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 (va) of tube to string is :

A)  \[{{v}_{0}}\]                                     

B)  2v0

C)  \[\frac{2}{3}{{v}_{0}}\]                                

D) \[\frac{3}{2}{{v}_{0}}\]

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\[0.25\,l\] \[\therefore \]  \[{{v}_{c}}=\frac{v}{4(0.25l)}=\frac{v}{2l}\times 2=2{{v}_{o}}\]