Surface Tension PDF

60 48 Surface Tension 10.1 Intermolecular Force. E3 The force of attraction or repulsion acting between the molecules are known as intermolecular force. The nature of intermolecular force is electromagnetic. The intermolecular forces of attraction may be classified into two types. Cohesive force Adh...

60 48 Surface Tension 10.1 Intermolecular Force. E3 The force of attraction or repulsion acting between the molecules are known as intermolecular force. The nature of intermolecular force is electromagnetic. The intermolecular forces of attraction may be classified into two types. Cohesive force Adhesive force The force of attraction between the molecules of the different substances is called the force of adhesion. Ex. (i) Two drops of a liquid coalesce into one when brought in mutual contact. Ex. (i) Adhesive force enables us to write on the blackboard with a chalk. (ii) It is difficult to separate two sticky plates of glass welded with water. (ii) A piece of paper sticks to another due to large force of adhesion between the paper and gum molecules. D YG U ID The force of attraction between molecules of same substance is called the force of cohesion. This force is lesser in liquids and least in gases. (iii) It is difficult to break a drop of mercury into small droplets because of large cohesive force between the mercury molecules. (iii) Water wets the glass surface due to force of adhesion. Note :  Cohesive or adhesive forces are inversely proportional to the eighth power of distance U between the molecules. 10.2 Surface Tension. ST The property of a liquid due to which its free surface tries to have minimum surface area and behaves as if it were under tension some what like a stretched elastic membrane is called surface tension. A small liquid drop has Imaginary line spherical shape, as due to surface tension the liquid surface tries to have minimum surface area and for a given volume, the sphere has minimum surface area. Surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of liquid, the direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if F is the force acting on one side of imaginary line of length L, then T = (F/L) (1) It depends only on the nature of liquid and is independent of the area of surface or length of line considered. Surface Tension 49 (2) It is a scalar as it has a unique direction which is not to be specified. (3) Dimension : [MT –2 ]. (Similar to force constant) (4) Units : N/m (S.I.) and Dyne/cm [C.G.S.] (5) It is a molecular phenomenon and its root cause is the electromagnetic forces. 60 10.3 Force Due to Surface Tension. If a body of weight W is placed on the liquid surface, whose surface tension is T. If F is the Body Figure (Length = l ) Hollow disc D YG (Inner radius = r1 U T T F = T F × 2l ID F Needle E3 minimum force required to pull it away from the water then value of F for different bodies can be calculated by the following table. Force F = 2l T + W F = 2 (r1 + r2)T + W Outer radius = r2) F Thin ring ST U (Radius = r) Circular plate or disc (Radius = r) F = 2 (r + r)T + W F = 4rT + W F F = 2rT + W 50 Surface Tension F Square frame F = 8l T + W 60 (Side = l ) F Square plate E3 F = 4l T + W ID 10.4 Examples of Surface Tension. (2) When a greased iron needle is placed gently on the surface of water at rest, so that it does not prick the water surface, the needle floats on T T the surface of water despite it being heavier because the weight of needle is balanced by the vertical mg components of the forces of surface tension. If the water surface is pricked by one end of the needle, the needle sinks down. (4) Take a frame of wire and dip it in soap solution and take it out, a soap film will be formed in the frame. Place a loop of wet thread gently on the film. It will remain in the form, we place it on the film according to figure. Now, piercing the film Thread loop with a pin at any point inside the loop, It immediately takes the circular form as shown in figure. D YG U (1) When mercury is split on a clean glass plate, it forms globules. Tiny globules are spherical on the account of surface tension because force of gravity is negligible. The bigger globules get flattened from the middle but have round shape near the edges, figure (3) When a molten metal is poured into water from a suitable height, the falling stream of metal breaks up and the detached portion of the liquid in small quantity acquireMolten the spherical shape. U metal ST Wate r (5) Hair of shaving brush/painting brush when dipped