Water is filled up to a height $h$ in a beaker of radius $R$ as shown in the figure. The density of water is $\rho$, the surface tension of water is $T$ and the atmospheric pressure is $P_0$. Consider a vertical section $ABCD$ of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude

Step 1: Determine the dimensions of the vertical section.

The vertical section $ABCD$ passes through the diameter of the beaker.

• Width of the section: $2R$
• Height of the section: $h$

Step 2: Calculate the force due to pressure.

At a depth $z$ below the free surface of the water, the pressure is given by:

$$P(z) = P_0 + \rho g z$$

The force $F_p$ due to this pressure is found by integrating over the area of the section:

$$F_p = \int_0^h (P_0 + \rho g z)(2R) \, dz$$ $$F_p = 2R \left[ P_0 z + \frac{1}{2} \rho g z^2 \right]_0^h$$ $$F_p = 2P_0Rh + \rho g R h^2$$

Step 3: Calculate the force due to surface tension.

Surface tension acts along the top edge of the section of length $2R$ at the free surface, pulling the two halves of the water column together.

$$F_T = 2RT$$

Since surface tension pulls the surface together, this force acts in the direction opposite to the pressure force.

Step 4: Find the magnitude of the net force.

The net force $F$ is the difference between the pressure force and the surface tension force.

$$F = F_p - F_T$$ $$F = 2P_0Rh + \rho g R h^2 - 2RT$$

Taking the magnitude of this net force:

$$F = |2P_0Rh + \rho g R h^2 - 2RT|$$

Answer: The magnitude of the force is $|2P_0Rh + \rho g R h^2 - 2RT|$.