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JEE Main 20198 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Morning Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2019 · 8 April · Morning Shift · Physics · Center of Mass, Momentum and Collision

If 10 ^ 22 gas molecules each of mass 10 ^ - 26 kg collides with a...

If

{10}^{22}

gas molecules each of mass

{10}^{- 26}

kg collides with a surface (perpendicular to it) elastically per second over an area

1{m}^{2}

with a speed

{10}^{4}m / s,

the pressure exerted by the gas molecules will be of the order of:

Options

A $2Pa$
B $4Pa$
C $6Pa$
D $8Pa$

Correct answer

A. $2Pa$

Step-by-step solution

Total momentum transferred in one second

= n\left(2mv\right) = {10}^{22}\left(2 \times {10}^{- 26} \times {10}^{4}\right) = 2

Force experienced by surface

= 2N

Pressure on surface

= \dfrac{Force}{Area} = \dfrac{2}{1} = 2Pa

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