Two large glass plates are placed vertically and parallel to one...

Let the length of the plates be $L$ and the separation between them be $d$.

The upward force due to surface tension acts along the length of the two plates. Assuming the angle of contact for water and glass to be $\theta = 0^{{}^{{}^{\circ}}}$, the upward force is given by:

$F = 2TL \cos\theta = 2TL$

The weight of the water column raised to a height $h$ is:

$W = \text{volume} \times \rho \times g = (Ldh)\rho g$

At equilibrium, the upward force balances the weight of the water column:

$2TL = Ldh\rho g$

$h = \dfrac{2T}{\rho g d}$

Given $T = 0.075 \text{ N/m}$, $d = 1 \text{ mm} = 10^{-3} \text{ m}$, $\rho = 1000 \text{ kg/m}^3$, and taking $g = 10 \text{ m/s}^2$:

$h = \dfrac{2 \times 0.075}{1000 \times 10 \times 10^{-3}}$

$h = \dfrac{0.15}{10} = 0.015 \text{ m} = 1.5 \text{ cm}$