6.Surface Tension (Marks: 04/05)

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1 Q. Q.1 Define the following terms. 1) Intermolecular force of attraction. 2) Cohesive force 3) Adhesive force 4) Molecular range. 5) Sphere of influence. Ans:- 1)Intermolecular force of attraction:- The force of attraction between any two molecules is called as intermolecular force of attraction. There are two types of intermolecular force of attraction: i) Cohesive forces ii) Adhesive forces 2)Cohesive forces:- The intermolecular force of attraction between two molecules of same substance is known as cohesive fore. Exa:- i) Force of attraction between two water molecules. ii) Force of attraction between two glass molecules. 2)Adhesive forces:- The intermolecular force of attraction between two molecules of different substances is known as adhesive force. Exa:- force of attraction between one glass and one air molecule. 3)Molecular Range:- The maximum distance up to which the molecule can exert force of attraction on another molecule is called as molecules range. 4)Sphere of Influence:- An imaginary sphere having molecule as the centre and molecular range as the radius, is known as sphere of influence. Q.2.Explain the phenomenon of surface Tension on the basis of Molecular Theory. Ans: Surface film: A film on the surface of the liquid having a thickness equal to the molecular range of the liquid is called surface film. 6.Surface Tension (Marks: 04/05) Consider a beaker filled with water. Let A, B and C are the three molecules of water with their sphere of influence are shown in the above figure. The molecules A is completely inside the liquid. So it is equally attracted by another water molecules. Therefore, the resultant cohesive force on molecule A is zero. The molecule B is just touching the liquid surface. The very small part of its sphere of influence lies outside the liquid and large part lies inside the liquid. Hence, the large number of water molecule attracts it in downward direction. Therefore, the resultant cohesive force on molecule B is in downward direction. The molecule C is on the surface of the liquid. Its upper half sphere of influence lies outside the liquid. Therefore, the resultant cohesive force on molecule C is in downward direction. If we try to bring the molecule from inside to the surface of the liquid, the work is done against the downward force. This work done is stored in the form of potential energy. Thus, the potential energy increases with increase in surface area and potential energy decreases with decrease in surface area. But in nature any system tries to have minimum potential energy. So the surface of the liquid has a tendency to contract and minimize its surface area like a stretched elastic membrane. This behavior gives surface tension of the liquid. Q.3.Define surface tension. Give its unit & dimension. Ans: Surface Tension:- The force per unit length acting at right angle to imaginary line drawn on the surface of the liquid is called surface tension. It l is length of an imaginary line, F is force acting at right angle, then T = F l S. I. unit: N/m CGS unit: dyne/cm Dimension: [ M 1 L 0 T -2 ] 1

2 Q.4.Define the term surface tension & surface energy. Obtain the relation between them. OR Show that surface tension is equal to surface energy. Ans: Surface tension: The force per unit length acting at right angle to imaginary line drawn on the surface of the liquid is called surface tension. Surface energy: potential energy per unit surface area is called surface energy. Relation between surface tension & surface Energy: Let ABCD is an open rectangular frame of wire on which wire PQ can move without any friction. If it is dipped in a soap solution, then the soap film APQD is formed as shown in above figure due to the surface tension of the soap solution. Let F be the force exerted on wire PQ, l be the length of wire PQ and T be surface tension of soap solution, F = 2 Tl Here, 2 indicates that the soap film has the two surfaces which are in contact with the wire PQ. If the wire PQ is displaced through a very small displacement dx, then the work done is given by Work done = force displacement dw = F. dx dw = T. 2l. dx But, 2l. dx = da = increase in surface area dw = T. da This work done is stored in the form of potential energy. But, P.E da P. E. = T. da P.E da = T = surface energy. T = surface energy. Surface tension = surface energy This is the relation between surface tension & surface energy. Q.5.Define the term angle of contact. State is character is tics. OR Define angle of contact, when it is acute & when is obtuse. Ans: Angle of contact: The angle between the surface of the solid & the tangent drawn to the surface of the liquid at the point of contact, measured from the side of the liquid is called the angle of contact. Characteristics of angle of contact: 1)Angle of contact is 0 o when the liquid completely wets the solid. Exa: pure water in contact with glass. [In this case, the tangent to the liquid surface is almost along the surface of the solid. 2)Angle of contact is acute (θ < 90 0 ) when the liquid partially wets the solid. Exa:- kerosene in contact with glass. 3)Angle of contact is obtuse (θ > 90 0 ) when the liquid does not wet the solid. Exa:- mercury in contact with glass. 4)The angle of contact is constant for a given solid liquid pair. 5)If we add impurity in a liquid, the angle of contact changes. 6)Angle of contact changes due to the temperature change. 7)Angle of contact depends upon the free surface or the medium of the free surface which is in contact with the liquid. Explaination of Angle of Contact: Q.6.Explain why free surface of some liquid in contact with solid is not horizontal. 2

