6_Properties of Matter _ Fluid Mechanics.docx

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PHYSICSSerialNo.MODULE - 2PageNo.1.Work, Power & Energy2.Circular Motion3.Centre of Mass & Collisions4.Rotational Motion5.Gravitation6.Properties of Matter & Fluid Mechanics PROPERTIES OF MATTER & FLUID MECHANICSS. No.CONTENTSPage No.1.Elasticity(Some definitions, Stress, Strain, Stress -StrainGraph, Hooke's Law, Types of ElasticityCoefficients, Work done in stretching a wire(Potential energy of a stretched wire), FactorEffecting Elasticity)2.Hydro-statics(Density, Pressure, Types of Pressure, Pascal's Law, Buoyancy and Archimede's Principle, Floatation)3.Hydro-Dynamics(Types of fluid flow, Equation of continuity,Bernoulli's Theorem, Applications of Bernoulli'stheorem)4.ViscosityNewton's law of viscosity, Stoke's Law andTerminal velocity, Reynolds Number,Dependency of viscosity of fluids, Steady flowin capillary tube )5.Surface tension(Intermolecular forces, Explanation of SurfaceTension (Molecular theory of surface Tension),Definition of surface tension, Surface Energy,Excess pressure inside a curved liquid surface,Angle of contact, capillary tube and capillarity)6.Exercise-I (Conceptual Questions)7.Exercise-II (Previous Years Questions)8.Exercise-III (Analytical Questions)9.Exercise-IV (Assertion & Reason)1. ELASTICITY1. ELASTICITY A body is said to be rigid if the relative positions of its constituent particles remain unchanged when external deforming forces are applied to it. The nearest approach to a rigid body is diamond or carborundum. Actually no body is perfectly rigid and every object can be deformed to some extent or other by the application of suitable forces. All these deformed bodies, however, regain their original shape or size, when the deforming forces are removed. The property of matter by virtue of which a body tends to regain its original shape and size after the removal of deforming forces is called elasticity.1.1 Some Definitions● Deforming Force External force which tends to bring about a change in the length, volume or shape of a body is called deforming force.● Elasticity Elasticity is that property of a material of a body by virtue of which it opposes any change in its shape or size when deforming forces are applied on it, and recover its original state as soon as the deforming force are removed.● Perfectly Elastic Body A body which perfectly regains its original form on removing the external deforming force, is defined as a perfectly elastic body. Example : quartz. - It is quite dose to a perfect elastic body.● Plastic Body(a) A body which does not have the property of opposing the deforming forces, is known as a plastic body.(b) All bodies which remain in the deformed state even after the removed of the deforming forces are known as plastic bodies.● Restoring force When an external force acts at any object then an internal resistance produced in the substance due to the intermolecular forces which is called restoring force. At equilibrium the numerical value of internal restoring force is equal to the external deforming force.1.2 Stress The restoring force acting per unit area of cross-section of the deformed body is called stress. Stress = (at equilibrium). SI UNIT : N/m2 Dimensions : [M1L–1T–2] Type of stress – (i) Longitudinal Stress When the stress is normal to the surface of body, then it is known as longitudinal stress. There are two types of longitudinal stress : (a) Tensile Stress The longitudinal stress, produced due to increase in the length of a body, is defined as tensile stress. (b) Compressive Stress The longitudinal stress, produced due to the decrease in the length of a body, is defined as compressive stress. (ii) Volume Stress If equal normal forces are applied over every unit surface of a body, then it undergoes a certain change in volume. The force opposing this change in volume per unit area is defined as volume stress. (iii) Shear Stress When the stress is tangential or parallel to the surface of a body then it is known as shear stress. Due to this stress, the shape of the body changes or it gets twisted but not its volume. (iv) Breaking Stress The stress required to cause the actual fracture of a material is called the breaking stress or ultimate strength. Breaking stress = ; F = force required to break the body.Dependence of breaking stress : (i) Nature of material (ii) Temperature (iii) Impurities.Independence of breaking stress : (i) Cross sectional area or thickness (ii) Applied force. Maximum load (force) which can applied on the wire depends on(i) Cross sectional area or thickness (ii) Nature of material (iii) Temperature (iv) Impurities.1.3 Strain Strain = There are three types of strains : Types of strains depend upon the directions of applied force. (i) Longitudinal strain = Longitudinal strain is possible only in solids. (ii) Volume strain = (iii) Shear strain When a deforming force is applied to a body parallel to its surface its shape (not size) changes. The strain produced in this manner is known as shear strain. The stain produced due to a change in shape of the body is known as shear strain. tanφ = (Here φ is very small) Shear strain φ = φ = ● Relation Between angle of twist and Angle of shear When a cylinder of length '' and radius 'r' is fixed at one end and tangential force is applied at the other end, then the cylinder gets twisted. Figure shows the angle of shear ABA' and angle of twist AOA'. Arc AA' = rθ and Arc AA' = φ  so rθ = φ ⇒ φ = θ = angle of twist, φ = angle of shear.1.4 Stress – Strain Graph● In the region between O to A the curve is linear. Hooke's law is obeyed. In this region the solid behaves as an elastic body. ● In the region A to B stress and strain and not proportional but body regains its original shape and size when the load is removed.● Point B is known as the elastic limit or yield point.● If the load is increased further the stress developed than strains increases rapidly.● In the region from B to D, when load is removed the body does not regain its original dimensions. Even when stress is zero the strain is not zero. The material is said to have a permanent set. The region beyond point B is known as the plastic region.● Point D corresponds to the tensile strength; beyond this point additional strain is produced by even a reduced applied force and fracture occurs at point E.● If plastic region is large then material will be ductile. ● If plastic region is small then material will be brittle.● For some materials elastic region is very large and the material does not obey Hooke's law over most of the region. These are called elastomers e.g. Tissu of Aorta, rubber, etc.1.5 Hooke's Law If the deformation is small, the stress in a body is proportional to the corresponding strain; this fact is known , as Hooke's Law. Within elastic limit : stress ∝ strain ⇒  = constant. This constant is known as coefficient of elasticity or modulus of elasticity. The modulus of elasticity depends on the type of material temperature and impurity. It does not depend upon the values of stress and strain.GOLDEN KEY POINTS● When a material is under tensile stress restoring force is generated due to the intermolecular attraction while under compressive stress, it is due to the intermolecular repulsion.