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 :

<p>We know, for closed pipe,</p> <p>${f_n} = {{nv} \over {4l}},\,n = 1,3,5,7$</p> <p>for open pipe, ${f_n} = {{nv} \over {2L}},\,n = 1,2,3,4$</p> <p>So, 9$^{th}$ harmonic of closed pipe $= {{9{v_1}} \over {4{l_1}}}$</p> <p>4$^{th}$ harmonic of open pipe $= {{4{v_2}} \over {2{l_2}}} = {{2{v_2}} \over {{l_2}}}$</p> <p>$\therefore$ $${{9{v_1}} \over {4{l_1}}} = {{2{v_2}} \over {{l_2}}} \Rightarrow {{{l_2}} \over {{l_1}}} = {8 \over 9}{{{v_2}} \over {{v_1}}}$$</p> <p>We know, $v = \sqrt {{\beta \over \rho }}$</p> <p>So, ${{{v_2}} \over {{v_1}}} = \sqrt {{{{\rho _1}} \over {{\rho _2}}}}$ (for same $\beta$)</p> <p>hence, $${{{l_2}} \over {{l_1}}} = {8 \over 9}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} = {8 \over 9}\sqrt {{1 \over {16}}} = {8 \over 9} \times {1 \over 4}$$</p> <p>$$ \Rightarrow {l_2} = {2 \over 9} \times {l_1} \Rightarrow {l_2} = {2 \over 9} \times 10 = {{20} \over 9}cm$$</p>