A source of frequency 340 ,Hz is kept above a | vedclass

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(D) The frequency of the source is $f = 340 \,Hz$ and the velocity of sound is $v = 340 \,m/s$.The wavelength of the sound wave is $\lambda = \frac{v}

Vedclass · Accepted Answer

$0.45$

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(D) The frequency of the source is $f = 340 \,Hz$ and the velocity of sound is $v = 340 \,m/s$.The wavelength of the sound wave is $\lambda = \frac{v}{f} = \frac{340}{340} = 1 \,m = 100 \,cm$.For a tube closed at one end, resonance occurs when the length of the air column $l$ satisfies $l = (2n-1) \frac{\lambda}{4}$, where $n = 1, 2, 3, \dots$.For $n=1$, $l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25 \,cm$.For $n=2$, $l_2 = \frac{3\lambda}{4} = \frac{3 	imes 100}{4} = 75 \,cm$.For $n=3$, $l_3 = \frac{5\lambda}{4} = \frac{5 	imes 100}{4} = 125 \,cm$.Since the total length of the tube is $120 \,cm$, the possible air column lengths are $25 \,cm$ and $75 \,cm$.The height of the water level $h$ from the bottom is given by $h = L - l$, where $L = 120 \,cm$ is the total length of the tube.For $l_1 = 25 \,cm$, $h_1 = 120 - 25 = 95 \,cm = 0.95 \,m$.For $l_2 = 75 \,cm$, $h_2 = 120 - 75 = 45 \,cm = 0.45 \,m$.The minimum height of the water level is $0.45 \,m$.

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