Question

an open tube in resonance with string frequency of vibration of tube is n0 tube is dipped in water so that 75% of length of tube is inside the water then the ratio of frequency of tube to string now will be​

Answer 1

Explanation:

   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 \[\frac{n}{{{n}_{0}}}=2\]