3 Ans: The angle of contact can be explained on the basis of molecular forces. Therefore, the liquid creeps upward on the solid surface. Hence, the liquid surface in contact with solid is concave upward and the angle of contact is acute. Case II: Consider a liquid molecule A in contact with the solid as shown in the adjacent figure. The sphere of influence of this molecules is partly in solid, liquid and air. There are four forces acting on this molecule A: 1)Resultant adhesive forces by the molecules of solid along vector AP. 2)Resultant cohesive force by the liquid molecules along AQ. 3)Resultant adhesive force by the air molecules which can be neglected. 4)The gravitational force which is very small and it is also neglected. Therefore the behavior of the molecule depends upon the two forces AP and AQ. In case of liquid which does not wet the solid (e.g. mercury), the resultant adhesive force AP is less than the resultant cohesive force AQ. Therefoere, their resultant force AR lies inside the liquid as shown in the fig. (2). In equilibrium state, the tangent AT to the liquid surface which is perpendicular to the resultant force AR is drawn. Therefore, the liquid creeps downward on the solid surface. Hence, the liquid surface in contact with the solid is convex upward and the angle of contact is obtuse. Case III: Case I:- In case of liquid which partially wets the solid(e.g. kerosene), the resultant adhesive force AP is greater then the resultant cohesive force AQ. Therefore the resultant(ar) of AP & AQ lies outside the liquid (inside the solid) as shown in the fig. (1). In equilibrium state, the tangent AT to the liquid surface which is perpendicular to the resultant force AR is drawn. In case of liquid e.g. pure water in contact with clear glass, which completely wets the solid; the resultant adhesive force AP acting on molecule A is very strong & the resultant cohesive force AQ can be neglected. Therefore, the resultant of these two forces is along AP. In equilibrium state, the tangent AT to the liquid surface is always perpendicular to AR. Therefore, the angle of contact is 0 o for pure water & clean glass. Shape of Molecule (Liquid Drop): 3

4 When a small quantity of liquid is dropped on the plane solid surface, force of surface tension acts along a surface separating any two areas. There is a formation of liquid drop, when it is in equilibrium. It depends upon force of interface media. Thus, there is a Surface tension along a surface between a)liquid and air b)solid and air c)liquid and solid Let θ be the angle of contact of given solid liquid pair, T 1 be surface tension at solid -liquid interface, T 2 be surface tension at solid-air interface, T 3 be surface tension at liquid-air interface. For equilibrium of the drop, T 2 = T 1 + T 3 cos θ T 2 T 1 = T 3 cos θ cos θ = T 2 T 1 T 3 From this equation, we get following cases: 1)If T 2 >T 1 and T 2 T 1 <T 3, then cos θ is positive, θ < 90 o i.e. θ is acute. 2)If T 2 < T 1 and T 2 T 1 < T 3, then cos θ is negative, = θ > 90 o i.e. θ is abtase. 3)If T 2 T 1 = T 3, then cos θ = 1, θ = 0 o. 4)If T 2 T 1 > T 3, then cos θ > 1 which is impossible. Then the drop spread over the solid and drop shall not be formed. Excess Pressure inside the liquid drop and A Bubble:- a)inside a liquid drop:- Let us consider a drop of liquid of surface Tension T and radius R. Let P i and P o be the values of pressures of inside and outside the drop of the liquid (as shown in fig.). Then, Excess pressure inside the liquid drop = P i - P o Suppose that the radius of the drop is increased from R to R+ R under the pressure difference (P i - P o ). The outward force acting on the drop = excess of pressure surface area = (P i - P o ) 4πR 2 The small amount of workdone to increase its radius by R, dw= (P i - P o ) 4πR 2 R = 4πR 2 (P i - P o ) R (1) This work (dw) is done by the excess of pressure against the force of surface tension and is stored in the form of its potential energy (du). Also, increase in the potential energy of the liquid drop, du= surface tension increase in surface area = T (4π(R + R) 2 4πR 2 ) = T (4πR 2 + 8πR R + 4π R 2 4πR 2 ) As R is very small, the term containing R 2 can be neglected. Therefore increase in the potential energy of the drop, du=8πr R T (2) From equations (1) and (2),we get, 4πR 2 (P i - P o ) R = 8πR R T (P i - P o ) = 2T R Excess pressure inside the liquid drop = 2T R a)inside a liquid bubble:- 4

