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Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both ends and is filled with a monatomic gas of molar mass MA. Pipe B is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass MB. 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 MA/MB.
(b) Now the open end 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.

Hrishant Goswami , 10 Years ago
Grade 10
anser 1 Answers
Jitender Pal

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
(a) Second harmonic in pipe A is
2 (v0)A = 2 [v/2ℓ] = 1/ ℓ √yART/MA
The harmonic in pipe B is
3(v0) B = 3[v/4 ℓ] = 3/4 ℓ √y BRT/MB
Given v A = v B
1/ ℓ √y ART/MA = 3/4 ℓ √y BRT/MB
Or, MA/MB = yA/yB x (4/3)2 = 5/3 / 7/5 x 16/9 = 400/189
235-954_6.png
Now, (v0)A / (v0)B = √yA/ yA x MB/ MB = 3/4

Thanks
Jitender Pal
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