MHT CET202521 Apr 2025Morning ShiftPhysicsMechanical Properties of FluidsActual
A liquid drop of volume V is placed on the surface of glass plate. Then another glass plate is placed on it such that liquid forms a thin layer of area A between the surfaces of two plates. To separate the plates a force F has to be applied normal to the surfaces. The surface tension of the liquid is
Options
- AFV 2 ~A
- BF V 2 A^2
- CF V A^2
- DF V A
Correct answer
B. F V 2 A^2
Step-by-step solution
Separating glass plates with liquid film. The force F required to overcome surface tension attraction depends on the liquid volume V , plate area A , and surface tension T . The film thickness d relates to volume and area as d = V/A . The separation force for parallel plates with surface tension T at distance d is given by F = 2AT/d , accounting for two liquid-air interfaces. Substituting the thickness expression yields F = 2AT/(V/A) = 2A^2T/V . Solving for surface tension gives T = FV/(2A^2) . The result correspon