`AB` is a cylinder of length `1m` fitted with a thin flexible diaphragm `C` at the middle and other thin flexible diaphragms `A` and `B` at the ends. The portions `AC` and `BC` contain hydrogen and oxygen gases respectively. The diaphragms `A` and `B` are set into vibrations of same frequency. What is the minimum frequency of these vibrations for which diaphragms `C` is a node? (Under the conditions of experiment `v_(H_(2) = 1100m//s`, `v_(0_(2)` = `300m//s`).

Options

A. `1100Hz`
B. `3300Hz`
C. `1650Hz`
D. `1500Hz`

Answer

C

Explanation

As diaphrgm C is a node, A and B will be antinodes. Thus each part behave as closed organ pipe. Their fundamental frequency are
`n_(H)=(V_(H))/(4l_(H))=(1100)/(4xx0.5)=550z` `n_(0)=(V_(0))/(4l_(0))=(330)/(4xx0.5)=150Hz`
As `n_(H) ne n_(0)`, then common frequency of system is written as `n=pn_(H)=qn_(0)implies(p)/(q)=(3)/(11)`