JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2018
- Medium
- MCQ
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:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2.35 \times 10^3\) \( N/m^2\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Change\,in\,momentum\\
\Delta P = \frac{P}{{\sqrt 2 }}\hat j + \frac{P}{{\sqrt 2 }}\hat j + \frac{P}{{\sqrt 2 }}\hat i - \frac{P}{{\sqrt 2 }}\hat i\\
\Delta P = \frac{{2P}}{{\sqrt 2 }}\hat j = {I_H}\,molecule\\
\Rightarrow \,{I_{wall}} = - \frac{{2P}}{{\sqrt 2 }}\hat j
\end{array}\)
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 }}\hat j
\end{array}\)
\(\begin{array}{l}
pressure,P\\
= \frac{F}{A} = \frac{{\sqrt 2 P}}{A}n\,\,\,\,\,\left( {\,n = no.of\,particles} \right)\\
= \frac{{\sqrt 2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}}}{{2 \times {{10}^{ - 4}}}}\\
= 2.35 \times {10^3}N/{m^2}
\end{array}\)
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