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Physics · Kinetic Theory of Gases

JEE Main 2026 — 28 January, Evening Shift — Question 29

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

  1. Option A:

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

  2. Option B:

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

  3. Option C:

    22×10−82 \sqrt{2} \times 10^{-8}

    Correct
  4. Option D:

    2×10−82 \times 10^{-8}

Answer: C

Step-by-step solution

λ=kBT2πσ2P\lambda=\frac{k_{B} T}{\sqrt{2} \pi \sigma^{2} P} =1.38×10−23×(273+41)×1002×3.14×(5×10−10)2×1.38×105=22×10−8=\frac{1.38 \times 10^{-23} \times(273+41) \times 100}{\sqrt{2} \times 3.14 \times\left(5 \times 10^{-10}\right)^{2} \times 1.38 \times 10^{5}}=2 \sqrt{2} \times 10^{-8}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Pressure of Gas and Different speeds of Gas Molecules
The mean free path of a molecule of diameter 5 × 10 -10 m at the… | JEE Main 2026 PYQ with Solution · DhiX AI