in water spread out, but as soon as it is taken out, its hair stick together. (6) If a small irregular piece of camphor is floated on the surface of pure water, it does not remain steady but dances about on the surface. This is because, irregular shaped camphor dissolves unequally and decreases the surface tension of the water locally. The unbalanced forces make it move haphazardly in different directions. (7) Rain drops are spherical in shape because each drop tends to acquire minimum surface area (8) Oil drop spreads on cold water. Whereas it may remain as a drop on hot water. This is due to Surface Tension 51 due to surface tension, and for a given volume, the surface area of sphere is minimum. the fact that the surface tension of oil is less than that of cold water and is more than that of hot water. 10.5 Factors Affecting Surface Tension. tension of liquid is zero at its boiling point and it vanishes at critical temperature. At temperature, intermolecular forces for liquid and gases becomes equal and liquid can without any restriction. For small temperature differences, the variation in surface with temperature is linear and is given by the relation E3 surface critical expand tension 60 (1) Temperature : The surface tension of liquid decreases with rise of temperature. The Tt  T0 (1   t) where Tt , T0 are the surface tensions at t o C and 0 o C respectively and  is the temperature ID coefficient of surface tension. Examples : (i) Hot soup tastes better than the cold soup. U (ii) Machinery parts get jammed in winter. D YG (2) Impurities : The presence of impurities either on the liquid surface or dissolved in it, considerably affect the force of surface tension, depending upon the degree of contamination. A highly soluble substance like sodium chloride when dissolved in water, increases the surface tension of water. But the sparingly soluble substances like phenol when dissolved in water, decreases the surface tension of water. 10.6 Applications of Surface Tension. (1) The oil and grease spots on clothes cannot be removed by pure water. On the other hand, when detergents (like soap) are added in water, the surface tension of water decreases. As a result of this, wetting power of soap solution increases. Also the force of adhesion between U soap solution and oil or grease on the clothes increases. Thus, oil, grease and dirt particles get mixed with soap solution easily. Hence clothes are washed easily. ST (2) The antiseptics have very low value of surface tension. The low value of surface tension prevents the formation of drops that may otherwise block the entrance to skin or a wound. Due to low surface tension, the antiseptics spreads properly over wound. (3) Surface tension of all lubricating oils and paints is kept low so that they spread over a large area. (4) Oil spreads over the surface of water because the surface tension of oil is less than the surface tension of cold water. (5) A rough sea can be calmed by pouring oil on its surface. (6) In soldering, addition of ‘flux’ reduces the surface tension of molten tin, hence, it spreads. 52 Surface Tension 10.7 Molecular Theory of Surface Tension. The maximum distance upto which the force of attraction between two molecules is appreciable is called molecular range ( 10 9 m). A sphere with a molecule as centre and radius equal to molecular range is called the sphere of influence. The liquid enclosed between free 60 surface (PQ) of the liquid and an imaginary plane (RS) at a distance r (equal to molecular range) from the free surface of the liquid form a liquid film. To understand the tension acting on the free surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure. E3 (1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid. ID (2) Molecule B is little below the free surface of the liquid and it is also attracted equally in all directions. Hence the resultant force on it is also zero. D P R C B Q S A U (3) Molecule C is just below the upper surface of the liquid film and the part of its sphere of D YG influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force. (4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force. Thus all molecules lying in surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane. U Sample problems based on Surface tension Problem 1. A wooden stick 2m long is floating on the surface of water. The surface tension of water ST 0.07 N/m. By putting soap solution on one side of the sticks the surface tension is reduced to 0.06 N/m. The net force on the stick will be (a) 0.07 N (b) 0.06 N Solution : (d) Force on one side of the stick F1  T1  L (c) 0.01 N (d) 0.02 N  0. 