2. $P Q$ is any chord passing through the focus of the parabola $y^{2}=2 p x$, $M N$ is the projection of $P Q$ on the directrix $l$, the surface area of the solid of revolution obtained by rotating $P Q$ around $l$ once is $S_{1}$, and the surface area of the sphere with $M N$ as its diameter is $S_{2}$. Among the following conclusions, the correct one is
A. $S_{1}>S_{2}$
B. $S_{1}S_{2}$, sometimes $S_{1}=S_{2}$, sometimes $S_{1}<S_{2}$
Let $|P F|=x,|Q F|=y \Rightarrow|P Q|=x+y \Rightarrow|M N|=\sqrt{(x+y)^{2}-(x-y)^{2}}$ $=2 \sqrt{x y} \Rightarrow S_{1}=\pi(x+y)^{2}, S_{2}=4 \pi x y$, so the answer is $C$.