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The frequency of the first overtone of a closed pipe of length $l_{1}$ is equal to that of the first. overtone of an open pipe of length $l_{2}$. The ratio of their lengths $\left(l_{1}: l_{2}\right)$ is
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The correct answer is:
$3: 4$
Here, $\quad v_{1}=\frac{(2 n-1)}{4 l_{1}} v$
$$
v_{2}=\frac{n}{2 l_{2}} v
$$
Hence, $y_{1}=v_{2}$
$$
\frac{l_{1}}{l_{2}}=\frac{3}{4}
$$
$$
v_{2}=\frac{n}{2 l_{2}} v
$$
Hence, $y_{1}=v_{2}$
$$
\frac{l_{1}}{l_{2}}=\frac{3}{4}
$$
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