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Physics · Kinetic Theory of Gases

JEE Main 2024 — 5 April, Shift 2 — Question 36

If n is the number density and d is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by :

  1. Option A:

    12nπd2\frac{1}{\sqrt{2 n \pi d^{2}}}

  2. Option B:

    2nπd2\sqrt{2} n \pi d^{2}

  3. Option C:

    12nπd2\frac{1}{\sqrt{2} n \pi d^{2}}

    Correct
  4. Option D:

    12n2π2d2\frac{1}{\sqrt{2} n^{2} \pi^{2} d^{2}}

Answer: C

Step-by-step solution

n=\mathrm{n}= number of molecule per unit volume

d=\mathrm{d}= diameter of the molecule

λ=12πd2n\lambda=\frac{1}{\sqrt{2} \pi d^{2} n} (By Theory)

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 2
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems
If n is the number density and d is the diameter of the molecule… | JEE Main 2024 PYQ with Solution · DhiX AI