JEE Main On 08 April 2018 Question 2
Question: The mass of a hydrogen molecule is $ 3.32\times {10^{-27}}kg $ . If $ 10^{23} $ hydrogen molecules strike, per second, a fixed wall of area $ \text{2 c}{m^{2}} $ at an angle of $ \text{45 }{}^\circ $ to the normal, and rebound elastically with a speed of $ 1{0^{3}}\text{ m/s} $ , then the pressure on the wall is nearly: [JEE Main Online 08-04-2018]
Options:
A) $ 2\text{.35}\times 1{0^{2}}N/m^{2} $
B) $ 4.70\times 10^{2}N/m^{2} $
C) $ 2.35\times 10^{3}N/m^{2} $
D) $ 4.70\times 10^{3}N/m^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Force = rate of change of momentum (Perpendicular to area) $ =n(2mu\cos \theta ) $
Pressure $ =\frac{Force}{Area}=\frac{n(2mu\cos \theta )}{A} $
$ =\frac{3.32}{\sqrt{2}}\times \frac{{10^{-1}}}{{10^{-4}}}=\frac{3.32}{1.41}\times 10^{3}=2.35\times 10^{3}N/m^{2} $