5 A liquid bubble has air both inside and outside it and therefore it has two free surfaces. If a liquid bubble increases in size from radius R to R + R,then area of its inner surface as well as that of outer surface will increase. Therefore, increase in potential energy of the bubble, du= surface tension increase in surface area 2)When the capillary tube is dipped in mercury, mercury level fails in the capillary tube as shown in the following figure. = 2T (4π(R + R) 2 4πR 2 ) = 2T (4πR 2 + 8πR R + 4π R 2 4πR 2 ) = 16πR R T (3) The small amount of workdone to increase its radius by R, dw= (P i - P o ) 4πR 2 R = 4πR 2 (P i - P o ) R (4) From equations (3) and (4),we get, 4πR 2 (P i - P o ) R = 16πR R T (P i - P o ) = 4T R Excess pressure inside the liquid bubble = 4T R Q.7.Define the term capillary tube and capillary Action. Explain capillary Action with example. Ans: Capillary tube: A glass tube having a bore of very small diameter is called capillary tube. Capillary Action: The phenomenon of rise or fall of the liquid inside a capillary tube when the capillary tube is dipped in the liquid is called capillarity or capillary action. 1)When the capillary tube is dipped in water, the water rises in the capillary tube as shown in following figure. Examples (Applications of Capillarity):- 1)Ink rises in a pen. 2)Blotting paper absorb ink as water. 3)Oil rises up through wick in oil lamps. 4)Water and sap rises from the roots to the top of the tree. 5)Towel soaks the water. Pressure Difference Across the Surface of the Liquid:- Consider that AB is the section of small element kept on the liquid surface. Suppose that it has unit length perpendicular to the plane of liquid surface. 1)If the surface of the liquid is plane, the forces due to the surface tension acting on two sides of the element AB are equal & opposite. So that the resultant force is zero. Therefore, there is no difference of pressure on the two sides of plane surface (fig. (1)). 2)If the surface of the liquid is concave, the forces due to the surface tension acting on the element AB produce a resultant force acting in vertically upward direction (fig.(2)). To balance the effect of resultant force, the pressure on the concave surface must be greater than the pressure on the other side. 5

6 3)If the surface of liquid is convex, the forces due to the surface tension acting on the element AB produce resultant force acting in vertically downward direction. (fig. (3)). To balance the effect of resultant force, the pressure on the concave side (from below) must be greater than the pressure on convex surface. Therefore, we conduce that 1)If the liquid surface is plane there is no difference of pressure on the two sides of the liquid surface. 2)If the liquid surface is curved (convex or concave), there is a difference of pressure on the two sides of the surface. The pressure on the concave side is greater than the pressure on the convex side. capillary and point C is just above the plane surface outside the capillary and point D is just below the plane surface outside the capillary. Let P A, P B, P C and P D be the pressure at the four points A, B, C and D respectively. The pressure on the concave side of the liquid surface is greater than that on the convex side. Therefore, P A > P B. The pressure is same on both sides of the plane surface outside the capillary. P C = P D Now, P A = P C = atmospheric pressure Therefore, we can writs P D > P B ( P D = P A = P C ) Therefore, the liquid cannot remain in equilibrium and it flows into the capillary and rises above the point B till the pressure at B becomes same as that at point D. Therefore, the liquid rises inside the capillary tube. For a liquid like mercury, which does not wet the solid (tube), the liquid surface inside the capillary tube is convex. In case, there is fall of liquid inside the capillary. Q.9.Explain the rise of liquid in a capillary tube. Derive an expression for the surface tension of the liquid. Ans: Rise of liquid in capillary tube:- Q.8.Explain the cause of capillary Action. Ans: Suppose that a capillary tube is dipped into a liquid when wets the capillary consider the situation before the movement of water inside the capillary. Shape or the surface of the water in the capillary is concave. Let us consider the four points, A, B, C and D as shown in the above figure. The point A is just above the curved surface inside the capillary. The point B is just below the curved surface inside the 6 Consider a capillary tube of radius r dipped vertically in water as shown in figure. Due to the surface tension of water, the water rises inside the capillary tube. Let h is the height of water inside the capillary tube above the free surface of water in the vessel. The meniscus is concave as shown in the figure. The force of surface tension is the force per unit length which acts along the tangent to the liquid, making an angle θ with the wall of the capillary,

7 where θ is the angle of contact for the liquid glass pair. By Newton s third law, the capillary tube exerts an equal and opposite force of reaction on the liquid. The force of reaction T can be resolved into two components:- 1)T cos θ acting in vertically upward direction. 