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(C) Let the length of both pipes be $L$. The speed of sound is $v$.For an open pipe,the frequencies are $f_{open, n} = n(v/2L)$. The first overtone $(

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$14$

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(C) Let the length of both pipes be $L$. The speed of sound is $v$.For an open pipe,the frequencies are $f_{open, n} = n(v/2L)$. The first overtone $(n=2)$ is $f_{open, 1st} = 2(v/2L) = v/L$.For a closed pipe,the frequencies are $f_{closed, n} = (2n-1)(v/4L)$. The first overtone $(n=2)$ is $f_{closed, 1st} = 3(v/4L)$.Given that the beat frequency is $4$ Hz: $|v/L - 3v/4L| = 4 \implies v/4L = 4 \implies v/L = 16$.Now,the new length of the open pipe is $L' = L/2$ and the new length of the closed pipe is $L'' = 2L$.The fundamental frequency of the new open pipe is $f'_{open} = v/(2L') = v/(2(L/2)) = v/L = 16$ Hz.The fundamental frequency of the new closed pipe is $f''_{closed} = v/(4L'') = v/(4(2L)) = v/8L = 16/8 = 2$ Hz.The number of beats produced is $|f'_{open} - f''_{closed}| = |16 - 2| = 14$ Hz.

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