11 SL CH 10 Mechanical Properties of Fluids PDF

## CHAPTER 10 MECHANICAL PROPERTIES OF FLUIDS ### 10.1 WHAT IS A FLUID ? 1. What are fluids? Give their important characteristics. > Fluid. A fluid is a substance that can flow. It ultimately assumes the shape of the containing vessel because it cannot withstand shearing stress. Thus, both liquid...

## CHAPTER 10 MECHANICAL PROPERTIES OF FLUIDS ### 10.1 WHAT IS A FLUID ? 1. What are fluids? Give their important characteristics. > Fluid. A fluid is a substance that can flow. It ultimately assumes the shape of the containing vessel because it cannot withstand shearing stress. Thus, both liquids and gases are fluids. > Important characteristics of fluids: > (i) The atoms or molecules in a fluid are arranged in a random manner. > (ii) A fluid cannot withstand tangential or shearing stress for an indefinite period. It begins to flow when a shearing stress is applied. > (iii) A fluid has no definite shape of its own. It ultimately assumes the shape of the containing vessel. So a fluid has no modulus of rigidity. > (iv) A fluid can exert/withstand a force in a direction perpendicular to its surface. So a fluid does have a bulk modulus of rigidity. 2. Both liquids and gases are fluids. What is the main difference between them? > Difference between liquid and gas. A liquid is incompressible and has a definite volume and a free surface of its own. A gas is compressible and it expands to occupy all the space available to it. 3. Distinguish between the terms fluid statics and fluid dynamics. > Fluid statics. The branch of physics that deals with the study of fluids at rest is called fluid statics or hydrostatics. Its study includes hydrostatic pressure, Pascal's law, Archimedes' principle, floatation of bodies and surface tension. > Fluid dynamics. The branch of physics that deals with the study of fluids in motion is called fluid dynamics or hydrodynamics. Its study includes equation of continuity, Bernoulli's theorem, Toricelli's theorem, viscosity, etc. ### 10.2 THRUST OF A LIQUID 4. Define the term thrust. Give its SI unit.. > Thrust. A liquid in equilibrium has a fundamental property that it exerts a force on any surface in contact with it and this force acts perpendicular to the surface. The total force exerted by a liquid on any surface in contact with it is called thrust. It is because of this thrust that a liquid flows out through the holes of the containing vessel. > As thrust is a force, so its SI unit is newton (N). 5. Show that a liquid at rest exerts force perpendicular to the surface of the container at every point. > Liquid in equilibrium. > Consider a liquid contained in a vessel in the equilibrium state of rest. As shown in Fig. 10.1, suppose the liquid exerts force F on the bottom surface in an inclined direction OA. The surface exerts an equal reaction R to water along OB. > The reaction R along OB has two rectangular components: > (i) Tangential component, OC = R cos 0 > (ii) Normal component, OD = R sin 0 > Since a liquid cannot resist any tangential force, so the liquid near O should begin to flow along OC. But the liquid is at rest, the force along OC must be zero. > .. R cos 0=0 > As R≠0, so cos 0=0 or 0=90° > Hence a liquid always exerts force perpendicular to the surface of the container at every point. ### 10.3 PRESSURE 6. Define the term pressure. Is it a scalar or a vector? Give its units and dimensions. > Pressure. The pressure at a point on a surface is the thrust acting normally per unit area around that point. If a total force Facts normally over a flat area A, then the pressure is > F > P = > A > If the force is not distributed uniformly over the given surface, then pressure will be different at different points. If a force AF acts normally on a small area AA surrounding a given point, then pressure at that point will be > AF dF > P = lim > ΔΑΟ ΔΑ dA > Pressure is a scalar quantity, because fluid pressure at a particular point in fluid has same magnitude in all directions. This shows that a definite direction is not associated with fluid pressure. Though force is a vector quantity, only the magnitude of the normal component of the force appears in the above equations for pressure. Units and dimensions of pressure : SI unit of pressure = Nm or Pascal (Pa) CGS unit of pressure = dyne cm Dimensional formula of pressure is [ML -1T-2]. 