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. 1
B. 2
C. `2/3`
D. `3/2`

Answer

B

Explanation

For open tube `n_0 = (v)/(2l)` tube is partially dipped in water resonating length is `l_2=l/4`
The fundamental frequency of water filled tube is ,
`n=(v)/(4l_2) = (v)/(4l/4)`
`n=(2v)/(2l)=2xx(v)/(2l) = 2n_0`
`(n)/(n_0) = 2`