Step-by-step Derivation

1. Identify the forces: We need to find the total force exerted by the water on one side of the vertical section ABCD on the water on the other side. This force is an internal force within the fluid. We can calculate it by considering all the interactions across this imaginary dividing plane.

2. Interactions across the plane: There are two main interactions across the vertical plane ABCD: a. A force due to the pressure of the water, acting on the entire rectangular area of the section. b. A force due to the surface tension of the water, acting along the line where the section ABCD intersects the free surface of the water.

3. Calculate the force due to pressure ($F_P$): The pressure in the water varies with depth. At a depth z from the surface, the pressure is given by the hydrostatic pressure formula: $P(z) = P_0 + \rho g z$ where $P_0$ is the atmospheric pressure, $\rho$ is the density of water, and $g$ is the acceleration due to gravity.

   The section ABCD is a rectangle with width equal to the diameter of the beaker, $2R$, and height $h$. To find the total force due to pressure, we integrate the pressure over the area of this rectangle. Consider a small horizontal strip of height dz at depth z. Its area is $dA = 2R \cdot dz$.

   The force on this strip is $dF_P = P(z) \cdot dA = (P_0 + \rho g z) (2R \cdot dz)$.

   Integrating from the surface ($z=0$) to the bottom ($z=h$): $F_P = \int_0^h (P_0 + \rho g z) (2R) dz$ $F_P = 2R \int_0^h (P_0 + \rho g z) dz$ $F_P = 2R \left[ P_0 z + \frac{1}{2} \rho g z^2 \right]_0^h$ $F_P = 2R \left( P_0 h + \frac{1}{2} \rho g h^2 \right)$ $F_P = 2P_0 R h + R \rho g h^2$ This force acts perpendicularly to the plane, pushing the water on one side away from the other.

4. Calculate the force due to surface tension ($F_T$): Surface tension acts at the free surface of the water. The dividing plane ABCD cuts the surface along a line segment which is the diameter of the beaker. The length of this line is $2R$.

   Surface tension ($T$) is defined as force per unit length. The molecules on one side of this line pull the molecules on the other side. This force acts along the surface and perpendicular to the line. $F_T = T \times \text{length} = T \times (2R) = 2RT$ This force tends to minimize the surface area, so it acts to pull the two halves of the water surface together. Therefore, it acts in the opposite direction to the pressure force.

5. Calculate the total force: The total force is the resultant of the pressure force and the surface tension force. Since they act in opposite directions, the magnitude of the net force is the absolute difference between their magnitudes: $F_{\text{net}} = |F_P - F_T| = |2P_0 R h + R \rho g h^2 - 2RT|$

6. Compare with options: Let's examine the given options: A: $|2 P_0 Rh + \pi R^2 \rho g h - 2RT|$ - Incorrect hydrostatic term. B: $|2 P_0 Rh + \pi R \rho gh^2 - 2RT|$ - This option has the correct terms for atmospheric pressure and surface tension. However, the hydrostatic term is $\pi R \rho g h^2$, which differs from our derived result $R \rho g h^2$ by a factor of $\pi$. C: $|P_0 \pi R^2 + R \rho g h^2 - 2RT|$ - Mixes vertical and horizontal forces. D: $|P_0 R^2 + R \rho g h^2 + 2RT|$ - Incorrect terms and signs.

7. Conclusion: Our derived expression, based on fundamental principles, is $|2P_0 R h + R \rho g h^2 - 2RT|$. None of the options exactly match this result. However, option (B) has the correct structure, correctly identifying all the contributing forces and their signs. The terms for atmospheric pressure force ($2P_0Rh$) and surface tension force ($2RT$) are correct. The hydrostatic force term ($R\rho gh^2$) appears to have a typo, with an extra factor of $\pi$. In multiple-choice questions with errors, the intended answer is often the one that is structurally and conceptually closest to the correct one. Therefore, option (B) is the most likely intended answer, despite the apparent typo.