● If the deforming force is inclined to the surface at an angular θ such that θ ≠ 0 and θ ≠ 90° then both tangential and normal stress are developed.● Linear strain in the direction of force is called longitudinal strain while in a direction perpendicular to the force it is lateral strain.● Breaking stress· also measures the tensile strength.IllustrationsIllustration 1. The ratio of radii of two wires of same materials is 2 : 1. Find if they are stretched by the same force, the ratio of stress :Solution: Stress = ⇒ = .Illustration 2. A bar of cross-section A is subjected to equal and opposite tensile forces F at its ends. Consider a plane through the bar making an angle θ with a plane at right angles to the bar length. (a) What is the tensile stress at this plane in terms of F, A and θ ? (b) What is the shearing stress at this plane, in terms of F, A and θ ?Solution: (a) As tensile stress = (normal force / area) here AN = area = and normal force FN = F cosθ So tensile stress = . (b) As shear stress = (tangential force / area) here Area = and tangential force = F sinθ So shear stress = Illustration 3. The upper end of a wire 1 meter long and 2 mm radius is clamped. The lower end is twisted through an angle of 45°. The angle of shear is .....Solution: φ = θ = × 45º = 0.09º.Illustration 4. The stress versus strain graphs for two materials A and B are shown below. Explain the following (a) Which material has greater Young's modulus ? (b) Which material is more ductile ? (c) Which material is more brittle ? (d) Which of the two is more stronger material ?Solution: (a) Material A has greater value of Young's modulus, because slope of A is greater than that of B. (b) Material A is more ductile because there is a large plastic deformation range between the elastic limit and the breaking point. (c) Material B is more brittle because the plastic region between the elastic limit and breaking point is small. (d) Strength of a material is determined by the stress required to cause fracture. Material A is stronger than material B.Illustration 5. A body of mass 10 kg is attached to a 30 cm long wire whose breaking stress is 4.8 × 107 N/m2. The area of cross section of the wire is 10–6 m2. What is the maximum angular velocity with which it can be rotated in a horizontal circle ?Solution: = breaking stress (BS) ⇒ ω = = 4 rad/sIllustration 6. The breaking stress of aluminium is 7.5 × 108 dyne/cm2. Find the maximum length of aluminium wire that can hang vertically without getting broken. Density of aluminium is 2.7 g/cm3. Given : g = 980 cm/s2,Solution: Let  be the maximum length of the wire that can hang vertically without getting broken. Mass of the wire, m = cross-sectional area (A) × length () × density (ρ) Weight of the wire = mg = Aρg This is equal to the maximum force that the wire can withstand. ∴ Breaking stress = = ρg or 7.5 × 108 =  × 2.7 × 980 ⇒  = = 2.834 × 105 cm 2.83 km.BEGINNER’S BOX - 11. Find out the longitudinal stress and tangential stress on the fixed block shown in figure.2. A 2m long rod of radius 1 cm which is fixed at one end is given a twist of 0.8 radians at the other end. Find the shear strain developed.3. The maximum stress that can be applied to the material of a wire employed to suspend an elevator is × 108 N/m2. If the mass of the elevator is 900 kg and it moves up with an acceleration of 2.2 m/s2 then calculate the minimum radius of the wire.4. A human bone is subjected to a compressive force of 5.0 × 105N. The bone is 25 cm long and has an approximate cross sectional area of 4.0 cm2. If the ultimate compressive strength of the bone is 1.70 × 108 N/m2, will the bone be compressed or will it break under this force?5. The breaking stress of steel is 7.9 × 108 N/m2 and density is 7.9 × 103 kg/m3. What should be the maximum length of a steel wire so that it may not break under its own weight?6. A wire can bear a weight of 20 kg before it breaks. If the wire is divided into two equal parts, then each part will support a maximum weight .....1.6 Types of Elasticity Coefficients1. Young’s Modulus of Elasticity 'Y' Within elastic limit, the ratio of longitudinal stress to longitudinal strain is called Young’s modulus of elasticity. Y = . Within elastic limit, the normal force acting on a unit cross-sectional area of a wire due to which the length of the wire becomes double, is equivalent to the Young’s modulus of elasticity of the material of the wire. If L is the original length of the wire, r is its radius and  the increase in its length as a result of suspending a weight Mg at its lower end then Young's modulus of elasticity of the material of the wire is Y = Unit of Y : N/m2 or pascal Dimensions of Y : [M1L–1T–2]● Increment of length due to own weight Consider a rope of mass M and length L hanging vertically. As the tension at different points on the rope is different, stress as well as strain will be different at different points. (i) maximum stress will be at the point of suspension (ii) minimum stress will be at the lower end. Consider an element of rope of length dx at x distance from the lower end, then tension there T = x g So stress = Let increase in length of this element be dy then strain = So, Young modulus of elasticity Y = ⇒ dx = Ydy Summing up the expression for full length of the rope, ⇒ = Y∆ ⇒ ∆ = [Since the stress is varying linearly we may apply the average method to evaluate strain.] Alternate Method : Since the, weight acts at the centre of gravity, therefore ∴ the original length will be taken as ∴ Y = ⇒ ∆ = But M = (A)ρ ∴∆ = or ∆ =  2. Bulk’s modulus of elasticity 'K' or 'B' Within elastic limit, the ratio of the volume stress (i.e., change in pressure) to the volume strain is called bulk’s modulus of elasticity. K or B = The minus sign indicates a decrease in volume with an increase in stress and vice-versa. Unit of K : M/m2 or pascal Compressibility 'C' The reciprocal of bulk’s modulus of elasticity is defined as compressibility. C = ; SI unit of C : m2/N or pascal–13. Modulus of Rigidity 'η' Within elastic limit, the ration of shearing stress to shearing strain is called modulus of rigidity of a material η = Note : Angle of shear 'φ' is always taken in radians4. Poisson’s Ratio (σ) Within elastic limit, the ratio of lateral strain to the longitudinal strain is called Poisson’s ratio. σ = β = and α = –1 ≤ σ ≤ 0.5 (theoretical limit) σ ≈ 0.2 – 0.4 (experimental limit)● Relation between Y, K, η and σ : (To be remembered) Y = 3K (1 – 2σ), Y = 2η (1 + σ), .1.7 Work done in stretching a wire (Potential energy of a stretched wire) For a wire of length Lo stretched by a length x, the restoring elastic force is : F = stress × area = A The work required to be done against the elastic restoring forces to elongate it further by a length dx is, dW = F.dx = x.dx The total work done in stretching the wire from x = 0 to x = ∆ is, W = or W = AL0 W = ½ × Y × (strain)2 × volume W = ½ (stress) (strain) (volume).1.8 Factor Affecting Elasticity● Effect of Temperature T ↑ ⇒ Y ↓ Due to weakness of intermolecular force. When temperature is increased, the elastic properties in general decreases i.e. elastic constants decrease. Plasticity increases with temperature. For a special kind of steel, elastic constants do not vary appreciably with temperature. This steel is called INVAR steel.