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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Chapter 10, Mechanical Properties of Fluids, Liquids and gases can flow and are therefore, called fluids., The fluid does not have any resistance to change of its shape. Thus, the, shape of a fluid is governed by the shape of its container., Basic difference between Liquids and Gases, A liquid is incompressible and has a free surface of its own. A gas is, compressible and it expands to occupy all the space available to it. Gas has, no free surface., , Pressure, The normal force(F) exerted by a fluid on an area A is called pressure., Pressure, P =, , 𝐅, , 𝐀, , Pressure is a scalar quantity., Its SI unit is Nm−2 or pascal (Pa), Dimensional formula is ML−1 T −2, A common unit of pressure is the atmosphere (atm). It is the pressure, exerted by the atmosphere at sea level., 1 atm = 1.013 × 𝟏𝟎𝟓 Pa., , Density, Density ρ for a fluid of mass m occupying volume V is given by, , ρ=, , 𝐦, 𝐕, , It is a positive scalar quantity., Its SI unit is kg m−3 ., The dimensions of density are [ML−3 ]., The density of water at 𝟒𝟎 C (277 K) is 1000 kg 𝐦−𝟑 ., A liquid is incompressible and its density is therefore, nearly constant at all, pressures. Gases, on the other hand exhibit a large variation in densities, with pressure.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Relative Density, The relative density of a substance is the ratio of its density to the density of, water at 40 C., Relative density=, , 𝐃𝐞𝐧𝐬𝐢𝐭𝐲 𝐨𝐟 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜𝐞, 𝐃𝐞𝐧𝐬𝐢𝐭𝐲 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 𝐚𝐭 𝟒𝟎 𝐂, , It is a dimensionless positive scalar quantity., , Variation of Pressure with Depth, , A fluid is at rest in a container. Consider a cylindrical element of fluid having, area of base A and height h., In equilibrium, the resultant vertical forces should be balanced., P2 A = P1 A+mg, P2 A - P1 A =mg, (P2 - P1 )A =mg, But m= ρV, V = hA, m=ρhA, (P2 - P1 )A = ρhA g, , P2 - P1 = ρ gh, If the point 1 at the top of the fluid , which is open to the atmosphere, P1 may, be replaced by atmospheric pressure (Pa ) and we replace P2 by P, Gauge pressure, 𝐏 - 𝐏𝐚 = ρ gh, The excess of pressure, P - Pa , at depth h is called a gauge pressure at that, point., , Absolute Pressure, 𝐏 = 𝐏𝐚 + ρ gh, Thus, the absolute pressure P, at depth below the surface of a liquid open to, the atmosphere is greater than atmospheric pressure by an amount ρgh.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Hydrostatic paradox., , The absolute pressure depends on the height of the fluid column and not on, cross sectional or base area or the shape of the container. The liquid, pressure is the same at all points at the same horizontal level (same depth)., The result is appreciated through the example of hydrostatic paradox., , Problem, What is the pressure on a swimmer 10 m below the surface of a lake?, h = 10 m, ρ = 1000 kg m−3, , Take g = 10 m s−2, P = Pa + ρ gh, = 1.01 × 105 + 1000 × 10 × 10, = 1.01 × 105 + 1 × 105, = 2.01 × 105 Pa, ≈ 2 atm, , (This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is, 100 atm. Submarines are designed to withstand such enormous pressures.), , Atmospheric Pressure, It is the pressure exerted by the atmosphere at sea level., 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., 1 atm = 1.013 × 𝟏𝟎𝟓 Pa

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Mercury barometer, Mercury barometer is used to measure Atmospheric Pressure. Italian, scientist Evangelista Torricelli devised mercury barometer., , (The space above the mercury column in the, tube contains only mercury vapour whose, pressure P is so small that it may be neglected.), , The pressure inside the column at point A =The pressure at point B, which is, at the same level., Pressure at B = Pa (atmospheric pressure), Pressure at A = ρgh, , Pa = ρgh, where ρ is the density of mercury and h is the height of the mercury column, in the tube., At sea level h= 76 cm and is equivalent to 1 atm., , Open-tube manometer, An open-tube manometer is a used for measuring Guage pressure or, pressure differences.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , 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., One end of the tube is open to the atmosphere and other end is connected to, the system whose pressure we want to measure ., The pressure at A = pressure at point B, P= Pa + ρ gh, , P - 𝐏𝐚 = ρ gh, The gauge pressure is proportional to manometer height h., , Problem, 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, Pa = ρ gh, h=, h=, , Pa, ρg, 1.01×105, 1.29 × 9.8, , h = 7989 m ≈ 8 km, , Problem, 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 = 10m s −2 ), h = 1000 m, (a), , ,, , ρ = 1.03 × 103 kg m−3, , Absolute pressure, P = Pa + ρgh, = 1.01 × 105 + 1.03 × 103 × 10 × 1000, = 1.01 × 105 + 103 × 105, = 104.01 × 105 Pa, ≈ 104 atm

