A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is ( x /3) L √(ρ1/ρ2) where x is . (Round off to the Nearest Integer)

Question

A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is ( x /3) L √(ρ1/ρ2) where x is . (Round off to the Nearest Integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/609b5877d310e9549408348f7f644604-.png" /> f c = f 0 (3 V C /4 L )=(2 V 0/2 L prime) (3 V C /4 L )=( V 0/ L prime) L prime=(4 L /3) ( V 0/ V C )=(4 L /3) √( B ⋅ ρ1/ρ2 ⋅ B ) ( B is bulk modulus) =(4 L /3) √(ρ1/ρ2) x =4

fc​=f0​ 4L3VC​​=2L′2V0​​ 4L3VC​​=L′V0​​ L′=34L​VC​V0​​=34L​ρ2​⋅BB⋅ρ1​​​( B is bulk modulus) =34L​ρ2​ρ1​​​ x=4

$f_{c} = f_{0}$
$\frac{3 V _{C}}{4 L} = \frac{2 V _{0}}{2 L ^{'}}$
$\frac{3 V _{C}}{4 L} = \frac{V _{0}}{L ^{'}}$
$L^{'} = \frac{4 L}{3} \frac{V _{0}}{V _{C}} = \frac{4 L}{3} \frac{B \cdot ρ _{1}}{ρ _{2} \cdot B}$ ( B is bulk modulus)
$= \frac{4 L}{3} \frac{ρ _{1}}{ρ _{2}}$
$x = 4$