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(C) The frequency of the first overtone of a closed organ pipe of length $L_1 = L$ is given by $f_1 = \frac{3v_1}{4L_1}$,where $v_1 = \sqrt{\frac{1}{\

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$\frac{4L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$

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(C) The frequency of the first overtone of a closed organ pipe of length $L_1 = L$ is given by $f_1 = \frac{3v_1}{4L_1}$,where $v_1 = \sqrt{\frac{1}{\beta \rho_1}}$ and $\beta$ is the compressibility.The frequency of the first overtone of an open organ pipe of length $L_2$ is given by $f_2 = \frac{2v_2}{2L_2} = \frac{v_2}{L_2}$,where $v_2 = \sqrt{\frac{1}{\beta \rho_2}}$.Given that the frequencies are equal $(f_1 = f_2)$: $\frac{3}{4L} \sqrt{\frac{1}{\beta \rho_1}} = \frac{1}{L_2} \sqrt{\frac{1}{\beta \rho_2}}$Canceling the common term $\sqrt{\frac{1}{\beta}}$ from both sides:$\frac{3}{4L\sqrt{\rho_1}} = \frac{1}{L_2\sqrt{\rho_2}}$Solving for $L_2$:$L_2 = \frac{4L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$

$A$ closed organ pipe of length $L$ and an open organ pipe contain gases of densities ${\rho _1}$ and ${\rho _2}$ respectively. The compressibility of gases is equal in both the pipes. Both the pipes are vibrating in their first overtone with the same frequency. The length of the open organ pipe is

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