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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.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))` |
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Answer» Correct Answer - B Frequency of first overtone in closed pipe, `v = (3)/(4l_(1)) sqrt((B)/(rho_(1)))`…(i) Frequency of the first overtone in open pipe, `v = (1)/(l_(2)) sqrt((B)/(rho_(2))) rArr l_(2) =(4)/(3) l_(1) sqrt((rho_(1))/(rho_(2)))`….(ii). |
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