Physics · 2026
JEE Main · 28 January 2026, Shift 2 · Q34
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 ×…
The mean free path of a molecule of diameter $\displaystyle 5 \times 10^{-10} \mathrm{~m}$ at the temperature $\displaystyle 41$ °C and pressure $\displaystyle 1.38 \times 10^5 \mathrm{~Pa}$, is given as $\displaystyle \_\_\_\_$ m. (Given $\displaystyle k_B=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ ).
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 2 \sqrt{2} \times 10^{-8}$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.