A tube of diameter `d` and of length `l` unit is open at both ends. Its fundamental frequency of resonance is found to be `v_(1)`. The velocity of sound in air is `330 m//sec`. One end of tube is now closed. The lowest frequency of resonance of tube is now `v_(2)`. Taking into consideration the end correction, `(v_(2))/(v_(1))` is

Options

A. `((l + 0.6d))/((l + 0.3d))`
B. `(1(l + 0.3d))/(2(l + 0.6d))`
C. `(1(l + 0.6d))/(2(l + 0.3d))`
D. `(1(d + 0.3l))/(2(d + 0.6l))`

Answer

C

Explanation

`(lambda_(1))/(2) = l ++ .6d, upsilon_(1) = (V)/(lambda_(1))`

`(lambda_(2))/(4) = l + 0.3d, upsilon = (V)/(lambda_(2))`

`(upsilon_(2))/(upsilon_(1)) = (2(l + 6d))/(4(l + .3d)) = (l + .6d)/(2(l + .3d))`