Page 1 : Chapter 10, , Mechanical Properties of Fluids, Prepared by:- Shri S. S. Sitre M.Sc.,B.Ed., 10.1 Introduction :Fluids: 1) Fluid is a substance that flows under the action of an applied force and does not have, a shape of its own. There are two types of fluids one is liquid and another is gaseous., 2) The study of fluids at rest us known as hydrostatic or fluid statics. The study of fluid, in motion is known as hydrodynamics., 10.2 Pressure: a) Definition:-The magnitude of force acting perpendicular to the surface per unit area., F, i. e., P=, A, b) Pressure-measuring device:1) It consists of an evacuated chamber with a spring that is, calibrated to measure the force acting on the piston., 2) This device is placed at a point inside the fluid., 3) The inward force exerted by the fluid on the piston is, balanced by the outward spring force and is thereby, measured, 4) S.I. unit of pressure is N/m2 and C.G.S. unit is dynes/cm2., 5) The dimensional formula for pressure is, [ M L−1 T −2 ]., 6) A common unit of pressure is the atmosphere (atm), i.e. the, pressure exerted by the atmosphere at sea level (1 atm =, 1.013 × 105 Pa)., 7) Density ρ for a fluid of mass m occupying volume V, ρ = m/V, 10.2.1 Pascal’s Law: a) Statement :-“The pressure applied to an enclosed fluid is equally distributed to every, point of the fluid and the walls of the containing vessel”., b) Diagram :-, , c) Explanation :1) An element in the interior of a fluid at rest., PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 1 of 20

Page 2 : 2) The fluid exerts pressures Pa , Pb and Pc on this element of area corresponding to the, normal forces Fa , Fb and Fc on the faces BEFC, ADFC and ADEB denoted by Aa , Ab, and Ac respectively., 3) Then, Fbsinθ = Fc , Fbcosθ = Fa (by equilibrium), Absinθ = Ac , Ab cosθ = Aa (by geometry), 4) Thus,, 𝐹𝑏, 𝐹𝑐, 𝐹𝑎, =, =, 𝐴𝑏 𝐴𝑐 𝐴𝑎, Pb=Pc=Pa, 5) Hence, pressure exerted is same in all directions in a fluid at rest., d) Applications of Pascal’s Law, Hydraulic lift is a device used to lift a heavy load or to press the things., Hydraulic break system is used in auto-mobile to reduce the motion., , 10.2.2 Variation of pressure with depth :a) Gauge Pressure: 1) It is the pressure difference between two points inside fluid separated by a depth., Consider a vessel filled with fluid of density ‘ρ’ and a small volume element of, thickness ‘dh’ and cross section ‘A’ as shown in figure., 2) According to equilibrium condition of concurrent forces,, (, , downward force acting, weight of the, upward force acting, )+ (, )− (, )=0, on volume element, voulme element, on volume element, F1 + W − F2 = 0, P1 A + m g − P2 A = 0, m g = ( P2 − P1 ) A, , 3) density of fluid, ρ =, , m, V, , m=V ρ, m=h A ρ, h A ρ g = ( P2 − P1 ) A, ( P2 − P1 ) = ρ g h, 4) P1 may be replaced by atmospheric pressure (Pa ) and we replace P2 by P. Then, (P − Pa ) = h ρ g, 5) The excess of pressure, P − Pa , at depth h is called a gauge pressure at that point., , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 2 of 20