● Effect of Impurities Y slightly increases with impurities. The inter molecular attraction strengthens impurities consequently, external deformation can be more effectively opposed.● Interatomic Force Constant : k or ka = Y. r0 Y = Young's modulus ; r0 = interatomic distance under normal circumstancesGOLDEN KEY POINTS● The value of K is maximum for solids and minimum for gases.● Maxwell was the first to define bulk's modulus.● For liquid and gases Young's modulus and modulus of rigidity are each equal to zero.● For any ideal rigid body all the three elastic modulii are infinite .● Modulus of rigidity (η) is the characteristic of solid material only as the fluids do not have a fixed shape.● W = (Load) (extension) = ∆ [where ∆ is the extension in length]● This work done in elongating a wire is stored in the wire as elastic potential energy. Thus, the elastic potential energy density u = (stress) (strain) ● Potential energy density = area under the stress-strain curve.● Young's modulus = Slope of the stress-strain curve. ● Application of Elastic Behaviour of Materials :(1) Crane Cross-sectional area A ≥ Sy = yield strength or breaking stress; W = weight of the object being lifted. e.g. for 10 metric tone load Sy for steel being 300 × 106 N/m2 A ≥ 3.3 × 10–4 m2(2) Girder (3) Max height of mountain Hρg = breaking stress / tensile strength for a rock = 30 × 107 N/m2IllustrationsIllustration 7. Two wires are made of the same metal. The length of the first wire is half that of the second wire and its diameter is double that of the second wire. If equal loads are applied on both the wires, find the ratio of increase in their lengths.Solution: Y = ⇒ ∆ = = .Illustration 8. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Young's modulus of steel is 2.0 × 1011 Pa and that of brass is 0.91 × 1011 Pa. Calculate the elongations of the steel and brass wires. (1 Pa = 1 N/m2) Solution:  The elongation in steel wire ∆S = = 1.50 × 10–4 m The elongation in brass wire ∆B = = 1.32 × 10–4 mIllustration 9. A copper wire of negligible mass, length 1 m and cross-sectional area 10–6 m2 is kept on a smooth horizontal table with one end fixed. A ball of mass 1kg is attached to the other end. The wire and the ball are rotating with an angular velocity of 20 rad/s. If the elongation in the wire is 10–3 m, obtain the Young's modulus of copper.Solution:  Centripetal force F = mωrr ∴ Stress in the wire = = and Strain in the wire = Young’s modulus Y = = 4 × 1011 N/m2.Illustration 10. By how much will a 3.0 m long copper wire elongate if a weight of 10 kg is suspended from the lower end with the upper end keeping fixed? The diameter of the wire is 0.4 mm. Given: Y for copper = 1011 N/m2 and g = 9.8 m/s2.Solution: ∆ = ∴ ∆ = = 2.34 cm.Illustration 11. Calculate the force required to increase the length of a steel wire of cross-sectional area l0–6 m2 by 0.5% given : Y(for steel) = 2 × 1011 N/m2.Solution: × 100 = 0.5% ⇒ = 5 × 10–3 so F = YA = 2 × 1011 × 10–6 × 5 × 10–3 = 103 N.Illustration 12. The graph shows the extension of a wire of length l m suspended from a roof at one end and with a load W connected to the other end. If the cross sectional area of the wire is 1 mm2, then the Young's modulus of the material of the wire isSolution: Y == slope ⇒ Y = (slope) == 2 × 1010 N/m2.Illustration 13. A rubber rope of length 8 m is hung from the ceiling of a room. What is the increase in length of the rope due to its own weight? (Given: Young's modulus of elasticity of rubber = 5 × 106 N m–2 and density of rubber = 1.5 × 103 kg/m3 and g = 10 m/s2).Solution: ∆ = = 9.6 × 10–2 m = 9.6 × 10–2 × 103 mm = 96 mm.Illustration 14. A sphere contracts in volume by 0.01% when taken to the bottom of sea 1 km deep. Find the bulk's modulus of the material of the sphere. Given: density of sea water is 1 g/cm3, g = 980 cm/s2.Solution: , h = 1 km = 105 cm, ρ = 1g/cm3 ; ∆P = 105 × 1 × 980 dyne/cm2, K = ? K = dyne/cm2 = 9.8 × 1011 dyne/cm2.Illustration 15. A rubber cord has a cross-sectional area 1 mm2 and total unstretched length 10 cm. It is stretched to 12 cm and then released to project a mass of 5 g. Young's modulus for rubber is 5 × 108 N/m2. Find the tension in the cord and velocity of the mass.Solution: Tension in the cord = = 100 N. When the mass is released, elastic energy stored = Kinetic energy of mass ⇒ F × ∆= mv2 v = = 20 m/s.Illustration 16. Young modulus of elasticity of steel is 2 × 1011 N/m2. If interatomic distance for steel is 3.2 Aº, then find the interatomic force constant.Solution: k = Y × r0 = 2 × 1011 × 3.2 × 10–10 = 64 N/m.BEGINNER’S BOX - 21. A stress of 20 × 108 N/m2 is developed when the length of a wire is doubled. Its Young's modulus of elasticity in N/m2 will be ............2. The ratio of lengths of two wires made up of the same material is 3 : 1 and the ratio of their radii is 1 : 3. The ratio of increments of lengths on account of suspending the same weight will be...... 3. The following four wires are made of same material. Which one will have the largest elongation when subjected to the same tension ? (1) Length 500 cm, diameter 0.05 mm. (2) Length 200 cm, diameter 0.02 mm. (3) Length 300 cm, diameter 0.03 mm. (4) Length 400 cm, diameter 0. 01 mm.4. The bulk's modulus of copper is 138 × 109 Pa. The additional pressure generated in an explosion chamber is 345 × 106 Pa. Then the percentage change in the volume of a piece of copper placed in this chamber will be ............5. The compressibility of water per unit atmospheric pressure is 4 × 10–5. Decrease in the volume of 100 cm3 water of volume at 100 atmospheric pressure will be ............6. Two parallel and opposite forces, each of magnitude 4000 N, are applied tangentially to the upper and lower faces of a cubical metal block of side 25 cm. If the shear modulus for the metal is 8 × 1010 Pa, then the displacement of the upper surface relative to the lower surface will be......7. Young modulus of elasticity of brass is 1011 N/m2. The increase in its energy on pressing a rod of length 0.1 m and cross-sectional area 1 cm2 made of brass with a force of 10 kg along its length, will be ............2. HYDRO-STATICS Fluids are the substances that can flow or deforms. Therefore liquids and gases both are fluids. Study of fluids at rest is called fluid statics or hydrostatics and the study of fluid in motion is called fluid dynamics or hydrodynamics. Fluid statics and fluid dynamics collectively known as fluid mechanics. The intermolecular force in liquids are comparatively weaker than in solids. Therefore, their shapes can be changed easily. When external force (shear stress) are present, liquid can flow until it conforms to the boundaries of its container. Most liquids resist compression. Unlike a gas, a liquid does not disperse to fill every space of a container and it forms a free surface. The intermolecular forces are weakest in gases, so their shapes and sizes can be changed much easily. Gases are highly compressible and occupy the entire space of the container