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 :

Options

A. ` 1 m`
B. `1.5 m`
C. ` 2 m`
D. `3 m`

Answer

C

Explanation

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`