A water drop of 0.05cm³ is squeezed between two glass plates and spreads into area of 40cm² . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be

The correct option/answer is 45 N.

Let us find the normal force required to separate the two glass plates with a thin film of water between them. Let the surface tension of water be $T$, the volume of the water drop be $V$, the area of the water film be $A$, and the thickness of the water film (distance between the plates) be $d$.

First, we express the volume of the water drop in terms of the area and thickness of the film:
$V=A·d$
From this, the thickness $d$ of the film is given by:
$d=\frac{V}{A}$

The pressure difference (excess pressure) inside the thin water film due to the curved surface (concave meniscus of radius of curvature $R\approx \frac{d}{2}$) is given by:
$\Delta P=\frac{2T}{d}$

Substituting the expression for $d$ into the excess pressure equation, we get:
$\Delta P=\frac{2T}{V/A}=\frac{2TA}{V}$

The normal force $F$ required to separate the glass plates is equal to the pressure difference multiplied by the area of contact $A$:
$F=\Delta P·A=\frac{2T{A}^2}{V}$

Now, let us convert the given values into consistent SI units:
Volume of the water drop, $V=0.05{\text{cm}}^3=0.05·{10}^{-6}{\text{m}}^3=5·{10}^{-8}{\text{m}}^3$
Area of the film, $A=40{\text{cm}}^2=40·{10}^{-4}{\text{m}}^2=4·{10}^{-3}{\text{m}}^2$
Surface tension of water, $T=70\text{dyne/cm}=70·{10}^{-3}\text{N/m}=0.07\text{N/m}$

Substituting these values into the force formula:
$F=\frac{2·(7·{10}^{-2})·{(4·{10}^{-3})}^2}{5·{10}^{-8}}$

Simplifying the numerator and denominator:
$F=\frac{2·7·{10}^{-2}·16·{10}^{-6}}{5·{10}^{-8}}$
$F=\frac{224·{10}^{-8}}{5·{10}^{-8}}$
$F=\frac{224}{5}=44.8\text{N}\approx 45\text{N}$

Thus, the normal force required to separate the glass plates is approximately 45 N.