Answer:
Option A
Explanation:
Momentum imparted due to first collision
$\Rightarrow 2mv\sin45$ =$\sqrt{2mv}$ \[[\therefore sin45^{0}=\frac{1}{\sqrt{2}}]\]
$\therefore$ Pressure on surface
= $\frac{n\sqrt{2}mv}{Area}=\frac{10^{23}\times \sqrt{2}\times 3.32\times 10^{-27} \times10^{3}}{(2\times 10^{-2})^{2}}$
$p =2.35\times 10^{3} N/m^{2}$