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 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 :

Two pipes are submerged in sea water , arranged as shown in figure . P

1.	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 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 :
Answer» ` 1 m` `1.5 m` ` 2 m` `3 m` Solution :For pipe `A`, second RESONANT frequency is third harmonic thus ` f= ( 3V)/( 4 L_(A))` For pipe `B` , second resonant frequency is second harmonic thus `f = ( 2V)/( 2 L_(B))` Equating , `(3 V)/( 4L_(A)) = ( 2 V)/( 2L_(B))` `rArr L_(B) = (4)/(3) L_(A) = (4)/(3) . (1.5) = 2 m`