Zagazig University Physics PDF

Zagazig University Faculty of Oral and Dental Medicine Physics  CONTENTS page Chapter 1 Dimensions and units………………………………………. 1 Chapter 2 Elasticity…………………………………………………………… 20 Chapter 3 Surface tension……………………………...

Zagazig University Faculty of Oral and Dental Medicine Physics  CONTENTS page Chapter 1 Dimensions and units………………………………………. 1 Chapter 2 Elasticity…………………………………………………………… 20 Chapter 3 Surface tension…………………………………………………. 47 Chapter 4 Viscosity…………………………………………………….......… 64 Chapter 5 Atomic Structure and Chemical Bonding …………. 82 Chapter 6 Physics of Diagnostic X-rays …………………………….. 106 Chapter 7 Heat………………………………………………………...……….. 132 Chapter 1 Dimensions and Units Chapter (1) Dimensions and Units Fundamental concepts of the Physics start from this chapter. Basically the terms & concepts which are illustrated in this topic will be used in so many ways because all Physical quantities have units. It is must to measure all Physical quantities so that we can use them. In this chapter we will have an over view of different units of different Physical quantities. We will learn the dimension and dependence of the unit of any Physical quantity on fundamental quantities or unit. Three “Fundamental” Physical Quantities 1. Length 2. Mass 3. Time 1 Chapter 1 Dimensions and Units These three form a “fundamental set” in the sense that they can be used to derive the dimensions of (some) other physical quantities. By the dimensions of a physical quantity, we mean, “what kind of physical quantity is it (length, mass, time, some combination of these three, etc.).” Systems of Standard Units There are three systems of units: 1. SI (“mks system”) Length → meter (m) Mass → kilogram (kg) Time → second (s) 2. cgs system Length → centimeter (cm) Mass → gram (g) Time → second (s) 2 Chapter 1 Dimensions and Units 3. British system Length → foot (ft) Mass → slug* Time → second (s) Abbreviations for “three fundamental” dimensions: Length = L Mass = M Time = T These Abbreviations are called the dimensional equation of the physical quantity. The following EXAMPLES show the dimensional equation of some known physical quantities: 1. The area In general, the area is obtained by multiplying two lengths together. This means that the dimensional equation of area L2. 3 Chapter 1 Dimensions and Units Its units: (MKS system) → m2 (cgs system) → cm2 2. The density The density of a substance is defined as the mass of a substance divided by its volume, In other words, But since the dimension of the mass is M that of the volume is L3, we can easily see that the dimensional equation of density is ML- 3. Its units: (MKS system) → Kg.m-3 (cgs system) → gm.cm-3 3. The velocity Velocity is defined as the rate of change of distance with respect to time. i.e. Hence the dimensional equation of velocity is LT-1. Its units: (MKS system) →m.s-1, (cgs system) →cm.s-1 4 Chapter 1 Dimensions and Units 4. The acceleration Acceleration is defined as the rate of change of velocity with respect to time. Hence the dimensional equation of acceleration is LT-2. Its units: (MKS system) →m.s-2 (cgs system) →cm.s-2 5. The work Work is defined as the force multiplied by the distance through which it acts. This means that: But force is equal to acceleration multiplied by mass M. Also the dimensional equation of acceleration is LT-2. Hence the dimensional equation of work is given by: M.LT-2.L = ML2T-2 Its units: (MKS system) → kg.m2.s-2 (cgs system) → gm.cm2.s-2 5 Chapter 1 Dimensions and Units The following Table gives the dimensions of some physical quantities which are usually dealt with in studying Physics or Mechanics. Physical Definition Dimensions Cgs units quantity Area L2 cm2 Volume ( ) L3 cm3 Density mass/volume ML-3 gm.cm-3 Velocity distance/time LT -1 cm.sec-1 Acceleration velocity/time LT -2 cm.sec-2 MLT -2 Force gm.cm.sec-2 Work M L2 T -2 gm.cm2.sec-2 Power work/time M L2 T -3 gm.cm2.sec-3 Pressure force/area M L -1 T -2 gm.cm-1.sec-2 Momentum MLT -1 gm.cm.sec-1 Modulus of stress/strain or, M L -1 T -2 gm.cm-1.sec-2 elasticity force/area Moment of M L2 T -2 gm.cm2.sec-2 force 6 Chapter 1 Dimensions and Units Notes: 1) There are quantities which have no dimensions (or their dimensional equation is equal to 1). They are called non-dimensional quantities such as angles and ratios. 2) In the above table one can easily see that some physical quantities have the same dimensions, such as work and moment of force, both have dimensions ML2T -2. Also both pressure and modulus of elasticity have the same dimensions, ML-1T -2. Applications of Dimensional Equation Dimensional equations are found to be useful in studying physics and mechanics. The following examples may illustrate the usefulness of such dimensional equations: 7 Chapter 1 Dimensions and Units A. Testing the correctness of physical laws: The correctness of any physical relation or law can be tasted by obtaining the dimensional equation of both sides. If the dimensions of both sides agree, the physical relation under test is considered to be correct. The following examples may illustrate the above fact: (1) The relation which gives the period of oscillation of a simple pendulum: The period of complete oscillation is given by the following relation: √ where, T: is the time of one complete oscillation (sec.), π is 22/7, L: is the length of the pendulum (cm), g is the earth gravity acceleration (cm.sec-2). 