Page 3 : b) Hydrostatic paradox:Consider three vessels A, B and C of different, shapes. They are connected at the bottom by a, horizontal pipe. On filling with water, the level in, the three vessels is the same, though they hold, different amounts of water. This is so because water, at the bottom has the same pressure below each, section of the vessel., 10.2.3 Atmospheric Pressure and Gauge Pressure:a) Measurement of atmospheric pressure:1) The pressure of the atmosphere at any point is equal to the, weight of a column of air of unit cross-sectional area, extending from that point to the top of the atmosphere. At, sea level, it is 1.013 × 105 Pa (1 atm), 2) A long glass tube closed at one end and filled with mercury, is inverted into a trough of mercury as shown in Fig. This, device is known as ‘mercury barometer’., 3) The space above the mercury column in the tube contains, only mercury vapour whose pressure P is so small that it, may be neglected., 4) Thus, the pressure at Point A=0., 5) The pressure inside the coloumn at Point B must be the same as the pressure at Point, C, which is atmospheric pressure, Pa., Pa = ρgh, where ρ is the density of mercury and h is the height of the mercury column in the, tube., 6) A pressure equivalent of 1 mm is called a torr (after Torricelli). 1 torr = 133 Pa., 7) The mm of Hg and torr are used in medicine and physiology., 8) In meteorology, a common unit is the bar and millibar. 1 bar = 105 Pa, b) Measurement of gauge pressure:1) An open tube manometer is a useful instrument for measuring, pressure differences., 2) It consists of a U-tube containing a suitable liquid i.e., a low, density liquid (such as oil) for measuring small pressure, differences and a high density liquid (such as mercury) for large, pressure differences., 3) One end of the tube is open to the atmosphere and the other end, is connected to the system whose pressure we want to measure., 4) The pressure P at A is equal to pressure at point B. What we normally measure is the, gauge pressure, which is P − Pa, and is proportional to manometer height h., PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 3 of 20

Page 4 : 10.2.4 Hydraulic Machines:a) Pascal’s law:Whenever external pressure is applied on any part of a fluid contained in a vessel, it is, transmitted undiminished and equally in all direction., b) Explanation:1) A number of devices, such as hydraulic lift and hydraulic brakes, are based on the, Pascal’s law., 2) A piston of small cross-section A1 is used to exert a force F1 directly on the liquid., 3) The pressure P = F1 /A1 is transmitted throughout the, liquid to the larger cylinder attached with a larger, piston of area A2 , which results in an upward force, of P × A2 ., 4) Therefore, the piston is capable of supporting a large, force (large weight of, say a car, or a truck,, F2 = PA2 = F1A2/A1, 5) By changing the force at A1 , the platform can be, moved up or down. Thus, the applied force has been increased by a factor of A2/A1, and this factor is the mechanical advantage of the device., 10.3 Streamline Flow: a) Definition:-The flow of fluid in which the velocity of every particle of the fluid at any, point is constant such flow is known as streamline flow or steady flow., The path travelled by the particles of a fluid in a streamline flow is known as streamline., b) Characteristics of Streamline flow:,  A tangent at any point at the streamline gives the direction of the velocity of the fluid, particle at that point.,  No two streamlines can cross each other., c) Equation of Continuity: 1) Consider a fluid of density ‘ρ’ flowing through a tube of varying cross-section area as, shown in figure., 2) Let ‘a1 ’ & ‘v1 ’ are the cross-section area and velocity of flow at ‘P’. Also ‘a2 ’ & ‘v2 ’, are the cross-section and velocity of flow at ‘Q’., , 3) The mass of fluid crossing the area ‘a1 ’ in time ‘∆t’ is,, ∆m1 = volume × density, ∆m1 = a1 v1 ∆t ρ, PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 4 of 20

Page 5 : 4), , 5), , 6), , 7), 8), , 9), , ∆m1, = a1 v1 ρ, ∆t, This is mass flux at ‘P’, similarly the mass flux at ‘Q’ is,, ∆m2, = a 2 v2 ρ, ∆t, But, no fluid is leaving or entering in between ‘P’ and ‘Q’., ∆m1 ∆m2, =, ∆t, ∆t, a1 v1 ρ = a2 v2 ρ, a1 v1 = a2 v2, In general, a v = constant, 1, ∴v ∝, a, Thus above equation is called equation of continuity. It states that, “the rate of flow of, fluid is inversely proportional to area of cross-section”., Laminar Flow: If the fluid flows over a horizontal surface in the form of layers of different velocities, then the flow of the fluid is called laminar flow., Turbulent Flow: The flow of fluid in which velocity of all particles crossing a given a point is not same, and the motion of the fluid becomes irregular is called turbulent flow., , 10.4 Bernoulli’s Principle: a) Statement:-The sum of the pressure (P), the kinetic energy per unit volume and the, potential energy per unit volume (ρgh) remains a constant., b) Expression :1) Suppose v1 is the speed at B and v2 at D, then fluid initially at B has moved a distance, v1 ∆t to C., 2) In the same interval ∆t the fluid initially at D moves to E, a distance equal to v2 ∆t., 3) The work done on the fluid at left end (BC) is, W1 = P1A1 (v1 ∆t) = P1 ∆V., , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 5 of 20

