Answer: 106

Solution

Given: $V_{\text{air}} = 300$ m/s, $\rho_{\text{gas}} = 2\rho$ air

Using, $V = \sqrt{\frac{B}{\rho}}$

$\frac{V_{\text{gas}}}{V_{\text{air}}} = \sqrt{\frac{B/2\rho_{\text{air}}}{B/\rho_{\text{air}}}}$

$\Rightarrow V_{\text{gas}} = \frac{V}{\sqrt{2}} = \frac{300}{\sqrt{2}} = 150\sqrt{2}$ m/s

And $f_{n\text{th}}$ harmonic $= \frac{nv}{2L}$ (in open organ pipe)

($L = 1$ metre given)

$\therefore f_{2\text{nd}}$ harmonic $- f_{\text{fundamental}} = \frac{2v}{2 \times 1} - \frac{v}{2 \times 1} = \frac{v}{2}$

$\therefore f_{2\text{nd}}$ harmonic $- f_{\text{fundamental}} = \frac{150\sqrt{2}}{2} = \frac{150}{\sqrt{2}} \approx 106$ Hz