An open pipe of sufficient length is dipping in water with a speed V vertically if at any instant l is length of tube above water then the rate at which fundamental frequency of pipe changes is (C is the speed of sound in air )

Options

A. `(CV)/2l^(2)`
B. `(CV)/4l^(2)`
C. `(CV)/(2v^(2)t^(2))`
D. `(CV)/(4v^(2)t^(2))`

Answer

B

Explanation

At time t `l=(l-vt)=lambda//4`
fundamental frequency `f_(0)=C/(4l)`
`f_(0)=C/(4(l-vt))`
`:. (df)/(dt)=-C/(4(l-vt)^(2))(-v)=(CV)/(4l^(2))`