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 $l$ is $S_{1}$, and the surface area of the sphere with diameter $M N$ is $S_{2}$. Among the following conclusions, the correct one is
(A) $S_{1}>S_{2}$;
(B) $S_{1}<S_{2}$;
(C) $S_{1}=S_{2}$;
(D) $S_{1}>S_{2}$, sometimes $S_{1}=S_{2}$, sometimes $S_{1}<S_{2}$.
2. (C)

Among the four options (A), (B), (C), and (D), (C) includes (A), meaning if (A) is correct, then (C) must also be correct, so (A) does not need to be considered.

When the chord $P Q$ passing through the focus has a very small inclination angle (the angle with the $x$-axis), the generatrix of the solid of revolution (a frustum) formed by rotating $P Q$ around $l$ is very long, while the corresponding height is very short, meaning the diameter of the sphere is very small, so (B) does not hold.

We know that the lateral surface area of the frustum is determined by the generatrix $P Q$ and the radii of the two bases $P M$ and $Q N$. According to the definition of a parabola,
$$
|P M|=|P F|,|Q N|=|Q F| \quad \text{(where } F \text{ is the focus)}
$$

If we take the inclination angle $\alpha$ of $P Q$ as a parameter, and set
$$
|P F|=\rho_{1},|Q F|=\rho_{2}
$$

then $|P M|=\rho_{1},|Q N|=\rho_{2},|P Q|=\rho_{1}+\rho_{2}$,
thus
$$
\begin{array}{c}
S_{1}=\pi\left(\rho_{1}+\rho_{2}\right)^{2}, \\
S_{2}=\pi \cdot|M N|^{2}=\pi \sin ^{2} \alpha\left(\rho_{1}+\rho_{2}\right)^{2},
\end{array}
$$

Clearly, $S_{1} \geqslant S_{2}$, and equality holds when $\alpha=\frac{\pi}{2}$.