A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities ρ1 and ρ2 respectively. The frequency of 9th harmonic of closed tube is identical with 4th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1: ρ2 = 1:16, then the length of the open tube is:

Question

As we know the velocity of sound in gases:

v

=

β

ρ

v = \sqrt{\frac{\beta}{\rho}}

v

1

v

2

=

ρ

2

ρ

1

=

16

1

=

4

1

\frac{v_{1}}{v_{2}} = \sqrt{\frac{\rho_{2}}{\rho_{1}}} = \sqrt{\frac{16}{1}} = \frac{4}{1}

For closed pipe in 9th Harmonics: For open organ pipe in 4th Harmonics:

L

1

=

9

4

λ

1

L_{1} = \frac{9}{4}\lambda_{1}

given frequency for both is same and using

f

=

v

λ

f = \frac{v}{\lambda}

v

1

λ

1

=

v

2

λ

2

\frac{v_{1}}{\lambda_{1}} = \frac{v_{2}}{\lambda_{2}}

L

2

=

4

λ

2

L_{2} = \frac{4\lambda_{2}}{2}

f

1

=

f

2

f_{1} = f_{2}

v

1

v

2

=

λ

1

λ

2

\frac{v_{1}}{v_{2}} = \frac{\lambda_{1}}{\lambda_{2}}

PYQ · Accepted Answer

B