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Q.

The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length LA =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. The length of the pipe B is :    

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a

3 m

b

1.5 m

c

2 m

d

1 m

answer is C.

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Detailed Solution

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For pipe A, second resonant frequency is third harmonic thus f =  3V4LA
For pipe B, second resonant frequency is second harmonic thus f =  2V2LB
Equating   3V4LA=2V2LB
       LB=43LA=43.(1.5)    =  2m.
 

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