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\[0.25\,l\] \[\therefore \] \[{{v}_{c}}=\frac{v}{4(0.25l)}=\frac{v}{2l}\times 2=2{{v}_{o}}\]You need to login to perform this action.
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