Page 1 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , 9, Notes, , PROPERTIES OF FLUIDS, , In the previous lesson, you have learnt that interatomic forces in solids are, responsible for determining the elastic properties of solids.Does the same hold, for liquids and gases? (These are collectively called fluids because of their nature, to flow in suitable conditions). Have you ever visited the site of a dam on a river, in your area / state/ region? If so, you would have noticed that as we go deeper,, the thickness of the walls increases. Did you think of the underlying physical, principle? Similarly, can you believe that you can lift a car, truck or an elephant by, your own body weight standing on one platform of a hydraulic lift? Have you, seen a car on the platform of a hydraulic jack at a service centre? How easily is it, lifted? You might have also seen that mosquitoes can sit or walk on still water,, but we cannot do so. You can explain all these observations on the basis of, properties of liquids like hydrostatic pressure, Pascal’s law and surface tension., You will learn about these in this lesson., Have you experienced that you can walk faster on land than under water? If you, pour water and honey in separate funnels you will observe that water comes out, more easily than honey. In this lesson we will learn the properties of liquids which, cause this difference in their flow., You may have experienced that when the opening of soft plastic or rubber water, pipe is pressed, the stream of water falls at larger distance. Do you know how a, cricketer swings the ball? How does an aeroplane take off? These interesting, observations can be explained on the basis of Bernoulli’s principle. You will learn, about it in this lesson., , OBJECTIVES, After studying this lesson, you would be able to :, z, , calculate the hydrostatic pressure at a certain depth inside a liquid;, , PHYSICS, , 233

Page 2 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, z, , describe buoyancy and Archimedes Principle;, , z, , state Pascal’s law and explain the functioning of hydrostatic press , hydraulic, lift and hydraulic brakes.;, , z, , z, , explain surface tension and surface energy ;, derive an expression for the rise of water in a capillary tube;, differentiate between streamline and turbulent motion of fluids;, , z, , define critical velocity of flow of a liquid and calculate Reynold’s number;, , z, , define viscosity and explain some daily life phenomena based on viscosity of, a liquid; and, , z, , state Bernoulli’s Principle and apply it to some daily life experiences., , z, , Notes, , 9.1 HYDROSTATIC PRESSURE, While pinning papers, you must have experienced that it is easier to work with a, sharp tipped pin than a flatter one. If area is large, you will have to apply greater, force. Thus we can say that for the same force, the effect is greater for smaller, area. This effect of force on unit area is called pressure., Refer to Fig. 9.1. It shows the shape of the side wall of a dam. Note that it is, thicker at the base. Do we use similar shape for the walls of our house. No, the, walls of rooms are of uniform thickness. Do you know the basic physical, characteristic which makes us to introduce this change?, , Fig. 9.1 : The structure of side wall of a dam, , From the previous lesson you may recall that solids develop shearing stress when, deformed by an external force, because the magnitude of inter-atomic forces is, very large. But fluids do not have shearing stress and when an object is submerged, in a fluid, the force due to the fluid acts normal to the surface of the object (Fig., 9.2). Also, the fluid exerts a force on the container normal to its walls at all, points., 234, , PHYSICS

Page 3 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , Notes, , Fig. 9.2 : Force exerted by a fluid on a submerged object, , The normal force or thrust per unit area exerted by a fluid is called pressure. We, denote it by P :, P=, , Thrust, area, , (9.1), , The pressure exerted by a fluid at rest is known as hydrostatic pressure, The SI Unit of pressure is Nm–2 and is also called pascal (Pa) in the honour of, French scientist Blaise Pascal., 9.1.1 Hydrostatic Pressure at a point in side, a liquid, , A, , P2, , Consider a liquid in a container and an imaginary, h, right circular cylinder of cross sectional area A, and height h, as shown in Fig. 9.3. Let the pressure, P1, exerted by the liquid on the bottom and top faces, of the cylinder be P 1, and P 2, respectively., Therefore, the upward force exerted by the liquid, Fig. 9.3 : An imaginary cylinder, on the bottom of the cylinder is P1A and the, of height h in a liquid., downward force on the top of the cylinder is P2 A., ∴, , The net force in upward direction is (P1A – P2A)., , Now mass of the liquid in cylinder = density × volume of the cylinder, = ρ. A. h where ρ is the density of the liquid., ∴, , Weight of the liquid in the cylinder = ρ. g. h. A, , Since the cylinder is in equilibrium, the resultant force acting or it must be equal, to zero, i.e., PHYSICS, , 235

Page 4 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , P1 A – P2A – ρ g h A = 0, ⇒, , P1 – P 2 = ρ g h, , (9.2), , So, the pressure P at the bottom of a column of liquid of height h is given by, P = ρgh, Notes, , That is, hydrostatic pressure due to a fluid increases linearly with depth. It is for, this reason that the thickness of the wall of a dam has to be increased with increase, in the depth of the dam., If we consider the upper face of the cylinder to be at the open surface of the, liquid, as shown in Fig.(9.4), then P2 will have to be replaced by Patm (Atmospheric, pressure). If we denote P1 by P, the absolute pressure at a depth below the, surface will be, P – Patm + ρ g h, or, , P = Patm + ρ g h, , (9.3), free surface of, the liquid, , P2, h, P1, , Fig. 9.4 : Cylinder in a liquid with one face at the surface of the liquid, , Note that the expression given in Eqn. (9.3) does not show any term having area, of the cylinder It means that pressure in a liquid at a given depth is equal,, irrespective of the shape of the vessel (Fig 9.5)., , Fig. 9.5 : Pressure does not depend upon shape of the versel., , 236, , PHYSICS

Page 5 :
MODULE - 2, , Properties of Fluids, , Example 9.1: A cemented wall of thickness one metre can withstand a side pressure, of 105 Nm–2. What should be the thickness of the side wall at the bottom of a, water dam of depth 100 m. Take density of water = 103 kg m–3 and g = 9.8 ms–2., , Mechanics of Solids, and Fluids, , Solution: The pressure on the side wall of the dam at its bottom is given by, P=hdg, = 100 × 103× 9.8, , Notes, , = 9.8 × 105 Nm–2, Using unitary method, we can calculate the thickness of the wall, which will, withstand pressure of 9.8×105 Nm–2. Therefore thickness of the wall, 9.8 × 10 5 Nm, –2, 10 5, –2, , t=, , Nm, , = 9.8 m, 9.1.2 Atmospheric Pressure, We know that the earth is surrounded by an atmosphere upto a height of about, 200 km. The pressure exerted by the atmosphere is known as the atmospheric, pressure. A German Scientist O.V. Guericke performed an experiment to, demonstrate the force exerted on bodies due to the atmospheric pressure. He, took two hollow hemispheres made of copper,, having diameter 20 inches and tightly joined them, Vacuum, with each other. These could easily be separated, 76 cm = h, when air was inside. When air between them was, exhausted with an air pump, 8 horses were, P, B, A, required to pull the hemispheres apart., Toricelli used the formula for hydrostatic pressure, to determine the magnitude of atmospheric, Fig: 9.6 : Toricelli’s Barometer, pressure., He took a tube of about 1 m long filled with mercury of density 13,600 kg m–3, and placed it vertically inverted in a mercury tub as shown is Fig. 9.6. He observed, that the column of 76 cm of mercury above the free surface remained filled in the, tube., In equi1ibrium, atmospheric pressure equals the pressure exerted by the mercury, column. Therefore,, Patm = h ρ g = 0.76 × 13600 × 9.8 Nm–2, = 1.01 × 10 5 Nm–2, = 1.01 × 105 Pa, PHYSICS, , 237

Page 6 :
MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , Properties of Fluids, , 9.2 BUOYANCY, It is a common experience that lifting an object in water is easier than lifting it in, air. It is because of the difference in the upward forces exerted by these fluids on, these object. The upward force, which acts on an object when submerged in a, fluid, is known as buoyant force. The nature of buoyant force that acts on objects, placed inside a fluid was discovered by. Archimedes Based on his observations,, he enunciated a law now known as Archimedes principle. It state that when an, object is submerged partially or fully in a fluid, the magnitude of the buoyant, force on it is always equal to the weight of the fluid displaced by the object., The different conditions of an object under buoyant force is shown in Fig 9.7., , W, , B, , B, , W, , W, , B, , Fig. 9.7:, (a) : The magnitude of, buoyant force B on the, object is exactly equal, to its weight, in equilibrium., , (b) : A totally submerged, object of density less than, that of the fluid, experiences a net upward, force., , (c) : A totally submerged, object denser than the fluid, sinks., , Another example of buoyant force is provided by the motion of hot air balloon, shown in Fig. 9.8. Since hot air has less density than, cold air, a net upward buoyant force on the balloon makes, it to float., Floating objects, You must have observed a piece of wood floating on the, surface of water. Can you identify the forces acting on it, when it is in equilibrium? Obviously, one of the forces is, due to gravitational force, which pulls it downwards., However, the displaced water exerts buoyant force which, acts upwards. These forces balance each other in, equilibrium state and the object is then said to be floating, on water. It means that a floating body displaces the, fluid equal to its own weight., 238, , Fig. 9.8: Hot air, balloon floating in air, , PHYSICS

