AB is a cylinder of length 1.0 m fitted with a thin flexible diaphragm C at the middle and two other thin flexible diaphragms A and B at the ends. The portions AC and BC contain hydrogen and oxygen gases, respectively. The diaphragms A and B are set into vibrations of same frequency. Then the minimum frequency in Hz of these vibrations for which the diaphragm C is a node is (Under the conditions of the experiment, the velocity of sound in hydrogen is 1100 m/s and in oxygen is 300 m/s.)

When diaphragms A and Bare set in oscillations, antinodes are formed at A and B while a  node is formed at C (given) Also AB= 1.0 m and AC= CB= l (say) = 1/2 = 0.5 m. The  portions AC and BC behave as closed pipes. In a closed pipe, the modes of vibration are given by
                       $l=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4}\mathrm{......}$
 i.e.,              $l=(2r+1)\frac{\lambda }{4} ,r=0,1,2,3,\mathrm{.....}$
or                  $\lambda =\frac{4l}{2r+1}r=0,1,2,3,\mathrm{.....}$
for hydrogen   ${\lambda }_1=4l/(2r_1+1)$
for oxygen ${\lambda }_2=4l/(2r_2+1)$ In both gases the frequency is same
As     $n_1=n_2$
Or    $\frac{v_1}{{\lambda }_1}=\frac{v_2}{{\lambda }_2}$
  $\frac{v_1}{v_2}=\frac{{\lambda }_1}{\lambda }=\frac{l_1}{l_2}\times \frac{(2r_2+1)}{(2r_1+1)}$
As    $l_1=l_2=0.5m$
i.e.,    $\frac{v_1}{v_2}=\frac{2r_2+1}{2r_1+1}=\frac{1100}{300}=\frac{2r_2+1}{2r_1+1}$
i.e.,    $\frac{2r_1+1}{2r_2+1}=\frac{3}{11}$
For minimum frequency, the integers $r_1$  and $r_2$  should be least. 
Therefore by inspection  $r_1=1$ and  $r_2=5$
Therefore, the frequency of oscillations is given by
   $n_{\min }=\frac{v_1}{{\lambda }_1}=(2r_1+1)\frac{v_1}{4l}$
  $=(2\times 1+1)\times \frac{1100}{4\times 0.5}=\frac{3\times 1100}{2}=1650Hz$