2)T sin θ acting in the horizontal direction. All the horizontal components cancel each other, while all the vertical components are added to each other. If r is the radius of the capillary tube the liquid surface inside the capillary, in contact with capillary is along the length 2πr. Therefore, Total upward force = 2πr T cos θ (1) The upward force pulls the liquid till the weight mg of the liquid acting downward balances it. The volume of the liquid column (V) = πr 2 h Let ϱ is the density of the liquid. Density = mass volume mass = density volume m = ϱ πr 2 h Total downward force = weight of the liquid column = mg = ϱ πr 2 hg Now, for equilibrium of the liquid in the tube, Total upward force = Total downward force 2πr. T cos θ = ϱ πr 2 hg T = ϱ πr2 hg 2πr cos θ T = ϱ rhg 2 cos θ T = hr ϱ g 2 cos θ This is an expression for surface tension of the liquid. If θ = 0 o, (for pure water & clean glass), cos θ = 1 T = hr ϱ g 2 Also, height of the liquid column inside the capillary tube h = 2 T cos θ rρ g Q.10.Explain the effect of impurity and temperature on surface tension. Ans: 1)Effect of Impurity on surface Tension:- If surface of the liquid contains impurities of any kind, there is a change in the surface tension of the liquid. i)if the soluble impurity is added in the liquid, there is increase in the surface tension of the liquid.e.g. when common salt (NaCl) is dissolved in water, surface tension of the solution is greater than that of pure water. ii)if the insoluble impurity is added in the liquid, there is decrease in the surface tension of the solution. e.g. when soap or detergent is dissolved in water, the surface tension of the solution decreases to large extent. This is the reason why the soap is used for washing clothes. That s why, a soap bubble in air remains stable for reasonable time. By sprinkling oil in the sea, the waves calm down, due to decrease in surface tension of water and thereby the height of water waves is also reduced. 2)Effect of Temperature on surface Tension:- For most liquids, the surface tension decreases with increase in temperature of liquid. With increase in temperature the distance between the liquid molecules increases. This decreases the cohesive force of attraction between the molecules & hence, the surface tension of liquid decreases. Only in case of molten copper and molten cadmium, the surface tension increases with increase in temperature. The temperature at which surface tension of liquid becomes zero is called the critical temperature of the liquid. The surface tension of almost all liquids decrease with rise in temperature. If T and T 0 are the values of surface tension of liquid at θ and 0 respectively, then over a certain range of temperature T=T 0 (1-αθ), where α is constant which depends on the nature of the liquids. 3)Effect of contamination of surface tension:- 7

8 The presence of dust particles or lubricating materials on the liquid surface decreases its surface tension. Important Formulae:- Type I 1)Surface Tension of liquid:- T = F l SI unit : N/m CGS unit : dyne / cm 2)Force due to surface tension:- F = T l 3)Additional formulae (Forces) in different cases:- i)a needle (or stick) of length l : F = T. 2 l ii)a ring of radius r : F = T. 2. 2πr F = 4Tπr iii)a ring of inner radius r 1 and outer radius r 2 : F = T. (2πr 1 + 2πr 2 ) F = 2πr T (r 1 + r 2 ) iv)a disc of radius r : F = T 2πr F = T π D v)frame of length l and breadth b : F = T. (4l + 4b) F = 4T (l + b) vi)glass plate of length l and breadth b : F = T. (2l + 2b) F = 2T (l + b) Type II 1)Surface energy:- Surface energy = work done = T. da = surface tension increase or decrease in surface area. 1)Surface area of liquid drop = 4πR 2 2)Surface area of soap bubble = 8πR 2 3)Volume of the spherical drop = 4 3 πr2 4)Work done in increase in the radius of soap bubble for R 1 to R 2 is given by Work done = surface tension increase in surface area = T (final surface area - initial surface area) = T (8πR 2 2 8πR 1 2 ) = 8π T(R 2 2 R 1 2 ) = 2π T(D 2 2 D 1 2 ) Type III 1)Surface Tension:- T = rh ϱg 2 cos θ Where, h = height of liquid column r = radius of the capillary tube ϱ = density of the liquid g = acceleration due to gravity θ = angle of contact For a given liquid r h = constant r 1 h 1)Ratio of surface Tension: T 1 T 2 = h 1 h 2 r 1 r 2 ϱ 1 ϱ 2 g 1 g 2 cos θ 2 cos θ 1 Problems:- Type I 1.A needle, 5cm long can just rest on the surface of water without wetting. Find the weight of needle. Surface tension of water=0.07 N/m(Ans: 7 x 10-3 N) 2.A light square wire frame each side of which is 10 cm long hanges vertically in water with one side just touching the water surface.find the additional force necssary to pull the frame clear of water.(t=0.074n/m) (Ans: N) 3.A thin wire is bent in the form of a rectangle of length 4 cm and bredth 3 cm.what force due to the surface tension do the sides experience when a soap film is formed in the frame? S.T. of soap solution =0.030 N/m. (Ans: 8.4 x 10-3 N) 4.A square glass plate 9.5cm long and 0.5cm thick is suspended so that its lower edge is in contact with 8

9 water in a vessel. Calculate the force due to surface tension acting on the plate. Surface tension of water =72 dynes/cm. (Ans: 1440 dynes) 5.Compute the downward pull due to surface tension on a tube of internal diameter 10 cm and 4 mm thickness when it is partly submerged in water.s.t. of water=72 x 10-3 N/m. (Ans: 4.7 x 10-3 N) 6.A glass tube of internal diameter 3.5 cm and thickness 0.5cm is held vertically with its lower end immersed in water.find the downward pull on the tube due to surface tension. S. T. of water is N/m. (Ans: N) 7. Calculate the force required to take away a flat circular plate of radius 0.01 m from the surface of water. The surface tension of water is 0.075N/m. (Ans: N) 8. A circular loop of thin wire of radius 7/ cm is suspended from one arm of a balance. The plane of the loop is in contact with the surface of soap solution. The pull on the loop due to S.T. is found to be 0.6 x 10-3 kg wt. Calculate the S.T. of soap solution, assuming its of contact to be 0. (Ans:21 x 10-3 N/m) Type II 9. Calculate the work done in blowing a soap bubble of radius 2 cm in air. Surface tension of soap solution is 30 dyne/cm. (Ans:3.014 x 10-4 J) 10. A soap bubble in air is slowly expanded so that its radius increases from 4 cm to 10 cm. Determine the increase in its surface energy, if the S.T. of soap solution is 0.03 N/m.