Page 7 :
MODULE - 2, , Properties of Fluids, , Archimedes, (287- 212 B.C), A Greek physicist, engineer and mathematician was perhaps, the greatest scientist of his time. He is well known for, discovering the nature of buoyant forces acting on objects., The Archimedes screw is used even today. It is an inclined, rotating coiled tube used originally to lift water from the hold of ships. He also, invented the catapult and devised the system of levers and pulleys., , Mechanics of Solids, and Fluids, , Notes, , Once Archimedes was asked by king Hieron of his native city Syracuse to, determine whether his crown was made up of pure gold or alloyed with other, metals without damaging the crown. While taking bath, he got a solution,, noting a partial loss of weight when submerging his arm and legs in water. He, was so excited about his discovery that he ran undressed through the streets, of city shouting “ Eureka, Eureka’’, meaning I have found it., , 9.3 PASCAL’S LAW, While travelling by a bus, you must have observed that the driver stops the bus by, applying a little force on the brakes by his foot. Have you seen the hydraulic jack, or lift which can lift a car or truck up to a desired height? For this purpose you, may visit a motor workshop. Packing of cotton bales is also done with the help of, hydraulic press which works on the same principle., These devices are based on Pascal’s law, which states that when pressure is applied, at any part of an enclosed liquid, it is transmitted undiminished to every point of, the liquid as well as to the walls of the container., This law is also known as the law of transmission of liquid pressure., 9.3.1 Applications of Pascal’s Law, (A) Hydraulic Press/Balance/Jack/Lift, It is a simple device based on Pascal’s law and is used to lift heavy loads by, applying a small force. The basic arrangement is shown in Fig.9.9. Let a force F1, be applied to the smaller piston of area A1. On the other side, the piston of large, area A2 is attached to a platform where heavy load may be placed. The pressure, on the smaller piston is transmitted to the larger piston through the liquid filled, in-between the two pistons. Since the pressure is same on both the sides, we have, Pressure on the smaller piston, P =, , PHYSICS, , force, F1, =, A1, area, , 239

Page 8 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , Car, , F1, A2, , Heavy, Load, , Large, piston, , F2, , Notes, , Fig. 9.9: Hydraulic lift, , Compressed, air, , A1, , Inlet valve, Release valve, Air-line, , Platform, , Oil, , Fig. 9.10: Hydraulic jack, , According to Pascal’s law, the same pressure is transmitted to the larger cylinder, of area A2., Hence the force acting on the larger piston, F2 = pressure × area =, , F1, × A2, A1, , (9.4), , It is clear from Eqn. ( 9.4) that force F2 > F1 by an amount equal to the ratio, (A2/A1). With slight modifications, the same arrangement is used in hydraulic, press, hydraulic balance, and hydraulic Jack, etc., (B) Hydraulic Jack or Car Lifts, At automobile service stations, you, would see that cars, buses and trucks, are raised to the desired heights so that, a mechanic can work under them (Fig, 9.10). This is done by applying pressure,, which is transmited through a liquid to, a large surface to produce sufficient, force needed to lift the car., (C) Hydraulic Brakes, , 1 kg, , A, , 100 kg, , 0.01 m2, , 1 m2, B, P, , Liquid, , C, , Fig. 9.11(a) : Hydraulic balance, While traveling in a bus or a car, we see, how a driver applies a little force by his, foot on the brake paddle to stop the, vehicle. The pressure so applied gets, L, Cotton M, Bales, transmitted through the brake oil to the, Pump, Plunger, piston of slave cylinders, which, in turn,, P, a, Q, pushes the break shoes against the break, S, Press, drum in all four wheels, simultaneously. Plunger, The wheels stop rotating at the same, b, time and the vehicle comes to stop, Fig. 9.11(b) : Hydraulic press, instantaneously., , 240, , PHYSICS

Page 9 :
Properties of Fluids, , MODULE - 2, Mechanics of Solids, and Fluids, , INTEXT QUESTIONS 9.1, 1. Why are the shoes used for skiing on snow made big in size?, 2, , Calculate the pressure at the bottom of an ocean at a depth of 1500 m. Take, the density of sea water 1.024 × 103 kg m–3, atmospheric pressure=1.01 × 105, Pa and g = 9.80 ms–2., , Notes, , 3. An elephant of weight 5000 kg f is standing on the bigger piston of area 10, m2 of a hydraulic lift. Can a boy of 25 kg wt standing on the smaller piston of, area 0.05m2 balance or lift the elephant?, 4. If a pointed needle is pressed against your skin, you are hurt but if the same, force is applied by a rod on your skin nothing may happen. Why?, 5. A body of 50 kg f is put on the smaller piston of area 0.1m2 of a big hydraulic, lift. Calculate the maximum weight that can be balanced on the bigger piston, of area 10m2 of this hydraulic lift., , 9.4 SURFACE TENSION, It is common experience that in the absence of external forces, drops of liquid are, always spherical in shape. If you drop small amount of mercury from a small, height, it spreads in small spherical globules. The water drops falling from a tap, or shower are also spherical. Do you know why it is so? You may have enjoyed, the soap bubble game in your childhood. But you can not make pure water bubbles, with same case? All the above experiences are due to a characteristic property of, liquids, which we call surface tension. To appreciate this, we would like you to, do the following activity., , ACTIVITY 9.1, 1. Prepare a soap solution., 2. Add a small amount of glycerin to it., 3. Take a narrow hard plastic or glass tube. Dip its one end in the soap solution, so that some solution enters into it., 4. Take it out and blow air at the other end with your mouth., 5. Large soap bubble will be formed., 6. Give a jerk to the tube to detach the bubble which then floats in the air., To understand as to how surface tension arises, let us refresh our knowledge of, intermolecular forces. In the previous lesson, you have studied the variation of, intermolecular forces with distance between the centres of molecules/atoms., PHYSICS, , 241

Page 10 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , The intermolecular forces are of two types: cohesive and adhesive. Cohesive, forces characterise attraction between the molecules of the same substance,, whereas force of adhesion is the attractive force between the molecules of two, different substances. It is the force of adhesion which makes it possible for us to, write on this paper. Gum, Fevicol etc. show strong adhesion., We hope that now you can explain why water wets glass while mercury does not., , Notes, , ACTIVITY 9.2, To show adhesive forces between glass and water molecule., 1. Take a clean sheet of glass, 2. Put a few drops of water on it, 3. Hold water containing side downward., 4. Observe the water drops., Glass sheet, , Water drops, , Fig. 9.12: Water drops remain stuck to the glass sheet, , The Adhesive forces between glass and water molecules keep the water drops, sticking on the glass sheet, as shown in Fig. 9.12., 9.4.1 Surface Energy, The surface layer of a liquid in a container exhibits a property different from the, rest of the liquid. In Fig. 9.13, molecules, are shown at different heights in a liquid. A Spheres of molecular attraction, B Surface film, molecule, say P, well inside the liquid is A, S, C, D, R, attracted by other molecules from all sides., Q, However, it is not the case for the molecules, P, at the surface., Molecules S and R, which lie on the surface, layer, experience a net resultant force, downward because the number of molecules, , 242, , Fig 9.13 : Resultant force acting on, P and Q is zero but molecules R and, S experience a net vertically, downward force., , PHYSICS