8 Chapter 1 Dimensions and Units (a) The dimension of the left hand side of the relation is T. (b) the dimensional equation of the right-hand side of the relation is: ( ) Hence, the dimensions of L.H.S. = the dimensions of R.H.S = T. This proves that, the above relation who governs the period of oscillation of the simple pendulum is correct. (2) The velocity of a transverse wave: The velocity of a transverse vibration of a string is given by the relation: √ where: 9 Chapter 1 Dimensions and Units V: is the velocity of the transverse wave (cm. sec-1), T: the tension of the string (dynes), m: the mass per unit length of the string. Deriving the dimensional equation for each side of the relation, we get: (a) For the left hand side: LT-1 (b) For the right hand side, we get: Tension is force, whose dimensional equation can be written as: Also the mass per unit length has a dimensional equation equal to ML-1. Hence, the dimensional equation of the right hand side is: ( ) So, the dimensions of L.H.S. = the dimensions of R.H.S. 10 Chapter 1 Dimensions and Units This proves that the relation, who governs the velocity of transverse vibrations given by a string, is correct. B. Deciding the nature of any constant contained in a physical quantity: Usually physical quantities contain a constant whose nature may not be known. Such a constant may be a pure number (with no dimensions), or may be itself a physical quantity which has dimensions. In the case, we apply the theory of the dimensional equations to discover the nature of such constant. The following examples illustrate the significance of such application: (1) The nature of the constant G included in Newton's general law of gravitation. This law can be written as: where, 11 Chapter 1 Dimensions and Units F: the force of attraction between the two masses (dynes), m1 and m2, the two masses (gm),d, the distance which lies between the centers of the two masses (cm),G: the gravitational constant. The constant G can be written as: So, the dimensions of G equal to the dimensions of R.H.S. The dimensions of R.H.S can be written as: This means that: The dimensions of G = and it is units are: (MKS system) → kg-1.m3.s-2 (cgs system) → gm-1.cm3.s-2 Hence, the gravitational constant G is not a pure number. 12 Chapter 1 Dimensions and Units It is a physical quantity and has dimensions. (2) The constant K included in the law which governs the relation between the periods of oscillation of the simple pendulum: √ where, T is the time of one complete oscillation (sec.), L: is the length of the pendulum (cm), g is the earth gravity acceleration (cm.sec-2). Find the nature of constant k????? The constant k is given by: ( ) the dimensions of k equal to the dimensions of R.H.S which can be written as: ( ) 13 Chapter 1 Dimensions and Units ( ) This means that the constant K is a pure number C. The determination of dimensions of an unknown physical quantity: This can be illustrated by the following example: By experiment, Coulomb could discover the following relation: Where, F, the force of attraction or repulsion between the two point-charges (dynes),K, the dielectric constant of the medium which lies between the two charges, Q1, the first electric charge,Q2, the second electric charge, d, the distance between the two charges (cm). Obtaining the dimensions of both sides of the above equation, we get: 14 Chapter 1 Dimensions and Units [ ] [ ] Hence, we can write: [ ] This gives the dimensions of the electrostatic charge. D. The discovery of a new physical law: The application of dimensional equations may lead to the discovery of a new physical relation. D: gm. Cm-3., the destiny of its material, and E, Young's modulus of the material it is made of. Obtaining the dimensional equation of each side of the above relation, we may write: i) The dimension of the left hand-side is T. ii) The dimensions of the right hand-side is given by: [ ] [ ] [ ] Hence, the dimensional equation of the above relation can be written as follows: 15 Chapter 1 Dimensions and Units [ ] [ ] [ ] [ ] Equation the powers of M, L, and T in each one of the two sides of the dimensional equation, we get: 1) Comparing the powers of T in both sides we get: or 2) Comparing the powers of M in both sides we get: or 3) Comparing the powers of L in both sides we get: or Putting the numerical value obtained for each of a, b, and c, in the original equation which was assumed, we get: √ 16 Chapter 1 Dimensions and Units Which gives the exact law which governs the relation between the period of oscillation of the tuning fork and both its density and elasticity. 