7. Briefly explain a method for measuring fluid pressure at any point inside the fluid. ### 10.4 PRACTICAL APPLICATIONS OF PRESSURE 8. Describe some practical applications from daily life which make use of the concept of pressure. > Practical applications based on the concept of pressure : > (i) A sharp knife cuts better than a blunt one. The area of a sharp edge is much less than the area of a blunt edge. For the same total force, the effective force per unit area (or pressure) is more for the sharp edge than the blunt edge. Hence a sharp knife cuts better. > (ii) Railway tracks are laid on wooden sleepers. This spreads force due to the weight of the train on a larger area and hence reduces the pressure considerably. This, in turn, prevents the yielding of the ground under the weight of the train. > (iii) It is difficult for a man to walk on sand while a camel walks easily on sand inspite of the fact that a camel is much heavier than a man. This is because camel's feet have a larger area than the feet of man. Due to larger area, pressure is less. > (iv) Pins and nails are made to have pointed ends. Their pointed ends have very small area. When a force is applied over head of a pin or a nail, it transmits a large pressure (= force/area) on the surface and hence easily penetrate the surface. ### Examples based on Thrust and Pressure FORMULAE USED 1. Thrust = Total force exerted by a liquid on the surface in contact 2. Pressure = = F Thrust Area or P = A UNITS USED Thrust is in newton and pressure in Nm or pascal (Pa). EXAMPLE 1. The two thigh bones (femurs), each of cross-sectional area 10 cm2 support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs. Take g = 10 ms-2. [NCERT] > Solution. Total cross-sectional area of the femurs, > A = 2 x 10 cm2 = 20 × 10-4 m2 > Force acting vertically downwards and hence normally on femurs, > F = mg = 40 x 10 = 400 N > = > = 2 × 105 Nm-2. > > F > P = > A > 400 > 20x10-4 EXAMPLE 2. How much pressure will a man of weight 80 kgf exert on the ground when (i) he is lying and (ii) he is standing on his feet? Given that the area of the body of the man is 0.6 m2 and that of a foot is 80 cm2. > Solution. Force, F = 80 kgf = 80 × 9.8 N > (i) When the man is lying on the ground, > = 1. > = > > A= 0.6 m2 > F > P= > A > 80 x 9.8 > 0.6 > 1.307 x 103 Nm-2. > > (ii) When the man is standing on his feet, > = 4. > = > > A = 2 x 80 cm2 = 160 x 10-4 m2 > 80 x 9.8 > P = > 160 x 10-4 > 4.9 x 104 Nm-2. #### PROBLEMS FOR PRACTICE 1. Find the pressure exerted at the tip of a drawing pin if it is pushed against a board with a force of 20 N. Assume the area of the tip to be 0.1 mm2. (Ans. 2 x 108 Pa) 2. Atmospheric pressure is nearly 100 kPa. How large the force does the air in a room exert on the inside of a window pan that is 40 cm x 80 cm? (Ans. 32 kN) 3. The force on a phonograph needle is 1.2 N. The point has a circular cross-section whose radius is 0.1 mm. Find the pressure (in atm) it exerts on the records. Given 1 atm = 1.013 × 105 Pa. (Ans. 377 atm) 4. A cylindrical vessel containing liquid is closed by a smooth piston of mass m. The area of cross-section of the piston is A. If the atmospheric pressure is Po, find the pressure of the liquid just below the piston. (Ans. P + mg /A) ## 10.3 DENSITY 9. Define the term density. Give its units and dimensions. > Density. The density of any material is defined as its mass per unit volume. If a body of mass M occupies volume V, then its density is > M > p = > V > i.e., Density = > Mass > Volume > Density is a positive scalar quantity. As liquids are incompressible, their density remains constant at all pressures. Density of a gas varies largely with pressure. Units and dimensions of density: SI unit of density = kg m CGS unit of density = g cm Dimensional formula of density is [ML³]. 