Question

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 diatomic gas of molar mass `M_(B)`. Both gases are at the same temperature.
(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 1

Correct Answer - A::B::C::D
Frequency of second harmonic in pipe
`A=` frequency of third harmonic in pipe `B`
`:. 2((v_(A))/(2l_(AP))=3((v_(B))/(4l_(B))`
or `(v_(A))/(v_(B))=(3)/(4)implies(sqrt(gamma_(A)RT_(A))/(M_(A)))/(sqrt((gamma_(B)RT_(B))/(M_(B))))=(3)/(4)`
or `sqrt((gamma_(A))/(gamma_(B)))sqrt((M_(B))/M_(A))=(3)/(4)` (as `T_(A)=T_(B)`)
`:. (M_(A))/(M_(B))=((25)/(21))((16)/(9))=(400)/(189)`
(b) Ratio of fundamental frequency in pipe `A` and in pipe `B` is
`(f_(A))/(f_(B))=(v_(A)//2l_(A))/(v_(B)//2l_(B))=(v_(A))/(v_(B) )` (as `T_(A)=T_(B)`)
`=sqrt((((gamma_(A)RT_(A))/(M_(A)))/(gamma_(B)RT_(B)))/(M_(B)))=sqrt((gamma_(A))/(gamma_(B)).(M_(B))/(M_(A)))` (as `T_(A)=T_(B)`)
Substituting `(M_(B))/(M_(A))=(189)/(400)` from paert (a). we get
`(f_(A))/(f_(B))=sqrt((25)/(21)xx(189)/(400))=(3)/(4)`