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A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities…

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A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities \(\rho_1\) and \(\rho_2\), respectively. The frequency of \({9}^{\text{th }}\) harmonic of closed tube is identical with \({4}^{\text{th }}\) harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is \({\rho }_{1}:{\rho }_{2}=1:16\), then the length of the open tube is :

[JEE Main 2025, 22 Jan (Shift 1)]

a

\(\frac{15}{9}\mathrm{cm}\)

b

\(\frac{20}{9}\mathrm{cm}\)

c

\(\frac{20}{7}\mathrm{cm}\)

d

\(\frac{15}{7}\mathrm{cm}\)

✓ Correct answer: b)

\(\frac{20}{9}\mathrm{cm}\)

Explanation

\({9}^{th}\) harmonic of closed pipe \(=\frac{9{V}_{1}}{4{ℓ}_{1}}\)

\({4}^{\text{th }}\) harmonic of open pipe \(=\frac{2{V}_{2}}{{ℓ}_{2}}\)
Since, the frequency of the 9th harmonic of the closed pipe equals the frequency of the 4th harmonic of the open pipe, therefore:

\(\frac{9{V}_{1}}{4{ℓ}_{1}}=\frac{2{V}_{2}}{{ℓ}_{2}}\)

\(∴\frac{9}{4{ℓ}_{1}}\sqrt{\frac{B}{{\rho }_{1}}}=\frac{2}{{ℓ}_{2}}\sqrt{\frac{B}{{\rho }_{2}}}\Rightarrow \frac{{ℓ}_{2}}{{ℓ}_{1}}=\frac{8}{9}\sqrt{\frac{{\rho }_{1}}{{\rho }_{2}}}\\ \Rightarrow {ℓ}_{2}==\frac{20}{9}cm\)

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