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , (b) Gauge pressure ,, , P −Pa = ρgh, = 1.03 × 103 × 10 × 1000, = 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, ρgh., Since the area of the window is A = 0.04 m2 , the force acting on it is, F = Gauge Pressure x A, = 103 × 105 × 0.04, = 4.12 ×10 N, , Pascal’s law for transmission of fluid pressure, Whenever external pressure is applied on any part of a fluid contained in a, vessel, it is transmitted undiminished and equally in all directions., , Applications of Pascal’s law, 1.Hydraulic lift, , The pressure on smaller piston, P=, , F1, A1, , --------------(1), , This pressure is transmitted equally to the larger cylinder with a larger, piston of area A2 producing an upward force F2 ., P=, From eq(1) and (2), , F1, A1, , =, , F2, A2, F2, , --------------(2), , A2, , 𝐅𝟐 = 𝐅𝟏, , 𝐀𝟐, 𝐀𝟏, , Thus, the applied force has been increased by a factor of, the mechanical advantage of the device., , A2, A1, , and this factor is

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Problem, 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?, F2 = F1, , A2, A1, , π x (1.5 x 10−2 )2, F2 = 10 x, π x (0.5 x 10−2 )2, = 10 x9, =90 N, (b) Volume covered by the smaller piston is equal to volume moved by the, larger piston., L1 A1 = L2 A2, L2 = L1, , A1, A2, 2, , = 6 x10, , −2, , x, , π x (0.5 x 10−2 ), , π x (1.5 x 10−2 )2, , =0.54m, , = 6 x10−2 x0.111, = 0.67 x10−2 m, = 0.67cm, , Problem, 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 )., A1, , F1 = F2 (, , A2, , ), F2 = mg = 1350 x 9.8, =13230N, 2, , F1 = 13230 x, = 13230 x, =1470 N, , π x (5 x 10−2 ), , πx (15 x 10−2 )2, 25, 225

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , The air pressure that will produce this force is, P=, P=, , F1, A1, , 1470, 3⋅14 x (5 x 10−2 )2, , =1.9 x105 Pa, , 2.Hydraulic brakes, When we apply a force on the pedal with our foot the master piston moves, inside the master cylinder, and the pressure caused is transmitted through, the brake oil to act on a piston of larger area. A large force acts on the piston, and is pushed down expanding the brake shoes against brake lining. In this, way a small force on the pedal produces a large retarding force on the wheel., The pressure set up by pressing pedal is transmitted equally to all cylinders, attached to the four wheels so that the braking effort is equal on all wheels., , Streamline Flow (Steady Flow), The study of the fluids in motion is known as fluid dynamics., The flow of the fluid is said to be steady if at any given point, the velocity of, each passing fluid particle remains constant in time., The velocity of a particular particle may change as it moves from one point, to another., , The path taken by a fluid particle under a steady flow is a streamline., Streamline is defined as a curve whose tangent at any point is in the, direction of the fluid velocity at that point., No two streamlines can cross, for if they do, an oncoming fluid particle can, go either one way or the other and the flow would not be steady.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Equation of Continuity, , Consider a region of streamline flow of a fluid. The points P, R and Q are, planes perpendicular to the direction of fluid flow . The area of crosssections at these points are AP , AR , AQ and speeds of fluid particles are vP ,, vR and vQ ., The mass of fluid crossing at P in a small interval of time Δt = ρP AP vP Δt, The mass of fluid crossing at Q in a small interval of time Δt = ρQ AQ vQ Δt, The mass of fluid crossing at R in a small interval of time Δt = ρR AR vR Δt, The mass of liquid flowing out = The mass of liquid flowing in, ρP AP vP Δt = ρQ AQ vQ Δt = ρR AR vR Δt, If the fluid is incompressible ρP = ρQ = ρR, 𝐀𝐏 𝐯𝐏 = 𝐀𝐐 𝐯𝐐 = 𝐀𝐑 𝐯𝐑, , Av = constant, This is called the equation of continuity and it is a statement of conservation, of mass in flow of incompressible fluids., Thus, at narrower portions where the streamlines are closely spaced,, velocity increases and its vice versa., , Turbulent Flow, Steady flow is achieved at low flow speeds. Beyond a limiting value, called, critical speed, the flow of fluid loses steadiness and becomes turbulent., , A jet of air striking a flat plate placed perpendicular to it is an example of, turbulent flow.