Page 11 :
MODULE - 2, , Properties of Fluids, , in the upper half of sphere of influence attracting these molecules is less than, those in the lower half. If we consider the molecules of liquid on the upper half of, the surface of the liquid or liquid-air interface, even then the molecules will, experience a net downward force because of less number of molecules of liquid., Therefore, if any liquid molecule is brought to the surface layer, work has to be, done against the net inward force, which increases their potential energy. This, means that surface layer possesses an additional energy, which is termed as surface, energy., , Mechanics of Solids, and Fluids, , Notes, , For a system to be in equilibrium, its potential energy must be minimum. Therefore,, the area of surface must be minimum. That is why free surface of a liquid at rest, tends to attain minimum surface area. This produces a tension in the surface,, called surface tension., Surface tension is a property of the liquid surface, due to which it has the tendency to decrease its, surface area. As a result, the surface of a liquid acts, like a stretched membrane You can visualise its, existence easily by placing a needle gently on water, surface and see it float., , B, F, , F, A, , Let us now understand this physically. Consider an, imaginary line AB drawn at the surface of a liquid at Fig. 9.14 : Direction of, surface tension on a liquid, rest, as shown in Fig 9.14. The surface on either side, surface, of this line exerts a pulling force on the surface on the, other side., The surface tension of a liquid can be defined as the force per unit length in, the plane of liquid surface :, T = F/L, , (9.5), , where surface tension is denoted by T and F is the magnitude of total force acting, in a direction normal to the imaginary line of length L, (Fig 9.14) and tangential, to the liquid surface. SI unit of surface tension is Nm–1 and its dimensions are, [MT–2]., Let us take a rectangular frame, as shown in Fig. 9.15 having a sliding wire, on one of its arms. Dip the frame in a soap solution and take out. A soap, film will be formed on the frame and have two surfaces. Both the surfaces, are in contact with the sliding wire, So we can say that surface tension acts, on the wire due to both these surfaces., Let T be the surface tension of the soap solution and L be the length of the wire., , PHYSICS, , 243

Page 12 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , Dx, , L, , F, , F, , F=T×2, , Notes, Fig. 9.15: A Film in equilibirum, , The force exerted by each surface on the wire will be equal to T × L. Therefore,, the total force F on the wire = 2TL., Suppose that the surfaces tend to contract say, by Δx. To keep the wire in, equilibrium we will have to apply an external uniform force equal to F. If we, increase the surface area of the film by pulling the wire with a constant speed, through a distance Δx, as shown in Fig. 9.15b, the work done on the film is given, by, W = F × Δx = T × 2L × Δx, where 2L × Δx is the total increase in the area of both the surfaces of the film. Let, us denote it by A. Then, the expressopm for work done on the film simplifies to, W =T×A, This work done by the external force is stored as the potential energy of the new, surface and is called as surface energy. By rearranging terms, we get the required, expresion for surface tension :, T = W/A, , (9.6), , Thus, we see that surface tension of a liquid is equal to the work done in, increasing the surface area of its free surface by one unit. We can also say, that surface tension is equal to the surface energy per unit area., We may now conclude that surface tension, z, , is a property of the surface layer of the liquid or the interface between a, liquid and any other substance like air;, , z, , tends to reduce the surface area of the free surface of the liquid;, , z, , acts perpendicular to any line at the free surface of the liquid and is tangential, to its meniscus;, , z, , has genesis in intermolecular forces, which depend on temperature; and, , z, , decreases with temperature., , A simple experiment described below demonstrates the property of surface tension, of liquid surfaces., 244, , PHYSICS

Page 13 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , ACTIVITY 9.3, Take a thin circular frame of wire and dip it in a soap solution. You will find that, a soap film is formed on it. Now take a small circular loop of cotton thread and, put it gently on the soap film. The loop stays on the film in an irregular shape as, shown in Fig. 9.16(a). Now take a needle and touch its tip to the soap film inside, the loop. What do you observe?, Thread, , Thread, A, , T, , T, , T, , Soap film, , Soap film, , A, , T, , Fig 9.16 (a) : A soap film with, closed loop of thread, , Notes, , T, , Fig. 9.16 (b) : The shape of the thread, without inner soap film, , You will find that the loop of cotton thread takes a circular shape as shown in Fig, 9.16(b). Initially there was soap film on both sides of the thread. The surface on, both sides pulled it and net forces of surface tension were zero. When inner side, was punctured by the needle, the outside surface pulled the thread to bring it into, the circular shape, so that it may acquire minimum area., 9.4.2 Applications of Surface Tension, (a) Mosquitoes sitting on water, In rainy reason, we witness spread of diseases like dengue, malaria and, chickungunya by mosquito breeding on fresh stagnant water. Have you seen, mosquitoes sitting on water surface? They do not sink in water due to surface, tension. At the points where the legs of the mosquito touch the liquid surface, the, surface becomes concave due to the weight of the mosquito. The surface tension, , T cos q, T sin q, , q, , Leg of mosquito, T cos q, q T, T sin q, , Leg of mosquito, , mg, , (a), , (b), , Fig. 9.17 : The weight of a mosquito is balanced by the force of surface tension, = 2π rT cos θ (a) Dip in the level to form concave surface, and (b) magnified image, PHYSICS, , 245

Page 14 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , acting tangentially on the free surface, therefore, acts at a certain angle to the, horizontal. Its vertical component acts upwards. The total force acting vertically, upwards all along the line of contact of certain length balances the weight of the, mosquito acting vertically downward, as shown in Fig 9.17., (b) Excess pressure on concave side of a spherical surface, , Notes, , Consider a small surface element with a line PQ of unit length on it, as shown in, Fig. 9.18. If the surface is plane, i.e. θ = 900, the surface tension on the two sides, tangential to the surface balances and the resultant tangential force is zero [Fig., 9.18 (a)]. If, however, the surface is convex, [Fig. (9.18 (b)] or concave [Fig., 9.18 (c)], the forces due to surface tension acting across the sides of the line PQ, will have resultant force R towards the center of curvature of the surface., Thus, whenever the surface is curved, the surface tension gives rise to a pressure, directed towards the center of curvature of the surface. This pressure is balanced, by an equal and opposite pressure acting on the surface. Therefore, there is, always an excess pressure on the concave side of the curved liquid surface, [Fig. (9.18 b)]., , T, , P, , Q, , P, , T, , R (Resultant, force), r, Q, , Q, T, , T, , T, P, R, , (a) plane surface, , (b) convex surface, , (c) concave surface, , Fig. 9.18, , (i) Spherical drop, A liquid drop has only one surface i.e. the outer surface. (The liquid area in, contact with air is called the surface of the liquid.) Let r be the radius of a small, spherical liquid drop and P be excess pressure inside the drop (which is concave, on the inner side, but convex on the outside). Then, P = (Pi – P0), where Pi and P0 are the inside and outside pressures of the drop, respectively, (Fig 9.19a), If the radius of the drop increases by Δr due to this constant excess pressure P,, then increase in surface area of the spherical drop is given by, 246, , PHYSICS

Page 15 :
MODULE - 2, , Properties of Fluids, , ΔA = 4π (r + Δr)2 – 4πr2, , P0, , r + Dr, , = 8π r Δr, where we have neglected the term containing second, power of Δr., The work done on the drop for this increase in area is, given by, W = Extra surface energy, = TΔA = T. 8π r Δr, , Mechanics of Solids, and Fluids, , air, , Notes, , Fig. 9.19 (a) : A, spherical drop, , (9.7), , If the drop is in equilibrium, this extra surface energy is equal to the work done, due to expansion under the pressure difference or excess pressure P:, Work done = P ΔV = P. 4π r2 Δr, , (9.8), , On combining Eqns. (9.7) and (9.8), we get, P. 4π r2 Δr = T.8 π r Δr, Or, , P = 2 T/ r, , (9.9), P0, , (ii) Air Bubble in water, An air bubble also has a single surface, which is the, inner surface (Fig. 9.19b). Hence, the excess of pressure, P inside an air bubble of radius r in a liquid of surface, tension T is given by, P = 2T/ r, , (9.10), , Pc, air, , Fig. 9.19 b : Air Bubble, , (iii) Soap bubble floating in air, The soap bubble has two surfaces of equal surface area (i.e. the, outer and inner), as shown in Fig. 9.19(c). Hence, excess pressure, inside a soap bubble floating in air is given by, P = 4T/ r, where T is suface tension of soap solution., , (9.11), , P0, Pc, air, air, , Fig. 9.19 (c), , This is twice that inside a spherical drop of same radius or an air, bubble in water. Now you can understand why a little extra pressure is needed to, form a soap bubble., Example 9.3: Calculate the difference of pressure between inside and outside of, a (i) spherical soap bubble in air, (ii) air bubble in water, and (iii) spherical drop of, water, each of radius 1 mm. Given surface tension of water = 7.2 × 10–2 Nm–1 and, surface tension of soap solution = 2.5 × 10–2 Nm–1., PHYSICS, , 247

