Magnitude of Force Across Diametrical Vertical Section of Water Column

To find the magnitude of the force exerted on the water on one side of the vertical section $ABCD$ by the water on the other side, we analyze the two interactions acting across the vertical plane $ABCD$:

1. Force due to Hydrostatic Pressure:
   The vertical section $ABCD$ is a rectangle of width equal to the diameter of the beaker, $2R$, and height $h$.
   The absolute pressure at a depth $y$ below the free surface of the water is given by:
   $P(y) = P_0 + \rho g y$

The force exerted on an infinitesimal horizontal strip of height $dy$ and length $2R$ is:
$dF_P = P(y) \, dA = (P_0 + \rho g y)(2R \, dy)$

Integrating from the surface ($y = 0$) to the bottom of the beaker ($y = h$):
$F_P = \int_0^h (P_0 + \rho g y)(2R) \, dy = 2R \left[ P_0 y + \frac{1}{2}\rho g y^2 \right]_0^h$
$F_P = 2 P_0 R h + R \rho g h^2$

This pressure force is compressive and acts perpendicularly outward from the section, pushing the water on this side away.

1. Force due to Surface Tension:
   Along the line of the water's free surface at the top of the section (length $AB = 2R$), surface tension exerts an attractive (tensile) force that acts to pull the two halves of the water surface together.
   The magnitude of this force is:
   $F_T = T \cdot (2R) = 2RT$

Because surface tension acts as a tensile force, its direction is opposite to that of the compressive pressure force.

1. Net Force:
   The net force acting across the section $ABCD$ from the other side of the water is the vector sum of these two collinear opposing forces:
   $F_{\text{net}} = F_P - F_T = 2 P_0 R h + R \rho g h^2 - 2 R T$

Thus, the magnitude of the total force is:
$|F_{\text{net}}| = \left| 2 P_0 R h + R \rho g h^2 - 2 R T \right|$

Therefore, the correct option is B.