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MCQEasyJEE Main 2026 · 28 January, Shift 2Mean Free Path & Avogadro's Number

Physics Question from JEE Main 2026 · 28 January, Shift 2

The mean free path of a molecule of diameter 5×10−10 m5\times10^{-10}\,m at temperature 41∘C41^\circ C and pressure 1.38×105 Pa1.38\times10^5\,Pa is given as _____ mm. (Given kB=1.38×10−23 J/Kk_B=1.38\times10^{-23}\,J/K)

  • A

    22×10−82\sqrt{2}\times10^{-8}

  • B

    102×10−810\sqrt{2}\times10^{-8}

  • C

    2×10−82\times10^{-8}

  • D

    22×10−102\sqrt{2}\times10^{-10}

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