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Venturimeter principle is used in,, • The carburetor of automobile has a Venturi channel (nozzle) through, which air flows with a large 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., • Filter pumps or aspirators, • Bunsen burner, • Atomisers, • Sprayers used for perfumes or to spray insecticides, , The spray gun. Piston forces air at high speeds causing a lowering of, pressure at the neck of the container., , 3.Blood Flow and Heart Attack, The artery may get constricted due to the accumulation of plaque on its, inner walls. 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. The heart exerts further pressure to open this artery and forces, the blood through. 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., , 4.Dynamic Lift, (i)Ball moving without spin:, , 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.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , (ii)Ball moving with spin:Magnus Effect, , The ball is moving forward and relative to it the air is moving backwards., Therefore, the relative velocity of air above 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 spining is called Magnus effect., (iii)Aerofoil or lift on aircraft wing, , Aerofoil is a solid piece shaped to provide an upward dynamic lift when it, moves horizontally through air., When the aerofoil moves against the wind, the orientation of the wing, relative to flow direction causes the streamlines to crowd together above the, wing more than those below it. The flow 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., , Viscosity, The internal frictional force that acts when there is relative motion between, layers of the liquid is called viscosity., , When liquid flows between a fixed and moving, glass plates, the layer of the liquid in contact with, top surface moves with a velocity v and the layer, of the liquid in contact with the fixed surface is, stationary. The velocities of layers increase, uniformly from bottom to the top layer.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , When a fluid is flowing in a pipe or a tube, then velocity of, the liquid layer along the axis of the tube is maximum and, decreases gradually as we move towards the walls where, it becomes zero., , Coefficient of viscosity( 𝛈), , Due to viscous force, a portion of liquid, which at some instant has the shape, ABCD, take the shape of AEFD after short interval of time (Δt)., Shearing stress =, Shearing strain =, Strain rate=, , F, A, 𝛥𝑥, 𝑙, 𝛥𝑥, (𝑙), , 𝛥𝑡, , =, , 𝛥𝑥, 𝑙 𝛥𝑡, , =, , 𝑣, 𝑙, , The coefficient of viscosity( η)for a fluid is defined as the ratio of shearing, stress to the strain rate., η=, , Shearing stress, , 𝛈=, , Strain rate, , =, , F, A, 𝑣, 𝑙, , 𝐅𝒍, 𝐯𝐀, , The SI unit of coefficient viscosity is poiseiulle (Pl)., Its other units are N s m−2 or Pa s., The dimensions are [ML−1 T −1], Generally thin liquids like water, alcohol etc. are less viscous than thick, liquids like coal tar, blood, glycerin etc., The viscosity of liquids decreases with temperature while it increases in the, case of gases.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Stokes’ Law, Stokes’ law states that the viscous drag force F on a sphere of radius a, moving with velocity v through a fluid of coefficient of viscosity η is,, , F = 6πηav, Terminal velocity, When an object falls through a viscous medium (raindrop in air), it, accelerates initially due to gravity. As the velocity increases, the retarding, force also increases. Finally when viscous force plus buoyant force becomes, equal to the force due to gravity(weight of the body), the net force and, acceleration become zero. The sphere (raindrop) then descends with a, constant velocity clalled terminal velocity., , Expression for Terminal velocity, , Consider a raindrop in air. The forces acting on the drop are, 4, , 1. Force due to gravity(weight,mg) acting downwards , FG = πa3 ρg, 4, , 3, , 2. Buoyant force acting upwards, FB = πa3 σg, 3, , 3. Viscous force, FV = 6πηav, In equilibrium,, 4, , 4, , 3, , 3, 4, , 6πηav + πa3 σg = πa3 gρ, 6πηav = πa3 (ρ −σ)g, 3, , Terminal velocity ,, , 𝐯𝐭 =, , 𝟐𝐚𝟐 (𝛒 −𝛔)𝐠, 𝟗𝛈, , So the terminal velocity vt depends on the square of the radius of the sphere, and inversely on the viscosity of the medium.