A closed organ pipe of length `L` and an open organ pipe contain gass of densities `rho_(1)` and `rho_(2)`, respectively. The compressibility of gass are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency . The length of the open orange pipe is
(a) `(L)/(3)`
`(4l) / (3)`
(c ) `(4l)/(3)sqrt((rho_(1))/(rho_(2)))`
(d) `(4l)/(3)sqrt((rho_(2))/(rho_(1)))`

Explanation

Frequency of first overtone in closed pipe = `3((v_(c))/(4L))`
Let length of open pipe = `l_(0)`
`therefore` Frequency of first overtone in open pipe = `2((v_(0))/(2l_(0)))`
The two frequencies are given to be equal.
`therefore (3v_(c))/(4L)=(2v_(o))/(2l_(o))orl_(o)=(4)/(3)((v_(o))/(v_(c)))L`
Now, velocity of sound in a gas `v=sqrt((K)/(rho))`
`therefore (v_(o))/(v_(c))=sqrt((rho_(c))/(rho_(o)))`
`therefore l_(0)=(4L)/(3)sqrt((rho_(c))/(rho_(0)))=(4L)/(3)sqrt((rho_(1))/(rho_(2)))`
`therefore` Length of open organ pipe = `(4L)/(3)sqrt((rho_(1))/(rho_(2)))`.