10. What do you mean by specific gravity or relative density of a substance? > Specific gravity or relative density. The specific gravity or relative density of a substance is defined as the ratio of the density of the substance to the density of water at 4°C. The density of water at 4°C is 1.0 × 103 kg m³. Specific gravity = Density of substance Density of water at 4°C. Specific gravity is a dimensionless positive scalar quantity. Clearly, Density of a substance = Specific gravity × Density of water at 4°C some common fluids at STP table 10.1 Densities of Fluid Density (kg m³) Water 1.00 × 103 Sea water 1.03 x 103 Mercury 13.6 x 103 Ethyl alcohol 0.806 x 103 Whole blood 1.06 x 103 Air 1.29 Oxygen 1.43 Hydrogen 9.0×10-2 Intersteller space ≈ 10-20 ### 10.6 PASCAL'S LAW 11. State and prove Pascal's law of transmission of fluid pressure. > Pascal's law. This law tells as how pressure can be transmitted in a fluid. It can be stated in a number of equivalent ways as follows: > (i) The pressure exerted at any point on an enclosed liquid is transmitted equally in all directions. > (ii) A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel. > (iii) The pressure in a fluid at rest is same at all points if we ignore gravity. Proof of Pascal's law. Pascal law can be proved by using two principles: (i) The force on any layer of a fluid at rest is normal to the layer and (ii) Newton's first law of motion. As shown in Fig. 10.3, consider a small element ABC - DEF in the form of a right angled prism in the interior of a fluid at rest. The element is so small that all its parts can be assumed to be at same depth from the liquid surface and, therefore, the effect of gravity is same for all of its points. Suppose the fluid exerts pressure P, P, and P on the faces BEFC, ADFC and ADEB respectively of the this element and the corresponding normal forces on these faces are F, Fo and F. Let A, A₁ and A be the respective areas of the three faces. In right ∆ABC, let ∠ACB = 0. As the prismatic element is in equilibrium with remaining fluid, by Newton's law, the fluid force should balance in various directions. Along horizontal direction, F, sin 0 = F Along vertical direction, and F cos 0 = Fa From the geometry of the figure, we get A sin 0 = A A cos 0 = Aa From the above equations, we get F, sin F A sin 0 F₁ cos A and Fa A cos a or = = FaFF A A A P = P = P Hence, pressure exerted is same in all directions in a fluid at rest. This proves Pascal's law of transmission of fluid pressure. The above discussion again shows that pressure is not a vector quantity. No direction can be assigned to it. 12. How will you experimentally verify the Pascal's law of transmission of fluid pressure? > Experimental verification of Pascal's law. As shown in Fig. 10.4, take a vessel having three openings A, B and C and provided with frictionless and water tight pistons. Let their cross-sectional areas be A, 2 A and > A > 2 > respectively. Fill the vessel with water and apply an additional force F on piston A. To keep the pistons B and C in their positions, forces equal to 2 F and > F > 2 > respectively have to be applied on them. This shows that the pressure P is transmitted equally in all directions because > F 2F F/2 > P= > A 2A A/2 > = ### 10.7 APPLICATIONS OF PASCAL'S LAW 13. Explain how is Pascal's law applied in a hydraulic lift. > Hydraulic lift. Hydraulic lift is an application of Pascal's law. It is used to lift heavy objects. It is a force multiplier. > It consists of two cylinders C₁ and C₂ connected to each other by a pipe. The cylinders are fitted with water-tight frictionless pistons of different cross-sectional areas. The cylinders and the pipe contain a liquid. Suppose a force f is applied on the smaller piston of cross-sectional area a. Then > Pressure exerted on the liquid, > P= > = > f a > Load > Area = a > F > Area = A > C₁ > C₂ > Liquid > Fig. 10.5 Hydraulic lift. > According to Pascal's law, the same pressure P is also transmitted to the larger piston of cross-sectional area A. > .. Force on larger piston is > F=PxA=xA= > F=Px A=1x A=Ax f > a > As A > a, therefore, F > f. > a > Hence by making the ratio A/ a large, very heavy loads (like cars and trucks) can be lifted by the application of a small force. However, there is no gain of work. The work done by force f is equal to the work done by F. The piston P₁ has to be moved down by a larger distance compared to the distance moved up by piston P₂. 14. With the help of a labelled diagram, explain the working of hydraulic brakes. > Hydraulic brakes. The hydraulic brakes used in automobiles are based on Pascal's law of transmission of pressure in a liquid. > Construction. As shown in Fig. 10.6, a