Page 7 : 𝑣1 = √2𝑔ℎ +, , 2(P − Pa), ρ, , 7) When P >>Pa and 2 g h may be ignored, the speed of efflux is determined by the container, pressure. Such a situation occurs in rocket propulsion., 8) If the tank is open to the atmosphere, then P = Pa Then 𝑣1 = √2𝑔ℎ, This is also the speed of a freely falling body and Equation 𝑣1 = √2𝑔ℎ represents, Torricelli’s law., 10.4.2 Venturi-meter:1) The Venturi-meter is a device to measure, the flow speed of incompressible fluid., 2) It consists of a tube with a broad diameter, and a small constriction at the middle., 3) The speed v1 of the liquid flowing through, the manometer in the form of a U-tube at the, broad neck area A., 𝐴, , 4) The speed at the constriction 𝑣2 = 𝑣1, 𝑎, , 5) Using Bernoulli’s equation for (h1 =h2),, 1, , 𝐴 2, , 1, , P1 + ρ 𝑣12 = P2 + ρ 𝑣12 ( ), 2, 2, 𝑎, P1 - P2 =, 𝜌𝑚 𝑔ℎ =, 𝑣1 = √, , 1, , 𝐴 2, , 1, , 𝐴 2, , ρ 𝑣12 [( ) − 1], 2, 𝑎, ρ 𝑣12 [( ) − 1], 2, 𝑎, , 2𝜌𝑚 𝑔ℎ, 𝜌, , 𝐴 2, , −, , [(𝑎) − 1], , 1, 2, , 6) The principle behind this meter has many applications., 7) The carburetor of automobile has a Venturi, channel (nozzle) through which air flows, with a high speed. The pressure is then, lowered at the narrow neck and the petrol, (gasoline) is sucked up in the chamber to, provide the correct mixture of air to fuel, necessary for combustion., 8) Filter pumps or aspirators, Bunsen burner,, atomisers and sprayers used for perfumes or, to spray insecticides work on the same, principle., , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 7 of 20

Page 8 : 10.4.3 Blood Flow and Heart Attack:1) The artery may get constricted due to the, accumulation of plaque on its inner walls., 2) The speed of the flow of the blood in this region is, raised which lowers the pressure inside and the artery, may collapse due to the external pressure., 3) The heart exerts further pressure to open this artery, and forces the blood through., 4) As the blood rushes through the opening, the internal, pressure once again drops due to same reasons, leading to a repeat collapse. This may result in heart, attack., 10.4.4 Dynamic Lift :1) Ball moving without spin: From the symmetry of streamlines it is clear that the, velocity of fluid (air) above and below the ball at corresponding points is the same, resulting in zero pressure difference. The air therefore, exerts no upward or downward, force on the ball., 2) Ball moving with spin: The velocity, of air above the ball relative to the ball, is larger and below it is smaller. This, difference in the velocities of air, results in the pressure difference, between the lower and upper faces, and there is a net upward force on the ball. This dynamic lift, due to spinning is called Magnus effect., 3) Aerofoil or lift on aircraft wing: The flow streamlines to, crowd together speed on top is higher than that below it., There is an upward force resulting in a dynamic lift of the, wings and this balances the weight of the plane., 10.5 Viscosity: a) Definition:-The property of a liquid by virtue of which it opposes relative motion, between its different layers is called viscosity., b) Viscous force: In case of a liquid having relative motion between the layers internal, forces are developed, which retard the relative motion. These retarding forces are called, viscous force., c) Viscous drag or drag force: When an object moves relative to a fluid, the fluid exerts a, friction like retarding force on the object. This force is called viscous drag or drag force., d) Co-efficient of viscosity (𝜼): The coefficient of viscosity of a fluid is defined as the, ratio of shearing stress to the strain rate., PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 8 of 20

