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:

Heat and Thermodynamics PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q60 with Solution

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Correct answer: B

For a gas, the collision frequency per molecule is

z = \sqrt{2}\,\pi d^2 n \bar{v}

where:

Assuming the gas and temperature remain the same, $d$ and $\bar{v}$ are constant. Hence,

z \propto n

Initial number density:

n_1 = 3\times 10^{19}\;\text{cm}^{-3}

Final number density:

n_2 = 12\times 10^{19}\;\text{cm}^{-3}

The question asks for the ratio of collision frequency before and after the increase, respectively:

\frac{z_1}{z_2} = \frac{n_1}{n_2}

Substitute values:

\frac{z_1}{z_2} = \frac{3\times 10^{19}}{12\times 10^{19}} = \frac{3}{12} = \frac{1}{4} = 0.25

$0.25$ corresponds to Option B.

Stored correct answer: B

Our derived answer is also B. So they agree.