17 Chapter 1 Dimensions and Units Problems 1. Check dimensionally the following equations: i. , where, v: velocity and a: acceleration. ii. , where F: force, x: distance, vo: initial velocity, v: final velocity 2. Find the dimensions of the following physical quantities: Kinetic energy – Potential energy – Power – Relative density. 3. Name the physical quantities represented by the following dimensions: LT-1 – ML2T-2 – ML-1T-2 4. Mention two pairs of physical quantities which have the same dimensions. 5. The centripetal force (F) acting on a particle moving uniformly in a circle depends upon its mass (m), velocity (v) and radius of circle (r). Derive the 18 Chapter 1 Dimensions and Units expression for centripetal force using method of dimensions. 19 Chapter 2 Elasticity Chapter (2) Elasticity Elasticity is the property of solid materials to return to their original shape and size after the forces deforming them have been removed. Recall Hooke's law — first stated formally by Robert Hooke. Hooke’s Law — Stress and Strain Hooke’s Law The extension of an elastic object (like a spring) is directly proportional to the force applied, provided the limit of proportionality is not exceeded. F = − k Δx where, F: is the applied force. Δx: is the extension. K: a constant of proportionality. 20 Chapter 2 Elasticity This is Hooke's law for a spring — a simple object that's essentially one-dimensional. Hooke's law can be generalized to… Stress is proportional to strain Stress refers to the cause of the change (a force applied to a surface) Strain refers to a change in some spatial dimension (length, angle, or volume) compared to its original value. The coefficient (proportionality constant) that relates a particular type of stress to the strain that results is called an elastic modulus. Elastic moduli are properties of materials, not objects. There are three basic types of stress and three associated moduli. 1. Young Modulus (E or Y). 2. Bulk Modulus (K or B). 3. Shear Modulus (G or S). 21 Chapter 2 Elasticity 1. Young's Modulus For the description of the elastic properties of linear objects like wires, rods, columns which are either stretched or compressed, a convenient parameter is the ratio of the stress to the strain, a parameter called the Young's modulus of the material. Young's modulus can be used to predict the elongation or compression of an object as long as the stress is less than the yield strength of the material. 22 Chapter 2 Elasticity Let’s imagine a rod of length ℓ0 and cross sectional area A being stretched by a force F to a new length ℓ0 + Δℓ. Tensile stress (σ) is the outward normal force per unit cross-sectional area A Its dimensions = ML-1T-2 Its units: (MKS system) → kg.m-1.s-2 ≡ N.m-2 (cgs system) → gm.cm-1.s-2 ≡ dyne.cm-2 23 Chapter 2 Elasticity Tensile strain (ε) is the ratio between the change in length to original length of the rod. Tensile strain dimensions = 1 Tensile strain units: has no unit. The proportionality constant that relates these two quantities together is the ratio of tensile stress to tensile strain —Young's modulus. Its dimensions = the dimensions of stress = ML-1T-2 Its units: (MKS system) → kg.m-1.s-2 ≡ N.m-2 (cgs system) → gm.cm-1.s-2 ≡ dyne.cm-2 24 Chapter 2 Elasticity  Stress-Stain Curve The relation between the stress and the strain for a given material under tensile stress can be found tensile properties, a test cylinder or a wire is experimentally. In a standard test of tensile properties, a test cylinder or a wire is stretched by an applied force. The fractional change in length (the strain) and the applied force needed to cause the strain are recorded. A graph is plotted between the stress (which is equal in magnitude to the applied force per unit area) and the strain produced. The stress-strain curves vary from material to material. These curves help us to understand how a 25 Chapter 2 Elasticity given material deforms with increasing loads. From the graph, we can see that:  In the region between O to A, the curve is linear and Hooke’s law is obeyed (when the applied stress is removed, the material return to its original shape).  In the region from A to B, stress and strain are not proportional.  The point B in the curve is known as yield point (also known as elastic limit) and the corresponding stress is known as yield strength (Sy) of the material.  The portion of the curve between B and D shows, if the load is increased further, the stress developed exceeds the yield strength and strain increases rapidly even for a small change in the stress.  