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Two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length $L_{A} = 1.5 m$ and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. Find the length of the pipe B in meters

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answer is B.

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Detailed Solution

For pipe A, second resonant frequency is third harmonic thus $f = \frac{3 V}{4 L_{A}}$

For pipe B, second resonant frequency is second harmonic thus $f = \frac{2 V}{2 L_{B}}$

Equating $\frac{3 V}{4 L_{A}} = \frac{2 V}{2 L_{B}}$

$L_{B} = \frac{4}{3} L_{A} = \frac{4}{3} . (1.5) = 2 m$

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