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Problem, The terminal velocity of a copper ball of radius 2.0 mm falling through a, tank of oil at 200 C is 6.5 cm s−1 . Compute the viscosity of the oil at 200 C., Density of oil is 1.5 × 103 kg m−3 , density of copper is 8.9 × 103 kg m−3 ., vt = 6.5 × 10−2 ms −1, , ρ = 8.9 × 103 kg m−3, , a = 2 × 10−3 m, g = 9.8m s −2 ,, , σ =1.5 ×103 kg m−3, , vt =, η=, , 2a2 (ρ −σ)g, 9η, , 2a2 (ρ −σ)g, 9vt, 2, , η=, , 2 x (2 × 10−3 ) (8.9 × 103 −1.5 ×103 )x9.8, 9 x 6.5 × 10−2, , η =9.9 × 10−1 kg m−1 s−1, , Reynolds Number, Osborne Reynolds defined a dimensionless number, whose value gives one, an approximate idea whether the flow would be turbulent . This number is, called the Reynolds number( R e ), , 𝐑𝐞 =, , 𝛒𝐯𝐝, 𝛈, , where ρ is the density of the fluid ,v is the speed of fluid, d stands for the, dimension of the pipe, and η is the viscosity of the fluid., 𝐑 𝐞 <1000 – The flow is streamline or laminar., 𝐑 𝐞 > 𝟐𝟎𝟎𝟎 − The flow is turbulent ., 𝐑 𝐞 between 1000 and 2000-The flow becomes unsteady ., The critical value of Reynolds number at which turbulence sets, is known as, critical Reynolds number., • Turbulence dissipates kinetic energy usually in the form of heat. Racing, cars and planes are engineered to precision in order to minimise, turbulence., • Turbulence is sometimes desirable. The blades of a kitchen mixer, induce turbulent flow and provide thick milk shakes as well as beat, eggs into a uniform texture.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Surface Tension, The free surface of a liquid possess some additional energy and it behaves, like a stretched elastic membrane. This phenomenon is known as surface, tension., Surface tension is concerned with only liquid as gases do not have free, surfaces., , Surface Energy, , For a molecule well inside a liquid the net force on it is zero. But the, molecules on the surface have a net downward pull. So work has to be done, against this downward force and this work is stored as energy in suface, molecules. Thus, molecules on a liquid surface have some extra energy in, comparison to molecules in the interior,which is termed as surface energy., A liquid thus tends to have the least surface area inorder to reduce surface, enegy., , Surface Energy and Surface Tension, , Consider a horizontal liquid film ending in a movable bar. Due to surface, tension the bar is pulled inwards ., Inorder to keep the bar in its original position some work has to be done, against this inward full., W = F x d----------(1), This work done increases surface energy.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , If the surface energy of the film is S per unit area, the extra area is 2d l (film, has two sides),, The extra surface energy = S x 2d l ------------(2), The extra surface energy = work done, S x2dl =Fd, 𝐅, , S = 𝟐𝒍, This quantity S is the magnitude of surface tension., , Definition of Surface tension, Surface tension is a force per unit length (or surface energy per unit area), acting in the plane of the interface between the plane of the liquid and any, other substance. It is the extra energy that the molecules at the interface, have as compared to the interior, Force, , Surface Tension, S= Length, The SI Unit is Nm−1, Dimensional formula is MT −2, The value of surface tension depends on temperature., The surface tension of a liquid decreases with temperature., Some effects of surface Tension, Oil and water do not mix., Water wets you and me but not ducks., Mercury does not wet glass but water sticks to it., Oil rises up a cotton wick, inspite of gravity., Sap and water rise up to the top of the leaves of the tree., Hairs of a paint brush do not cling together when dry and even when dipped, in water but form a fine tip when taken out of it.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Angle of Contact, The angle between tangent to the liquid surface at the point of contact and, solid surface inside the liquid is termed as angle of contact(θ), The value of θ determines whether a liquid will spread on the surface of a, solid or it will form droplets on it., When Angle of contact is Obtuse:, , When θ is an obtuse angle(greater than 90) then molecules of liquids are, attracted strongly to themselves and weakly to those of solid, and liquid, then does not wet the solid., Eg: Water on a waxy or oily surface, Mercury on any surface., When Angle of contact is Acute:, , When θ is an acute angle (less than 90), the molecules of the liquid are, strongly attracted to those of the solid and liquid then wets the solid., Eg: Water on glass or on plastic, Kerosene oil on virtually anything ., Action Soaps and detergents, Soaps, detergents and dying substances are wetting agents. When they are, added the angle of contact becomes small so that these may penetrate well, and become effective., Action of Water proofing agents, Water proofing agents are added to create a large angle of contact between, the water and fibres.

