Question

Let $T_1$ be surface tension between solid and air, $T_2$ be the surface tension between solid and liquid and $T$ be the surface tension between liquid and air. Then in equilibrium, for a drop of liquid on a clean glass plate, the relation is.

• A. $\cos \theta =\frac{T}{T_1+T_2}$
• B. $\cos \theta =\frac{T}{T_1-T_2}$
• C. $\cos \theta =\frac{T_1+T_2}{T}$
• D. $\cos \theta =\frac{T_1-T_2}{T}$

Answer 1

Answer: (D)

$T_1+T\cos (\pi -\theta )=T_2$

$\therefore \cos (\pi -\theta )=\frac{T_2-T_1}{T}$
$\therefore \cos \theta =\frac{T_1-T_2}{T}$