Page 16 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , Solution:, (i) Excess pressure inside a soap bubble of radius r is, P = 4T/r, =, , Notes, , 4 × 2.5 × 10 –2, 1 × 10 –3 m, , Nm–1, , = 100 Nm–2, (ii) Excess pressure inside an air bubble in water, = 2T ′/ r, =, , 2 × 7.2 × 10 –2 Nm –1, 1 × 10 –3 m, , = 144 Nm–2, (iii) Excess pressure inside a spherical drop of water =2T ′/ r, = 144 Nm–2, (c) Detergents and surface tension, You may have seen different advertisements highlighting that detergents can, remove oil stains from clothes. Water is used as cleaning agent. Soap and detergents, lower the surface tension of water. This is desirable for washing and cleaning, since high surface tension of pure water does not allow it to penetrate easily, between the fibers of materials, where dirt particles or oil molecules are held up., Soap, molecules, , water, Soap molecules with head, attracted to water, Platter with particles of greasy dirt, , Insert ends surround dirt and the, platter dirt can now be dislodged, say by moving water., , Water is added; dirt is not dislooged, , Detergent is added the inert waxy, ends of molecules are attracted to, boundary where water meals dirt., , Dirt is held suspended, surrounded, by soap molecules., , Fig: 9.20 : Detergent action, , 248, , PHYSICS

Page 17 :
Properties of Fluids, , You now know that surface tension of soap solution is smaller than that of pure, water but the surface tension of detergent solutions is smaller than that of soap, solution. That is why detergents are more effective than soap. A detergent dissolved, in water weakens the hold of dirt particles on the cloth fibers which therefore, get, easily detached on squeezing the cloth., The addition of detergent, whose molecules attract water as well as oil, drastically, reduces the surface tension (T) of water-oil. It may even become favourable to, form such interfaces, i.e. globes of dirt surrounded by detergent and then by, water. This kind of process using surface active detergents is important for not, only cleaning the clothes but also in recovering oil, mineral ores etc., , MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , (d) Wax-Duck floating on water, You have learnt that the surface tension of liquids decreases due to dissolved, impurities. If you stick a tablet of camphor to the bottom of a wax-duck and float, it on still water surface, you will observe that it begins to move randomly after a, minute or two. This is because camphor dissolves in water and the surface tension, of water just below the duck becomes smaller than the surrounding liquid. This, creates a net difference of force of surface tension which makes the duck to, move., Now, it is time for you to check how much you have learnt. Therefore, answer, the following questions., , INTEXT QUESTIONS 9.2, 1. What is the difference between force of cohesion and force of adhesion?, 2. Why do small liquid drops assume a spherical shape., 3. Do solids also show the property of surface tension? Why?, 4. Why does mercury collect into globules when poured on plane surface?, 5. Which of the following has more excess pressure?, (i), , An air bubble in water of radius 2 cm. Surface tension of water is 727 ×, 10–3 Nm–1 or, , (ii) A soap bubble in air of radius 4 cm. Surface tension of soap solution is, 25 × 10–3 Nm–1., , 9.5 ANGLE OF CONTACT, You can observe that the free surface of a liquid kept in a container is curved. For, example, when water is filled in a glass jar, it becomes concave but if we fill water, PHYSICS, , 249

Page 18 :
MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , Properties of Fluids, , in a paraffin wax container, the surface of water becomes convex. Similarly, when, mercury is filled in a glass jar, its surface become convex. Thus, we see that shape, of the liquid surface in a container depends on the nature of the liquid, material of, container and the medium above free surface of the liquid. To characterize it, we, introduce the concept of angle of contact., It is the angle that the tangential plane to the liquid surface makes with the, tangential plane to the wall of the container, to the point of contact, as, measured from within the liquid, is known as angle of contact., Fig. 9.21 shows the angles of, contact for water in a glass jar and, paraffin jar. The angle of contact is, acute for concave spherical, meniscus, e.g. water with glass and, obtuse (or greater than 900) for, convex spherical meniscus e.g., water in paraffin or mercury in, glass tube., , q, , q, , Various forces act on a molecule, in the surface of a liquid contained, Fig 9.21 : Nature of free surface when water is, in a vessel near the boundary of the filled in (a) glass jar, and (b) paraffin wax jar, menisus. As the liquid is present, only in the lower quadrant, the resultant cohesive force acts on the molecule at P, symmetrically, as shown in the Fig.9.22(a). Similarly due to symmetry, the resultant, adhesive force Fa acts outwards at right angles to the walls of the container vessel., The force Fc can be resolved into two mutually perpendicular components Fc cos, Fa, , P, , Fc sin q, , Fa, q, , Fc, Fc, , Fc cos q, , Fa, , Fc, , Fa, Fc, , F, , F, (a), , Fc, , (b), , (c), , Fig. 9.22 : Different shapes of liquid meniscuses, , 250, , PHYSICS

Page 19 :
Properties of Fluids, , θ acting vertically downwards and Fc sin θ acting at right angled to the boundary,, The value of the angle of contact depends upon the relative values of Fc and Fa., CASE 1: If Fa > Fc sin θ, the net horizontal force is outward and the resultant of, (Fa – Fc sin θ) and Fc cos θ lies outside the wall. Since liquids can not sustain, constant shear, the liquid surface and hence all the molecules in it near the boundary, adjust themselves at right angles to Fc so that no component of F acts tangential, to the liquid surface. Obviously such a surface at the boundary is concave spherical, ( Since radius of a circle is perpendicular to the circumference at every point.), This is true in the case of water filled in a glass tube., , MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , Case 2 : If Fa < Fc sin θ the resultant F of (Fc sin θ – Fa) acting horizontally and, Fc cos θ acting vertically down wards is in the lower quadrant acting into the, liquid. The liquid surface at the boundary, therefore, adjusts itself at right angles, to this and hence becomes convex spherical. This is true for the case of mercury, filled in the glass tube., Case 3 :When Fa = Fc sinθ, the resultant force acts vertically downwards and, hence the liquid surface near the boundary becomes horizontal or plane., , 9.6 CAPILLARY ACTION, You might have used blotting paper to absorb extra ink from your notebook. The, ink rises in the narrow air gaps in the blotting paper. Similarly, if the lower end of, a cloth gets wet, water slowly rises upward. Also water given to the fields rises in, the innumerable capillaries in the stems of plants and trees and reaches the branches, and leaves. Do you know that farmers plough their fields only after rains so that, the capillaries formed in the upper layers of the soil are broken. Thus, water, trapped in the soil is taken up by the plants. On the other hand, we find that when, a capillary tube is dipped into mercury, the level of mercury inside it is below the, outside level. Such an important phenomenon of the elevation or depression of a, liquid in an open tube of small cross- section (i.e., capillary tube) is basically due, to surface tension and is known as capillary action., The phenomenon of rise or depression of liquids in capillary tubes is known, as capillary action or capillarity., 9.6.1 Rise of a Liquid in a Capillary Tube, Let us take a capillary tube dipped in a liquid, say water. The meniscus inside the, tube will be concave, as shown in Fig. 9.23 (a). This is essentially because the, forces of adhesion between glass and water are greater than cohesive forces., Let us consider four points A, B, C and D near the liquid-air interface Fig. 9.23(a)., We know that pressure just below the meniscus is less than the pressure just, above it by 2T/R, i.e., PHYSICS, , 251

Page 20 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , PB = PA – 2T/R, , (9.12), , where T is surface tension at liquid-air interface and R is the radius of concave, surface., , A, B, , Notes, , h, E, D, , A D, B C, , C, , (a), , (b), Fig. 9.23 : Capillary action, , But pressure at A is equal to the pressure at D and is equal to the atmospheric, pressure P (say). And pressure at D is equal to pressure at C . Therefore, pressure, at B is less than pressure at D. But we know that the pressure at all points at the, same level in a liquid must be same. That’s why water begins to flow from the, outside region into the tube to make up the deficiency of pressure at point B., Thus liquid begins to rise in the capillary tube to a certain height h (Fig 9.23 b) till, the pressure of liquid column of height h becomes equal to 2T/R..Thereafter,, water stops rising. In this condition, h ρ g = 2 T/R, , (9.13), , where ρ is the density of the liquid and g is the, acceleration due to gravity. If r be radius of capillary T, tube and θ be the angle of contact, then from Fig., 9.24, we can write, R = r /cosθ, , h p g = 2T/ r /cos θ, h = 2T cosθ / r ρ g, , R, q, O, , P, , Substituting this value of R in Equation (9.13), , or, , C, , (9.14), , q, Q, S, Fig. 9.24 : Angle of, contact, , It is clear from the above expression that if the radius, of tube is less (i.e. in a very fine bore capillary), liquid rise will be high., , 252, , PHYSICS

