JEE Main 2026 — 22 January, Evening Shift — Question 28

Collision frequency $(z)=\sqrt{2} \pi d^{2} N \sqrt{\frac{8 R T}{\pi M}}$ Temp, N are same $\mathrm{Z} \propto \frac{\mathrm{d}^{2}}{\sqrt{\mathrm{M}}}$ $\mathrm{d}_{\mathrm{A}}=\frac{\mathrm{d}_{\mathrm{B}}}{2}$ $\mathrm{M}_{\mathrm{A}}=4 \mathrm{M}_{\mathrm{B}}$ $\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\mathrm{d}_{\mathrm{A}}^{2}}{\sqrt{\mathrm{M}_{\mathrm{A}}}} \times \frac{\sqrt{\mathrm{M}_{\mathrm{B}}}}{\mathrm{d}_{\mathrm{B}}^{2}}=\left(\sqrt{\frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}}\right)\left(\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}\right)^{2}$ $=\left(\sqrt{\frac{1}{4}}\right)\left(\frac{1}{2}\right)^{2}$ $\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$ $\Rightarrow \mathrm{Z}_{\mathrm{A}}=\frac{32 \times 10^{8}}{8}=4 \times 10^{8} / \mathrm{s}$