Question

Two glass plates are separated by water. If surface tension of water is $75$ dyne/cm and area of each plate wetted by water is $8c{m}^2$ and the distance between the plates is $0.12mm$, then the force applied to separate the two plates is

• A. $10^2$ dyne
• B. $10^4$ dyne
• C. $10^5$ dyne
• D. $10^6$ dyne

Answer 1

Answer: (C)

The shape of water layer between the two plates is shown in the figure.
Thickness $d$ of the film $=0.12mm$
$=0.012cm$
Radius $R$ of the cylindrical face $=\frac{d}{2}$
Pressure difference across the surface
$=\frac{T}{R}=\frac{2T}{d}$
Area of each plate wetted by water $=A$
Force $F$ required to separate the two plates is given by
$F=$ pressure difference $\times$ area
$=\frac{2T}{d}A$
Putting the given values, we get
$F=\frac{2\times 75\times 8}{0.012}=10^5$ dyne