In a closed organ pipe,the frequency of the f | vedclass

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(A) For a closed organ pipe,the fundamental frequency is given by $f = \frac{v}{4L}$.Initially,$f_1 = 30 \,Hz$ and $v = 330 \,m/s$,so $L_1 = \frac{330

Vedclass · Accepted Answer

$400$

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(A) For a closed organ pipe,the fundamental frequency is given by $f = \frac{v}{4L}$.Initially,$f_1 = 30 \,Hz$ and $v = 330 \,m/s$,so $L_1 = \frac{330}{4 	imes 30} = \frac{330}{120} = 2.75 \,m$.After pouring water,the new length of the air column is $L_2$. The new frequency is $f_2 = 110 \,Hz$.$L_2 = \frac{330}{4 	imes 110} = \frac{330}{440} = 0.75 \,m$.The change in the length of the air column is $\Delta L = L_1 - L_2 = 2.75 - 0.75 = 2.0 \,m = 200 \,cm$.The volume of water poured is $V = A 	imes \Delta L = 2 \,cm^2 	imes 200 \,cm = 400 \,cm^3$.Since the density of water is $1 \,g/cm^3$,the mass of water is $400 \,g$.

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