When the load is removed say at some point C between B and D, the body does not regain its original dimension. In this case, even when the stress is zero, the strain is not zero. The material is said to have a permanent set. The deformation is said to be plastic deformation. 26 Chapter 2 Elasticity  The point D on the graph is the ultimate tensile strength (Su) of the material beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at point E. If the ultimate strength and fracture points D and E are close, the material is said to be brittle. If they are far apart, the material is said to be elastic. Example 1 A structural steel rod has a radius of 10 mm and a length of 1.0 m. A100 k N force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is 2.0 x10 11 N m -2. Solution We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. The other end, parallel to the length of the rod then the stress on the rod is given by 27 Chapter 2 Elasticity The strain is given by Example 2 A steel wire 1.00 m long with a diameter d = 1.00 mm has a 10.0-kg mass hung from it. (a) How much will the wire stretch? (b) What is the stress on the wire? (c) What is the strain? (Y = 21 × 1010 N/m2) Solution a. The cross-sectional area of the wire is given by 28 Chapter 2 Elasticity The force stretching the wire is the weight of the 10.0-kg mass, that is, The elongation of the wire is given by, b. The stress acting on the wire is c. The strain of the wire is 29 Chapter 2 Elasticity Example3 A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied. Solution The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A. From Eq. (3.7) we have Stress= strain x Young’s modulus Therefore ( ) ( ) Where the subscripts c and s refer to copper and stainless steel respectively Or, Lc/ Ls = (Ys/Yc).(Lc/Ls) 30 Chapter 2 Elasticity Given Lc = 2.2 m, Ls = 1.6 m, Yc = 1.1 x10 11 N.m-2, Ys = 2.0 x 10 11 N.m-2 Lc/ Ls = (2.0 x 1011/1.1 x 1011) x (2.2/1.6) = 2.5. The total elongation is given to be Lc + Ls = 7.0 x10 -4 m Solving the above equations, Lc = 5.0 x 10 -4 m, Ls = 2.0 x 10 -4 m. Therefore [ ] 31 Chapter 2 Elasticity 2. Bulk Modulus If a uniform force is exerted on all sides of an object, as in the figure, such as a block under water, each side of the block is compressed (the volume of the object compress without changing its shape). Thus, the entire volume of the block decreases. The compression (bulk stress) is defined as: where F is the magnitude of the normal force acting on the cross-sectional area A of the block. The bulk strain is 32 Chapter 2 Elasticity measured by the change in volume per unit volume, that is, Since the stress is directly proportional to the strain, by Hooke’s law, we have To obtain equality, we introduce a constant of proportionality B, called the bulk modulus, The minus sign is introduced because an increase in the stress (F/A) causes a decrease in the volume, leaving ∆V negative. The bulk modulus is a measure of how difficult it is to compress a substance. 33 Chapter 2 Elasticity In dealing with liquids and gases it is convenient to deal with the pressure exerted by the liquid or gas. The pressure is defined as the force that is acting over a unit area of the body, that is, For the case of volume elasticity, the stress F/A, acting on the body by the fluid, can be replaced by the pressure of the fluid itself. Thus, Hooke’s law for bulk modulus can also be written as The bulk modulus of a gas which obeys Boyle's law at constant temperature is given by the pressure of the gas, for we may state Boyle's law as PV = constant and, taking differentials of this equation, we find PdV + V dP = 0, And, solving for P, we find 34 Chapter 2 Elasticity If we replace the differentials by small increments so that the differential dP is replaced by P and the differential dV is replaced by V, we have From the above analysis it can be seen that, if the pressure of a gas is changed by any amount while the temperature is kept constant, the bulk modulus will vary and, at any stage of the process, will be equal to the pressure of the gas at that stage. This is the reason that tires or basketballs inflated to high pressures seem hard, or difficult to deform, while the same object inflated to low pressure is easy to deform, or soft. 35 Chapter 2 Elasticity Example 4 A solid copper sphere of 0.5m3 volume is placed 30.5 m below the ocean surface where the pressure is 3.00 × 105 N/m2. What is the change in volume of the sphere? The bulk modulus for copper is 14 × 1010 N/m2. Solution The change in volume is given by, Example 4 The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, V/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 x10 9 N m -2. (Take g = 10 m s -2) Solution 36 Chapter 2 Elasticity The pressure exerted by a 3000 m column of water on the bottom layer p =h ρ g = 3000 m.1000 kg m -3.10 m s -2 = 3.107 kg m -1 s-2 = 3.107 N m-2 Fractional compression V/V, is 3. Shear Modulus Consider a cube fixed to the surface as in the figure. A tangential force F is applied at the top of the cube, a distance h above the bottom. 