Consider two boxes containing… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

Consider two boxes containing ideal gases $A$ and $B$ such that their temperatures, pressures and number densities are same. The molecular size of $A$ is half of that of $B$ and mass of molecule $A$ is four times that of $B$ . If the collision frequency in gas $B$ is $32 \times 10^{18} / \mathrm{s}$ then collision frequency in gas $A$ is $\_\_\_\_$ $/ \mathrm{s}$ .

Official previous-year question

Held on 22 Jan 2026 · Verified 6 Jul 2026.

Options

A
$2 \times 10^{8}$
B
$8 \times 10^{8}$
C
$4 \times 10^{8}$
D
$32 \times 10^{8}$

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Solution

Collision frequency $(z) = \sqrt{2} \pi d^2 N \sqrt{\dfrac{8RT}{\pi M}}$

Temp, N are same

$Z \propto \dfrac{d^2}{\sqrt{M}}$

$d_A = \dfrac{d_B}{2}$

$M_A = 4M_B$

$\dfrac{Z_A}{Z_B} = \dfrac{d_A^2}{d_B^2} \times \dfrac{\sqrt{M_B}}{\sqrt{M_A}} = \left(\sqrt{\dfrac{M_B}{M_A}}\right) \left(\dfrac{d_A}{d_B}\right)^2$

$= \left(\sqrt{\dfrac{1}{4}}\right) \left(\dfrac{1}{2}\right)^2$

$\dfrac{Z_A}{Z_B} = \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$

$\Rightarrow Z_A = \dfrac{32 \times 10^8}{8} = 4 \times 10^8 /s$

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Solution

Collision frequency $(z) = \sqrt{2} \pi d^2 N \sqrt{\dfrac{8RT}{\pi M}}$

Temp, N are same

$Z \propto \dfrac{d^2}{\sqrt{M}}$

$d_A = \dfrac{d_B}{2}$

$M_A = 4M_B$

$\dfrac{Z_A}{Z_B} = \dfrac{d_A^2}{d_B^2} \times \dfrac{\sqrt{M_B}}{\sqrt{M_A}} = \left(\sqrt{\dfrac{M_B}{M_A}}\right) \left(\dfrac{d_A}{d_B}\right)^2$

$= \left(\sqrt{\dfrac{1}{4}}\right) \left(\dfrac{1}{2}\right)^2$

$\dfrac{Z_A}{Z_B} = \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$

$\Rightarrow Z_A = \dfrac{32 \times 10^8}{8} = 4 \times 10^8 /s$

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Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics