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The mean free path of a molecule of diameter 5 × 10^-10 ~m at the temperature 41° C and pressure 1.38 × 10^5 ~Pa, is given as ____ m. (Given k_B=1.38…

JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

The mean free path of a molecule of diameter 5×10−10 m5 \times 10^{-10} \mathrm{~m}5×10−10 m at the temperature 41∘C41^{\circ} \mathrm{C}41∘C and pressure 1.38×105 Pa1.38 \times 10^{5} \mathrm{~Pa}1.38×105 Pa, is given as ____\_\_\_\_____ m. (Given kB=1.38×10−23 J/Kk_{B}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}kB​=1.38×10−23 J/K).

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

  1. A

    102×10−810 \sqrt{2} \times 10^{-8}102​×10−8

  2. B

    22×10−82 \sqrt{2} \times 10^{-8}22​×10−8

  3. C

    22×10−102 \sqrt{2} \times 10^{-10}22​×10−10

  4. D

    2×10−82 \times 10^{-8}2×10−8

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