Answer

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**Hint:**This question is based on momentum conservation. That is the initial momentum of a body will be as same as the final momentum of the same body. This can be shown in an expression like,

${{m}_{i}}{{u}_{i}}\cos {{\theta }_{i}}={{m}_{f}}{{u}_{f}}\cos {{\theta }_{f}}$

**Complete step by step answer:**

First of all let us discuss the momentum. Momentum is the impact happening to a body due its mass and its velocity. Momentum is the product of its mass and the velocity. Momentum is a vector quantity in which it is having both direction and magnitude.

Newton's second law of motion states that the time rate of change of momentum is the same as the force acting on the particle. Here in the question it is given that,

$\begin{align}

& u={{10}^{3}}m{{s}^{-1}} \\

& m=3.2\times {{10}^{-27}}kg \\

& n={{10}^{23}} \\

\end{align}$

Therefore the change in momentum in y direction of one \[{{H}_{2}}\] molecule is given by the equation,

\[\Delta P=mu\cos 45{}^\circ -\left( -mu\cos 45{}^\circ \right)=2mu\cos 45{}^\circ \]

Now the change in momentum in y direction of total \[{{H}_{2}}\]will be

\[\Delta P=2mnu\cos 45{}^\circ \]

Area of the wall at which it is incident,

\[A=2c{{m}^{2}}=2\times {{10}^{-4}}{{m}^{2}}\]

Therefore the resultant pressure on the wall will be

\[\begin{align}

&P=\dfrac{\Delta {{P}_{T}}}{A}=\dfrac{2\times {{10}^{23}}\times 3.32\times {{10}^{-27}}\times {{10}^{3}}\times 0.707}{2\times {{10}^{-4}}} \\

& =2.35\times {{10}^{3}}N{{m}^{-2}} \\

\end{align}\]

\[\begin{align}

& P=\dfrac{\Delta {{P}_{T}}}{A}=\dfrac{2\times {{10}^{23}}\times 3.32\times {{10}^{-27}}\times {{10}^{3}}\times 0.707}{2\times {{10}^{-4}}} \\

& =2.35\times {{10}^{3}}N{{m}^{-2}} \\

\end{align}\]

**Hence, the correct answer is option C.**

**Note:**

There are two kinds of momentum. Angular momentum and linear momentum. Angular momentum is defined as the inertia of rotation motion. Linear momentum is defined as inertia of translation motion. The significant difference is that the type of motion which is related to each momentum is different. It is important to notice the place where the force related to rotation applies, which appears as 'r' in the equation.

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