A closed organ and an open organ tube filled by two different gases having same bulk modulus but different densities and respectively. The frequency of 9th harmonic of closed tube is identical with 4th harmonic of open tube. If the length of the closed

Question

A closed organ and an open organ tube filled by two different gases having same bulk modulus but different densities $ρ_{1}$ and $ρ_{2}$ 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 $ρ_{1} : ρ_{2} = 1 : 16$ , then the length of the open tube is [2025]

Smart Achievers · Accepted Answer

(3)

9th harmonic of closed pipe = $\frac{9 v_{1}}{4 l_{1}}$

4th harmonic of open pipe = $\frac{2 v_{2}}{l_{2}}$

$∴ \frac{9 v_{1}}{4 l_{1}} = \frac{2 v_{2}}{l_{2}}$

$∴ \frac{9}{4 l_{1}} \sqrt{\frac{B}{ρ_{1}}} = \frac{2}{l_{2}} \sqrt{\frac{B}{ρ_{2}}}$

$\Rightarrow \frac{l_{2}}{l_{1}} = \frac{8}{9} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

$l_{2} = l_{1} \times \frac{8}{9} \times \frac{1}{4} = \frac{20}{9} cm$

Solution

1 $\frac{20}{7} cm$

2 $\frac{15}{7} cm$

3 $\frac{20}{9} cm$

4 $\frac{15}{9} cm$

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Solution

1 207 cm

2 157 cm

3 209 cm

4 159 cm

Ans. (3) 9th harmonic of closed pipe = 9v14l1 4th harmonic of open pipe = 2v2l2 ∴ 9v14l1=2v2l2 ∴ 94l1Bρ1=2l2Bρ2 ⇒ l2l1=89ρ1ρ2 l2=l1×89×14=209 cm

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1 $\frac{20}{7} cm$

2 $\frac{15}{7} cm$

3 $\frac{20}{9} cm$

4 $\frac{15}{9} cm$