quite rapidly. Unlike liquid, gases can't form free surface.2.1 Density (ρ) Mass per unit volume of a substance is defined as density. So density at a point of a fluid is expressed as ρ = SI UNIT : kg/m3 ; CGS UNIT : g/cc Dimensions : [ML–3] Density is a positive scalar quantity.2.2 Specific Weight or Weight Density (w) It is defined as the ration of the weight of the substance to its volume or the weight acting per unit volume of the fluid. w = g = ρg. SI UNIT : N/m3 Dimensions : [ML–2T–2] Specific weight of pure water is 9.81 kN/m3 at 4ºC.2.3 Relative Density It is defined as the ratio of the density of the given fluid to the density of pure water at 4ºC. Relative density (R.D.) = Relative density is a unitless and dimensionless positive scalar quantity. Being a dimensionless/unitless quantity R.D. of a substance is same in both SI and CGS system.2.4 Specific Gravity It is defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at 4ºC. Specific gravity = = = R.D. of liquid. Thus the specific gravity of a liquid is numerically equal to the relative density of that liquid and for calculation purposes they are used interchangeably.2.5 Pressure Pressure P is defined as the magnitude of the normal force acting per unit surface area. P = , here ∆F = normal force on a surface of area ∆A SI UNIT : Pascal (Pa) ; 1 Pa = 1 N/m2 Dimensions : [ML–1T–2] Practical units : atmospheric pressure (atm), bar and torr. 1 atm = 1.01325 × 105 Pa = 1.01325 bar = 760 torr = 760 mm of Hg = 10.33 m of water 1 bar = 105 Pa; 1 torr = pressure exerted by 1 mm of mercury column= 133 Pa. Pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions when force is applied, which shows that pressure is not associated with a definite direction. Consequences of pressure(i) Railway tracks are laid on wide wooden or iron sleepers. This is because the weight (force) of the train is spread over a large area of the sleeper. This reduces the pressure acting on the ground and hence prevents the yielding of ground under the weight of the train.(ii) A sharp knife is more effective in cutting the objects than a blunt knife. The pressure exerted = Force / area. The sharp knife transmits force over a small area as compared to the blunt knife. Hence the pressure exerted in case of a sharp knife is more than that in case of a blunt knife.(iii) A camel walks easily on sand but a horse cannot inspite of the fact that a camel is heavier than horse. This is because the area of camel's feet is large as compared to horse's feet. So the pressure exerted by camel on the sand is very small as compared to the pressure exerted by horse. Due to large pressure, sand under the feet of horse yields and hence it cannot walk easily on sand. Type of Pressures In our day to day activity we commonly encounter the following three types of pressures : Pressure is of three types (i) Atmospheric pressure (Po) (ii) Gauge pressure (Pgauge) (iii) Absolute pressure (Pabs.) (1) Atmospheric pressure and Torricelli’s experiment :- Force exerted by atmospheric column on unit cross-sectional area at mean sea level is called atmospheric pressure (Po) Po = 101.3 kN/m2 ∴ Po = 1.013 × 105 N/m2 At tube of length 1 m and uniform cross section is taken. It is filled with mercury and inverted into a mercury tray. The height of the mercury column in equilibrium inside the tube is 76 cm. ∴ atmospheric pressure Po = ρgh = 13.6 × 103 × 9.81 × 76 × 10–2 = 1.013 × 105 N/m2 Note : The above apparatus is known as a barometer. Barometer is used to measure the atmospheric pressure. (2) Gauge Pressure :- Excess Pressure over the atmospheric pressure (P – Patm) measured with the help of pressure measuring instruments is called gauge pressure. Pgauge = Pgauge = hρg or Pgauge ∝ h Pressure due to liquid PA = PB = PC Px' = Px Px' = Po + hρg Gauage pressure = Px' – Po =hρg Note : Gauge pressure is always measured with the help of a “manometer”. (3) Absolute Pressure :- Sum of the atmospheric and gauge pressure is called absolute pressure. Pabs = Patm + Pgauge Pabs = Po + hρg Pressure exerted by a liquid (Effect of gravity) : Consider a vessel containing liquid. As the liquid is in equilibrium, so every volume element of the fluid is also in equilibrium. Consider one such volume element in the form of a cylindrical column of liquid of height h and of area of cross section A. The various forces acting on the cylindrical column of liquid are :(i) Force F1 = P1 A, acting vertically downward on the top face of the column. P1 is the pressure of the liquid on the top face of the column.(ii) Force F2 = P2 A, acting vertically upward at the bottom face of the cylindrical column. P2 is the pressure of the liquid on the bottom face of the column.(iii) W = mg, weight of the cylindrical column of the liquid acting vertically downward. Since the cylindrical column of the liquid is in equilibrium, so the net force acting on the column is zero. F1 + W – F2 = 0 ⇒ P1A + mg – P2A = 0 ⇒ P1A + mg = P2A ∴ P2 = P1 +…(i) Now, mass of the cylindrical column of the liquid is, m = volume × density of the liquid = Area of cross section × height × density = Ahρ ∴ equation (i) becomes P2 = P1 + , P2 = P1 + hρg …..(ii) P2 is the absolute pressure at a depth h below the free surface of the liquid. Equation (ii), shows that the absolute pressure at a depth h is greater than the absolute pressure (P1) by an amount equal to hρg. Equation (ii) can also be written as (P2 – P1) = hρg which is the difference of pressures between two points separated by a depth h.2.6 Pascal's Law Pascal's law is stated in following ways- • A liquid exerts equal pressures in all directions.• If the pressure in an enclosed fluid is changed at a particular point, the change is transmitted to every point of the fluid and to the walls of the container without being diminished in magnitude. Applications of pascal's law: hydraulic jacks, hydraulic lifts, hydraulic press, hydraulic brakes, etc Hydraulic lift Pressure applied = ∴ Pressure transmitted =  Pressure is equally transmitted ∴ ∴ Upward force on A2 is F2 = × A2 = × F12.7 Buoyancy and Archimede's Principle Buoyant Force If a body is partially or fully immersed in a fluid, it experiences an upward force due to the fluid surrounding it. This phenomenon of force exerted by fluid on the body is called buoyancy and force is called buoyant force or force of up thrust. Archimede's Principle It states that the upward buoyant force on a body that is partially or totally immersed in a fluid is equal to the weight of the fluid displaced by it. Consider a body immersed in a liquid of density σ. Top surface of the body experiences a downward force F1 = AP1 = A[h1.σ.g + P0]…….(i) Lower face of the body will experience an upward force F2 = AP2 = A[h2.σ.g + P0]…….(ii) As h2 > h1 so F2 is greater than F1 so net upward force F = F2 – F1 = Aσg[h2 – h1] ∴ F = A.σ.g.L. = Vin.σ,g [ Vin = volume of the body submerged in the fluid = AL] Principle of Floatation When a body of density (ρ) and volume (V) is completely immersed in a liquid of density (σ), the forces acting on the body are :(i) Weight of the body W = Mg = Vρg directed vertically downwards through the Centre of gravity of the body.