(Ans:6.33 x 10-3 J) 11. Compare the amounts of work done in blowing two soap bubbles of radii in the ratio 4 : 5. (Ans:16:25) 12. A drop of mercury, 2 mm in diameter, is broken up into 1000 small droplets, all of the same size, Calculate the work done in the process, if the surface tension of mercury is 0.46 N/m.(Ans:5.2 x 10-5 J) 13. Calculate the amount of energy needed to break a drop of water of radius 1 mm into 10 9 droplets of equal size. The surface tension of water is 75 x 10-3 N/m. (Ans:9.144 x 10-4 J) 14. Calculate the amount of energy evolved when 27 droplets of water, each of radius 0.5 mm, combine to form one drop. S.T. of water is 70 x10-3 N/m. (Ans:3.956 x 10-6 J) A drop of mercury of radius 0.5 cm falls on a plane surface from a certain height and breaks into a million droplets all of the same size. If the surface tension of mercury is 0.5 N/m. find the height from which the drop falls. (Density of mercury = 13,600 kg/m 3, acceleration due to gravity = 9.8 m/s 2 ) (Ans:22.28 cm) 16. Eight droplets of mercury each of radius 1 mm coalesce into a single drop. Find change in surface energy. Surface tension of mercury is J/m 2. (Ans:2.33 x 10-5 J) 17. The total energy of the free surface of a liquid drop is 2 times the surface tension of the liquid. What is the diameter of the drop? (Assume all terms in SI unit.)[hint: Numerically, T (4 r 2 ) = 2 T] (Ans m) 18. A soap bubble has a radius of 3.5 cm.if ergs of work is done to blow it further, find the new radius of the bubble. (The surface tension of soap solution is 40 dynes/cm)(ans.7cm) 19. A drop oil of diameter 2 mm is sprayed into one million droplets, all of same size. If surface tension of oil is 32 dyne/cm. Calculate energy required. (Ans J) 20. Sixty four drop of mercury each of diameter 0.1 mm coalesce to form a single drop. Find the energy released in this process. (Surface tension of mercury = 0.49 N/m) (Ans J) Type III 21. Capillary tube of radius 0.5 mm is dipped vertically into water. The capillary rise is of 2.94 cm. If density of liquid is 1 gm/cm 3. Find surface tension of water. (Ans N/m) 22. The rise of water in a capillary tube of radius 0.1 mm is 5.5 cm. What is the rise of water in a capillary tube of 0.1 mm diameter? (Ans:11.0cm) 23. A liquid rises to a height of 9 cm in a glass capillary of radius 0.02 cm. What will be the height of liquid column in a glass capillary of radius 0.03 cm? (Ans:6 cm) 24. A capillary tube of uniform bore is dipped vertically in water which rises by 7 cm in the tube. Find the radius of the capillary if the S.T. of water is 70 dyne/cm. (g = 980 cm/s 2 ) (Ans: cm) 25. Water rises to a height of 5 cm in a certain capillary tube. In the same capillary tube, mercuryis

10 depressed by 1.54 cm. Compare the surface tensions of water and mecury. (Given : Density of water = 1000 kg/m 3 Density of mercury = kg/m 3 Angle of contact for water = 0 Angle of contact for mercury=130 ) (Ans:0.153) 26. A capillary tube of 0.5mm in radius is immersed in a beaker of mercury. The mercury level inside the tube is found to be 0.80cm below the level of reservoir. Determine the angle of contact between mercury and glass. Surface tension of mercury = N/m and density is 13.6 x 10 3 kg/m 3,g=9.8 m/s 2 (Ans: ) 27. Water rises to a height of 4cm in a certain capillary tube.if the same capillary tube is dipped into mercury,the level of mercury decrease to 3cm,compare the surface tension of water and mercury, if densities of mercury and water are 13.6 x 10 3 kg/m 3 and 10 3 kg/m 3 respectively. Angle of contact for water is 0 0 and that of mercury is (Ans : :1) 28. A capillary tube of radius 0.5 mm is dipped vertically in a liquid of surface tension 0.04 N/m and relative density 0.8 gm/cc. Calculate the height of capillary rise, if the angle of contact is 10 0.(g=9.8 m/s 2 ) (Ans :2 cm) 29. The surface tension of water at 0 0 c is 70 dyne/cm. Find surface tension of water at 25 0 c ( for water = / 0 c) (Ans: dyne/cm) 30. Calculate the density of paraffin oil, if glass capillary of diameter 0.25mm dipped in paraffin oil of the suface tension N/m rises a height of 4 cm.