hydraulic brake consists of a tube T containing brake oil. One end of this tube is connected to a master cylinder fitted with piston P. The piston P is attached to the brake pedal through a lever system. The other end of the tube is connected to the wheel cylinder having two pistons P₁ and P₂. The pistons P and P₂ are connected to the brake shoes S₁ and S₂ respectively. The area of cross-section of the wheel cylinder is larger than that of master cylinder. > To other > wheels > Lever system > P > Tube T > P > P2 > Wheel > cylinder > Master > cylinder - > Brake > paddle > 0000 > Brake > shoe > S₁ > S₂ > Fig. 10.6 Hydraulic brakes. > Working. When the pedal is pressed, its lever system pushes the piston P into the master cylinder. The pressure is transmitted through the oil to the pistons P₁ and P₂ in the wheel cylinder, in accordance with Pascal's law. The pistons P₁ and P₂ are pushed outwards. The brake shoes get pressed against the inner rim of the wheel, retarding the motion of the wheel. As the cross-sectional area of wheel cylinder is larger than that of master cylinder, a small force applied to the pedal produces a large retarding force. When the paddle is released, a spring pulls the brake shoes away from the rim. The pistons in both cylinders move towards their normal positions and the oil is forced back into the master cylinder. > Advantages of hydraulic brakes:. > (i) The master cylinder transmits equal retarding force on each wheel. So a hydraulic brake operates uniformly and hence prevents skidding. > (ii) A small force applied to the pedal exerts a much larger force on the wheel drums. It enables the driver to keep the vehicle under control. ## 10.5 DENSITY 9. Define the term density. Give its units and dimensions. > Density. The density of any material is defined as its mass per unit volume. If a body of mass M occupies volume V, then its density is > M > p = > V > i.e., Density = > Mass > Volume > Density is a positive scalar quantity. As liquids are incompressible, their density remains constant at all pressures. Density of a gas varies largely with pressure. Units and dimensions of density: SI unit of density = kg m CGS unit of density = g cm Dimensional formula of density is [ML³]. ## 10.13 DIFFERENT UNITS OF PRESSURE 22. Name the various units used for measuring pressure > Various units for pressure: > (i) SI unit of pressure = Nm > or Pascal (Pa). > (ii) CGS unit of pressure = dyne cm > (iii) Atmosphere (atm). It is the pressure exerted by 76 cm of Hg column (at 0°C, 95° latitude and mean sea level). > 1 atm = 1.013 × 105 Pa = 1.013 x 106 dyne cm > (iv) In meteorology, the atmospheric pressure is measured in bar and millibar. > 1 bar = 105 Pa = 106 dyne cm > 1 millibar = 10-3 bar = 100 Pa > (v) Atmospheric pressure is also measured in torr, a unit named after Torricelli. > 1 torr = 1 mm of Hg > 1 atm 1.013 bar = 760 torr 23. In what units is the blood pressure measured? > Units for blood pressure. The blood pressure is measured in mm of Hg. When the heart is contracted to its smallest size, the pumping is hardest and the pressure of blood flowing in major arteries is nearly 120 mm of Hg. This is known as systolic pressure. When the heart is expanded to its largest size, the blood pressure is nearly 80 mm of Hg. This is known as diastolic pressure. ## 10.14 BUOYANCY 24. What do you understand by buoyancy and centre of buoyancy ? > Buoyancy. When body is immersed in a fluid, the fluid exerts pressure on all faces of the body. But the fluid pressure increases with depth. The upward thrust at the bottom is more than the downward thrust on the top because the bottom is at the greater depth than the top. Hence a resultant upward force acts on the body. The upward force acting on a body immersed in a fluid is called upthrust or buoyant force and the phenomenon is called buoyancy. For example, a cork taken inside water experiences an upward thrust and comes to the surface. Similarly, while drawing water from a well, a bucket is found too much lighter when it is inside water than when it comes out of it. > The force of buoyancy acts through the centre of gravity of the displaced fluid which is called centre of buoyancy. ## 10.15 ARCHEMEDES' PRINCIPLE 25. State Archemedes' principle and prove it mathematically. Deduce