A) 1
B) 2
C) \[\frac{2}{3}\]
D) \[\frac{3}{2}\]
Correct Answer: B
Solution :
For open tube, \[{{n}_{0}}=\frac{v}{2l}\] For closed tube length available for resonance is \[l'=l\times \frac{25}{100}=\frac{l}{4}\] \ Fundamental frequency of water filled tube \[\,n=\frac{v}{4l'}=\frac{v}{4\times (l/4)}=\frac{v}{l}=2{{n}_{0}}\]Þ \[\frac{n}{{{n}_{0}}}=2\]You need to login to perform this action.
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