illustrations . . . Liquid drops are always spherical The bristles of a paint brush spread out when the brush is immersed in water. However, when the brush is taken out, the bristles cling together. The film of water formed between adjacent bristles is under tension, therefore, it shrinks and draws the bristles together. Some insects, like water spiders walk on water surface as if they are walking on a stretched elastic membrane. Their feet produce 'dimples on the surface film without rupturing the film.

called and the force of attraction is called the ‘Cohesive Force’. Intermolecular forces Let us understand this concept by considering water molecules in a beaker. COHESIVE FORCE : Attraction between two molecules of the is called and the force of attraction is called the ‘Cohesive Force’. ADHESIVE FORCE : Attraction between two molecules of is called and the force of attraction is called the ‘Adhesive Force’. n Water Molecules same substance Oil Molecules ‘cohesion’ Now let us consider a layer of oil molecules above the water surface different substances ‘adhesion’ Cohesion Adhesion

F1 > F2 > F3 F1 F2 F3 F3 F2 F1

Surface tension on the basis of molecular theory Sphere of influence C P Q B Surface Film R R = 10–8 m A Q’ P’ FR FR

l F F F F F A B O1 O2 O3 O4 O5 F F F F F

Relation between surface energy and surface tension.  Inward force due to film F'  External force applied on wires CD F F' F F' l dw = F' dx A D dx D' dw = 2 T l dx = = T (2l dx) T (2l dx) But, dw = T (dA) F l T = l dx = dA 2 2 l dx dw = = Surface energy dA (Increase in surface area due to both sides of film ) (Increase in surface area of film ) Potential energy But, F = F' F = T l 2 F = 2 T l (Force due to both sides of film)  Surface energy = T dA F' = 2 T l

Analysis of molecular forces Solid Air Fa A P Fc Q Liquid

Angle of contact () CASE : 1 Air Liquid surface is Concave upward A P R Q Liquid (kerosene) T  Solid ↑ Angle of contact is acute

CASE : 2 Air Solid Liquid surface is convex upward T A P A Angle of contact is obtuse  R Q Liquid (mercury)

CASE : 3 Air Surface of water is perpendicular to resultant force AR P A solid R  = 00 Liquid (water)

Pressure on two sides of a liquid surface PA PA A A PA B B B PB PB PB PA  PB PA > PB PA < PB

Rise in level of liquid. 1) PA < PB, (The liquid inside the capillary is concave) 2) PA = PC, (Both point A and C are at the same horizontal level. They exert only atmospheric pressure.) A C 3) PC = PD (The liquid surface is plane.) B D  PA = PD  PD < PB

Fall in level of liquid. 1) PA > PB, (The liquid inside the capillary is concave) 2) PA = PC, (Both point A and C are at the same horizontal level. They exert only atmospheric pressure.) A C 3) PC = PD (The liquid surface is plane.) B D  PA = PD  PD > PB

Rise of liquid in a capillary tube : T cos T cos R T R T Resolution of Tension According to Newton’s 3rd law   T sin T sin Horizontal Component R = T = T sin   T T Total upward Force = Weight of liquid column in capillary tube h Vertical Component = T cos Total effect of Horizontal Component of Force (M = V  r ) W 2r  T cos  = M =  g 2T cos  = r h r g (V = A  h ) Total effect of Vertical Component of Force Total effect of Vertical Component of Force = T cos  2r = V  r  g 2 T cos = T cos  2r (A =  r2) h = = A  h r r g  r  g = =  r2  h  r g  r2  h  r g T cos  2r r Mg sin  Mg sin 

dW A1 = 4 r2  (i) (A) T = dA A2 = 4 ( ) r + ∆r 2  (ii) Excess pressure inside drop P0 dW A1 = 4 r2  (i) (A) T = dA A2 = 4 ( ) r + ∆r 2  (ii) r + ∆r Since Pi > P0 Pi  dW dW = = dF T dA ∆r (B) = 4 (r2 + 2r∆r + ∆r2) dW dW = T (8r ∆r) = = (Pi – P0) 4r2 ∆r = (Pi – P0) T (8r ∆r) 4r2 ∆r = 4r2 + 8r ∆r + 4 ∆r2 From (A) and (B) = = 4r2 + 8r ∆r 4r2 + 8r ∆r + As ∆r is very small  ∆r2 is neglected dA = 8r ∆r Excess force Excess pressure T (8 r ∆r) = (Pi – P0) 4r2 ∆r i.e. 4  area = ∆r2  dA = A2 – A1 8T = 4 (Pi – P0) r = 4r2 + 8r ∆r – 4 r2 dF = (Pi – P0) 4r2 = 8r ∆r = 8r ∆r 2T (Pi – P0) = r

Excess pressure inside bubble In case of bubble, there are two free surfaces in contact with air.  dA = 2 (8  r ∆r ) = 16  r ∆r dw = T . dA dw = 16  r ∆rT  (A) Also, dw = (Pi – Po)  4r2 ∆r  (B) From A & B  16  r ∆r T = (Pi – Po)  4  r2 ∆r 4T r  (Pi – Po) =