(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

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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 $L_1 = \frac{330}{4 \times 30} = \frac{330}{120} = 2.75 \,m$.After pouring water,the new length of the air column is $L_2$. The new frequency is $f_2 = 110 \,Hz$.$L_2 = \frac{330}{4 \times 110} = \frac{330}{440} = 0.75 \,m$.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 $V = A \times \Delta L = 2 \,cm^2 \times 200 \,cm = 400 \,cm^3$.Since the density of water is $1 \,g/cm^3$,the mass of water is $400 \,g$.

Class 11 Physics Waves Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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In a closed organ pipe,the frequency of the fundamental note is $30 \,Hz$. $A$ certain amount of water is now poured into the organ pipe so that the fundamental frequency increases to $110 \,Hz$. If the organ pipe has a cross-sectional area of $2 \,cm^2$,the amount of water poured into the organ tube is . . . . . . $g$. (Take the speed of sound in air as $330 \,m/s$)

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$400$
B
$200$
C
$600$
D
$800$

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Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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$A$ 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 $= 340 \ m/s$)

An open organ pipe of length $l$ is sounded together with another organ pipe of length $l + x$ in their fundamental tones $(x << l)$. The beat frequency heard will be (speed of sound is $v$):

The fundamental note produced by a closed organ pipe is of frequency $f$. The fundamental note produced by an open organ pipe of the same length will be of frequency .... $(f)$

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