A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is __________ Hz.

We begin with the well-known formula for the speed of sound in an ideal gas, $$v=\sqrt{\dfrac{\gamma P}{\rho}}$$, where $$\gamma$$ is the ratio of specific heats, $$P$$ is the pressure and $$\rho$$ is the density. For two gases at the same pressure and temperature having the same $$\gamma$$, the speed of sound varies inversely as the square root of the density, so $$v\propto\dfrac1{\sqrt{\rho}}$$.

We are told that the given gas has double the density of air at STP. Hence, if the speed of sound in air is $$300\ \text{m s}^{-1}$$, then in the denser gas we have

$$v_{\text{gas}} = \dfrac{300}{\sqrt{2}}\ \text{m s}^{-1} = 300 \times \dfrac{1}{1.414}\ \text{m s}^{-1} = 212.1\ \text{m s}^{-1}\ (\text{approximately}).$$

Now consider the organ pipe. Because both ends are open, the standing-wave pattern has antinodes at each end. For an open-open pipe of length $$L$$, the allowed wavelengths are given by

$$\lambda_n = \dfrac{2L}{n},\qquad n = 1,2,3,\ldots$$

Therefore, the corresponding frequencies are

$$f_n = \dfrac{v}{\lambda_n} = \dfrac{v}{2L}\,n.$$

For the fundamental (first harmonic) we set $$n=1$$:

$$f_1 = \dfrac{v}{2L}.$$

For the second harmonic we set $$n=2$$:

$$f_2 = \dfrac{2v}{2L} = \dfrac{v}{L}.$$

The question asks for the difference between these two frequencies. Substituting the expressions we just obtained, we find

$$f_2 - f_1 = \dfrac{v}{L} - \dfrac{v}{2L} = \dfrac{v}{2L}.$$

The pipe length is given as $$L = 1\ \text{m}$$, so

$$f_2 - f_1 = \dfrac{v}{2(1)} = \dfrac{v}{2}.$$

Finally, insert the speed of sound in the gas, $$v = 212.1\ \text{m s}^{-1}$$:

$$f_2 - f_1 = \dfrac{212.1}{2}\ \text{Hz} \approx 106\ \text{Hz}.$$

Hence, the correct answer is Option 106 Hz.