Q 11-12-077JEE MainJEE Main 2023 (6 Apr, Shift 1)Easy
The number of air molecules per $\text{cm}^3$ is 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
Answer: (D) 0.25
Collision frequency $=\dfrac{v}{\lambda}$ with $\lambda\propto\dfrac1n$, so it is proportional to $n$: $\dfrac{3}{12}=0.25$.
Solution by Sreeraj P, M.Sc Physics