JEE Main 2019PhysicsCenter of Mass Momentum and CollisionMediumMCQ

JEE Main 2019 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

If 1022 gas molecules each of mass 10-26 kg collides with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 104m/s, the pressure exerted by the gas molecules will be of the order of:

Choose an option

Show full solutionCorrect option: A
Correct answer
A 2 Pa

Step-by-step explanation

Total momentum transferred in one second =n2mv=10222×10-26×104=2
Force experienced by surface =2 N
Pressure on surface =ForceArea=21=2 Pa

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About this question

This is a previous-year question from JEE Main 2019, covering the Center of Mass Momentum and Collision chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.