37 Chapter 2 Elasticity The magnitude of this force F times the height h of the cube would normally cause a torque to act on the cube to rotate it. However, since the cube is not free to rotate, the body instead becomes deformed and changes its shape. The tangential force thus causes a change in the shape of the body that is measured by the angle φ, called the angle of shear. We can also relate φ to the linear change from the original position of the body by noting from the figure that Because the deformations are usually quite small, as a first approximation the tan φ can be replaced by the angle φ itself, expressed in radians. Thus, this equation represents the shearing strain of the body. 38 Chapter 2 Elasticity The proportionality constant that relates these two quantities together is the ratio of shear stress to shear strain —Shear modulus or modulus of rigidity (R). Shear modulus dimensions = the dimensions of stress = ML-1T-2 Shear modulus units: (MKS system) → kg.m-1.s-2 ≡ N.m-2 (cgs system) → gm.cm-1.s-2 ≡ dyne.cm-2 Example 5 A sheet of copper 1.00 m high, base length 0.750 m, and 0.500 cm thick is acted on by a tangential force of 50,000 N, as shown in the below figure. The value of R for copper is 4.20 × 1010 N/m2. Find (a) the shearing stress, (b) the shearing strain, and (c) the linear displacement ∆x. 39 Chapter 2 Elasticity Solution a. The area that the tangential force is acting over is A = bt = (0.750 m)(5.00 × 10 m) = 3.75 × 10−3 m2 where b is the length of the base and t is the thickness of the copper sheet shown in the figure. The shearing stress is b. The shearing strain is = 3.17 × 10−4 rad c. The linear displacement ∆x is 40 Chapter 2 Elasticity ∆x = hφ = (1.00 m)(3.17 × 10−4 rad) = 3.17 × 10−4 m = 0.317 mm Poisson’s ratio One more important physical property is Poisson’s relationship. When an object is compressed (or extended) it is observed that it expands (or contracts) perpendicular to the direction of the applied stress. The ratio of the transverse strain to the axial strain is Poisson’s ratio (σ). Consider a rectangular block in the following figure with original length (l) and diameter (d) is subjected to a tensile force (axial force) then, the length is increased by (Δl) and the diameter is decreased by (Δd). There is exist two strains 41 Chapter 2 Elasticity Poisson’s ratio is given by: σ σ If the force is compresses Poisson’s ratio it is given by: σ Poisson’s ratio dimensions = dimensionless Poisson’s ratio units: has no unit. 42 Chapter 2 Elasticity Example 6 A steel wire 10 m long and 2 mm in diameter is attached to the ceiling and a 200-N weight is attached to the end. What is the applied stress? Solution L=2m Diameter =d=2mm=2x10 -3m r=d/2=1x10-3m A= = F=200 N 43 Chapter 2 Elasticity Stress=? Stress=F/A N.m-2 Example 7 A 10 m steel wire stretches 3.08 mm due to the 200 N loads. What is the is the longitudinal strain Solution Given =10m =3.08mm F=200N Longitudinal strain = (3.08mm) x /10m=0.308 44 Chapter 2 Elasticity Problems 1. A coil spring stretches 4.00 cm when a mass of 500 g is suspended from it. What is the force constant of the spring? 2. A coil spring stretches by 2.50 cm when a mass of 750 g is suspended from it. (a) Find the force constant of the spring. (b) How much will the spring stretch if 800 gm is suspended from it? 3. An aluminum wire has a diameter of 0.850 mm and is subjected to a force of 1000 N. Find the stress acting on the wire. 4. A copper wire experiences a stress of 5.00 × 103 N/m2. If the diameter of the wire is 0.750 mm, find the force acting on the wire. 5. A steel wire, 1.00 m long, has a diameter of 1.50 mm. If a mass of 3.00 kg is hung from the wire, by how much will it stretch? 6. A copper wire, 1.00 m long, has a diameter of 0.750 mm. When an unknown weight is suspended from the 45 Chapter 2 Elasticity wire it stretches 0.200 mm. What was the load placed on the wire? 7. A copper block, 7.50 cm on a side, is subjected to a tangential force of 3.5 × 103 N. Find the angle of shear. 8. A cube of lead 15.0 cm on a side is subjected to a uniform pressure of 5.00 × 105 N/m2. By how much does the volume of the cube change? 46 Chapter 3 Surface tension Chapter (3) Surface Tension One of most important phenomena in nature which describes: - Water climbing up trees. - Shape of rain drops. -Some insects moving above water. These observations suggest that the surface of a liquid acts like an elastic membrane covering the surface liquid or in a state of tension. As we shall soon show, surface tension is due to intermolecular attraction. Molecular basis for surface tension The fact that a liquid surface is in a state of tension can be explained by the intermolecular forces. The figure illustrates the molecular basis for surface tension by considering the attractive forces that molecules in a liquid exert on one another. 