An open tube is in resonance with string (frequency of vibration of tube is n 0 ). If tube is dipped in water so that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be

Options

A. `v_(0)`
B. `2v_(0)`
C. `(2)/(3)v_(0)`
D. `(3)/(2)v_(0)`

Answer

B

Explanation

When open tube is dipped in water, it becomes a tube closed at one end. Fundamental frequency for open tube is

Length available for resonance of closed tube is 0.25 l.
`:." "v_(c)=(v)/(4(0.25l))=(v)/(2l)xx2=2v_(0)`