The 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 towards resonant freiquency of that pipe. Sound from pipe `A` set 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` in meter is :

Answer

2

Explanation

For pipe `A`, second resonant frequency is third harmonic thus `f = (3V)/(4L_(A))`
For pipe `B`, second resonant frequency is third harmonic thus `f = (2V)/(2L_(B))`
Equating `(3V)/(4L_(A)) = (2V)/(2L_(B)) rArr L_(B) = (4)/(3)L_(A) , = (4)/(3).(1.5)`