac{3\lambda}{4} = \frac{300}{4} = 75 \, cm$.For $n=3$,$l_3 = \frac{5\lambda}{4} = 125 \, cm$ (which is greater than the tube length $120 \, cm$).The tube is initially resonant at $l_2 = 75 \, cm$ (since $l_1 = 25 \, cm$ is also possible,but the tube length is $120 \, cm$,so the air column length must be $75 \, cm$ to be resonant).When filling with water,the air column length decreases. The next resonance occurs at $l_1 = 25 \, cm$.The length of the water column required is $L - l_1 = 120 \, cm - 25 \, cm = 95 \, cm$ or $L - l_2 = 120 \, cm - 75 \, cm = 45 \, cm$.The minimum length of the water column to get resonance again is $45 \, cm$.

$A$ cylindrical tube $(L = 120 \, cm)$ is resonant with a tuning fork of frequency $330 \, Hz$. If it is being filled with water,then to get resonance again,the minimum length of the water column is ...... $cm$ $(v_{air} = 330 \, m/s)$.

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