(ii) Buoyant force or Upthrust Th = Vσg directed vertically upwards through Centre of buoyancy. The following three cases are possible : Case I : Density of the body is greater than that of liquid (ρ > σ) In this case W > Th So the body will sink to the bottom of the liquid. WApp = W – Th = Vρg – Vσg = Vρg (1 – σ/ρ) = W (1 – σ/ρ). Case II : Density of the body is equal to the density of liquid (ρ = σ) In this case W = Th So the body will float fully submerged in the liquid. It will be in neutral equilibrium. WApp = W – Th = 0 Case III : Density of the body is lesser than that of liquid (ρ < σ) In this case W < Th So the body will float partially submerged in the liquid. In this case the volume of liquid displaced by the body (Vin) will be less than the volume of body (V). This ensures that Th equally to W ∴ WApp = W – Th = 0 The above three cases constitute the laws of floatation which states that a body will float in a liquid if weight of the liquid displaced by the immersed part of the body is at least equal to the weight of the body.GOLDEN KEY POINTS● For a solid body volume and density will be same as that of its constituent substance of equal mass i.e. if Mbody = Msub then Vbody = Vsub and ρbody = Psub. But for a hollow body or body with air gaps or avities, Mbody = Msub and Vbody > Vsub then ρbody < ρsub● If m1 mass of liquid of density ρ1 and m2 mass of an immiscible liquid of density ρ2 are mixed then Mmix = m1 + m2 and Vmix = V1 + V2 = ∴ ρmix. = If liquids with same masses are mixed i.e. m1 = m2 = m then ρmix. = (Harmonic mean of individual densities) ● If V1 volume of liquid of density ρ1 and V2 volume of liquid of density ρ2 are mixed then Vmix. = V1 + V2 and Mmix. = m1 + m2 = ρ1V1 + ρ2V2 ∴ ρmix. = If liquids with same volumes are mixed i.e. V1 = V2 = V then ρmix. = (arithmetic mean of individual densities)● Pressure due to liquid on a vertical wall is different at different depths, so average fluid pressure on side wall a of container = mean pressure = (h = height of wall)● Buoyance force acts vertically upward through the centre of gravity (C.G.) of the displaced fluid. This point is called the centre of buoyancy (C.B.). Thus the centre of buoyancy is the point through which the force of buoyancy is supposed to act.● Buoyant force or force of up thrust does not depend upon the characteristics of the body such as its mass, size, density, etc. However it depends upon the volume of the body inside the liquid. Th ∝ Vin● It depends upon the nature of the fluid as it is proportional to the density of the fluid. ⇒ Th ∝ σ .  This is the reason that force of upthrust on a fully submerged body is more in sea water than in pure water. ( σsea > σpure)● The effective weight of a body decreases due to upthrust WApp = W – Th (W is the true weight of the body) Decrease in weight = W – WApp = Th = Weight of the fluid displaced ● Using Archimede's principle we can determine the relative density (R D) of a body as R.D. = = =● If a body is weighed in air (WA), in water (W­A) and in a liquid (WL), then Specific gravity of liquid = ● In case of W = Th, the equilibrium of a floating body does not depend upon the variation in g though both thrust and weight depends upon g.● The weight of a plastic bag full of airs is same as that of empty bag because the force of upthrust is equal to the weight of the air enclosed. IllustrationsIllustration 17. A hollow metallic sphere has inner and outer radii, as 5 cm and 10 cm respectively. If the mass of the sphere is 2.5 kg. Find the (a) density of the materia1, (b) relative density of the material of the sphere.Solution: Volume of the material of the sphere is V = × 3.14 ×× 3.14 × [0.001 – 0.000125] = × 3.14 × 0.000875 m3 = 0.00367 m3 (a) Therefore, density of the material of the sphere is ρ = kg/m3 = 681.2 kg/m3 (b) Relative density of the material of the sphere ρr = = 0.6812Illustration 18. Two immiscible liquids of densities 2.5 g/cm3 are taken in the ratio of their masses as 2.3 respectively. Find the average density of the liquid combination.Solution: Let masses be 2M & 3M then V = V1 + V2 = cm3 Total mass = 2M + 3M = 5M Therefore, the average density ρav = = g/cm3 = 1.09 g/cm3 Illustration 19. During a blood transfusion a needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein ? [Density of blood= 1.06 × 103 kg/m3]. (1) 0.192 m (2) 0.182 m (3) 0.172 m (4) 0.162 mAns. (1)Solution: Pressure P = hρg ⇒ h = = 0.192 m.Illustration 20. Calculate the depth of a well if the pressure at its bottom is 15 times that at a depth of 3 meters. Atmospheric pressure is 10 m column of water.Solution: Let the depth of the well be h then according to the question, Patm + hρw g = 15 (Patm + 3ρw g) hρw g = 14 Patm + 45 ρw g = 14 (10 × ρw g) + 45 ρw g h = 185 m.Illustration 21. AU-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm column of water in one arm and 12.5 cm column of spirit in the other. What is the specific gravity of spirit ?Solution: Since level of mercury in the two arms of U-tube is same and if Pa is the atmospheric pressure then Pa + hwρwg = Pa + hsρsg ⇒ = 0.8 ∴ Specific gravity of spirit = 0.8 Illustration 22. Two liquids that do not mix are poured into a U-shaped tube as shown in fig. Find the difference H in these heights of liquids in terms of ρ1, ρ2 h.Solution: Starting from the point B, we apply the manometric equation as – PB + ρ2g (h + H) – ρ1gh = PA Since PA = PB = Patm therefore H = Illustration 23. A vertical U-tube of uniform cross-section contains mercury in both arms. A glycerine (relative density = 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg/m3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is 13.6 × 103 kg/m3.Solution: Pressure at A and B must be same Pressure at A= P0 + 0.1 × (l.3 × 1000) × g Pressure at B = P0 + h × 800 × g + (0.1 – h) × 13.6 × 1000 g ⇒ 0.1 × 1300 = 800 h + (0.1 – h) × 13600 ⇒ h = 0.096 m = 9.6 cmIllustration 24. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the piston have to bear? (taking g = 10 m/s2).Solution: According to Pascal’s law P1 = P2 = Pa = × 105 Pa = 7.06 × 105 Pa.Illustration 25. A cubical box of wood of side 30 cm weighing 21.6 kg floats on water with two faces horizontal. Calculate the depth of immersion of wood.Solution: For a floating body, the weight of displaced liquid should be equal to the weight of the block. Let x be the depth of immersion. Then (x×30×30)×1×g = 21.6×103 × g ⇒ x == 24 cm.Illustration 26. A block of brass of mass 0.5 kg and density 8 × 103 kg/m3 is suspended from a string. What will be the tension in the string if the block is completely immersed in water? ( g = 10 m/s2)Solution: Volume of the block V = m3 upthrust due to water (Th) = Vρwg = × 103 × 10 = = 0.625 N The tension in the string T = W – Th = mg – Th = 0.5 × 10 – 0.625 = 4.375 N.Illustration 27. A log of wood floats in water with of its volume above the surface. What is the density of wood?Solution: For a floating body, weight = force of upthrust ⇒ VBρB.g = Vin.ρg ⇒ V.ρB = V.