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Stuck here, help me understand: - Laws of motion - JEE Main

If 10^{22} gas molecules each of mass 10^{-26} kg collide 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 :

  • Option 1)

    2\; N/m^{2}

  • Option 2)

    10^{4}\; N/m^{2}

  • Option 3)

    10^{16}\; N/m^{2}

     

  • Option 4)

    10^{8}\; N/m^{2}

 
Answers (1)
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S solutionqc

   

\\V_{i}=10^{4}\; m/s\; \; (\widehat{l})\\\; \; \; \\V_{f}=10^{4}\; m/s\; \; (\widehat{-l})

\Delta V=2\times 10^{4}\; m/s

For 1 particle

Momentum  = \Delta P=m\Delta V

change for n particle

\Delta P=m.n.\Delta V

For unit second

Pressure=\frac{F}{A}=\frac{\Delta P}{A}=\frac{2mnV}{A}

Pressure=\frac{2\times10^{-26}\times10^{22}\times10^{4}}{1}

Pressure= 2\; N/m^{2}


Option 1)

2\; N/m^{2}

Option 2)

10^{4}\; N/m^{2}

Option 3)

10^{16}\; N/m^{2}

 

Option 4)

10^{8}\; N/m^{2}

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