Page 9 : e) Explanation:, 1) Consider the bottom plate is fixed and top, plate is moved with a constant velocity, 𝑣, with respect to the fixed plate., 2) The fluid in contact with the surface has same, velocity as the surface., 3) Therefore, the layer of the liquid in contact, with the top surface moves with a velocity, 𝑣, and the layer in contact with the fixed surface, remain at rest., 4) The velocities of the layers increase uniformly from bottom to the top., 5) Due to this motion, a portion of liquid initially having the shape ABCD takes the, shape AEFD after a short interval of time Δ𝑡., 6) If the distance between the plates is 𝑙 and plate at the top moves through a distance, Δ𝑥 in time Δ𝑡 then,, 𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑎𝑖𝑛 = Δ𝑥/𝑙, 𝑆𝑡𝑟𝑎𝑖𝑛 𝑟𝑎𝑡𝑒 = Δ𝑥Δ𝑡/l = 𝑣/𝑙, 7) The co − efficient of viscosity,, 𝜼=, , 𝑭, 𝑨, 𝒗, 𝒍, , =, , 𝑭𝒍, 𝒗𝑨, , SI unit of viscosity is 𝑁𝑠𝑚−2 It can be expressed also in 𝑝𝑎𝑠𝑐𝑎𝑙 𝑠𝑒𝑐𝑜𝑛𝑑., The dimensions are 𝑀𝐿−1𝑇−1., 10.5.1 Stokes Law: a) Stokes Law :-The viscous force acting on a spherical body moving in a fluid is directly, proportional to radius of spherical body, terminal velocity of body and coefficient of, viscosity of fluid., b) Expression:-The viscous drag(𝐹) acting on a spherical body of radius 𝑎 moving with, velocity 𝑣 in a fluid of co-efficient of viscosity 𝜂 is given by, 𝑭 = 𝟔𝝅𝜼𝒂𝒗, c) Terminal velocity (𝒗𝒕): When a body is dropped in a viscous fluid, it is first accelerated, and then its acceleration becomes zero and it attains a constant velocity. This constant, velocity is called terminal velocity., d) Expression for Terminal velocity:, 1) Consider a spherical body of radius 𝑎 falling through a viscous fluid having density 𝜎, and co-efficient of viscosity 𝜂., 2) Let 𝜌 be the density of the material of the body., 3) The viscous forces acting on that spherical body are, (i) its weight (𝑚𝑔) in downward, direction. (ii) upward thrust 𝑇 equal to the weight of the displaced fluid. (iii) viscous, drag 𝐹 in a direction opposite to the direction of motion of the body., 4) Net downward force acting on that body = 𝑚𝑔 − 𝑇 − 𝐹, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 9 of 20

Page 10 : 5) When the body attains terminal velocity, acceleration, 𝑎 = 0, 6) Net force acting on the body is zero. 𝑚𝑔 − 𝑇 − 𝐹 = 0, , 10.6 Reynolds Number: 1) Reynolds number: When the rate of flow of a fluid becomes very large, the flow loses, its orderliness and becomes turbulent., 2) In turbulent flow, the velocity of particles of the fluid at any point varies randomly with, time., 3) Osborne Reynolds defined a dimension less number which gives an approximate idea, about whether the flow would be turbulent or not. This number is called Reynolds number, denoted by 𝑅 ., 4) If 𝜂 is the viscosity and 𝜌 is the density of fluid flowing with a speed 𝑣 in a pipe of, diameter 𝑑, the value of 𝑅𝑒 is given by,, 𝑣ρD, 𝑅𝑒 =, η, 5) Classification of flow based on Reynolds number:, 1. For laminar or streamline flow, 𝑅𝑒 < 1000, 2. For the turbulent flow, 𝑅𝑒 > 2000, 3. For 1000 < 𝑅𝑒 < 2000, the flow becomes unsteady., 6) Critical Reynolds number: The value of Reynolds number at which the turbulence just, occurs is called critical Reynolds number., 7) Critical velocity: The maximum velocity of a fluid in a tube for which the flow remains, streamline is called Critical velocity., PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 10 of 20