Page 21 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , INTEXT QUESTIONS 9.3, 1. Does the value of angle of contact depend on the surface tension of the, liquid?, 2. The angle of contact for a solid and liquid is less than the 900. Will the liquid, wet the solid? If a capillary is made of that solid, will the liquid rise or fall in, it?, , Notes, , 3. Why it is difficult to enter mercury in a capillary tube, by simply dipping it, into a vessel containing mercury while designing a thermometer., 4. Calculate the radius of a capillary to have a rise of 3 cm when dipped in a, vessel containing water of surface tension 7.2 × 10–2 N m–1. The density of, water is 1000 kg m–3, angle of contact is zero, and g = 10 m s–2., 5. How does kerosene oil rise in the wick of a lantern?, , 9.7 VISCOSITY, If you stir a liquid taken in a beaker with a glass rod, in the middle, you will note that the motion of the, liquid near the walls and in the middle is not same, (Fig.9.25). Next watch the flow of two liquids (e.g., glycerin and water) through identical pipes. You will, find that water flows rapidly out of the vessel whereas, glycerine flows slowly. Drop a steel ball through each, liquid. The ball falls more slowly in glycerin than in, water. These observations indicate a characteristic, property of the liquid that determines their motion., This property is known as viscosity. Let us now learn, how it arises., , Fig. 9.25: Water being, stirred with a glass rod, , 9.7.1 Viscosity, We know that when one body slides over the other, a frictional force acts between, them. Similarly, whenever a fluid flows, two adjacent layers of the fluid exert a, tangential force on each other; this force acts as a drag and opposes the relative, motion between them. The property of a fluid by virtue of which it opposes the, relative motion in its adjacent layers is known as viscosity., Fig. 9.26 shows a liquid flowing through a tube. The layer of the liquid in touch, with the wall of the tube can be assumed to be stationary due to friction between, the solid wall and the liquid. Other layers are in motion and have different velocities, Let v be the velocity of the layer at a distance x from the surface and v + dv be the, velocity at a distance x + dx., PHYSICS, , 253

Page 22 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , Moving, , Q, , v + dv, , P, , v, , Notes, x, , x + dx, Rest, , Fig. 9.26 : Flow of a liquid in a tube: Different layers move with different velocities, , Thus, the velocity changes by dv in going through, a distance dx perpendicular to it. The quantity dv/, dx is called the velocity gradient., The viscous force F between two layers of the, fluid is proportional to, z, , area (A) of the layer in contact : F α A, , z, , velocity gradient (dv/dx) in a direction, perpendicular to the flow of liquid : F α dv/dx, , Table 10.1 : Viscosity of a, few typical fluids, Name T [0C] Viscosity η, of fluid, (PR), Water, , 20, , 1.0 × 10–3, , Water, , 100, , 0.3 × 10–3, , blood, , 37, , 2.7 × 10–3, , Air, , 40, , 1.9 × 10–5, , On combining these, we can write, F α A dv/dx, or, , F = – η A (dv/dx), , (9.15), , where η is constant of proportionality and is called coefficient of viscosity. The, negative sign indicates that force is frictional in nature and opposes motion., The SI unit of coefficient of viscosity is Nsm–2. In cgs system, the unit of viscosity, is poise., 1 poise = 0.1 Nsm–2, Dimensions of coefficient of viscosity are [ML–1 T–1], , 9.8 TYPES OF LIQUID FLOW, Have you ever seen a river in floods? Is it similar to the flow of water in a city, water supply system? If not, how are the two different? To discover answer to, such questions, let as study the flow of liquids., , 254, , PHYSICS

Page 23 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , 9.8.1 Streamline Motion, The path followed by fluid particles is called line of flow. If every particle passing, through a given point of the path follows the, same line of flow as that of preceding particles,, the flow is said to be streamlined. A streamline, can be represented as the curve or path whose, tangent at any point gives the direction of the, liquid velocity at that point. In steady flow, the, streamlines coincide with the line of flow, Fig. 9.27: Streamline flow, (Fig. 9.27)., , Notes, , Note that streamlines do not intersect each other because two tangents can then, be drawn at the point of intersection giving two directions of velocities, which is, not possible., When the velocity of flow is less than the critical velocity of a given liquid flowing, through a tube, the motion is streamlined. In such a case, we can imagine the, entire, thickness, of the stream of the liquid to be made up of a large number of plane layers (laminae), one sliding past the other, i.e. one flowing over the other. Such a flow is called, laminar flow., If the velocity of flow exceeds the critical velocity vc, the mixing of streamlines, takes place and the flow path becomes zig-zag. Such a motion is said to be, turbulent., 9.8.2 Equation of Continuity, If an incompressible, non-viscous fluid flows through a tube of non-uniform cross, section, the product of the area of cross section and the fluid speed at any point, in the tube is constant for a streamline flow. Let A1 and A2 denote the areas of, cross section of the tube where the fluid is entering and leaving, as shown in Fig., 9.28. If v1 and v2 are the speeds of the fluid at the ends A and B respectively, and, ρ is the density of the fluid, then the liquid entering the tube at A covers a distance, v1 in one second. So volume of the liquid entering per second= A1 × v1. Therefore, A2, B, , A1, A, Fig. 9.28: Liquid flowing through a tube, PHYSICS, , 255

Page 24 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , Mass of the liquid entering per second at point A = A1 v1 ρ, Similarly, mass of the liquid leaving per second at point B = A2 v2 ρ, Since there is no accumulation of fluid inside the tube, the mass of the liquid, crossing any section of the tube must be same. Therefore, we get, A1 v1 ρ = A2 v2 ρ, , Notes, , A1 v1 = A2 v2, , or, , This expression is called equation of continuity., 9.8.3 Critical Velocity and Reynolds’s Number, We now know that when the velocity of flow is less than a certain value, valled, critical velocity, the flow remains streamlined. But when the velocity of flow, exceeds the critical velocity, the flow becomes turbulent., The value of critical velocity of any liquid depends on the, z, , nature of the liquid, i.e. coefficient of viscosity ( η ) of the liquid;, , z, , diameter of the tube (d) through which the liquid flows; and, , z, , density of the liquid (ρ)., , Experiments show that vc α η ; vc α, , 1, 1, and vc α ., ρ, d, , Hence, we can write, vc = R.η/ρ d, , (9.16), , where R is constant of proportionality and is called Reynolds’s Number. It has no, dimensions. Experiments show that if R is below 1000, the flow is laminar. The, flow becomes unsteady when R is between 1000 and 2000 and the flow becomes, turbulent for R greater than 2000., Example 9.1: The average speed of blood in the artery (d =2.0 cm) during the, resting part of heart’s cycle is about 30 cm s–1. Is the flow laminar or turbulent?, Density of blood 1.05 g cm–3; and η = 4.0 × 10–2 poise., Solution: From Eqn. (9.16) we recall that Reynold’s number R = vc ρ d/η. On, substituting the given values, we get, (30 cm s –1 ) × 2cm × (1.05g cm –3 ), R =, (4.0 × 10 –2 g cm –1s –1 ), , = 1575, Since 1575 < 2000, the flow is unsteady., 256, , PHYSICS

Page 25 :
Properties of Fluids, , MODULE - 2, , 9.9 STOKES’ LAW, , Mechanics of Solids, and Fluids, , George Stokes gave an empirical law for the magnitude of the tangential backward, viscous force F acting on a freely falling smooth spherical body of radius r in a, highly viscous liquid of coefficient of viscosity η moving with velocity v. This is, known as Stokes’ law., According to Stokes’ law, , Notes, Fαη rv, F=Kηrv, , or, , where K is constant of proportionality. It has been found experimentally that K =, 6π., Hence Stokes’ law can be written as, F = 6π η r v, , (9.17), , Stokes’ Law can also be derived using the method of dimensions as follows:, According to Stokes, the viscous force depends on:, z, , coefficient of viscosity (η) of the medium, , z, , radius of the spherical body (r), , z, , velocity of the body (v), , Then, , F α ηa rb vc, , or, , F = K ηa rb vc, , where K is constant of proportionality, Taking dimensions on both the sides, we get, [MLT–2] = [ML–1T–1]a [L]b [LT–1]c, or, , [MLT–2] = [Ma L–a+b+c T–a-c], , Comparing the exponents on both the sides and solving the equations we get a =, b = c = 1., Hence, , F =Kηrv, , 9.9.1 Terminal Velocity, Let us consider a spherical body of radius r and, density ρ falling through a liquid of density σ., , B, , F, v, , The forces acting on the body will be, (i) Weight of the body W acting downward., (ii) The viscous force F acting vertically upward., (iii) The buoyant force B acting upward., , PHYSICS, , W, viscous liquid, Fig. 9.29 : Force acting on a, sphere falling in viscous fluid, , 257

