Question

A water drop of 0.01 cm³ is squeezed between two glass plates and spreads in to area of 10 cm². If surface tension of water is 70 dyne / cm then the normal force required to separate glass plates from each other will be ______.

Options

• 12 N

• 14 N

• 16 N

• 28 N

Answer 1

A water drop of 0.01 cm³ is squeezed between two glass plates and spreads in to area of 10 cm². If surface tension of water is 70 dyne / cm then the normal force required to separate glass plates from each other will be 14 N.

Explanation:

Force F = \[\frac {2TA}{d}\]

= \[\frac {2TA^2}{V}\]   (\[\because\]d = \[\frac {V}{A}\] = Thickness of the layer)

= \[\frac{2\times(70)\times(10)^2}{0.01}\] = 14 ×10⁵ dyne

= 14 N