The number of air molecules per cm$^3$ increased from $3\times10^{19}$ to $12\times10^{19}$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:

1. The collision frequency (f) is given by the formula : <br/><br/>$f = \sqrt{2} \pi d^2 v n_v$ <br/><br/> Where: <br/><br/>- $d$ is the diameter of the molecule, <br/><br/>- $v$ is the average velocity of the molecules, and <br/><br/>- $n_v$ is the number density (number of molecules per unit volume). <br/><br/>2. From this equation, we can see that the collision frequency (f) is directly proportional to the number density ($n_v$), because all the other variables ($d$ and $v$) are constant : <br/><br/>$f \propto n_v$ <br/><br/>3. Therefore, the ratio of two different collision frequencies ($f_1$ and $f_2$) is equal to the ratio of their corresponding number densities ($n_{v1}$ and $n_{v2}$): <br/><br/>$\frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}}$ <br/><br/>4. Given that the number density increased from $3 \times 10^{19}$ to $12 \times 10^{19}$, we can substitute these values into the equation to find the ratio of the collision frequencies: <br/><br/> $\frac{f_1}{f_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}}$ <br/><br/>5. Simplifying this equation gives: <br/><br/> $\frac{f_1}{f_2} = 0.25$ <br/><br/>So, the ratio of the collision frequency of air molecules before and after the increase in number is 0.25.