Page 26 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , Under the action of these forces, at some instant the net force on the body becomes, zero, (since the viscous force increases with the increase of velocity). Then, the, body falls with a constant velocity known as terminal velocity. We know that, magnitude of these forces are, F = 6π η r v0, , Notes, , where v0 is the terminal velocity., W = (4/3) π r3 ρg, and, , B = (4/3) π r3 σg, , The net force is zero when object attains terminal velocity. Hence, 6π η r v0 =, Hence, , v0 =, , 4, 4, π r3 ρg –, π r3 σg, 3, 3, 2r 2 (ρ – σ) g, 9η, , (9.18), , 9.9.2 Applications of Stokes’ Law, A. Parachute, When a soldier jumps from a flying aeroplane, he falls with acceleration due to, gravity g but due to viscous drag in air, the acceleration goes on decreasing till, he acquires terminal velocity. The soldier then descends with constant velocity, and opens his parachute close to the ground at a pre-calculated moment, so that, he may land safely near his destination., B. Velocity of rain drops, When raindrops fall under gravity, their motion is opposed by the viscous drag in, air. When viscous force becomes equal to the force of gravity, the drop attains a, terminal velocity. That is why rain drops reaching the earth do not have very high, kinetic energy., Example 9.2: Determine the radius of a drop of rain falling through air with, terminal velocity 0.12 ms–1. Given η = 1.8 × 10–5 kg m–1 s–1, ρ = 1.21 kg m–3, σ =, 1.0 × 103 kg m–3 and g = 9.8 m s–2., Solution: We know that terminal velocity is given by, v0 =, , 2r 2 (ρ – σ) g, 9η, , On rearranging terms, we can write, r =, 258, , 9η v 0, 2 (ρ – σ)g, PHYSICS

Page 27 :
MODULE - 2, , Properties of Fluids, , 9 × 1.8 × 10 × 0.12, m, 2 (1000 – 1.21) 9.8, –5, , =, , Mechanics of Solids, and Fluids, , = 10–5 m, , INTEXT QUESTIONS 9.4, , Notes, , 1. Differentiate between streamline flow and turbulent flow?, 2. Can two streamlines cross each other in a flowing liquid?, 3. Name the physical quantities on which critical velocity of a viscous liquid, depends., 4. Calculate the terminal velocity of a rain drop of radius 0.01m if the coeflicient, of viscosity of air is 1.8 × 10–5 Ns m–2 and its density is 1.2 kg m–3. Density of, water = 1000 kg m–3. Take g = 10 m s–2., 5. When a liquid contained in a tumbler is stirred and placed for some time, it, comes to rest, Why?, Daniel Bernoulli (1700-1782), Daniel Bernoulli, a Swiss Physicist and mathematician was born in a family of, mathematicians on February 8, 1700. He made important contributions in, hydrodynamics. His famous work, Hydrodyanamica was, published in 1738. He also explained the behavior of gases with, changing pressure and temperature, which led to the, development of kinetic theory of gases., He is known as the founder of mathematical physics. Bernoulli’s, principle is used to produce vacuum in chemical laboratories by, connecting a vessel to a tube through which water is running, rapidly., , 9.10 BERNOULLI’S PRINCIPLE, Have you ever thought how air circulates in a dog’s burrow, smoke comes quickly, out of a chimney or why car’s convertible top bulges upward at high speed? You, must have definitely experienced the bulging upwards of your umbrella on a, stormy- rainy day. All these can be understood on the basis of Bernoulli’s principle., Bernoulli’s Principle states that where the velocity of a fluid is high, the pressure, is low and where the velocity of the fluid is low, pressure is high., , PHYSICS, , 259

Page 28 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , 9.10.1 Energy of a Flowing Fluid, Flowing fluids possess three types of energy. We are familiar with the kinetic and, potential energies. The third type of energy possessed by the fluid is pressure, energy. It is due to the pressure of the fluid. The pressure energy can be taken as, the product of pressure difference and its volume. If an element of liquid of mass, m, and density d is moving under a pressure difference p, then, , Notes, , Pressure energy = p × (m/d) joule, Pressure energy per unit mass = (p/d) J kg–1, 9.10.2 Bernoulli’s Equation, Bernoulli developed an equation that expresses this principle quantitatively. Three, important assumptions were made to develop this equation:, 1. The fluid is incompressible, i.e. its density does not change when it passes, from a wide bore tube to a narrow bore tube., 2. The fluid is non-viscous or the effect of viscosity is not to be taken into, account., 3. The motion of the fluid is streamlined., , PA22, , U2, U2, PA11, , h2, , h2, U1, , Fig. 9.30, , We consider a tube of varying cross section shown in the Fig. 9.30. Suppose at, point A the pressure is P1, area of cross section A1, velocity of flow v1, height, above the ground h1 and at B, the pressure is P2 ,area of cross-section A2 velocity, of flow = v2 , and height above the ground h2., Since points A and B can be any two points along a tube of flow, we write, Bernoulli’s equation, 260, , PHYSICS

Page 29 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , P + 1/2 dv2 + h dg = Constant., That is, the sum of pressure energy, kinetic energy and potential energy of a fluid, remains constant in streamline motion., , ACTIVITY 9.4, , Notes, , 1. Take a sheet of paper in your hand., 2. Press down lightly on horizontal part of the paper as, shown in Fig. 9.31 so that the paper curves down., 3. Blow on the paper along the horizontal line., , Sheet of paper, , Watch the paper. It lifts up because speed increases and, pressure on the upper side of the paper decreases., , Fig. 9.31, , 9.10.3 Applications of Bernoulli’s Theorem, Bernoulli’s theorem finds many applications in our lives. Some commonly observed, phenomena can also be explained on the basis of Bernoulli’s theorem., A. Flow meter or Venturimeter, It is a device used to measure the rate of flow of liquids through pipes. The, device is inserted in the flow pipe, as shown in the Fig. 9.32, P1, H1, Main, pipe, , V, , V, , h, A2, , A, , P1, P2, h1, h2, V 2 A 2 A 1 r1, B, , V, , Main, pipe, , Venturimeter, , Fig. 9.32 : A Venturimeter, , It consists of a manometer, whose two limbs are connected to a tube having two, different cross-sectional areas say A1 and A2 at A and B, respectively. Suppose the, main pipe is horizontal at a height h above the ground. Then applying Bernoulli’s, theorem for the steady flow of liquid through the venturimeter at A and B, we can, write, Total Energy at A = Total Energy At B, 1, mp1, 1, mp2, m υ12 + mgh +, =, m υ22 + mgh +, 2, d, 2, d, PHYSICS, , 261

Page 30 :
MODULE - 2, Mechanics of Solids, and Fluids, , Properties of Fluids, , On rearranging terms we can write,, 2, ⎤, v12 d ⎡⎛ v 2 ⎞, d, 2, 2, ( p1 – p2) =, ( v2 – v1 ) = 2 ⎢⎜ v ⎟ − 1⎥, 2, ⎣⎝ 1 ⎠, ⎦, , Notes, , (9.19), , It shows that points of higher velocities are the points of lower pressure (because, of the sum of pressure energy and K.E. remain constant). This is called Venturi’s, Principle., For steady flow through the ventrurimeter, volume of liquid entering per second, at A = liquid volume leaving per second at B. Therefore, A1v1 = A2v2, , (9.20), , (The liquid is assumed incompressible i.e., velocity is more at narrow ends and, vice versa., Using this result in Eqn. (9.19), we conclude that pressure is lesser at the narrow, ends;, p1 – p2 =, , v12 d, 2, , ⎡ A12 ⎤, ⎢ A2 − 1⎥, ⎣ 2 ⎦, , 1 2, = dv1, 2, , ⎡⎛ A ⎞ 2 ⎤, ⎢⎜ 1 ⎟ − 1⎥, ⎢⎣⎝ A 2 ⎠, ⎥⎦, , 2( p1 – p2 ), ⎛ A2 ⎞, d ⎜ 12 ⎟ –1, ⎝ A2 ⎠, , v1 =, , (9.21), , If h denotes level difference between the two limbs of the venturimeter, then, p1 – p2 = h d g, and, , 2hg [(A12 / A 22 ) – 1], , v1 =, , From this we note that v1 ∝, , h since all other parameters are constant for a, , given venturimeter. Thus, v1 = K, , h;, , where K is constant., The volume of liquid flowing per second is given by, V = A1 v1 = A1 × K h, or, V = K′h, where K′ = K A1 is another constant., 262, , PHYSICS

