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(C) The given frequencies are $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$.Dividing these by their greatest common divisor $(85)$,we get the ratio: $425:59

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$1$

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(C) The given frequencies are $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$.Dividing these by their greatest common divisor $(85)$,we get the ratio: $425:595:765 = 5:7:9$.Since the frequencies are in the ratio of odd integers $(5:7:9)$,the pipe must be a closed organ pipe.The resonance frequencies for a closed pipe are given by $f_n = \frac{n V}{4l}$,where $n$ is an odd integer $(n = 1, 3, 5, ...)$.For the first observed frequency,$n = 5$,so $\frac{5 V}{4l} = 425 \, Hz$.Substituting $V = 340 \, m/s$:$l = \frac{5 	imes 340}{4 	imes 425} = \frac{1700}{1700} = 1 \, m$.

For a certain organ pipe,three successive resonance frequencies are observed at $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$ respectively. If the speed of sound in air is $340 \, m/s$,then the length of the pipe is ..... $m$.

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