Question

A drop of some liquid of volume $0.04c{m}^3$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20c{m}^2$ between the surfaces of the two slides. To separate the slides a force of $16\times {10}^5$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-$c{m}^{-1}$):

• A. $60$
• B. $70$
• C. $80$
• D. $90$

Answer 1

Answer: (C)

$80$

Liquid between the slides forms a cylindrical drop.

Excess pressure inside the cylindrical drop $P=\frac{T}{R}=\frac{2T}{d}$ $(\because R=d/2)$

Cross-section area of the liquid surface formed on the slides $A=20$ $c{m}^2$

Volume of the liquid $V=0.04$ $c{m}^3$

Thus separation between the slides $d=\frac{V}{A}=\frac{0.04}{20}=0.002$ $cm$

Using $P=\frac{2T}{d}$ where $T$ is the surface tension of liquid

Thus force required to separate the slides $F=PA=\frac{2T}{d}A$

$\therefore$ $16\times {10}^5=\frac{2T}{0.002}\times 20$ $⟹T=80$ $dyne/cm$