Question

A drop of water of mass 0.2 g is placed between two glass plates. The distance between them is 0.01 cm. The force of attraction between the plates is…. (surface tension of water = 0.07 $N{m}^{-1}$)

• A. 2.8 N
• B. 3.5 N
• C. 0.7 N
• D. 1.25 N

Answer 1

The correct option is A 2.8 N

Let R be the radius of circular layer of water. Then $\pi R^{2}d\rho =m$
Pressure at $A=P_0-\frac{2T}{d}$(meniscus is cylindrical in shape)
Pressure between the plates is less than tha atmospheric pressure and so the plates are pressed together.
F = Force of attraction = $\Delta P\times$ area
$F=\frac{2T}{d}\times \pi R^2$
$=\frac{2T}{d}\times \frac{m}{d.p}=\frac{2Tm}{d^2\rho }$
$=\frac{2\times 0.2\times {10}^{-3}\times 0.07}{{0.01}^2\times {10}^{-4}\times 1000}=2.8$ N