ρW ⇒ ρB = × 103 = 0.8 × 103 kg/m3.Illustration 28. A body weighs 160 g in air, 130 g in water and 136 g in oil. What is the specific gravity of oil?Solution: Specific gravity of oil = = 0.8.Illustration 29. An iceberg is floating partially immersed in sea-water. The density of sea-water is 1.03 g/cm3 and that of ice is 0.92 g/cm3. What is the fraction of the total volume of the iceberg above the level of sea-water?Solution: In case of floatation weight = upthrust i.e. mg = Vinσg or Vρg Vinσg or Vin­ = V so Vout = V – Vin­ = V ∴ fout = = 0.106Illustration 30. A piece of ice floats in a liquid. What will happen to the level of liquid after the ice melts completely?Solution: Consider a liquid of density ρL with level A in a beaker. Let a piece of ice of mass m float in the liquid. The increase in level of the liquid is AB. Suppose VD is the volume of liquid displaced. Then, weight of the ice = weight of liquid displaced mg = VD ρLg or VD = When the ice gets completely melted, let the level of (liquid + water) be at C. The difference levels A and C is due to the which got converted into water. Thus if volume of the molten ice (i.e., water) be V0. then, V0 = Where ρw = density of water. Here we consider the following three cases – (i) If ρL > ρw then V0 > VD i.e. the level of (liquid + water) will rise (ii) If ρL < ρw then V0 < VD the level of (liquid + water) will come down (iii) If ρL = ρw then V0 = VD then the level will remain unchanged.Illustration 31. A boat carrying a number of large stones is floating in water. What will happen to the water level if the stones are unloaded into the water ?Solution: Let M = mass of the boat, m = mass of the stones For floating condition weight = upthrust (M + m) g = VD ρw g ; where VD = volume of water displaced VD = …..(1) After the stones are unloaded into the water ( = volume of water displaced by boat) ( = volume of water displaced by stones) ∴ total volume of water displaced VD = + = ….(2) ∴ ⇒ VD > VD So the water level will fall.BEGINNER’S BOX - 31. When two metals with equal volumes are mixed together, the density of the mixture is 4 kg/m3. When equal masses of the same two metals are mixed together, the mixture density is 3 kg/m3. Calculate the densities of each metal.2. A mercury barometer reads 75 cm in a stationary lift. What reading does it show when the lift is moving downwards, with an acceleration of 1 m/s2 ?3. The diameter of a piston P2 is 50 cm and that of a piston P1 is 10 cm. What is the force exerted on P2 when a force of 1 N is applied on P1 ?4. An open U-tube of uniform cross-section contains mercury. If 27.2 cm of water column is poured into one limb of the tube, how high does the mercury surface rise in the other limb from its initial level? [ ρw = 1 g/cm3 and ρHg = 13.6 g/cm3 ]5. A certain block weighs 15 N in air, 12 N in water. When immersed in another liquid, it weighs 13 N. Calculate the relative density of (i) the block (ii) the other liquid.6. A block of wood floats in water with two-third of its volume submerged. The block floats in oil with 0.90 of its volume submerged. Find the density of (i) wood and (ii) oil. Density of water is 103 kg/m3.7. A 700 g solid cube having an edge of length 10 cm floats in water. What volume of the cube is outside water?8. If a block of iron of density 5 g/cm3 and size 5 cm × 5 cm × 5 cm was weighed whilst completely submerged in water, what would be the apparent Weight in g – f (gram-force) ?3. HYDRO-DYNAMICS When a fluid moves in such a way that there are relative motions among the fluid particles, the fluid is said to be flowing.3.1 Types of fluid flow Fluid flow can be classified as :● Steady Flow Steady flow is defined as that type of flow in which the fluid velocity at a point do not change with time. The fluid particle may have a different velocity at some other point. In steady flow all the particles passing through a given point follow the same path and hence a unique line, of flow. This line or path is called a streamline. Streamlines do not intersect each other, if they do so any particle at the point of intersection can move in either directions and consequently the flow cannot be steady.● Laminar and Turbulent Flow Laminar flow is the flow in which fluid particles move along well-defined streamlines which are straight and parallel. In laminar flow the velocities at different points in the fluid may have different magnitudes, but their directions are parallel. Thus the particles move in laminae or layers sliding smoothly over the adjacent layer. Turbulent flow is an irregular flow in which the particles move in zig-zag way due to which eddy formation take place which are responsible for high energy losses.● Compressible and Incompressible Flow In compressible flow the density of fluid varies from point to point i.e., the fluid density is not constant, whereas in an incompressible flow the density of the fluid remains uniform throughout. Liquids are practically incompressible while gases are highly compressible. ● Rotational and Irrotational Flow Rotational flow is the flow in which the fluid particles while flowing along different path-lines also rotate about their own axes. In irrotational flow the particles do not rotate about their axes.3.2 Equation of continuity● The continuity equation is the mathematical expression of the law of conservation of mass in fluid dynamics.● In the steady flow mass of the fluid entering into a tube of flow in a particular time interval is equal to the mass of fluid leaving the tube. or ρ1A1v1 = ρ2A2v2 for incompressible fluid ρ1 = ρ2 or A1v1 = A2v2 or Av = constant Volume flux = Rate of flow = Volume of liquid flowing per second Q = Av3.3 Bernoulli's Theorem● Bernoulli's theorem is the mathematical expression of the law of mechanical energy conservation in fluid dynamics.● Bernoullis theorem is applicable to ideal fluids. Characteristics of an ideal fluid are : (i) The fluid is incompressible. (ii) The fluid is non-viscous. (iii) The fluid flow is steady. (iv) The fluid flow is irrotational.● Every volume at a point in an ideal fluid flow is associated with three kinds of energies : (i) Kinetic Energy If a liquid of mass (m) and volume (V) is flowing with velocity (v) then K.E. = mv2 and kinetic energy per unit volume = (ii) Potential Energy If a liquid of mass (m) and volume (V) is at a height (h) above the surface of the earth then its P.E. = mgh and potential energy per unit volume = gh = ρgh (iii) Pressure Energy If liquid moves through a distance () due to pressure P on area A then Pressure energy = Work done = force × displacement = pressure × area × displacement = PA = PV [ A = volume V] Pressure energy per unit volume = = P.