Two narrow cylindrical pipes `A and B` have the same length. Pipe `A` is open at both ends and is filled with a monoatomic gas of molar mass `M_(A)`. Pipe `B` is open at one end and closed at the other end, and is filled with a distomic gas of molar mass `M_(B)`. Both gases are at the same tempreture.
(a) If the frequency of the second harmonic of the fundamental mode in pipe `A` is equal to the frequency of the third harmonic of the fundamental mode in pipe `B`, determine the value of `M_(B)//M_(B)`.
(b) Now the open end of pipe `B` is also closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe `A` to that in pipe `B`.

Answer

A, B, C, D

Explanation

(a) Second harmonic in pipe `A` is
`2(v_(0))_(A) = 2[(v)/(2l)] = (1)/(l)sqrt((gamma_(A)RT)/(M_(A))]`
Third harmonic in pipe `B` is
`3(v_(0))_(B) = 3[(v)/(4l)] = (3)/(4l)sqrt((gamma_(A)RT)/(M_(B))]`
Given `v_(A) = v_(B)`
`(1)/(l)sqrt((gamma_(A)RT)/(M_(A))) = (3)/(4l)sqrt((gamma_(A)RT)/(M_(B))`
or, `(M_(A))/(M_(B)) = (gamma_(A))/(gamma_(B)) xx ((4)/(3))^(2) = (5//3)/(7//5) xx (16)/(9) = (400)/(189)`


Now, `((v_(0))_(A))/((v_(0))_(B)) = sqrt((gamma_(A))/(gamma_(A)) xx (M_(B))/(M_(B))) = (3)/(4)`