Question

If two glasses plates have water between them and are separated by very small distance (see in figure), it is very difficult to pull them apart. It is because the water in between forms the cylindrical surface on the side that gives rise to lower pressure in the water in comparison to the atmosphere. If the radius of the cylindrical surface is R and surface tension of water is T then what is the lowered pressure between the plates?

(A) ${\displaystyle \frac{T}{4R}}$
(B) ${\displaystyle \frac{T}{2R}}$
(C) ${\displaystyle \frac{4T}{R}}$
(D) ${\displaystyle \frac{2T}{R}}$

Answer 1

Hint : Use the excess pressure formula to find the pressure lowered between the plates i.e.
 $\mathrm{\Delta }P=T[{\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}]$
Here, T is the surface tension of water
 $R_1$ and $R_2$ are the radii of the two surfaces due to which the pressure is lowered.

Complete step by step answer
The lowered pressure between the walls of the glass plates is due to the formation of the cylindrical surface on the sides. This lowered pressure is equal to the excess pressure which is given by:
 $\mathrm{\Delta }P=T[{\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}]$
This excess pressure is the same as the excess pressure inside a drop or air bubble.
So $R_1=R_2=R$ (Radius of the cylindrical surfaces in contact with the glass plates)
 $\begin{array}{rl}& \mathrm{\Delta }P=T[{\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}}]\\ & \mathrm{\Delta }P=T\left[{\displaystyle \frac{2}{R}}\right]\\ & \mathrm{\Delta }P={\displaystyle \frac{2T}{R}}\end{array}$
Therefore, option (D) is correct.

Note
 $R_1$ and $R_2$ are the radii of the curved surfaces which cause the lowering of the pressure in any medium. Here the curved surface is a cylinder with radius R. Since this behaves as an air bubble in water that is why we have taken the radii to be equal. Otherwise, one should calculate the excess pressure using the radii of the given shape which is not a cylinder or spherical as an air bubble.