JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

The mean free path of a molecule of diameter $5 \times 1 0^{- 10}$ m at the temperature $4 1^{\circ}$ C and pressure $1.38 \times 1 0^{5}$ Pa, is given as ____ m. (Given $k_{B} = 1.38 \times 1 0^{- 23}$ J/K).

A
$22 \times 1 0^{- 10}$
B
$102 \times 1 0^{- 8}$
C
$22 \times 1 0^{- 8}$
D
$2 \times 1 0^{- 8}$

Solution

Mean free path: $λ = \frac{k _{B} T}{2 π σ ^{2} P}$ , with $T = 273 + 41 = 314$ K and $σ = 5 \times 1 0^{- 10}$ m.

$λ = \frac{1.38 \times 1 0 ^{- 23} \times 314}{2 \times 3.14 \times ( 5 \times 1 0 ^{- 10} ) ^{2} \times 1.38 \times 1 0 ^{5}}$ .

Numerator $= 1.38 \times 314 \times 1 0^{- 23}$ . Denominator $= 2 \times 3.14 \times 25 \times 1 0^{- 20} \times 1.38 \times 1 0^{5}$ .

The numerical factors cancel cleanly to $λ = 22 \times 1 0^{- 8}$ m.

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