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InterviewSolution
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The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length `L_A=1.5m` 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. The length of the pipe B isA. 1mB. 1.5 mC. 2 mD. 3 m |
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Answer» Correct Answer - C For pipe A, second resonant frequency is third harmonic thus `f=(3V)/(4L_A)` For pipe B, second resonant frequency is second harmonic thus `f=(2V)/(2L_B)` Equating `(3V)/(4L_A)=(2V)/(2L_B)` `impliesL_B=(4)/(3)L_A=(4)/(3)xx(1.5)=2m` |
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