(angle of contact of paraffin oil with glass is 28 0 and g=9.8 m/s 2 ) (Ans:882.8 kg/m 3 ) 31. The tube of a mercury barometer is 1 cm in diameter. What correction due to capillarity is to be applied to the barometer reading if the surface tension of mercury is dynes/cm and the angle of contact of mercury with glass is 140? (Density of mercury = kg/m 3 ) (Ans. h = cm correction = cm) 32. The surface tension of water at 0 C is 70 dynes/cm. Find the surface tension of water at 20 C. ( for water = C) (Ans dynes/cm) 33.A vertical U-tube containing a liquid has one limb of diameter 2.4 mm and the other limb of diameter 0.8 mm. The density and surface tension of the liquid are 900 kg/m 3 and 0.05 N/m respectively. Find the difference between the liquid levels in the two limbs. Assume the angle of contact to be zero. (g = 9.8 m/s 2 ) (Ans:1.89 cm) 34. Find the difference in levels of mercury in two limbs of a U-tube if the diameter of the bore of one limb is 1 mmand the other is 4mm. The surface tension of mercury is 540 dynes/cm and density is 13.6 g/cm 3. Given: angle of contact of mercury = 135. (Ans cm i.e mm) 35. A capillary tube of radius 1 mm is immersed in a beaker of mercury. The mercury level inside the tube is found to be cm below the level in the reservoir. Determine the angel of contact between the mercury and glass. (Surface tension of mercury = N/m, density of mercury = kg/m 3 ) (Ans ) 36. A glass tube of diameter 0.05 cm is dipped in a liquid of density 0.8 g/cc. If S.T. of the liquid is 28 dynes/cm and angle of contact is 16 20,then find the height to which the liquid will rise in the tube. (Ans cm) 37. What should be the diameter of a capillary tube if water rises in it to a height of 2.4 cm. S.T. of water = 72 dynes/cm. (Ans mm) 38. Calculate the height to which an oil rises in a capillary tube of diameter 1 mm. Given: Density of oil = 800 kg/m 3, angle of contact = 28 27, surface tension of oil = N/m, g = 9.8 m/s 2 (Ans. 1.1 cm) 39. Find the height to which water will rise in capillary tube having a here of diameter 0.05 cm. Take surface tension of water = N/m. = 0, = 10 3 kg/m 3 (Ans cm) 40. By how much will liquid level depressed in capillary tube of diameter 0.04 cm. If angle of contact of liquid is 135 angle and surface tension of liquid is N/m. Density of liquid = 13.6 gm/cm 3. (Ans cm) 41. Radius of capillary tube is mm. The tube is dipped vertically in liquid of density 800 kg/m 3 and surface tension 0.03 N/m. Angle of contact 10

11 72.54.Calculate liquid rise in capillary g = 10 m/s 2 (Ans. 9 cm) 42. Water rises to height 10 cm in a capillary tube. If the area of cross section becomes 25% what will be capillary rise. (Ans. 20 cm) Problems For Practice:- 43. A needle,10cm long can just rest on the surface of water without wetting. Find the weight of needle. Surface tension of water=0.07 N/m(Ans: 14x10-3 N) 44.A glass plate 8cm long is 2mm thick. It is suspended with the long side horizontal and just touching the water surface in a trough full of water. Calculate the downward force on the plate due to surface tension.t=0.072 N/m(Ans:11.8x 10-3 N) 45. A glass plate 0.12 m long and 2 mm thick is suspended over the surface of water with its length just touching the surface. Calculate the downward force acting on the plate due to surface tenison.(s.t. of water = 75 x10-3 N/m) (Ans:1.83 x 10-3 N) 46. A capillary tube of internal diameter 1 mm and external diameter 5 mm is suspended vertically from one arm of a balance and its weight is determined. If the lower end of the tube is then kept dipping into a liquid of surface tensionn 39.2 x 10-3 N/m and angle of contact 0, find the change in the apparent weight of the tube due to surface tension. (g = 9.8 m/s 2 ).(Ans:7.536 x 10-5 kg wt) 47. A rectangular film of liquid is formed in the frame of wire and a movable rod of length 4 cm.what force must be applied to the rod to keep it in equilibrium if the surface tension of liquid is 40 x 10-3 N/m. (Ans:3.2 x 10-3 N) 48. A glass tube has inner diameter is 1mm outer diameter is 1.1mm,when it is kept vertical and partial dipped in water,calculate the downward pull due to surface tension.(surface tension of water = 75 dyne/cm) (Ans:103.6 dyne) 49. A horizontal circular loop of wire of radius 0.02 m is lowered into a liquid and form film.the force due to surface tension of the liquid is N.Calculate the surface tension of crude oil. (Ans:44.9 dyne/cm) 50. A glass plate 9.9 cm long and 1 mm thick is suspended in a trough containing water so that its length just touches the water surface. Calculate the downward force due to surface tension acting on the plate.(surface tension of water = 70 dynes/cm) (Ans.1400 dynes) 51. A needle, 3 cm long, can just rest on the surface of water without wetting. Find the weight of the needle(s T of water = 70 dynes/cm)(ans.420 dyne) 52. A glass plate 10cm long and 3 mm thick is suspended in a trough containing water so that its length just touches water surface. Calculate the downward force due to surface tension acting on the plate.s.t. of water = 74 dynes/cm.ans.1524 dynes 53. A glass plate of length 5 cm, breadth 3 cmand thickness 2 mm hold vertical with it s longest side just touching the surface of water what is the force due to surface tension acting on plate? (S.T. Of water = N/m (Ans N) 54. A soap bubble of radius 10 cm is blown. S.T. of soap solution is 30 dyne/cm. Calculate the workdone in blowing the bubble? (Ans:7.536 x 10 4 erg) 55. The diameter of a soap bubble is 5 cm. Determine the increase in its surface energy if its diameter is increased 3 times. The surface tension of soap solution is 25 x 10-3 N/m.(Ans:3.14 x 10-3 J) 56. Calculate the work done in blowing a soap bubble of radius 0.5 cm. Surface tension of soap solution is 25 x 10-3 N/m. (Ans:1.57 x 10-5 J) 57. A drop of mercury of radius 0.1 cm is broken into 27 droplets of same size. Find the work done if the S.T. of mercury is 540 dyne/cm.(ans:1.356 x 10-5 J) 58. Calculate the work done in breaking a mercury drop of radius 1 mm into one thousand droplets of the same size. Surface tension of mercury is 525 x 10-3 N/m. (Ans: x 10-5 J) 59. Calculate the work done in blowing a soap bubble of radius 0.5 cm. Surface tension of soap solution is 25 x 10-3 N/m. (Ans:7033 erg) 60. A mercury drop of radius 0.5 cm falls on a glass tube and breaks into a million droplets all of the same size. Find the height from which the drop must have fallen. (Density of mercury = 13,600 kg/m 3, S.T. of mercury = N/m,acceleration due to gravity = 9.8 m/s 2 ) (Ans: cm) 61. Calculate the work done in breaking a mercury drop of radius 1 mm into one thousad droplets of the same size. Surface tension of mercury = N/m. (Ans J) 11

