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. `3/2`
C. `2/3`
D. 2

Answer

D

Explanation

First case ` n_(o) = v/(2L) ` = frequency of the vibrating string ….(1)
In the second case, the tube becomes a closed tube of vibrating length ` L/4`.
` :. ` Frequency `n_(c) = v/(4L') = v/(4* L/4) = v/L ` ....(2)
` :. n_(c)/n_(o) = (v/L)/(v/(2L)) = 2`