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(C) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2L} = \frac{1}{2L} \sqrt{\frac{\gamma RT}{M}}$.Since $T$ is constant,$f

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$f_B/f_C = \sqrt{11/28}$

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(C) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2L} = \frac{1}{2L} \sqrt{\frac{\gamma RT}{M}}$.Since $T$ is constant,$f \propto \frac{\sqrt{\gamma}}{L \sqrt{M}}$.Let $K = \frac{\sqrt{RT}}{2}$. Then $f = K \frac{\sqrt{\gamma}}{L \sqrt{M}}$.For pipe $A$ $(H_2)$: $L_A = l$,$M_A = 2$,$\gamma_A = 1.4 = 7/5$. So,$f_A = K \frac{\sqrt{7/5}}{l \sqrt{2}}$.For pipe $B$ $(O_2)$: $L_B = l/2$,$M_B = 32$,$\gamma_B = 1.4 = 7/5$. So,$f_B = K \frac{\sqrt{7/5}}{(l/2) \sqrt{32}} = 2K \frac{\sqrt{7/5}}{l \sqrt{32}}$.For pipe $C$ $(N_2)$: $L_C = 2l/3$,$M_C = 28$,$\gamma_C = 1.4 = 7/5$. So,$f_C = K \frac{\sqrt{7/5}}{(2l/3) \sqrt{28}} = \frac{3}{2} K \frac{\sqrt{7/5}}{l \sqrt{28}}$.For pipe $D$ $(CO_2)$: $L_D = l/3$,$M_D = 44$,$\gamma_D = 7/5$. So,$f_D = K \frac{\sqrt{7/5}}{(l/3) \sqrt{44}} = 3K \frac{\sqrt{7/5}}{l \sqrt{44}}$.Calculating $f_B/f_C = \frac{2K \sqrt{7/5} / (l \sqrt{32})}{3K \sqrt{7/5} / (2l \sqrt{28})} = \frac{2}{\sqrt{32}} 	imes \frac{2 \sqrt{28}}{3} = \frac{4}{3} \sqrt{\frac{28}{32}} = \frac{4}{3} \sqrt{\frac{7}{8}} = \frac{4}{3} \frac{\sqrt{7}}{2\sqrt{2}} = \frac{2\sqrt{7}}{3\sqrt{2}} = \sqrt{\frac{28}{18}} = \sqrt{\frac{14}{9}}$.Checking $f_C/f_D = \frac{3K \sqrt{7/5} / (2l \sqrt{28})}{3K \sqrt{7/5} / (l \sqrt{44})} = \frac{1}{2 \sqrt{28}} 	imes \sqrt{44} = \frac{1}{2} \sqrt{\frac{44}{28}} = \frac{1}{2} \sqrt{\frac{11}{7}} = \sqrt{\frac{11}{28}}$.Thus,option $C$ is correct.

Four open organ pipes of different lengths and different gases at the same temperature are shown in the figure. Let $f_A$,$f_B$,$f_C$,and $f_D$ be their fundamental frequencies. Then: [Take $\gamma_{CO_2} = 7/5$]

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