A tuning fork of frequency 340 Hz is vibrate | vedclass

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(D) The tuning fork forms a resonance with the air column in a pipe closed at one end. The resonance condition is given by $n = \frac{(2N - 1)v}{4l}$,

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$45$

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(D) The tuning fork forms a resonance with the air column in a pipe closed at one end. The resonance condition is given by $n = \frac{(2N - 1)v}{4l}$,where $N = 1, 2, 3, \dots$ represents the mode of vibration.Given $n = 340 \ Hz$ and $v = 340 \ m/s$,the length of the air column $l$ is:$l = \frac{(2N - 1)v}{4n} = \frac{(2N - 1) 	imes 340}{4 	imes 340} = \frac{2N - 1}{4} \ m = (2N - 1) 	imes 25 \ cm$.For $N = 1, l_1 = 25 \ cm$.For $N = 2, l_2 = 75 \ cm$.For $N = 3, l_3 = 125 \ cm$ (not possible as the tube height is $120 \ cm$).The height of the water column $h$ is given by $h = H_{tube} - l$.For $l_1 = 25 \ cm$,$h_1 = 120 - 25 = 95 \ cm$.For $l_2 = 75 \ cm$,$h_2 = 120 - 75 = 45 \ cm$.The minimum height of water required for resonance is $45 \ cm$.

$A$ tuning fork of frequency $340 \ Hz$ is vibrated just above a tube of $120 \ cm$ height. Water is poured slowly into the tube. What is the minimum height of water necessary for resonance (in $cm$)? (Speed of sound in air $= 340 \ m/s$)

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