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Drops and Bubbles, Why are small drops and bubbles spherical?, Due to surface tension ,liquid surface has the tendency to reduce surface, area. For a given volume sphere has minimum surface area. So small drops, and bubbles are spherical., For large drops the effect of gravity predominates that of surface tension, and they get flattened., , Excess Pressure inside a spherical drop, , Due to surface tension the liquid surface experiences an inward pull and as, a result the pressure inside a spherical drop is more than the pressure, outside. Due to this excess pressure let the radius of drop increase by Δr, Work done in expansion= Force x Displacement, =Excess pressure x Area x Displacement, W = (𝐏𝐢 - 𝐏𝐨 ) x 4π𝐫 𝟐 x Δr ------------------(1), This workdone is equal to the increase in surface energy, Extra Surface energy = Surface tension x Increase in surface area, Increase in surface area of drop=4π(r + Δr)2 - 4πr 2, =4π (r 2 + 2rΔr + Δr 2 − r 2 ), =8πrΔr (neglecting higher order terms), Extra surface energy = 𝐒 𝐱𝟖𝛑𝐫𝚫𝐫 ---------------(2), The workdone = extra surface energy, (𝐏𝐢 - 𝐏𝐨 ) x 4π𝐫 𝟐 xΔr = 𝟖𝛑𝐫𝚫𝐫 𝐒 --------------(3), , (𝐏𝐢 - 𝐏𝐨 ) =, , 𝟐𝐒, 𝐫

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Excess Pressure Inside a Liquid Bubble, A bubble has two free surfaces., (Pi - Po ) = 2x, (𝐏𝐢 - 𝐏𝐨 ) =, , 𝟒𝐒, , 2S, r, , 𝐫, , Capillary Rise, Due to the pressure difference across a curved liquid-air interface, water, rises up in a narrow tube in spite of gravity. This is called capillary rise., , Consider a vertical capillary tube of circular cross section (radius a) inserted, into an open vessel of water., The excess pressure on the concave meniscus, (Pi - Po ) =, , 2S, r, , cosθ =, r=, (Pi - Po ) =, (Pi - Po ) =, , a, r, , a, , cosθ, , 2S, a, cosθ, , 2Scosθ, a, , ----------------(1), , Consider two points A and B in the same horizontal level i.e, the points are, at the same pressure., Pressure at A = Pi, Pressure at B = Po + h ρ g, Pi =Po + h ρ g, Pi - Po = h ρ g----------------(2), From eq(1) and (2), h ρ g=, , h=, , 2Scosθ, a, , 𝟐𝐒𝐜𝐨𝐬𝛉, 𝛒 𝐠𝐚

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Join Telegram Channel, https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Thus capillary rise is a consequence of surface tension., Capillary rise is larger, for capillary tube with smaller radius a., Note:, If the liquid meniscus is convex, as for mercury, angle of contact θ will be, obtuse . Then cos θ is negative and hence value of h will be negative. it is, clear that the liquid will lower in the capillary and this is called capillary fall, or capillary depression., , Problem, Find the capillary rise when a capillary tube of radius 0.05 cm is dipped, vertically in water. Surface tension for water is 0.073N𝑚−1 .Density of water, is 1000 kg𝑚−3 ., , h=, , 2Scosθ, ρ ga, , For water-glass angle of contact θ = 0, cos 0 =1, h=, h=, , 2S, ρ ga, 2𝑥 0.073, 1000 𝑥 9.8 𝑥 0.05 𝑥10−3, , h = 2.98 × 10−2 m = 2.98 cm., , Detergents and Surface Tension, Washing with water does not remove grease stains. This is because water, does not wet greasy dirt; i.e., there is very little area of contact between, them., The molecules of detergents are hairpin shaped, with one end attracted to, water and the other to molecules of grease, oil or wax .Thus water-oil, interfaces are formed which reduces the surface tension S (water-oil) and, dirt can be removed by running water., , Seema Elizabeth, MARM Govt HSS Santhipuram