A drop of water of volume V is pressed between the two glass plates so as to spread to an area A. If T is the surface tension, the normal force required to separate the glass plates is

To find the normal force required to separate the two glass plates, we can analyze the pressure difference across the liquid-gas interface of the water film between the plates.

Let $d$ be the distance of separation between the two plates, and $A$ be the area of the water film. The volume $V$ of the water drop is related to the area and separation distance by:
$V=A·d$
This gives the distance of separation as:
$d=\frac{V}{A}$

Due to surface tension, the water boundary between the plates forms a curved meniscus. Assuming complete wetting of glass by water (where the contact angle is approximately 0), the shape of the meniscus is semicircular with a radius of curvature:
$R=\frac{d}{2}$

According to the Laplace pressure formula, the excess pressure difference $\Delta P$ between the atmospheric pressure outside and the pressure inside the water film is given by:
$\Delta P=T\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$
Here, one radius of curvature is $R_1=\frac{d}{2}$ and the lateral radius of curvature $R_2$ along the perimeter of the flat film is extremely large compared to $d$ (so $\frac{1}{R_2}\approx 0$). Thus:
$\Delta P=\frac{2T}{d}$

The lower pressure inside the liquid film creates an attractive force holding the plates together. The normal force $F$ required to pull the plates apart is equal to the product of this pressure difference and the area $A$:
$F=\Delta P·A=\frac{2T}{d}·A$

Substituting the value of $d=\frac{V}{A}$ into the equation:
$F=\frac{2T}{\left(\frac{V}{A}\right)}$·A=2TA2V

Therefore, the normal force required to separate the glass plates is $\frac{2T{A}^2}{V}$.