Page 11 : Note :Cohesion / Cohesive Force: -The force of attraction between the molecules of same kinds, or same substance is called cohesive force., Adhesion / Adhesive Force: -The force of attraction between molecules of different kinds, or different substance is called adhesive force., 10.7 Surface Tension: a) Surface Tension: The property of a liquid at rest by virtue of which its free surface, behaves like a stretched membrane under tension and tries to minimise the surface area is, called surface tension. T =, , F, l, , b) Explanation:, 1) Consider three spheres indicating the spheres, of influence of the molecule at , 𝐵 and 𝐶., 2) The molecule 𝐴 is well inside the liquid, surface is attracted equally in all directions by, the neighbouring molecules. Hence the, resultant force on 𝐴 is zero., 3) The molecule 𝐵 has a part of its sphere of, influence above the liquid surface. Hence 𝐵, experiences a net force vertically downwards, in the liquid., 4) The molecule 𝐶 is just on the surface of the liquid, hence it experiences maximum, force pulling into the liquid as a result the surface of a liquid behaves as a stretched, membrane and the attractive inter molecular forces on the surface of the liquid tend to, compress the liquid surface, so that the liquid tries to minimise its surface area., 5) S.I. unit of surface tension is N m-1., 6) C.G.S. unit of surface tension is dynes cm-1., 7) Dimensional formula is [M L0 T-2], 10.7.1 Surface energy:, a) Definition The potential energy of the surface molecules per unit area of the surface is, called surface energy., b) Explanation:𝑺𝒖𝒓𝒇𝒂𝒄𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝒑𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚/𝒂𝒓𝒆𝒂, Unit of surface energy is 𝑗𝑜𝑢𝑙𝑒/𝑚𝑒𝑡𝑟𝑒2 and dimensions are 𝑀𝐿 0𝑇 −2, 10.7.2 Surface energy and surface tension:, a) Relation between surface energy and surface tension:1) Consider a thin horizontal film of liquid ending in bar, free to slide over parallel guide, as shown., , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 11 of 20

Page 12 : 2) Let the bar be moved along horizontal, through a small distance 𝑑 in order to, increase the surface area. Some work has, to be done against the internal force., 3) If the internal force is 𝐹, then work done, by the internal force is 𝐹𝑑., 4) The increase in the area of the film is 𝑙𝑑, since the film has two surfaces, the increase, in the area of the film should be 2𝑙𝑑., 5) If 𝑆 is the surface energy of the film per unit area, the extra energy supplied to the, film is,, 𝑆(2𝑙𝑑) = 𝐹𝑑, S=, , Fd, 2ld, , =, , F, 1l, , 6) The quantity 𝑆 is the magnitude of surface tension and it is equal to the surface energy, per unit area of the liquid surface., b) Measurement of surface tension:, 1) A flat vertical glass plate, below which a, vessel of some liquid is kept, forms one arm, of the balance. The plate is balanced by, weights on the other side., 2) The vessel is raised slowly till the liquid just, touches the glass plate and pulls it down a, little because of surface tension., 3) Weights are added till the plate just clears, water., 4) If the additional weight required is 𝑊 then, the surface tension of the liquid air interface is,, 𝑆𝑙𝑎 = 𝑊/2𝑙 = 𝑚𝑔/2𝑙, where 𝑙 →length of the plate., 10.7.3 Angle of Contact (𝜃): a) Angle of contact: The angle between tangent to the liquid surface at the point of contact, and solid surface inside the liquid is called as angle of contact., b) Explanation:1) Now let us consider three interfacial tensions at, all the three interfaces, liquid-air, solid-air and, solid-liquid denoted by 𝑆𝑙 , 𝑆𝑠𝑎 and 𝑆𝑠𝑙, respectively., Then, 𝑆𝑙𝑎𝑐𝑜𝑠 𝜃 + 𝑆𝑠𝑙 = 𝑆𝑠𝑎, 2) If 𝑆𝑠𝑙 > 𝑆𝑙 , the angle of contact is obtuse. When, 𝜃 is obtuse angle, the molecules of the liquid are, attracted strongly to themselves and weakly to, PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 12 of 20

