A pipe of length `l_(1)`, closed at one end is kept in a chamber of gas of density `rho_(1)`. A second pipe open at both ends is placed in a second chamber of gas of density `rho_(2)`. The compressibility of both the gases is equal. Calculate the length of the second pipe if frquency of first overtone in both the cases is equal

Options

A. `(4)/(3)l_(1)sqrt((rho_(2))/(rho_(1))`
B. `(4)/(3)l_(1)sqrt((rho_(1))/(rho_(2))`
C. `l_(1)sqrt((rho_(2))/(rho_(1))`
D. `l_(1)sqrt((rho_(1))/(rho_(2))`

Answer

B

Explanation

(b) Frequency of first overtone in closed pipe,
`v = (3v)/(4l_(1))sqrt((P)/(rho_(1))` ….(i)
Frequency of first overtone in open pipe,
`v = (1)/(4l_(2))sqrt((P)/(rho_(2))` …(ii)
From equation (i) and (ii)
rArr `l_(2) = (4)/(3)l_(1)sqrt((rho_(1))/(rho_(2))`