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(D) Fundamental frequency of a closed pipe is $n_{c} = \frac{v}{4L}$.Fundamental frequency of an open pipe is $n_{o} = \frac{v}{2L}$.Given that they p

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$7$

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(D) Fundamental frequency of a closed pipe is $n_{c} = \frac{v}{4L}$.Fundamental frequency of an open pipe is $n_{o} = \frac{v}{2L}$.Given that they produce $2$ beats per second:$n_{o} - n_{c} = 2$$\frac{v}{2L} - \frac{v}{4L} = 2$$\frac{v}{4L} = 2 \implies \frac{v}{L} = 8$.When the length of the open pipe is halved,the new frequency is:$n_{o}' = \frac{v}{2(L/2)} = \frac{v}{L} = 8 	ext{ Hz}$.When the length of the closed pipe is doubled,the new frequency is:$n_{c}' = \frac{v}{4(2L)} = \frac{v}{8L} = \frac{1}{8} 	imes \left(\frac{v}{L}ight) = \frac{1}{8} 	imes 8 = 1 	ext{ Hz}$.New beat frequency $= n_{o}' - n_{c}' = 8 - 1 = 7 	ext{ beats per second}$.

$A$ closed organ pipe and an open organ pipe of the same length produce $2$ beats per second when they are set into vibrations together in their fundamental mode. The length of the open pipe is now halved and that of the closed pipe is doubled. The number of beats produced per second will be:

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