question_answer

An open tube is in resonance with string. If tube is dipped in water, so that \[75%\]of length of tube is inside water, then the ratio of the frequency \[({{v}_{o}})\] of tube to string is

A)  \[{{v}_{o}}\]            

B)  \[2{{v}_{o}}\]

C)  \[\frac{2}{3}{{v}_{o}}\]          

D)  \[\frac{3}{2}{{v}_{o}}\]

Correct Answer: B

Solution :

When open tube is dipped in water, it becomes a tube closed at one end. Fundamental frequency for open tube is \[{{v}_{o}}=\frac{v}{2l}\] Length available for resonance of closed tube  is \[0.25\,\,l\]. \[\therefore \] \[{{v}_{c}}=\frac{v}{4(0.25\,l)}=\frac{v}{2l}\times 2=2{{v}_{o}}\]