07  2  0.14 N and force on other side of the stick F2  T2  L  0.06  2  0.12 N So net force on the stick  F1  F2  0.14  0.12  0.02 N Problem 2. A thin metal disc of radius r floats on water surface and bends the surface downwards along the perimeter making an angle  with vertical edge of disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight of metal disc is (a) 2 rT + W (b) 2 rT cos – W (c) 2 rT cos + W (d) W – 2 rT cos Surface Tension 53 Solution : (c) Weight of metal disc = total upward force T T = upthrust force + force due to surface tension   = weight of displaced water + T cos  (2 r) A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of 2  10 2 N to keep the wire in equilibrium. The surface tension in Nm–1 of water is [AMU (Med.) 1999] (a) 0.1 N/m (b) 0.2 N/m (c) 0.001 N/m Solution : (a) Force on wire due to surface tension F  T  2l T  There is a horizontal film of soap solution. On it a thread is placed in the form of a loop. ID Problem 4. F 2  10 2   0. 1 N /m 2l 2  10  10  2 (d) 0.002 N/m E3 Problem 3. 60 = W + 2 rT cos  The film is pierced inside the loop and the thread becomes a circular loop of radius R. If the surface tension of the loop be T, then what will be the tension in the thread (b) R 2 T (c) 2RT U (a) R 2 / T (d) 2RT Solution : (d) Suppose tension in thread is F, then for small part l of thread l  R  and 2 F sin  / 2  2T l  2TR  D YG TR  TR    2TR (sin  / 2   / 2) sin  / 2  / 2 F  Problem 5. F cos /2 /2 F l 2 × T × l F cos /2  /2 F  /2 /2 F sin/2 F sin/2 A liquid is filled into a tube with semi-elliptical cross-section as shown in the figure. The ratio of the surface tension forces on the curved part and the plane part of the tube in vertical position will be  (a  b) U (a) b 4b 2a b (c) a 4b ST (b) a (d)  (a  b) 4b Solution : (a) From the figure Curved part = semi perimeter   (a  b) 2 and the plane part = minor axis = 2b  Force on curved part = T   (a  b) 2 2b  (a+b) a 2 54 Surface Tension and force on plane part = T  2b  Ratio  Problem 6.  (a  b) 4b A liquid film is formed over a frame ABCD as shown in figure. Wire CD can slide without (a) 60 friction. The mass to be hung from CD to keep it in equilibrium is Tl g A 2 Tl (b) g B D C Liquid film E3 l g (c) 2 Tl (d) T  l m ID Solution : (b) Weight of the body hung from wire (mg) = upward force due to surface tension (2Tl )  2Tl g U 10.8 Surface Energy. D YG The molecules on the liquid surface experience net downward force. So to bring a molecule from the interior of the liquid to the free surface, some work is required to be done against the intermolecular force of attraction, which will be stored as potential energy of the molecule on the surface. The potential energy of surface molecules per unit area of the surface is called surface energy. Unit : Joule/m2 (S.I.) erg/cm2 (C.G.S.) Dimension : [MT–2] U If a rectangular wire frame ABCD, equipped with a sliding wire LM dipped in soap solution, a film is formed over the frame. Due to the surface tension, the film will have a tendency to ST shrink and thereby, the sliding wire LM will be pulled in inward direction. However, the sliding wire can be held in this position under a force F, which is equal and opposite to the force acting on the sliding wire LM all along its length due to surface tension in the soap film. If T is the force due to surface tension per unit length, then C l L L' D F T × 2l B M M x ' A F = T  2l Here, l is length of the sliding wire LM. The length of the sliding wire has been taken as 2l for the reason that the film has got two free surfaces. Suppose that the sliding wire LM is moved through a small distance x, so as to take the position L' M '. In this process, area of the film increases by 2l  x (on the two sides) and to do so, the work done is given by Surface Tension 55 W = F  x = (T  2l)  x = T  (2lx) = T  A  W = T  A [A = Total increase in area of the film from both the sides] If temperature of the film remains constant in this process, this work done is stored in the film as its surface energy. W or T = W A [If A = 1] 60 From the above expression T  E3 i.e. surface tension may be defined as the amount of work done in increasing the area of the liquid surface by unity against the force of surface tension at constant temperature. 