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 $(l_{1} : l_{2})$ is

2 : 3

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4 : 5

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3 : 5

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3 : 4

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Solution

The correct option is A $3 : 4$
$v_{1} = \frac{(2 n - 1)}{4 l_{1}} v$
$v_{2} = \frac{n}{2 l_{2}} v$
For the first overtone n = 2.
Since, $v_{1} = v_{2}$ $⟹ \frac{l_{1}}{l_{2}} = \frac{3}{4}$

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The frequency of the first overtone of a closed pipe of length l1 is equal to that of the first overtone of an open pipe of length l2. The ratio of their lengths (l1:l2) is

The correct option is A 3:4v1=(2n−1)4l1vv2=n2l2vFor the first overtone n = 2.Since, v1=v2 ⟹l1l2=34

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 $(l_{1} : l_{2})$ is

The correct option is A $3 : 4$
$v_{1} = \frac{(2 n - 1)}{4 l_{1}} v$
$v_{2} = \frac{n}{2 l_{2}} v$
For the first overtone n = 2.
Since, $v_{1} = v_{2}$ $⟹ \frac{l_{1}}{l_{2}} = \frac{3}{4}$