Explanation

(b) Let $A$ be the circular area over which the liquid wets the plate and $d$ be the distance between two plates. Mass of liquid drop, $m = Ad\rho$. If $S$ is the force of surface tension of water, then excess of pressure inside the liquid film in excess of atmospheric pressure is given by
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$p = \frac{S}{r} = \frac{S}{d/2} = \frac{2S}{d}$
Force of attraction between the plates,
$F = \frac{2S}{d}\ A\ \ \ \ \left\lbrack \because p = \frac{F}{A} \right\rbrack$
Or $F = \frac{2\ S}{\rho d^{2}} \times A\ \rho d = \frac{2\ S\ m}{\rho d^{2}}$
$= \frac{2 \times 0.07 \times (80 \times 10^{- 6})}{10^{3} \times (4 \times 10^{- 8})} = 0.28\ N$