an expression for the apparent weight of the immersed body. > Archemedes' principle. This principle was discovered by the great Greek scientist, Archimedes around 225 B.C. and it gives the magnitude of buoyant force on a body. > Archimedes' principle states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the displaced fluid. > Proof. As shown in Fig. 10.17, consider a body of height h lying inside a liquid of density p, at a depth x below the free surface of the liquid. Area of cross-section of the body is a. The forces on the sides of the body cancel out. ## 10.16 LAW OF FLOATATION 26. State and explain the law of floatation. Deduce an expression for the fraction of volume of the floating body submerged in the liquid. > Law of floatation. The law of floatation states that a body will float in a liquid if the weight of the liquid displaced by the immersed part of the body is equal to or greater than the weight of the body. ## 10.18 VISCOSITY 29. What is viscosity ? Explain the cause of viscosity. > Viscosity. Viscosity is the property of fluid by virtue of which an internal force of friction comes into play when a fluid is in motion and which opposes the relative motion between its different layers. The backward dragging force, called viscous drag or viscous force, acts tangentially on the layers of the fluid in motion and tends to destroy its motion. > Cause of viscosity. Consider a liquid moving slowly and steadily over a fixed horizontal surface. Each layer moves parallel to the fixed surface. The layer in contact with the fixed surface is at rest and the velocity of the every other layer increases uniformly upwards, as shown by arrows of increasing lengths in Fig. 10.21. ## 10.19 COEFFICIENT OF VISCOSITY 31. What is meant by coefficient of viscosity? Give its dimensions and units. > Coefficient of viscosity. As shown irt Fig. 10.23, suppose a liquid is flowing steadily in the form of parallel layers on a fixed horizontal surface. Consider two layers P and Q at distances x and x + dx from the solid surface and moving with velocities v and v + dv respectively. Then is the rate of change of velocity with distance in the direction of increasing distance and is called velocity gradient. > ## 10.20 COMPARISON BETWEEN VISCOUS FORCE AND SOLID FRICTION ## 10.21 VARIATION OF VISCOSITY WITH TEMPERATURE AND PRESSURE 33. Discuss the variation of fluid viscosity with temperature and pressure. ## 10.22 PRACTICAL APPLICATIONS OF THE KNOWLEDGE OF VISCOSITY ## Examples based on Coefficient of Viscosity ## 10.23 POISEUILLE'S FORMULA 35. State Poiseuille's formula. What are the assumptions used in the derivation of this formula? Derive this formula on the basis of dimensional considerations. ## Examples based on Poiseullie's Formula ## 10.24 VARIATION OF VISCOSITY WITH TEMPERATURE AND PRESSURE ## 10.25 TERMINAL VELOCITY 38. Explain how does a body attain a terminal velocity when it is dropped from rest in a viscous medium. Derive an expression for the terminal velocity of a small spherical body falling through a viscous medium. Also discuss the result. ## Examples based on Stokes' Law and Terminal Velocity ## 10.26 STREAMLINE AND TURBULENT FLOWS 39. Distinguish between streamline and turbulent flows. What do you understand by a streamline and tube of flow? Give important properties of streamlines. ## 10.27 LAMINAR FLOW 40. What is laminar flow of a liquid? Distinguish between the velocity profiles of non-viscous and viscous liquids. ## 10.28 CRITICAL VELOCITY 41. What do you mean by critical velocity of a liquid? Derive an expression for it on the basis of dimensional considerations. ## 10.29 REYNOLD'S NUMBER 42. What is Reynold's number? What is its importance? ## Examples based on Reynold's Number ## 10.30 IDEAL FLUID 44. What is meant by an ideal fluid ? ## 10.31 EQUATION OF CONTINUITY 45. Obtain the equation of continuity for the incompressible non-viscous fluid having a steady flow through a pipe. ## 10.32 ENERGY OF A FLUID IN A STEADY FLOW 47. What are different forms of energy possessed by a flowing liquid? Write expressions for them. ## 10.33 BERNOULLI'S PRINCIPLE 48. State and prove Bernoulli's principle for the flow of non-viscous fluids. Give its limitations. ## 10.34 TORRICELLI'S LAW OF EFFLUX 49. Apply Bernoulli's