47 Chapter 3 Surface tension  In the bulk of the liquid (part A), - A particular molecule is surrounded by an equal number of molecules on all sides. - The average distance apart of the molecules is such that the attractive forces balance the repulsive forces. - The average intermolecular force between A and the surrounding molecules is zero.  In the surface of the liquid (part B), - Since there are no molecules of the liquid above the surface, the molecule on the surface experiences a net attractive force pointing toward the liquid interior. This net attractive force causes the liquid surface to contract toward the interior. 48 Chapter 3 Surface tension Thus, if the particle on B is displaced very slightly upward, a resultant attractive force F on it, now has be overcome. It follows that if all molecules in the surface were removed to infinity, a definite amount of work is needed to bring molecules from the bulk to the surface. Consequently molecules in the surface have more potential energy than those in the bulk. Surface area (shape of drop) The potential energy of any system in stable equilibrium is a minimum. Thus under surface tension forces, the area of a liquid surface will have the least number of molecules in it, that is, the surface area of a given volume of liquid is a minimum. So, if the liquid is not acted upon by external forces, a liquid sample forms a sphere, which has the minimum surface area for a given volume. Nearly spherical drops of water are a familiar view. 49 Chapter 3 Surface tension Surface tension definition (units and dimensions) The surface tension is denoted by the Greek letter gamma ( ), sometimes called the coefficient of surface tension is defined as: The magnitude F of the force exerted parallel to the surface of a liquid divided by the length L of the line over which the force acts: - SI Unit of Surface Tension: N/m - Surface Tension dimensions The surface tension coefficient of depends on the temperature of the liquid and on the medium on the other side of the surface. 50 Chapter 3 Surface tension Capillary Action We have seen that surface tension arises because of the intermolecular forces of attraction that molecules in a liquid exert on one another. These forces, which are between like molecules, are called cohesive forces. A liquid, however, is often in contact with a solid surface, such as glass.Then additional forces of attraction come into play. They occur between molecules of the liquid and molecules of the solid surface and, being between unlike molecules, are called adhesive forces. Consider a tube with a very small diameter, which is called a capillary. When a capillary, open at both ends, is inserted into a liquid, the result of the competition between cohesive and adhesive forces can be observed. The figure shows a glass capillary inserted into water. In this case, - The adhesive forces are stronger than the cohesive forces, so that the water 51 Chapter 3 Surface tension molecules are attracted to the glass more strongly than to each other. - The result is that the water surface curves upward against the glass. - It is said that the water “wets” the glass. The surface tension leads to a force F acting on the circular boundary between the water and the glass. This force is oriented at an angle φ, which is determined by the competition between the cohesive and adhesive forces. The vertical component of F pulls the water up into the tube to a height h. At this height the vertical component of F balances the weight of the column of water of length h. The angle φ called “contact angle”, and is always measured through the liquid. The angle of contact between two given surfaces varies largely with their 52 Chapter 3 Surface tension freshness and cleanliness. The angle of contact between water and very clean glass is zero, but when the glass is not clear the angle of contact is < 90o. The figure, shows glass capillary inserted into mercury, - The adhesive forces are weaker than the cohesive forces. - The mercury atoms are attracted to each other more strongly than they are to the glass. As a result, the mercury surface curves downward against the glass and the mercury do not “wet” the glass. - The surface tension leads to a force F, the vertical component of which pulls the mercury down a distance h in the tube. - The angle of contact is > 90o. 53 Chapter 3 Surface tension Determination of surface tension coefficient by capillary tube: when capillary tube is immersed in liquid; there will be a raised or depressed column of liquid inside it. The case of a raised column is shown on the figure. The upward component of the surface tension force (upward force) will balance the weight of the liquid column (downward force). We know that: ; and the length of