12 drops of mercury, each of diameter 0.2 mm, coalesce to form a single drop. Find the energy released in this process. (Surface tension of mercury = 0.49 N/m) (Ans J OR ergs). 63. A drop of mercury of radius 0.1 cm is broken into 8 droplets of the same size.find the workdone if the S.T. of mercury is 490 dynes /cm(ans erg) 64. Eight droplets of water, each of radius 0.2 mm, coalesce into a single drop. Find the change in the total energy. (S.T. = N/m)(Ans J) 65. Calculate the work done when a spherical drop of mercury of radius 2 mm, falls from some height and breaks into a million drroplets, each of the same size. The S.T. of mercury is 0.5 N/m.(Ans J) 66. Calculate the work done in blowing a soap bubble of radius 0.5 cm.the S.T. of the soap solution is N/m.(Ans joule) 67. Compare the amount or work done in blowing soap bubbles of radii 5 cm to 7 cm. (Ans. 25 : 49) 68. Calculate work done required in blowing soap bubble from the diameter of 5 cm to 15 cm if S.T. of soap bubble is 125 dyne/cm. (Ans erg) 69. Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2cm. The surface tension of a soap solution is 30 dyne/cm. (Ans ergs) 70. A drop of mercury of radius 0.1 cm is broken into 8 droplets of same size. Find the work done if the surface tension of mercury is 540 dyne/cm. (Ans ergs.) 71. A liquid of density 900 kg/m 3 rises to a height of 9 mm in a capillary tube of 2.4 mm diameter. If the angle of contact is 25. find the surface tension of the liquid. (g = 9.8 m/s 2 ).(Ans:5.257 x 10-2 N/m) 72. A liquid rises to a height of 5 cm in a glass capillary of radius 0.02 cm. What will be the height of the same liquid column in a glass capillary of radius 0.04 cm? (Ans:2.5cm) 73. Water rises up in a capillary tube of radius 0.5 mm through height 3 cm. How far will it rise in a capillary tube of radius 1 mm? (Ans:1.5cm) 74. In determination of the surface tension of water by capillary rise method, the height of the water column in the glass capillary tube is found to be 4.9 cm. If the radius of the tube is 0.3 mm, find the surface tension of water. (Angle of contact for water and glass = 0 (Ans dynes/cm) 75. When a capillary tube of radius 0.4 mm is dipped into water, the level inside the tube rises to a height of 4 cm above the surface of water. Calculate the surface tension of water if its angle of contact is zero and density is 1 g/cm 3. (Ans dyes /cm) 76. The S.T. of water is N/m,. Find the height to which water rises in a capillary tube of glass having a bore of uniform radius 0.2 mm. (Density of water = 1000 kg/m 3 )(Ans cm) 77. A capillary tube of radius 0.5 mm is dipped vertically in a liquid of S.T N/m and density 0.8 g/cm 3. Calculate the height of the capillary rise if the angle of contact is 10. (Ans m) 78. When a capillary tube of radius 0.45 mm is dipped into water, the level inside the capillary tube rises to height of 3cm above the surface of water. Calculate the surface tension of the water.if its angle of contact is zero and its density is 1 gm/cm 3 (g = 9.8 m/s 2 ) (Ans: N/m) 79. A liquid rise to a height of 4.5 cm in a glass capillary tube of radius 0.01cm,what will be the height of liquid column in a glass tube of radius 0.02cm? (Ans:2.25cm) 80. Water rises to height 3.2cm in glass capillary tube. Find height of rise of same water in another capillary having half area of cross-section (Ans:4.525cm) 81. Calculate the radius of the capillary tube if water rises in it to a height of 4.5 cm. The surface tension of water is 72 dynes/cm. (Angle of contact for water = 0 ) (Ans cm) 82. Calculate the height to which paraffin oil rises in a capillary tube of radius 0.5 mm. (Given : Surface tension of paraffin oil = 24.5 x 10-3 N/m, density of paraffin = 800 kg/m 3, angle of contact = , g = 9.8 m/s 2 ) (Ans:1.1cm) 83. A certain liquid rises to a height of 6 cm in a capillary tube of radius 0.33 mm. The same liquid rises to height of 9 cm in another capillary tube. Calculate the radius of the other capillary tube. (Ans:0.22cm) 84. Water rises to height of 5 cm in a certain capillary tube. In the same capillary, mercury is depressed by 1.54 cm. Compare the surface tensions 12

