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If the seventh harmonic of a closed organ pipe is in unison with fourth harmonic of an open organ pipe, then the ratio of length of closed pipe to that of open pipe is
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Verified Answer
The correct answer is:
$7: 8$
For closed organ pipe
$$
\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{nv}}{4 \mathrm{I}_{\mathrm{c}}}=\frac{7 \mathrm{v}}{4 \mathrm{I}_{\mathrm{c}}}
$$
For open organ pipe
$$
\begin{aligned}
& f_o=\frac{n v}{21_o}=\frac{4 v}{21_o} \\
& \because f_c=f_o \\
& \frac{7 v}{41_c}=\frac{4 v}{21_o} \Rightarrow \frac{1_c}{1_o}=\frac{7}{8}
\end{aligned}
$$
$$
\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{nv}}{4 \mathrm{I}_{\mathrm{c}}}=\frac{7 \mathrm{v}}{4 \mathrm{I}_{\mathrm{c}}}
$$
For open organ pipe
$$
\begin{aligned}
& f_o=\frac{n v}{21_o}=\frac{4 v}{21_o} \\
& \because f_c=f_o \\
& \frac{7 v}{41_c}=\frac{4 v}{21_o} \Rightarrow \frac{1_c}{1_o}=\frac{7}{8}
\end{aligned}
$$
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