JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(2\) \(cm^2\) at an angle of \(45^o \) to the normal, and rebound elastically with a speed of \(10^3\ m/s\), then the pressure on the wall is nearly:
- A \(4.70 \times 10^3\) \(N/m^2\)
- B \(2.35 \times 10^2 \) \(N/m^2\)
- C \(4.70 \times 10^ 2\) \( N/m^2\)
- D \(2.35 \times 10^3\) \( N/m^2\)
Answer & Solution
Correct Answer
(D) \(2.35 \times 10^3\) \( N/m^2\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} Change\,in\,momentum\\ \Delta P = \frac{P}{{\sqrt 2 }}\hat j + \frac{P}{{\sqrt 2 }}\hat j + \frac{P}{{\sqrt 2 }}\hat i - \frac{P}{{\sqrt 2 }}\hat i\\ \Delta P = \frac{{2P}}{{\sqrt 2 }}\hat j = {I_H}\,molecule\\ \Rightarrow \,{I_{wall}} = - \frac{{2P}}{{\sqrt 2…
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