10.9 Work Done in Blowing a Liquid Drop or Soap Bubble. (1) If the initial radius of liquid drop is r1 and final radius of liquid drop is r2 then W = T  Increment in surface area W = T  4 [r22  r12 ] ID [drop has only one free surface] (2) In case of soap bubble W = T  8 [r22  r12 ] 10.10 Splitting of Bigger Drop. U [Bubble has two free surfaces] D YG When a drop of radius R splits into n smaller drops, (each of radius r) then surface area of liquid increases. Hence the work is to be done against surface tension. Since the volume of liquid remains constant therefore 4 4 R 3  n r 3  R 3  nr 3 3 3 Work done = T  A = T [Total final surface area of n drops – surface area of big drop] = U T[n4r 2  4R 2 ] Various formulae of work done 4T [nr  R ] 2 4R T [n 2 1/3  1] 4Tr n 2 2/3 [n 1/3  1] ST 2 R 1 1  4TR 3    r R  If the work is not done by an external source then internal energy of liquid decreases, subsequently temperature decreases. This is the reason why spraying causes cooling. By conservation of energy, Loss in thermal energy = work done against surface tension JQ = W 1 1  JmS   4TR 3    r R    J 4 1 1  R 3 d S   4R 3 T    3 r R  [As m = V  d = 4  R3 d ] 3 56 Surface Tension    Decrease in temperature 3T JSd 1 1  r  R    where J = mechanical equivalent of heat, S = specific heat of liquid, d = density of liquid. 10.11 Formation of Bigger Drop. 60 If n small drops of radius r coalesce to form a big drop of radius R then surface area of the liquid decreases. Amount of surface energy released = Initial surface energy – final surface energy Various formulae of released energy 4R 2T (n1 / 3  1) 4 T [nr 2  R 2 ] 4Tr 2n 2 / 3 (n1 / 3  1) E3 E  n 4r 2 T  4R 2 T R 1 1  4TR 3    r R  ID (i) If this released energy is absorbed by a big drop, its temperature increases and rise in 3 T 1 1  temperature can be given by    JSd  r R  v   1 4  1 1  R 3 d v 2  4R 3 T     2 3  r R  D YG 1 1 1  mv 2  4R 3 T    2 r R  U (ii) If this released energy is converted into kinetic energy of a big drop without dissipation then by the law of conservation of energy.  v2  6 T 1 1   d  r R  6T  1 1     d r R Sample problems based on Surface energy Problem 7. Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is (b) 21 / 3 : 1 U (a) 1 : 21 / 3 Solution : (b) As R  n1 / 3 r  2 1 / 3 r  R 2  2 2 / 3 r 2  ST  r2 Initial surface energy 2(4 r 2 T )    2  R2 Final surface energy (4R 2 T )  Problem 8. r 2 R2 (c) 2 : 1 (d) 1 : 2  22 / 3    2  2 2 / 3 = 21/3   Radius of a soap bubble is increased from R to 2R work done in this process in terms of surface tension is [CPMT 1991; RPET 2001; BHU 2003] (a) 24 R 2 S  (b) 48 R 2 S Solution : (a) W  8T R22  R12 Problem 9.   8S [(2 R) 2 (c) 12 R 2 S (d) 36 R 2 S  (R)2 ]  24 R 2 S The work done in blowing a soap bubble of 10cm radius is (surface tension of the soap solution is 3 N /m ) 100 Surface Tension 57 [MP PMT 1995; MH CET 2002] (b) 37.68  10 4 J (a) 75.36  10 4 J Solution : (a) W  8R 2 T  8 (10  10  2 ) 2 (c) 150.72  10 4 J (d) 75.36 J 3  75.36  10  4 J 100 surface energy. (Surface tension of mercury is 0.465 J/m2) (a) 23.4J (b) 18.5J (c) 26.8J 60 Problem 10. A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in (d) 16.8J E3 Solution : (a) Increase in surface energy  4R 2 T (n 1 / 3  1)  4 (2  10 3 )2 (0.465 )(8 1 / 3  1) = 23.4  10 6 J = 23.4 J Problem 11. The work done in increasing the size of a soap film from 10cm  6cm to 10cm  11cm is 3  10 4 J. The surface tension of the film is (b) 3.0  10 2 Nm 1 (c) 6.0  10 2 Nm 1 (a) 1.5  10 2 Nm 1 (d) 11.0  10 2 Nm 1 ID Solution : (b) A1  10  6  60 cm 2  60  10 4 m 2 , A 2  10  11  110 cm 2  110  10 4 m 2 As the soap film has two free surfaces W  T  2 A W 3  10 4   3  10  2 N/m 2  50  10  4 2  50  10  4 U  W  T  2  ( A2  A1 )  T  D YG Problem 12. A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water  7.2  10 2 N / m ) (a) 7.22  10 6 J (b) 1.44  10 5 J (d) 5.76  10 5 J (c) 2.88  10 5 J Solution : (b) As film have two free surfaces W  T  2A W  T  2l  x  7.2  10 2  2  0.1  1  10 3 F l U  1.44  10 5 J x ST Problem 13. If the work done in blowing a bubble of volume V is W, then the work done in blowing the bubble of volume 2V from the same soap solution will be (a) W/2 (b) 2W (c) 1/3 Solution : (d) As volume of the bubble V  4  3  R 3  R    3  4  3 2W  3  V1 / 3  R 2     4  Work done in blowing a soap bubble W  8R 2 T  W  R 2  V 2 / 3 V  W  2   2  W1  V1  2/3  2V     V  2/3  (2) 2 / 3  (4 )1 / 3  W2  3 4 W (d) 3 4 W 2/3 V 2/3  R2  V 2/3 58 Surface Tension Problem 14. Several spherical drops of a liquid of radius r coalesce to form a single drop of radius R. If T is surface tension and V is volume under consideration, then the release of energy is 1 1  (a) 3 VT    r R 1 1  (b) 3 VT    r R 1 1  (c) VT    r R 1   1 (d) VT  2  2  R  r 60 1 1  1 1  4  1 1  Solution : (b) Energy released = 4TR 3     3 R 3 T     3 VT    r R  r R   r R  3 10.12 Excess Pressure. Due to the property of surface tension a drop or bubble tries to contract and so compresses E3 the matter enclosed. This in turn increases the internal pressure which prevents further contraction and equilibrium is achieved. So in equilibrium the pressure inside a bubble or drop is greater than outside and the difference of pressure between two sides of the liquid surface is ID called excess pressure. In case of a drop excess pressure is provided by hydrostatic pressure of the liquid within the drop while in case of bubble the gauge pressure of the gas confined in the bubble provides it. Plane surface U Excess pressure in different cases is given in the following table : P D YG P= 0 P = 0 P  Convex surface 2T R U P  ST P P  4T R Bubble at depth h below the free surface of liquid of density d 2T R Drop P  P Bubble in air P Concave surface 2T R Bubble in liquid P P  2T R Cylindrical liquid surface Surface Tension 59 R P  h P Liquid surface of unequal radii Liquid film of unequal radii  1 1  P  T    R R 2   1 Note :  Excess  1 1  P  2T    R R 2   1 P E3 P T R 60 2T P   hdg R pressure is inversely proportional to the radius of bubble (or drop), i.e., U ID pressure inside a smaller bubble (or drop) is higher than inside a larger bubble (or drop). This is why when two bubbles of different sizes are put in communication with each other, the air will rush D YG from smaller to larger bubble, so that the smaller will shrink while the larger will expand till the smaller bubble reduces to droplet. Sample problems based on Excess pressure Problem 15. The pressure inside a small air bubble of radius 0.1mm situated just below the surface of water will be equal to (Take surface tension of water 70  10 3 Nm 1 and atmospheric pressure  1.013  10 5 Nm  2 ) U (a) 2.054  10 3 Pa Solution : (c) Pressure a bubble (c) 1.027  10 5 Pa when it (d) 2.054  10 5 Pa is in a liquid 3 70  10 2T  1. 013  10 5  2   1.027  10 5 Pa. R 0.1  10  3 ST  Po  inside (b) 1.027  10 3 Pa [AMU (Med.) 2002] Problem 16. If the radius of a soap bubble is four times that of another, then the ratio of their excess pressures will be (a) 1 : 4 [AIIMS 2000] (b) 4 : 1 Solution : (a) Excess pressure inside a soap bubble P  (c) 16 : 1 (d) 1 : 16 P1 r 4T  2 1:4  P2 r1 r Problem 17. Pressure inside two soap bubbles are 1.01 and 1.02 atmospheres. Ratio between their volumes is 60 Surface Tension [MP PMT 1991] (c) 8 : 1 Solution : (c) Excess pressure P  Pin  Pout  1.01 atm  1atm and volume of air bubble V   V  r 3   0.01 atm and similarly P2  0.02 atm 1 (P) 1 ] P V1  P2  V2  P1 3  8  0.02  2        0. 01 1 1      3 [as 3 3 P  1 r or E3 r 4 3 r 3 (d) 2 : 1 60 (b) (102 )3 : (101 )3 (a) 102 : 101 Problem 18. The excess pressure inside an air bubble of radius r just below the surface of water is P1. (a) P1  2P2 ID The excess pressure inside a drop of the same radius just outside the surface is P2. If T is surface tension then (c) P2  2P1 (b) P1  P2 (d) P2  0, P1  0 2T r U Solution : (b) Excess pressure inside a bubble just below the surface of water P1  2T r P1  P2 D YG and excess pressure inside a drop P2  10.13 Shape of Liquid Meniscus. We know that a liquid assumes the shape of the vessel in which it is contained i.e. it can not oppose permanently any force that tries to change its shape. As the effect of force is zero in a direction perpendicular to it, the free surface of liquid at rest adjusts itself at right angles to the resultant force. When a capillary tube is dipped in a liquid, the liquid surface becomes curved near the U point of contact. This curved surface is due to the resultant of two forces i.e. the force of ST cohesion and the force of adhesion. The curved surface of the liquid is called meniscus of the liquid. If liquid molecule A is in contact with solid (i.e. wall of capillary tube) then forces acting on molecule A are (i) Force of adhesion Fa (acts outwards at right angle to the wall of the tube). (ii) Force of cohesion Fc (acts at an angle 45o to the vertical). Resultant force FN depends upon the value of Fa and Fc. If resultant force FN make an angle  with Fa. Surface Tension 61 Then tan   Fc sin 135 o Fa  Fc cos 135 o  Fc 2 Fa  Fc Fc  2 Fa i.e. the   = 90o resultant force acts vertically downwards. Hence the liquid meniscus must be horizontal. Fa tan  = positive angle   is acute tan  = negative   is obtuse angle E3 tan =  Fc  2 Fa i.e. the resultant force directed i.e. the resultant force directed outside the liquid. Hence the liquid meniscus must be concave upward. inside the liquid. Hence the liquid meniscus must be convex upward. ID If Fc  2 Fa 60 By knowing the direction of resultant force we can find out the shape of meniscus because the free surface of the liquid adjust itself at right angle to this resultant force. Fa A A Fa   45 ° FN D YG FN 45 ° Example: Water capillary tube. in Fc FN glass Example: Mercury capillary tube. in glass U Example: Pure water in silver coated capillary tube. Fc U Fc A  10.14 Angle of Contact. ST Angle of contact between a liquid and a solid is defined as the angle enclosed between the tangents to the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of the liquid with the solid.  < 90o Fa  Fc 2  concave meniscus. Liquid wets the solid surface  = 90o Fa  Fc  > 90o  2 plane meniscus. Liquid does not wet the solid surface. Fa  Fc  2 convex meniscus. Liquid does not wet the solid 62 Surface Tension surface. Important points (i) Its value lies between 0o and 180o 60   0 o for pure water and glass,   8 o for tap water and glass,   90 o for water and silver   138 o for mercury and glass,   160 o for water and chromium E3 (ii) It is particular for a given pair of liquid and solid. Thus the angle of contact changes with the pair of solid and liquid. (iii) It does not depends upon the inclination of the solid in the liquid. (iv) On increasing the temperature, angle of contact decreases. ID (v) Soluble impurities increases the angle of contact. (vi) Partially soluble impurities decreases the angle of contact. U 10.15 Capillarity. If a tube of very narrow bore (called capillary) is dipped in a liquid, it is found that the D YG liquid in the capillary either ascends or descends relative to the surrounding liquid. This phenomenon is called capillarity. The root cause of capillarity is the difference in pressures on two sides of (concave and convex) curved surface of liquid. Examples of capillarity : (i) Ink rises in the fine pores of blotting paper leaving the paper dry. (ii) A towel soaks water. U (iii) Oil rises in the long narrow spaces between the threads of a wick. (iv) Wood swells in rainy season due to rise of moisture from air in the pores. ST (v) Ploughing of fields is essential for preserving moisture in the soil. (vi) Sand is drier soil than clay. This is because holes between the sand particles are not so fine as compared to that of clay, to draw up water by capillary action. 10.16 Ascent Formula. When one end of capillary tube of radius r is immersed into a liquid of density d which wets the sides of the capillary tube (water and capillary tube of glass), the shape of the liquid meniscus in the tube becomes concave upwards. R = radius of curvature of liquid meniscus. T = surface tension of liquid R  r A A(P) C(P B ) B D(P E(P ) (P – ) 2T/R) h  y Surface Tension 63 P = atmospheric pressure Pressure at point A = P, Pressure at point B = P  2T R Pressure at points C and D just above and below the plane surface of liquid in the vessel is 60 also P (atmospheric pressure). The points B and D are in the same horizontal plane in the liquid but the pressure at these points is different. In order to maintain the equilibrium the liquid level rises in the capillary tube upto height E3 h. Pressure due to liquid column = pressure difference due to surface tension hdg   h 2T R 2T cos  2T  rdg Rdg r    As R  cos     ID  U Important points (i) The capillary rise depends on the nature of liquid and solid both i.e. on T, d,  and R. D YG (ii) Capillary action for various liquid-solid pair. Meniscus Angle of contact Level  < 90o Rises Plane  = 90o No rise no fall Convex  > 90o Fall Glass Concave Water U Silver ST Water Glass Mercur y (iii) For a given liquid and solid at a given place h 1 r [As T, , d and g are constant] 64 Surface Tension i.e. lesser the radius of capillary greater will be the rise and vice-versa. This is called Jurin’s law. (iv) If the weight of the liquid contained in the meniscus is taken into consideration then h 2T cos  r  rdg 3 without changing nature such that : hr = Lr r r' h L E3 (v) In case of capillary of insufficient length, i.e., L < h, the liquid will neither overflow from the upper end like a fountain nor will it tickle along the vertical sides of the tube. The liquid after reaching the upper end will increase the radius of its meniscus 60 more accurate ascent formula is given by  L