Question

If ${10}^{26}$ gas molecules each of mass $2\times {10}^{-26}kg$ collides elastically per second on a surface of dimension $2m\times 4m$ at an angle ${37}^{\circ }$ with the normal to the surface with speed of ${10}^{4}m/s$, then pressure on surface will be

• A. 400 Pa
• B. 8000 Pa
• C. 4000 Pa
• D. 200 Pa

Answer 1

The correct option is C 4000 Pa


$\Delta p=2mv\cos \theta F=\frac{\Delta p}{\Delta t}=\frac{2Nmv\cos \theta }{1}\text{Pressure}=\frac{F}{A}=\frac{2Nmv\cos \theta }{A}PaP=\frac{2\times 2\times {10}^{-26}\times {10}^4}{2\times 4}\times \frac{4}{5}\times {10}^{26}=4\times {10}^{3}Pa$