Page 31 :
MODULE - 2, , Properties of Fluids, , Bernaulli’s principle has many applications in the design of many useful appliances, like atomizer, spray gun, Bunsen burner, carburetor, Aerofoil, etc., (i) Atomizer : An atomizer is shown in Fig. 9.33. When the rubber bulb A is, squeezed, air blows through the tube B and comes, B O, out of the narrow orifice with larger velocity creating A, N, a region of low pressure in its neighborhood. The liquid, C, (scent or paint) from the vessel is, therefore, sucked, Vessel, into the tube to come out to the nozzles N. As the, liquid reaches the nozzle, the air stream from the tube, Fig. 9.33 : Atomizer, B blows it into a fine spray., , Mechanics of Solids, and Fluids, , Notes, , (ii) Spray gun : When the piston is moved in, it blows the air out of the narrow, hole ‘O’ with large velocity creating a region of low, Piston, O, pressure in its neighborhood. The liquid (e.g. insecticide), is sucked through the narrow tube attached to the vessel, N, end having its opening just below ‘O’. The liquid on, reaching the end gets sprayed by out blown air from the, Fig. 9.34: Spray gun, piston (Fig. 9.34)., (iii) Bunsen Burner : When the gas emerges out of the, nozzle N, its velocity being high the pressure becomes, low in its vicinity. The air, therefore, rushed in through, the side hole A and gets mixed with the gas. The mixture, then burns at the mouth when ignited, to give a hot blue, flame (Fig.9.35)., , N, , Flame, Mixture of, gas and air, Air, A, Gas, , (iv) Carburetor : The carburetor shown in Fig. 9.36. is a Fig. 9.35: Bunsen, Burner, device used in motor cars for supplying a proper mixture, of air and petrol vapours to the cylinder of the engine., The energy is supplied by the explosion of this mixture inside the cylinders of the, engine. Petrol is contained in the float chamber. There is a decrease in the pressure, on the side A due to motion of the piston.This causes the air from outside to be, sucked in with large velocity. This causes a low pressure near the nozzle B (due, , B, Float, chamber, Air and petrol vapour-mixture, A, , Butterfly valve (throttle), Tube leading to cylinder, , Fig. 9.36: Carburettor, , PHYSICS, , 263

Page 32 :
MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , Properties of Fluids, , to constriction, velocity of air sucked is more near B) and, therefore, petrol, comes out of the nozzle B which gets mixed with the incoming. Air. The mixture, of vaporized petrol and air forming the fuel then enters the cylinder through the, tube A., (Sometimes when the nozzle B gets choked due to deposition of carbon or some, impurities, it checks the flow of petrol and the engine not getting fuel stops, working. The nozzle has therefore, to be opened and cleaned., (v) Aerofoil : When a solid moves in air , streamlines are formed . The shape of, the body of the aeroplane is designed specially as shown in the Fig. 9.37. When, the aeroplane runs on its runway, high velocity streamlines of air are formed. Due, to crowding of more streamlines on the upper side, it becomes a region of more, velocity and hence of comparatively low pressure region than below it. This, pressure difference gives the lift to the aeroplane., P, , v1, , Aerofoil, , P2, , v2, Low velocity region, (high pressure region), , Fig. 9.37 : Crowding of streamlines on the upper side., , Based on this very principle i.e., the regions of high velocities due to crowding of, steam lines are the regions of low pressure, following are interesting, demonstrations.1, (a) Attracted disc paradox : When air is blown through a narrow tube handle, into the space between two cardboard sheets [Fig. 9.38] placed one above the, other and the upper disc is lifted with the handle, the lower disc is attracted to, stick to the upper disc and is lifted with it. This is called attracted disc paradox,, Tube handle attached, with the upper disc, Low pressure in, between the, card-board discs, P1, P2, Fig. 9.38 : Attracted disc paradox, , 264, , PHYSICS

Page 33 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , (b) Dancing of a ping pong ball on a jet of water:, If a light hollow spherical ball (ping-pong ball or table, tennis ball) is gently put on a vertical stream of water, coming out of a vertically upward directed jet end of a, tube, it keeps on dancing this way and that way without, falling to the ground (Fig.9.39). When the ball shifts to, the lefts , then most of the jet streams pass by its right side, thereby creating a region of high velocity and hence low, pressure on its right side in comparison to that on the left, side and the ball is again pushed back to the center of the, jet stream ., , P1 v1, , v2 P2, , Jet end, , Notes, Fig. 9.39: Dancing, Pring Pongball, , (c) Water vacuum pump or aspirator or filter pump : Fig. 9.40 shows a filter, pump used for producing moderately low pressures., Water from the tap is allowed to come out of the narrow, Air, jet end of the tube A . Due to small aperture of the, nozzle, the velocity becomes high and hence a lowpressure region is created around the nozzle N. Air is,, Air, therefore, sucked from the vessel to be evacuated, through the tube B; gets mixed with the steam of water, and goes out through the outlet. After a few minutes., Fig. 9.40 : Filter Pump, the pressure of air in the vessel is decreased to about 1, cm of mercury by such a pump, (d) Swing of a cricket ball:, When a cricketer throws a spinning ball, it moves along a curved path in the air., This is called swing of the ball. It is clear from Fig. 9.41. That when a ball is, moved forward, the air occupies the space left by the ball with a velocity v (say)., When the ball spins, the layer of air around it also moves with the ball, say with, the velocity ‘u’. So the resultant velocity of air above the ball becomes (v – u), and below the ball becomes (v + u). Hence, the pressure difference above and, below the ball moves the ball in a curved path., v, u, , Curved path, of the ball, u, v, Fig. 9.41 : Swing of a cricket ball, , PHYSICS, , 265

Page 34 :
MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , Properties of Fluids, , Example 9.3: Water flows out of a small hole in, the wall of a large tank near its bottom (Fig. 942)., What is the speed of efflux of water when the, height of water level in the tank is 2.5m?, Solution: Let B be the hole near the bottom., Imagine a tube of flow A to B for the water to, flow from the surface point A to the hole B. We, can apply the Bernoulli’s theorem to the points A, and B for the streamline flow of small mass m., , A VA = 0, , hA, B, , vB, hB, , Fig. 9.42, , Total energy at B = Total energy at A, At A, vA= 0, pA= p = atmospheric pressure, h = height above the ground., At B, vB = v = ?, pB = p, hB = height of the hole above the ground., Let hA – hB = H = height of the water level in the vessel = 2.5m, and d = density of the water., Applying the Bernoulli’s Principle and substituting the values we get,, ½m v B2 = mg (hA – hB), or, , vB =, , 2g (hA – hB ), , =, , 2 × 9.8 × 2.5, , = 7 m s–1, , INTEXT QUESTIONS 9.5, 1. The windstorm often blows off the tin roof of the houses, How does, Bernoulli’s equation explain the phenomenon?, 2. When you press the mouth of a water pipe used for watering the plants,, water goes to a longer distance, why?, 3, , What are the conditions necessary for the application of Bernoulli’s theorem, to solve the problems of flowing liquid?, , 4. Water flows along a horizontal pipe having non-uniform cross section. The, pressure is 20 mm of mercury where the velocity is 0.20m/s. find the pressure, at a point where the velocity is 1.50 m/s?, 5. Why do bowlers in a cricket match shine only one side of the ball?, 266, , PHYSICS

Page 35 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , WHAT YOU HAVE LEARNT, z, , Hydrostatic pressure P at a depth h below the free surface of a liquid of, density is given by, P = hdg, , z, , The upward force acting on an object submerged in a fluid is known as buoyant, force., , z, , According to Pascal’s law, when pressure is applied to any part of an enclosed, liquid, it is transmitted undiminished to every point of the liquid as well as to, the walls of the container., , z, , The liquid molecules in the liquid surface have potential energy called surface, energy., , z, , The surface tension of a liquid may be defined as force per unit length acting, on a imaginary line drawn in the surface. It is measured in Nm–1., , z, , Surface tension of any liquid is the property by virtue of which a liquid surface, acts like a stretched membrane., , z, , Angle of contact is defined as the angle between the tangent to the liquid, surface and the wall of the container at the point of contact as measured from, within the liquid., , z, , The liquid surface in a capillary tube is either concave or convex. This curvature, is due to surface tension. The rise in capillary is given by, h=, , z, , Notes, , 2T cos θ, rd g, , The excess pressure P on the concave side of the liquid surface is given by, P=, , 2T, , where T is surface tension of the liquid, R, , P=, , 2T, , for air bubble in the liquid and, R, , P=, , 4T ′, , where T ′ is surface tension of soap solution, for soap bubble in air, r, , z, , Detergents are considered better cleaner of clothes because they reduce the, surface tension of water-oil., , z, , The property of a fluid by virtue of which it opposes the relative motion, between its adjacent layers is known as viscosity., , PHYSICS, , 267

