A water drop of 0.01 ~cm ^3 is squeezed between two glass plates and spreads in to area of 10 ~cm ^2 . If surface tension of water is 70 dyne / cm then the normal force required to separate glass plates from each other will be

Step-by-step solution

Normal force calculation for separating glass plates with water film The water drop of volume V = 0.01 cm ^3 spreads to cover area A = 10 cm ^2 , forming a film of thickness d = V A = 0.01 10 = 0.001 cm . The separation force counteracts surface tension, given by F = 2AT d for water with T = 70 dyne/cm . Substituting values: F = 2 10 70 0.001 = 1400 0.001 = 1.4 10^6 dyne Converting to Newtons using 1 N = 10^5 dyne : F = 1.4 10^6 10^5 = 14 N The required normal force is 14 N .