13 of water and mercury. (Density of water = 1000 kg/m 3, dennsity of mercury = kg/m 3, Angle of contact for water is 0 C and for mercury is 130. (Ans:1:6.516) 85. A certain liquid rises to a height of 9 cmin a capillary tube of radius 0.22 mm. The same liquid rises to a height of 15 cm in another capillary tube. Calculate the radius of the other capillary tube. (Ans mm) 86. A drop of mercury of diameter 4 mm falls on a plane surface from a certain height and breaks into 10 6 equal small spheres. If surface tension of mercury is 540 dyne/cm, find the height from which the drop falls. (Given : g = 980 cm/s 2, dennsity of mercury = 13.6 g/cm 3 ) (Ans cm) 87. In determination of the surface tension of a liquid by capillary rise method, two glass capillary tubes of radii 0.25 mm and 0.50 mm are held side by side. The difference in the levels of the liquid in the two tubes is found to be 2.60 cm. Find the surface tension of the liquid. (Density of the liquid = 0.9 gram/cm 3, angle of contact for the liquid and glass = 25 (Ans dynes/cm) 88. Calculate capillary rise from following data. radius of capillary tube = r = 0.5 mm, surface tension = T = 30 dyne/cm specific gravity = 0.8, angle of contact = =0.(Ans m) 89. Calculate the depression of mercury in a tube of diameter 0.4 cm when it is held vertically in mercury in a trough, with one end of the tube open to atmosphere. (Density of mercury = 13.6 grams/cm3, surface tension of mercury = 490 dynes/cm, angle of contact for mercury with glass = 140 ) [Hint : In this caase, cos is negative. Hence, h will be negative. The numberical value of h gives the depression of mercury.] (Ans cm) Board Book Problems 90. When a capillary tube of radius 0.45 mm is dipped into water, the level inside the capillary tube rises to height of 3 cm above the surface of water. Calculate the surface tension of water, if its angle of contact is zero and its density is 1g/cm 3 ( g = 9.8 m/s 2 ). 91. A liquid rises to a height of 4.5 cm in a glass capillary tube of radius 0.01 cm. What will be the height of liquid column in a glass capillary tube of radius 0.02 cm? 92. A capillary tube of radius 0.5 mm is immersed in a beaker of mercury. The mercury level inside the tube is found to be 0.80 cm below the level of reservoir. Determine the angle of contact between mercury and glass. Surface tension of mercury = N/m and density is 13.6 x 10 3 kg/m 3 m g = 9.8 m/s Water rises to a height of 4cm in a certain capillary tube. If the same capillary tube is dipped into mercury, the level of mercury decreases to 3cm. Compare the surface tension of water and mercury, if densities of mercury and water are 13.6 x 10 3 kg/m 3 and 10 3 kg/m 3 respectively. Angle of contact for water is 0 and that of mercury is A drop of mercury of radius 0.1 cm is broken into 27 droplets of the same size. Find the work done if the surface tension of mercury is 540 dyne/cm. 95. A rectangular film of liquid is formed in the a frame of wire and a movable rod of length 4 cm. What force must be applied to the rod to keep it in equilibrium if the surface tension of liquid is 40 x 10-3 N/m? 96. A soap bubble of radius 10 cm is blown. Surface tension of soap solution is 30 dyne/cm. Calculate the work done in blowing the bubble. 97. Water rises to height 3.2 cm in glass capillary tube. Find height to which same water rises in another capillary having half area of cross section. PROBLEMS FOR PRACTICE 98.Calculate the force required to take away a flat circular plate of radius 0.01 m from the surface of water. The surface tension of water is N/m. ( Ans : N) 99. There is a soap film on rectangular frame of wire of area 4 cm x 4cm. If the area of the frame is increased to 4 cm x 5 cm, find the work done in the process. (Surface tension of soap film = 3 x 10-2 N/m) ( Ans : 4.8 x 10-5 J ) 100. A capaillary tube of radius 0.5 mm is dipped vertically in a liquid of surface tension 0.04 N/m and relative density 0.8 g/cc. Calculate the height of capillary rise, if the angle of contact is 10. ( Given g = 9.8 m/s 2 ) ( Ans : cm) 101. A mercury drop of radius 0.5 cm falls from a height on a glass plate and breaks up into a million droplets, all of the same size, find the height from which the drop must have fallen. (Density of mercury = = kg/m 3. Surface tension of mercury = N/m) ( Ans : m) 102.The surface tension of water at 0 C is 70 dyne/cm. Find surface tension of water at 25 C (α 13

14 For water = / C). ( Ans : dyne/cm) 103. Eight droplets of water, each of radius 0.2 mm, coalesce into a single drop. Find the change in total surface energy. (Surface tension of water = N/m) ( Ans : x 10-7 J) 104. A horizontal circular loop of wire of radius 0.02 m is lowered into a crude oil from film. The force due to surface tension of the liquid is N. Calculate the surface tension of crude oil. ( Ans : dyne/cm) 105. The tube of a mercury barometer is 1 cm in diameter. What correction due to capillary with effect of minutes is to be applied to barometer reading if surface tension of mercury is dyne/cm and angle of contact of mercury with glass is 140? ( density of mercury = kg/m 3 ) ( Ans : cm) 106.A glass tube has inner diameter of 1 mm. outer diameter of 1.1 mm. It is kept vertical and partially dipped in water. Calculate the downward pull due to surface tension. (Surface tension of water = 75 dyne/cm) ( Ans : dyne) 107. The total energy of the free surface of a liquid drop is 2 π times the surface tension of the liquid. What is the diameter of the drop? (Assume all terms in SI units) ( Ans : m) 108.Calculate the density of paraffin oil, if glass capillary of diameter 0.25 mm dipped in paraffin oil of the surface tension N/m rises a height of 4 cm. (angle of contact of paraffin oil with glass is 28 and g = 9.8 m/s 2 ) ( Ans : kg/m 3 ) 109. What should be the diameter of a soap bubble, in order that the excess pressure inside it is 51.2 N/m 2? ( S.T. of soap solution = 3.2 x 10-2 N/m) ( Ans : 5 mm) 14