A uniform tube of length 60.5,cm is held vert | vedclass

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(D) For a resonance tube,the resonance condition is $L + e = (2n-1) \frac{\lambda}{4}$,where $e$ is the end correction.Given $L_1 = 16\,cm$ and $L_2 =

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$272, 544$

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(D) For a resonance tube,the resonance condition is $L + e = (2n-1) \frac{\lambda}{4}$,where $e$ is the end correction.Given $L_1 = 16\,cm$ and $L_2 = 50\,cm$ for $f = 500\,Hz$.$L_2 - L_1 = \frac{\lambda}{2} \implies 50 - 16 = 34\,cm = \frac{\lambda}{2} \implies \lambda = 68\,cm = 0.68\,m$.Speed of sound $v = f \lambda = 500 	imes 0.68 = 340\,m/s$.Now,$L_1 + e = \frac{\lambda}{4} \implies 16 + e = \frac{68}{4} = 17 \implies e = 1\,cm = 0.01\,m$.When the tube is taken out of water,it acts as an open organ pipe of length $L = 60.5\,cm + 2e = 60.5 + 2(1) = 62.5\,cm = 0.625\,m$.The resonant frequencies of an open pipe are $f_n = \frac{n v}{2L}$.For $n=1$,$f_1 = \frac{340}{2 	imes 0.625} = \frac{340}{1.25} = 272\,Hz$.For $n=2$,$f_2 = 2 	imes f_1 = 544\,Hz$.

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