Question:
If $10^{22}$ gas molecules each of mass $10^{-26} \mathrm{~kg}$ collide with a surface (perpendicular to it) elastically per second over an area $1 \mathrm{~m}^{2}$ with a speed $10^{4} \mathrm{~m} / \mathrm{s}$, the pressure exerted by the gas molecules will be of the order of:
$10^{4} \mathrm{~N} / \mathrm{m}^{2}$
$10^{8} \mathrm{~N} / \mathrm{m}^{2}$
$10^{3} \mathrm{~N} / \mathrm{m}^{2}$
$2\mathrm{~N} / \mathrm{m}^{2}$
Question of from chapter.
JEE Main Previous Year 8 April 2019 I
Correct Option: 4
Solution:
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