● Bernoulli's theorem According to Bernoulli's Theorem, in case of steady flow of incompressible and non-viscous fluid through a tube of non-uniform cross-section then the sum of the pressure, the potential energy per unit volume and the kinetic energy per unit volume is same at every point in the tube, i.e., P+ ρgh + ρv2 = constant. Consider a liquid flowing steadily through a tube of non- uniform cross-section as shown in figure. If P1 and P2 are the pressures at the two ends of the tube respectively, work done in pushing the volume ∆V of the incompressible liquid across sections B and C through the tube is W = (P1 – P2)∆V ……(i) This work is used by the liquid in two ways : (i) In changing the potential energy of mass ∆m (corresponding to the volume ∆V) ∆U = ∆mg (h2 – h1) …... (ii) (ii) In changing the kinetic energy ∆K = ∆m Now as the liquid is non-viscous, by work-energy theorem W = ∆U + ∆K i.e., (P1 – P2) ∆V = ∆mg (h2 – h1) + ∆m P1 – P2 = ρg(h2 – h1) + ρ  [as ρ = ∆m / ∆V] P1 + ρgh1 + = P2 + ρgh2 + ⇒ P + ρgh + ρv2 = constant This equation is the Bernoulli's equation and expresses principle of conservation of mechanical energy in case of moving fluids. The sum of pressure energy, kinetic energy and potential energy per unit volume remains constant along a streamline in an ideal fluid flow i.e., P + ρv2 + ρgh = constant (Energy per unit volume) or + gh = constant (Energy per unit mass) or + h = constant (Energy per unit weight) In the above equation is called the pressure head, is called the velocity head and h is called the gravitational / potential head.3.4 APPLICATIONS OF BERNOULLI'S THEOREM● Venturimeter or Venturi Tube or Flowmeter Venturimeter is used to measure the flow velocities in an incompressible fluid. As shown in figure if P1 and P2 are the pressures and v1 and v2 are the velocities of the fluid of density ρ at points 1 and 2 on the same horizontal level and A1 and A2 be the respective areas, then from equation of continuity A1v1 = A2v2 or v2 = v1 ……(1) From Bernoulli’s equation for horizontal flow, P1 + = P2 + or P1 + = P2 + [from equation (i)] or P1 – P2 = But P1 – P2 = ρgh [difference in heights between the liquid surfaces in the two arms is h] ∴ ρgh = ∴ v1 = = A2 If Q be the volume of liquid flowing per unit time  = A1v1 = A2v2 = A1A2 Thus at a point where the cross-sectional area is smaller velocity is greater and pressure is lower and vice versa.● Speed of efflux (Torricelli's Law) As shown in the figure the area of cross-section of the vessel A is very large as compared to that orifice B, therefore speed of liquid flow at A is zero i.e. vA ≈ 0. The fluid at sections A and B are at the same pressure P0 (atmospheric pressure). Applying Bernoulli's theorem at A and B. P0 + ρgH + = P0 + ρg(H – h) + or = ρgh or vB = This equation is same as that of the velocity acquired by a freely falling body after falling through h height and is known as Torricelli’s law. Writing the equation of uniformly accelerated motion in the vertical direction H – h = 0 + gt2 (from sy = uyt + ay t2) ⇒ t = , t = time of flight as in case of horizontal projection from the top of a tower. Horizontal range R = vx­t = or R = Range will be maximum when h = H – h or h = ∴ Rmax. = = H ● Magnus Effect (Observed in a Spinning Ball) Tennis and cricket players usually experience that when a ball is thrown spinning it moves along a curved path. This is called swing of the ball. This is due to the air which is being dragged round by the spinning ball. When the ball spins, the layer of the air around it also moves with the ball. So, as shown in figure the resultant velocity of air increases on the upper side and reduces on the lower side. Hence according to Bernoulli's theorem the pressure on the upper side becomes lower than that on the lower side. This pressure difference exerts a force on the ball due to which it moves along a curved path. This effect is known as Magnus-effect.● Aero foil This is a structure which is shaped in such a way so that its motion relative to a fluid produces a force perpendicular to the flow. As shown in the figure the shape of the aerofoil section causes the fluid to flow faster over the top surface than below the bottom i.e. the streamlines are closer above than below the aero foil. By Bernoulli's theorem the pressure at above reduced whereas that underneath it gets increased. Thus a resultant upward thrust is generated normal to the flow and it is this force which provides most of the upward lift for an aeroplane. Examples of aerofoils are aircraft wings, turbine blades and propellers.● Sprayer or Atomizer This is an instrument used to spray a liquid in the form of small droplets (fine spray). It consists of a vertical tube whose lower end is dipped in the liquid to be sprayed, filled in a vessel. The upper end opens in a horizontal tube. At one end of the horizontal tube there is a rubber bulb and the other end has is a fine bore (hole). When the rubber bulb is squeezed, air rushes out through the horizontal tube with very high velocity and thus the pressure reduces (according to Bernoulli's theorem). Consequently, the liquid in the vessel rises up and mixes with air in the form of small droplets which gets ejected in the form of a fine spray Example : paint guns, perfume or deodaurant sprayer, etc.● Motion of the Ping-Pong Ball When a ping-pong ball is placed on a vertical stream of water-fountain, it rises upto a certain height above the nozzle of the fountain and spins about its axis. The reason for this is that the streams of water rise up from the fountain with large velocity so that the air-pressure decreases. Therefore, whenever the ball tends to fall out from the stream, the outer air which is at atmospheric pressure pushes it back into the stream (in the region of low pressure). Thus the ball remains more or less stable in the fountain.● Blowing-off of Tin Roof Tops in Wind Storm l When wind blows with a high velocity above a tin roof, it causes lowering of pressure above the roof, while the pressure below the roof is still atmospheric. Due to this pressure-difference the roof is lifted up and is blown away during storms.● Pull-in or Attraction Force by Fast Moving Train If we are standing on a platform and a train passes through the platform with very high speed we are pulled towards the train. This is because as the train 'moves at high speed, the pressure close to the train decreases. Thus the air away from the train which is still at atmospheric pressure pushes us towards the train. The reason behind flying-off of small papers, straws and other light objects towards the train is also the same.GOLDEN KEY POINTS● At hills, where the river is narrow and shallow (i.e., small cross-section) the flow will be faster, while in planes where the river is wide and deep, (i.e., large cross section) the flow will be slower and so deep water appears to be still.● Which water falls from a tap, the velocity of falling water under the action of gravity will increase with distance from the tap (i.e.,v2 > v1). So in accordance with continuity equation the cross-section of the water stream will decrease (i.e., A2 < A1), i.e., falling stream of water becomes narrower.● Practically some energy of the fluid gets converted into heat energy and is lost. But Bernoulli's equation is derived without considering this loss of energy.● Speed of the liquid coming out of the orifice is independent of the nature and quantity of liquid in the container or the area of the orifice. (as long as the orifice is small)● Greater is the distance of the hole from the free surface of liquid greater will be the velocity of efflux. This is why liquid gushes out with maximum velocity from the orifice which is at maximum vertical distance below the free surface of the liquid.