Question

A drop of water of volume v is pressed between the two glasses placed so as to spread to an area
a. If t is the surface tension, the normal force required to separate the glass plates is

Answer 1

The normal force required to separate the glass plates is 2TA²/V

Given-

• Volume of water drop = V
• Spreading area = A
• Surface tension = T

We know that

Normal force required will be = 2 × (Surface Tension of liquid) × (Area)²/(Volume of liquid)

And we know that Volume V = A × t where t is the thickness

By substituting the values we get

F = 2AT/t × A/A = 2TA²/(A×T) = 2TA²/V