Page 13 : those of solids and the liquid will not wet the surface., 3) Ex: (i) water-leaf interface. (ii) water in waxy or oily surface. (iii) Mercury on any, surface. etc., 4) If 𝑆𝑠𝑙 < 𝑆𝑙 , the angle of contact is acute. When 𝜃 is acute angle, the molecules of the, liquid are attracted strongly to those of solids and the liquid will wet the surface, 5) Ex: (i) Water on glass. (ii) Water on plastic sheet., 10.7.3 Drops and Bubbles:, Due to surface tension, surface of liquid always has a tendency to have least surface area., For a given volume a sphere has the minimum surface., Excess pressure inside a liquid drop and Bubble, (i) For a liquid drop:, 1) Consider a liquid drop of radius 𝑟 having surface tension, 𝑆., 2) Let 𝑃𝑖 be the pressure inside the drop and 𝑃𝑜 be pressure, outside the drop., 3) The excess pressure inside the drop = (𝑃𝑖 − 𝑃𝑜 ), 4) Outward force acting on the drop = pressure × surface, area of the drop Outward force, = (𝑃𝑖 − 𝑃 ) × 4𝜋𝑟2, 5) Due to this force, the drop expends and radius increases by 𝑑𝑟., 6) Work done = force × change in radius work done, = (𝑃𝑖 − 𝑃 ) × 4𝜋𝑟 2 × 𝑑𝑟, 7) Increase in potential energy = surface tension × increase in surface area, = 𝑆 × [4(𝑟 + 𝑑𝑟) 2 − 4𝜋𝑟 2 ], = 𝑆 × [4(𝑟 2 + 𝑑𝑟2 + 2𝑟𝑑𝑟) − 4𝜋𝑟 2 ], = 𝑆 × [4𝜋𝑟 2 + 4𝜋𝑑𝑟2 + 8𝜋𝑟𝑑𝑟 − 4𝜋𝑟 2 ], = [4𝜋𝑑𝑟2 + 8𝜋𝑟𝑑𝑟] Since 𝑑𝑟 is very small 𝑑𝑟2 can be neglected., 8) Increase in potential energy = 8𝜋𝑆𝑟𝑑𝑟, 9) Therefore work done = 8𝜋𝑆𝑟𝑑𝑟 (𝑃𝑖 − 𝑃𝑜 ) × 4𝜋𝑟 2 × 𝑑𝑟, = 8𝜋𝑆𝑟𝑑𝑟 (𝑃𝑖 − 𝑃 ), = 8𝜋𝑆𝑟𝑑𝑟/4𝜋𝑟 2 × 𝑑𝑟, (𝑷𝒊 − 𝑷𝒐 ) = 𝟐𝑺/𝒓, (ii) For a bubble:, 1) In case of a liquid bubble, there are two, surfaces-inner and outer., 2) Consider a liquid bubble of radius 𝑟 having, surface tension 𝑆., 3) The excess pressure inside the bubble, = (𝑃𝑖 − 𝑃𝑜 ), , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 13 of 20

