A drop of volume \[V\] is pressed between two glasses placed as to spread to an area \[A\]. If \[T\] is the surface tension, the normal force required to separate the glass plates is
A. \[\dfrac{{T{A^2}}}{V}\]
B. \[\dfrac{{2T{A^2}}}{V}\]
C. \[\dfrac{{4T{A^2}}}{V}\]
D. \[\dfrac{{T{A^2}}}{{2V}}\]

Hint: In this question, when a drop is spread between two glass plates a surface tension is developed and a force is developed which makes the glass stick. To separate the glass plates, we need to apply a normal force equal to this force. We know excess pressure inside a drop is given by \[\Delta p = \dfrac{{2T}}{R}\].

Complete step by step answer:
The volume of the drop is \[V\]. The area over which the drop is spread between the two plates is \[A\]. The surface tension developed is given by \[T\]. Let the thickness of the plate be \[x\]. If the thickness of the plate is \[x\] the radius of the drop becomes \[\dfrac{x}{2}\].The excess pressure difference in the drop is given by,
\[\Delta p = \dfrac{{2T}}{R}\]

But we know \[\text{Pressure} = \dfrac{\text{Force}}{\text{Area}}\]
Substituting the values in the excess pressure difference equation,
\[\dfrac{F}{A} = \dfrac{{2T}}{{\dfrac{x}{2}}}\]
\[ \Rightarrow F = \dfrac{{4TA}}{x}\]
As the drop is spread between the plates, the volume between the plates will be equal to the volume of the drop. Therefore,
\[V = Ax\]
Substituting \[x = \dfrac{V}{A}\] we get,
\[F = \dfrac{{4T{A^2}}}{V}\]
Thus, the normal force required to separate the glass plates is \[\dfrac{{4T{A^2}}}{V}\].

Therefore, the correct option is C.

Note: If we have an air bubble inside a liquid, a single surface is formed, and the air is on the concave side and liquid is on the convex side. The pressure in the concave side is greater than the pressure in the convex side by an amount of \[\dfrac{{2T}}{R}\]. This statement can be generalized for a spherical liquid drop.