Page 36 :
MODULE - 2, Mechanics of Solids, and Fluids, , Notes, , Properties of Fluids, z, , The flow of liquid becomes turbulent when the velocity is greater than a, certain value called critical velocity (vc) which depends upon the nature of, the liquid and the diameter of the tube i.e. (η.P and d)., , z, , Coefficient of viscosity of any liquid may be defined as the magnitude of, tangential backward viscus force acting between two successive layers of, unit area in contact with each other moving in a region of unit velocity gradient., , z, , Stokes’ law states that tangential backward viscous force acting on a spherical, mass of radius r falling with velocity ‘v’ in a liquid of coefficient of viscosity, η is given by, F = 6π η r v., , z, , Bernoulli’s theorem states that the total energy of an element of mass (m) of, an incompressible liquid moving steadily remains constant throughout the, motion. Mathematically, Bernoullis’s equation as applied to any two points A, and B of tube of flow, 1, mPA, 1, mPB, m vA2 + m g h +, = m vB2 + m g hB +, 2, d, 2, d, A, , TERMINAL EXERCISES, 1. Derive an expression for hydrostatic pressure due to a liquid column., 2. State pascal’s law. Explain the working of hydraulic press., 3. Define surface tension. Find its dimensional formula., 4. Describe an experiment to show that liquid surfaces behave like a stretched, membrane., 5. The hydrostatic pressure due to a liquid filled in a vessel at a depth 0.9 m is, 3.0 N m2. What will be the hydrostatic pressure at a hole in the side wall of, the same vessel at a depth of 0.8 m., 6. In a hydraulic lift, how much weight is needed to lift a heavy stone of mass, 1000 kg? Given the ratio of the areas of cross section of the two pistons is 5., Is the work output greater than the work input? Explain., 7. A liquid filled in a capillary tube has convex meniscus. If Fa is force of adhesion,, Fc is force of cohesion and θ = angle of contact, which of the following, relations should hold good?, (a) Fa > Fc sinθ; (b) Fa < Fc sinθ; (c) Fa cosθ = Fc; (d) Fa sinθ > Fc, 8. 1000 drops of water of same radius coalesce to form a larger drop. What, happens to the temperature of the water drop? Why?, 268, , PHYSICS

Page 37 :
Properties of Fluids, , 9., , What is capillary action? What are the factors on which the rise or fall of a, liquid in a capillary tube depends?, , MODULE - 2, Mechanics of Solids, and Fluids, , 10. Calculate the approximate rise of a liquid of density 103 kg m–3 in a capillary, tube of length 0.05 m and radius 0.2 × 10–3 m. Given surface tension of the, liquid for the material of that capillary is 7.27 × 10–2 N m–1., 11. Why is it difficult to blow water bubbles in air while it is easier to blow soap, bubble in air?, , Notes, , 12. Why the detergents have replaced soaps to clean oily clothes., 13. Two identical spherical balloons have been inflated with air to different sizes, and connected with the help of a thin pipe. What do you expect out of the, following observations?, (i) The air from smaller balloon will rush into the bigger balloon till whole, of its air flows into the later., (ii) The air from the bigger balloon will rush into the smaller balloon till the, sizes of the two become equal., What will be your answer if the balloons are replaced by two soap bubbles of, different sizes., 14. Which process involves more pressure to blow a air bubble of radius 3 cm, inside a soap solution or a soap bubble in air? Why?, 15. Differentiate between laminar flow and turbulent flow and hence define critical, velocity., 16. Define viscosity and coefficient of viscosity. Derive the units and dimensional, formula of coefficient of viscosity. Which is more viscous : water or glycerine?, Why?, 17. What is Reynold’s number? What is its significance? Define critical velocity, on the basis of Reynold’s number., 18. State Bernoulli’s principle. Explain its application in the design of the body, of an aeroplane., 19. Explain Why :, (i) A spinning tennis ball curves during the flight?, (ii) A ping pong ball keeps on dancing on a jet of water without falling on, to either side?, (iii) The velocity of flow increases when the aperture of water pipe is, decreased by squeezing its opening., (iv) A small spherical ball falling in a viscous fluid attains a constant velocity, after some time., , PHYSICS, , 269

Page 38 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , (v) If mercury is poured on a flat glass plate; it breaks up into small spherical, droplets., 20. Calculate the terminal velocity of an air bubble with 0.8 mm in diameter, which rises in a liquid of viscosity of 0.15 kg m–1 s–1 and density 0.9 g m–3., What will be the terminal velocity of the same bubble while rising in water?, For water η = 10–2 kg m–1 s–1., , Notes, , 21. A pipe line 0.2 m in diameter, flowing full of water has a constriction of, diameter 0.1 m. If the velocity in the 0.2 m pipe-line is 2 m s–1. Calculate, (i) the velocity in the constriction, and, (ii) the discharge rate in cubic meters per second., 22. (i) With what velocity in a steel ball 1 mm is radius falling in a tank of, glycerine at an instant when its acceleration is one-half that of a freely, falling body?, (ii) What is the terminal velocity of the ball? The density of steel and of, glycerine are 8.5 gm cm–3 and 1.32 g cm–3 respectively; viscosity of, glycerine is 8.3 Poise., 23. Water at 20ºC flows with a speed of 50 cm s–1 through a pipe of diameter of, 3 mm., (i) What is Reynold’s number?, (ii) What is the nature of flow?, Given, viscosity of water at 20ºC as = 1.005 × 10–2 Poise; and, Density of water at 20ºC as = 1 g cm–3., 24. Modern aeroplane design calls for a lift of about 1000 N m–2 of wing area., Assume that air flows past the wing of an aircraft with streamline flow. If the, velocity of flow past the lower wing surface is 100 m s–1, what is the required, velocity over the upper surface to give a desired lift of 1000 N m–2? The, density of air is 1.3 kg m–3., 25. Water flows horizontally through a pipe of varying cross-section. If the, pressure of water equals 5 cm of mercury at a point where the velocity of, flow is 28 cm s–1, then what is the pressure at another point, where the, velocity of flow is 70 cm s–1? [Tube density of water 1 g cm–3]., , ANSWERS TO INTEXT QUESTIONS, 9.1, 1. Because then the weight of the person applies on a larger area hence pressure, on snow decreases., 270, , PHYSICS

Page 39 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , 2. P = Pa + ρ gh, P = 1.5 × 107 Pa, 3. Pressure applied by the weight of the boy =, Pressure due to the weight of the elephant =, , 2.5, = 500 N m–2., 0.05, 5000, = 500 N m–2., 10, , Notes, , ∴ The boy can balance the elephant., 4. Because of the larger area of the rod, pressure on the skin is small., 5., , w, 50, =, , w = 5000 kg wt., 0.1 10, , 9.2, 1. Force of attraction between molecules of same substance is called force of, cohesion and the force of attractive between molecules of different substance, is called force of adhesion., 2. Surface tension leads to the minimum surface area and for a given volume,, sphere has minimum surface area., 3. No, they have tightly bound molecules., 4. Due to surface tension forces., 5. For air bubble in water, , 2 × 727 × 10–3, 2T, P=, =, = 72.7 N m–2., r, 2 × 10–2, For soap bubble in air, P′ =, , 4 × 25 × 10–3, 4T ′, =, = 2.5 N m–2., r′, 4 × 10–2, , 9.3, 1. No., 2. Yes, the liquid will rise., 3. Mercury has a convex meniscus and the angle of contact is obtuse. The fall in, the level of mercury in capillary makes it difficult to enter., , PHYSICS, , 271

Page 40 :
MODULE - 2, , Properties of Fluids, , Mechanics of Solids, and Fluids, , 2T, , 2 × 7.2 × 10 –2, , 4. r = h ρ g =, 3 × 1000 × 10, = 4.8 × 10–6m., 5. Due to capillary action., Notes, , 9.4, 1. If every particle passing through a given point of path follows the same line, of flow as that of preceding particle the flow is stream lined, if its zig-zag, the, flow is turbulent., 2. No, otherwise the same flow will have two directions., 3. Critical velocity depends upon the viscous nature of the liquid, the diameter, of the tube and density of the liquid., 4. .012 ms–1, 5. Due to viscous force., 9.5, 1. High velocity of air creates low pressure on the upper part., 2. Decreasing in the area creates large pressure., 3. The fluid should be incompressible and non-viscous on (very less). The motion should be steamlined., 4. (P1 – P2) =, , 1, d ( v 22 – v12 ), 2, , 5. So that the stream lines with the two surfaces are different. More swing in the, ball will be obtained., Answers to the Terminal Exercises, 5. 2.67 N m–2., 6. 200 N, No., 20. 2.1 mm s–1, 35 cm s–1., 21. 8 m s–1, 6.3 × 10–2 m3 s–1., 22. 7.8 mm s–1, 0.19 m s–1., 23. 1500, Unsteady., 24. 2 cm of mercury., , 272, , PHYSICS