Page 14 : 4) Outward force acting on the bubble = pressure × surface area of the bubble Outward, force, = (𝑃𝑖 − 𝑃 ) × 4𝜋𝑟 2, 5) Due to this force, the bubble expends and radius increases by 𝑑𝑟., 6) Work done = force × change in radius work done, = (𝑃𝑖 − 𝑃 ) × 4𝜋𝑟 2 × 𝑑𝑟, 7) Increase in potential energy = surface tension × increase in surface area, = 𝑆 × 2[4(𝑟 + 𝑑𝑟) 2 − 4𝜋𝑟 2 ], = 𝑆 × 2[4(𝑟 2 + 𝑑𝑟2 + 2𝑟𝑑𝑟) − 4𝜋𝑟 2 ], = 𝑆 × 2[4𝜋𝑟 2 + 4𝜋𝑑𝑟2 + 8𝜋𝑟𝑑𝑟 − 4𝜋𝑟 2 ], = 2[4𝜋𝑑𝑟2 + 8𝜋𝑟𝑑𝑟] Since 𝑑𝑟 is very small 𝑑𝑟2 can be, neglected., 8) Increase in potential energy = 16𝜋𝑆𝑟𝑑𝑟, 9) Therefore work done = 16𝜋𝑆𝑟𝑑𝑟 (𝑃𝑖 − 𝑃𝑜 ) × 4𝜋𝑟 2 × 𝑑𝑟, = 16𝜋𝑆𝑟𝑑𝑟 (𝑃𝑖 − 𝑃 ), = 16𝜋𝑆𝑟𝑑𝑟/4𝜋𝑟 2 × 𝑑𝑟, (𝑷𝒊 − 𝑷𝒐 ) = 𝟒𝑺/𝒓, 10.7.5 Capillary rise: a) Capillarity: When a capillary tube is dipped in water, the water rises up in the tube. This, rise of liquid in a capillary tube is known as capillarity., b) Capillary rise:, 1) The surface of water in the capillary is concave., 2) Then there is a pressure difference between the, two sides of the top surface, which is given by,, (𝑃𝑖 − 𝑃𝑜) =, , 2 S cos θ, r, , where 𝑆 cos 𝜃 → vertical component of 𝑆, 3) Thus the pressure of the water inside the tube, just at the meniscus is less than the, atmospheric pressure., 4) Consider the two points 𝐴 and 𝐵.They must be at same height., Then, 𝑃𝑜 + 𝜌𝑔ℎ = 𝑃𝑖, 𝑃𝑖 − 𝑃𝑜 = 𝜌𝑔ℎ, where 𝜌 is the density of water and ℎ is called capillary rise, 2 S cos θ, ρgh =, r, 2 S cos θ, h=, r ρg, c) Practical applications of capillarity:, (i) The oil in the lamp rises in the wick to its top by capillary action., (ii) Sap and water rise up to the top of the leaves of the tree by capillary action., (iii) Ink is absorbed by the blotting paper due to capillarity., PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 14 of 20

Page 15 : (iv) The moisture rises in the capillaries of the soil to the surface, where it evaporates., To prevent this and preserve moisture in the soil, capillaries must be destroyed by, ploughing and levelling fields., 10.7.6 Detergents and Surface tension:, 1) We clean dirty clothes containing grease and oil stains by adding detergents or soap to, water., 2) Washing only with water does not remove grease stains., 3) This is because water does not wet grease dirt, because there is very little area of contact, between them., 4) If the water wet the grease, the flow of water could carry some grease away. This is, achieved by adding detergents., 5) The molecules of detergents are hairpin shaped, with one end attached to water and the, other to molecules of grease, oil or wax., 6) Thus they tend to form water-oil interfaces., 7) In other words, the addition of detergent reduces drastically the surface tension of water., 8) This kind of process using surface active detergent or surfactants is important not only, in cleaning but also in recovering oil, mineral ores etc., *****, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 15 of 20

Page 16 : Example 1, The two thigh bones (femurs), each of cross-sectional area10 cm2 support the upper part, of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs., Answer, Total cross-sectional area of the femurs is, A = 2 × 10 cm2, = 20 × 10–4 m2 ., The force acting on them is, F = 40 kg wt, = 400 N, g = 10 m s–2, This force is acting vertically down and hence, normally on the femurs., Thus, the average pressure is, Pav =, =, , 𝐹, 𝐴, 400, , 20 𝑋 10−4, 5, , = 2 𝑋10 𝑁𝑚−2, Example 2, What is the pressure on a swimmer 10 m below the surface of a lake?, Answer, h = 10 m, ρ = 1000 kg m-3., Take g = 10 m s–2, P = Pa + ρgh, = 1.01 × 105 + 1000 × 10 × 10, = 2.01 × 105 Pa, ≈ 2 atm, Example 3, The density of the atmosphere at sea level is 1.29 kg/m3 . Assume that it does not change, with altitude. Then how high would the atmosphere extend?, Answer, ρgh = 1.29 × 9.8 × h, 1.01 × 105 =1.29 × 9.8 × h, ℎ=, , 1.01 𝑥 105, 1.29 𝑥 9.8, , ℎ = 0.07989 𝑥 105, ∴ h = 7989 m, h ≈ 8 km, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 16 of 20