● The horizontal range is same for liquid coming out of holes at equidistant from the liquid surface and the base .IllustrationsIllustration 32. The cylindrical tube of a spray pump has a, cross-section of 8 cm2, one end of which has 40 fine holes each of area 10–8 m2. If the liquid flows inside the tube with a speed of 0.15 m/min, then find the speed with which the liquid is ejected through the holes.Solution: From equation of continuity A1v1 = A2v2 (8 × 10–4) × = (40 × 10–8) × v2 ⇒ v2 = 5 m/s.Illustration 33. A syringe containing water is held horizontally with its nozzle at a height h above the ground as shown in fig. The cross-sectional areas of the piston and the nozzle are A and a respectively. The piston is pushed with a constant speed v. Find the horizontal range R of the stream of water on the ground.Solution: let v' be the horizontal speed of water when it emerges from the nozzle then from equation of continuity Av = av' ⇒ v' = Let t be the time taken by the stream of water to strike the ground then h = gt2 ⇒ t = ⇒ horizontal distance R = v' = .Illustration 34. Water is flowing through two horizontal pipes of different diameters which are connected together. In the first pipe the speed of water is 4 m/s. and the pressure is 2 × 104 N/m2. Calculate the speed and pressure of water in the second pipe. The diameters of the pipes are 3 cm and 6 cm respectively ?Solution: If A is the area of cross-section of a pipe at a point and v is the velocity of flow of water at that point, then by the principle of continuity Av = constant ⇒ A1v1 = A2v2 ⇒   ⇒ v2 = v1 = × 4 = 1 m/s. From Bernoulli’s theorem : P1 + = P2 + ⇒ P2 = P1 + ∴ P2 = 2 × 104 + × (103) × (16 – 1) = 2 × 104 + 7.5 × 103 = 2.75 × 104 N/m2Illustration 35. The diagram (fig.) shows venturimeter through which water is flowing. The speed of water at X is 2 cm/s. Find the speed of water at Y (taking g = 1000 cm/s2).Solution: By using Bernoulli’s principle – P1 + = P2 + ⇒ P1 – P2 = ⇒ ρgh = putting the values in equation 1000 × 0.51 = ⇒ v2 = 32 cm/sIllustration 36. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m/s and 63 m/s respectively. What is the lift on the wing if its area is 2.5 m2? The density of air is 1.3 kg/m3.Solution: From Bernoulli’s equation P2 – P1 = Upward force on the wing, F = (P2 – P1)A, where A is area of the wing F = A = × 10.3 (702 – 632) × 2.5 = 1.5 × 103 N.Illustration 37. A cylindrical tank 1 m in radius rests on a platform 5 m high. Initially, the tank is filled with water to a height of 5 m. A small plug whose area is 10–4 m2 is removed from an orifice located on the side of the tank at the bottom. Calculate the : (i) initial speed with which water flows out from the orifice (ii) initial speed with which the water strikes the ground.Solution: (i) Applying Bernoulli’s theorem between the water surface and the orifice, P0 + ρ(0)2 + ρgh = P0 + ρv2 + ρg(0) ρgh = ρv2 ; v = = 10 m/s. (ii) Let v' be the initial velocity with which the water strikes the ground Then, applying Bernoulli’s theorem between the top of the tank and the ground level, we get v' = = = 14.1 m/sBEGINNER’S BOX - 41. An incompressible liquid flows as shown in the figure. calculate the speed v of the fluid in the lower branch. 2. A hole is made in a vessel containing water at a depth of 3.2 m below the free surface. What would be the velocity of efflux?3. Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 N/m2 between A and B where the area of cross-section are 30 cm2 and 15 cm2 respectively. Find the rate of flow of water through the tube.4. VILCOSITY Viscosity is the property of a fluid (liquid or gas) by virtue of which it opposes the relative motion between its adjacent layers. It is the fluid friction or internal friction. The internal tangential force which tends to retard the relative motion between the adjacent layers is called viscous force.4.1 Newton's law of viscosity. Suppose a liquid is flowing in a streamlined motion on a horizontal surface OX. The liquid layer in contact with the surface is almost at rest while the velocity of other layers increases with increasing distance from the surface OX. The uppermost layer flows with maximum velocity. Let us consider two parallel layers PQ and RS at distances y and y + ∆y from OX. Let the change in velocity over a perpendicular distance ∆y be ∆vx. The rate of change of velocity with distance perpendicular to the direction of flow i.e. , is called velocity gradient. According to Newton, the viscous force F acting between two adjacent layers of a liquid flowing in streamlined motion depends upon the following two factors : (i) F ∝ contact area of the layers i.e. F ∝ A (ii) F ∝ velocity gradient between the layers i.e. F ∝  Combining (i) and (ii) F µ A ⇒ F = ηA where η is constant called coefficient of viscosity of the liquid. Coefficient of viscosity η = SI UNIT : N-sm–2 = Pa-s = poiseuille (PI) = deca poise CGS UNITS : dyne-s/cm2 = poise ; 1 decapoise = 10 poise. Dimensions : [M1L–1T–1]4.2 Stoke’s Law and Terminal velocity● Stoke's Law Stoke showed that if a small sphere of radius r is moving with a velocity v through a homogeneous stationary medium (liquid or gas), of viscosity η then the viscous force acting on the sphere is Fv = 67πηrv.● Terminal Velocity When a solid sphere falls in a liquid, its accelerating velocity is controlled by the viscous force due to liquid and hence it attains a constant velocity which is known as the terminal velocity (vT). As shown in the figure when the body moves with constant velocity i.e. terminal velocity (with no acceleration) the net upward force (upthrust Th + viscous force Fv) balances the downward force (weight of the body W). Therefore Th + Fv = W ⇒ πr3σg + 6πηrv­T = πr3ρg ⇒ vT = where r = radius of body ρ = density of body a = density of medium η = coefficient of viscosity. Graph: The variation of velocity with time (or distance) is shown in the adjacent graph.● Some applications of terminal velocity :(a) Actual velocity of rain drops is very small in comparison to the velocity which would have acquired by a body falling freely from the height of clouds. (b) Descent of a parachute with moderate velocity. (c) Determination of electronic charge in Milikan's oil drop experiment.4.3 Reynold's Number (Re) The type of flow pattern (streamline, laminar or turbulent) is determined by a non-dimensional number called Reynold's number (Re). defined as Re = where ρ is the density of the fluid having viscosity η and flowing with a mean speed v. Here d denotes the lateral dimension of the obstacle or boundary of fluid flow. Although there is no perfect demarcation for the value of Re in case of laminar and turbulent flow but certain references take the value as : Re< 1000> 2000Between 1000 to 2000Type of flowStreamline or laminarturbulentunsteady Upon increasing the speed of flow gradually transition from laminar flow to turbulent flow takes place at certain speed. This speed is called critical speed. For fluids lower density and higher viscosity with laminar flow is more probable.4.4 Dependency of viscosity On Temperature of Fluid(a) Since cohesive forces decrease with increase in temperature. Viscosity of liquids decreases with a rise in temperature.(b) Viscosity of gases is the result of diffusion of gas molecule from one moving layer to other. With an increase in temperature, the rate of diffusion increases. Conseq