A tuning fork of frequency $340Hz$ is vibrated just above the tube of $120 cm$ height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance .... $cm$ (speed of sound in the air $= 340 m/sec$)
Diffcult
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(d) Because the tuning fork is in resonance with air column in the pipe closed at one end, the frequency is $n = \frac{{(2N - 1)v}}{{4l}}$ where $N = 1, 2, 3 ....$ corresponds to different mode of vibration
putting $n = 340Hz, v = 340 m/s,$ the length of air column in the pipe can be
$l = \frac{{(2N - 1)340}}{{4 \times 340}} = \frac{{(2N - 1)}}{4}m = \frac{{(2N - 1) \times 100}}{4}cm$
For $N = 1, 2, 3, ... $ we get $l = 25 cm, 75 cm, 125 cm ...$
As the tube is only $120 cm$ long, length of air column after water is poured in it may be $25 cm$ or $75 cm$ only, $125 cm$ is not possible, the corresponding length of water column in the tube will be $(120 -25) cm = 95 cm$ or $(120 -75) cm = 45 cm. $
Thus minimum length of water column is $45 cm.$
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