Page 17 : Example 4, At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the, gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a, submarine at this depth, the interior of which is maintained at sea level atmospheric, pressure. (The density of sea water is 1.03 × 103 kg m-3 , g = 10 m s–2.), Answer, h = 1000 m, ρ = 1.03 × 10 3 kg m-3 ., (a) P = Pa + ρgh, = 1.01 × 105 + 1.03 × 103 × 10 × 1000, = 104.01 × 105 Pa, ≈ 104 atm, (b) Gauge pressure is, P −Pa = ρgh, =PgPg, = 1.03 × 103 × 10 × 1000 m, = 103 × 105 Pa, ≈ 103 atm, (c) The pressure outside the submarine is, P =Pa + ρgh, and the pressure inside it is Pa ., Hence, the net pressure acting on the window is gauge pressure,, Pg = ρgh., Since the area of the window is A = 0.04 m2 ,, the force acting on it is, F = Pg A, = 103 × 105 × 0.04, = 4.12 × 105 N, , Example 5, Two syringes of different cross-sections (without needles) filled with water are, connected with a tightly fitted rubber tube filled with water. Diameters of the smaller, piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted, on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the, smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?, Answer, (a) Since pressure is transmitted undiminished throughout the fluid,, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 17 of 20

Page 18 : (b) Water is considered to be perfectly incompressible. Volume covered by the, movement of smaller piston inwards is equal to volume moved outwards due to the, larger piston., , Example 6, In a car lift compressed air exerts a force F1 on a small piston having a radius of 5.0 cm., This pressure is transmitted to a second piston of radius 15 cm . If the mass of the car to, be lifted is 1350 kg, calculate F1 . What is the pressure necessary to accomplish this, task? (g = 9.8 ms-2)., Answer, Since pressure is transmitted undiminished throughout the fluid,, , Example 7, Blood velocity: The flow of blood in a large artery of an anesthetised dog is diverted, through a Venturi meter. The wider part of the meter has a crosssectional area equal to, that of the artery. A = 8 mm2 . The narrower part has an area a = 4 mm2. The pressure, drop in the artery is 24 Pa. What is the speed of the blood in the artery?, Answer, density of blood = 1.06 × 103 kg m-3., 𝐴, , The ratio of the areas is = 2., 𝑎, , Using Eq. we obtain, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 18 of 20

Page 19 : Example 8, A fully loaded Boeing aircraft has a mass of 3.3 × 105 kg. Its total wing area is 500 m2 ., It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference, between the lower and upper surfaces of the wings (b) Estimate the fractional increase, in the speed of the air on the upper surface of the wing relative to the lower surface., [The density of air is ρ = 1.2 kg m-3], Answer, (a) The weight of the Boeing aircraft is balanced by the upward force due to the, pressure difference, ∆P × A = 3.3 × 105 kg × 9.8 ∆P, = (3.3 × 105 × 9.8 ) / 500, = 6.5 × 103 Nm-2, (b) We ignore the small height difference between the top and bottom sides in Eq., The pressure difference between them is then, , Example 9, A metal block of area 0.10 m2 is connected to a 0.010 kg mass via a string that passes, over an ideal pulley (considered massless and frictionless), as in Fig. A liquid with a, film thickness of 0.30 mm is placed between the block and the table. When released the, block moves to the right with a constant speed of 0.085 m s-1. Find the coefficient of, viscosity of the liquid., Answer, The metal block moves to the right because of the tension in the string., The tension T is equal in magnitude to the weight of the suspended mass m., Thus, the shear force F is, F = T = mg, = 0.010 kg × 9.8 m s–2, = 9.8 × 10-2 N, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 19 of 20

Page 20 : Example 10, The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at, 20oC is 6.5 cm s-1. Compute the viscosity of the oil at 20oC. Density of oil is 1.5 ×103 kg, m-3, density of copper is 8.9 × 103 kg m-3, Answer, vt = 6.5 × 10-2 ms-1 ,, a = 2 × 10-3 m, g = 9.8 ms-2 ,, ρ = 8.9 × 103 kg m-3 ,, σ =1.5 ×103 kg m-3., , Example 11, The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the, surface of water in a beaker. What is the pressure required in the tube in order to blow a, hemispherical bubble at its end in water? The surface tension of water at temperature of, the experiments is 7.30×10-2 Nm-1 . 1 atmospheric pressure = 1.01 × 105 Pa, density of, water = 1000 kg/m3 , g = 9.80 m s -2., Answer, , PUC I 10 MECHANICAL PROPERTIES OF FLUIDS, , Page 20 of 20