principle to determine the speed of efflux from the side of a container both when its top is closed and open. Hence derive Torricelli's law. ## 10.35 THE VENTURIMETER 50. What is a venturimeter? Describe its construction and working. ## 10.36 ATOMIZER OR SPRAYER 51. Briefly describe the working of an atomizer on the basis of Bernoulli's principle. ## 10.37 DYNAMIC LIFT 52. What is dynamic lift? If a ball is thrown and given a spin, then the path of the ball is curved more than in a usual spin free ball. Why? ## 10.38 BLOOD FLOW AND HEART ATTACK 54. How does Bernoulli's principle help in explaining vascular flutter and heart attack ? ## 10.39 BLOWING OFF THE ROOFS DURING WIND STORM 55. Why are the roofs of some houses blown off during a wind storm ? ## 10.40 EXAMPLES BASED ON EQUATION OF CONTINUITY AND BERNOULLI'S THEOREM ## 10.41 ANGLE OF CONTACT 67. Define the term angle of contact. On what factors does it depend? ## 10.42 SHAPE OF LIQUID MENISCUS IN A NARROW TUBE 68. Explain what determines the shape of liquid meniscus in a narrow tube. ## 10.43 MOLECULAR THEORY OF SURFACE TENSION 59. Explain surface tension on the basics of molecular theory. ## 10.44 SOME PHENOMENA BASED ON SURFACE TENSION 60. Explain some examples which illustrate the existence of surface tension. ## 10.45 SURFACE ENERGY 61. Define surface energy. Prove that it is numerically equal to the surface tension. ## 10.46 EXPERIMENTAL MEASUREMENT OF SURFACE TENSION 62. Describe a simple experiment for measuring the surface tension of a liquid. ## 10.47 PRESSURE DIFFERENCE ACROSS A CURVED LIQUID SURFACE 63. Show that a pressure difference exists between the two sides of a curved liquid surface . ## 10.48 EXCESS PRESSURE INSIDE A LIQUID DROP 64. Derive an expression for the excess pressure inside a liquid drop. ## 10.49 EXAMPLES BASED ON EXCESS PRESSURE IN DROPS & BUBBLES ## 10.50 PROBLEMS FOR PRACTICE ## 10.51 PROBLEMS FOR PRACTICE ## 10.52 PROBLEMS FOR PRACTICE ## 10.53 PROBLEMS FOR PRACTICE ## 10.54 FACTORS AFFECTING SURFACE TENSION 72. Describe the various factors which affect the surface tension of a liquid. ## 10.55 DETERGENTS AND SURFACE TENSION 73. Describe the cleansing action of detergents. ## 10.56 PROBLEMS FOR PRACTICE ## 10.57 PROBLEMS FOR PRACTICE ## 10.58 RISE OF LIQUID IN A CAPILLARY TUBE: ASCENT FORMULA 70. Derive an expression for the rise of liquid in a capillary tube and show that the height of the liquid column supported is inversely proportional to the radius of the tube. ## 10.59 RISE OF LIQUID IN A CAPILLARY TUBE OF INSUFFICIENT HEIGHT 71. Explain what happens when the length of a capillary tube is less than the height upto which the liquid may rise in it. ## 10.60 PROBLEMS FOR PRACTICE ## 10.61 PROBLEMS FOR PRACTICE ## 10.62 PROBLEMS FOR PRACTICE ## 10.63 PROBLEMS FOR PRACTICE ## 10.64 PROBLEMS FOR PRACTICE ## 10.65 PROBLEMS FOR PRACTICE ## 10.66 PROBLEMS FOR PRACTICE ## 10.67 PROBLEMS FOR PRACTICE ## 10.68 PROBLEMS FOR PRACTICE ## 10.69 PROBLEMS FOR PRACTICE ## 10.70 PROBLEMS FOR PRACTICE ## 10.71 PROBLEMS FOR PRACTICE ## 10.72 PROBLEMS FOR PRACTICE ## 10.73 PROBLEMS FOR PRACTICE ## 10.74 PROBLEMS FOR PRACTICE ## 10.75 PROBLEMS FOR PRACTICE ## 10.76 PROBLEMS FOR PRACTICE ## 10.77 PROBLEMS FOR PRACTICE ## 10.78 PROBLEMS FOR PRACTICE ## 10.79 PROBLEMS FOR PRACTICE ## 10.80 PROBLEMS FOR PRACTICE ## 10.81 PROBLEMS FOR PRACTICE ## 10.82 GLIMPSES 1. Fluid. A fluid is a substance that can flow. The term fluid refers to both liquids and gases. 2. Fluid statics. The branch of physics that deals with the study of fluids at rest is called fluid statics or hydrostatics. 3. Fluid dynamics. The branch of physics that deals with the study of fluids in motion is called fluid dynamics or hydrodynamics. 4. Thrust. The total force exerted by a liquid on any surface in contact with it is called thrust. A liquid always exerts force perpendicular to the surface of the container at every point. 5. Pressure. The thrust exerted by a liquid per unit area of the surface in contact with it is known as pressure. Pressure = Thrust Area or P = F A Pressure is a scalar quantity. 6. Units and dimensions of pressure. The CGS unit of pressure is dyne cm2 and its SI unit is Nm2 which is also called pascal (Pa). The dimensional formula of pressure is [ML-1T-2]. 