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

Answer

B

Explanation

Frequency of the first overtone in closed pipe,
`v=(3)/(4 1_1) sqrt((B)/(rho_1))`….(i)
Frequency of the first overtone in open pipe
`v=(1)/(1_2)sqrt((B)/(rho_2))implies 1_2=4/3 1_1sqrt((rho_1)/(rho_2))` ….(ii)