7. Density.The density of any material is defined as its mass per unit volume. Density = Mass Volume M or p = V Density is a positive scalar quantity. 8. Units and dimensions of density. The SI unit of density is kg m3 and the CGS unit is g cm-3. The dimensional formula of density is [ML3]. 9. Specific gravity. The relative density or specific gravity of a substance is defined as the ratio of the density of the substance to the density of water at 4°C. Specific gravity = Density of substance Density of water at 4°C Specific gravity is a dimensionless positive scalar quantity. 10. Pascal's law. It states that a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel. Or, the pressure exerted at any point on an enclosed liquid is transmitted equally in all directions. 11. Hydraulic lift. It is an application of Pascal's law. It is used to lift heavy objects. According to Pascal's law, Pressure applied on smaller piston = Pressure transmitted to larger piston As P=f=F or a A A > a, so F > f. F=PxA=LxA a Thus hydraulic lift acts as a force multiplier. 12. Hydraulic brakes. The hydraulic brakes used in automobiles are based on Pascal's law of transmission of pressure in a liquid. 13. Pressure exerted by a liquid. A liquid column of height h and density p exerts a pressure given by P=hpg 14. Effect of gravity on fluid pressure. The pressure in a fluid varies with depth h according to the expression P = P + hpg where p is the fluid density, assumed uniform. 15. Hydrostatic paradox. The pressure exerted by a liquid column depends only on the height of the liquid column and not on the shape of the containing vessel. 16. Atmospheric pressure. The pressure exerted by the atmosphere is called atmospheric pressure. At sea-level, we have Atmospheric pressure = Pressure exerted by 0.76 m of Hg = lpg = 0.76 x 13.6 x 103 x 9.8 = 1.013 × 105 Nm-2 Other units used for atmospheric pressure are as follows: 1 atm = 1.013 × 10º dyne cm-2 = 1.013 x 105 Nm-2 (or Pa) 1 bar = 10º dyne cm-2 = 105 Nm-2 1 millibar (m bar) = 103 bar = 103 dyne cm-2 = 102 Nm-2 1 torr = 1 mm Hg 1 atm = 101.3 kPa = 1.013 bar = 760 torr 17. Absolute pressure and gauge pressure. The total or actual pressure P at a point is called absolute pressure. Gauge pressure is the difference between the actual pressure (absolute pressure) at a point and the atmospheric pressure. Thus P = P - Pa P = P + Pa Absolute pressure = Atmospheric pressure + Gauge pressure. 18. Buoyancy and centre of buoyancy. The upward force acting on a body immersed in a fluid is called upthrust or buoyant force and the phenomenon is called buoyancy. The force of buoyancy acts though the centre of gravity of the displaced fluid which is called centre of buoyancy. 19. Archemedes' principle. It states that when a body is immersed partly or wholly in a fluid, it loses some weight. The loss in weight is equal to the weight of the fluid displaced. Apparent weight of a body in a fluid = True weight - Weight of fluid displaced Wapp = W - U = Vσg-Vpg =Vσ (1-2) where W = V σg is the weight of the body and σ its density. 20. Law of floatation. A body will float in a liquid if weight of the liquid displaced by the body is atleast equal to or greater than the weight of the body. When a body just floats Weight of the body = Weight of liquid displaced V'σg = V 'pg or V p or Volume of the immersed part Total volume of the body Density of the body = Density of liquid 21. Viscosity. It is the property of a fluid due to which an opposing force comes into play whenever there is relative motion between its different layers. 22. Newton's formula for viscous force. The viscous drag between two parallel layers each of area A and having velocity gradient dv/dx is given by F = - η A du dx where n is the coefficient of viscosity of the liquid. 23. Coefficient of viscosity. It may be defined as the tangential viscous force required to maintain a unit velocity gradient between two liquid layers each of unit area. Its dimensional formula is [ML-1T-1] 24. Units of η. The CGS unit of n is poise. The coefficient of viscosity of a liquid is 1 poise if a tangential force of 1 dyne cm-2 of the surface is required to maintain a relative velocity of 1cm s-1 between two layers of the liquid 1 cm apart. 1 poise = 1 dyne s cm-2 = 1g cm-1s-1. The SI unit of n is deca