liquid in contact with the glass is So: The upward component of the surface tension force is given by: 54 Chapter 3 Surface tension The downward force (weight of the liquid column) is given by: The volume of the liquid Thus, the weight of the liquid At balance Hence, the surface tension of liquid can be expressed as: Example 1 A capillary tube with an inside diameter of 250 μm can support a 100 mm column of liquid that has a density of 930 kg.m-3. The observed contact angle is 15°. Find the surface tension of the liquid. Solution 55 Chapter 3 Surface tension ( ) Pressure inside a soap bubble and a liquid drop Anyone who’s blown up a balloon has probably noticed that the air pressure inside the balloon is greater than on the outside. For instance, if the balloon is suddenly released, the greater inner pressure forces the air out, propelling the balloon much like a rocket. The reason for the greater pressure is that the tension in the stretched rubber tends to contract the balloon. To counteract this tendency, the balloon has a greater interior air pressure acting to expand the balloon. A soap bubble (see the figure below -a) has two spherical surfaces (inside and outside) with a thin layer of liquid in-between. Like a balloon, the pressure inside a soap bubble is greater than that on the outside. As we will see 56 Chapter 3 Surface tension shortly, this difference in pressure depends on the surface tension of the liquid and the radius R of the bubble. For the sake of simplicity, let’s assume that there is no pressure on the outside of the bubble (Po = 0). Now, imagine that the stationary soap bubble is cut into two halves. Being at rest, each half has no acceleration and so is in equilibrium. According to Newton’s second law of motion, a zero acceleration implies that the net force acting on each half must be zero (ΣF = 0). We will now use this equilibrium relation to obtain an expression relating the interior pressure to the surface tension and the radius of the bubble. 57 Chapter 3 Surface tension Figure -b shows a free-body diagram for the right half of the bubble, on which two forces act. 1. First, there is the force due to the surface tension in the film. This force is exerted on the right half of the bubble by the left half. The surface tension force points to the left and acts all along the circular edge of the hemispherical film. The magnitude of the force due to each surface of the film is the product of the surface tension and the circumference (2π R) of the circular edge, or (2π R). The total force due to the inner and outer surfaces is twice this amount or −2 (2π R). We have included the minus sign to denote that this force points to the left in the drawing. We have also assumed the film to be sufficiently thin enough that its inner and outer radii are nearly the same. 2. Second, there is a force caused by the air pressure inside the bubble. At each point on the surface of the bubble, the force due to the air pressure is 58 Chapter 3 Surface tension perpendicular to the surface and is directed outward. Figure 7b shows this force at six points on the surface. When these forces are added to obtain the total force due to the air pressure, all the components cancel, except those pointing to the right. The total force due to all the components pointing to the right is equal to the product of the pressure Pi inside the bubble times the circular cross-sectional area of the hemisphere, or Pi (π R2). Using these expressions for the forces due to the surface tension and air pressure, we can write Newton’s second law of motion as: 59 Chapter 3 Surface tension Solving this equation for the pressure inside the bubble gives Pi = 4 /R. In general, the pressure Po outside the bubble is not zero. However, this result still gives the difference between the inside and outside pressures, so that we have ( ) This result tells us that the difference in pressure depends on the surface tension and the radius of the sphere. What is surprising is that a greater pressure exists inside a smaller soap bubble (smaller value of R) than inside a larger one. A spherical drop of liquid, like a drop of water, has only one surface, rather than two surfaces, for there is no air within it. Thus, the force due to the surface tension is only one-half as large as that in a bubble. Consequently, the difference in pressure between the inside and outside of a liquid drop is one-half of that for a soap bubble: 60 Chapter 3 Surface tension ( ) This result also holds for a spherical bubble in a liquid, such as a gas bubble inside a glass of beer. However, the surface tension is that of the surrounding liquid in which the trapped bubble resides. Example 2 (a) A student, using a circular loop of wire and a pan of soapy water, produces a soap bubble whose radius is 1.0 mm. The surface tension of the soapy water is 2.5 × 10-2 N/m. Determine the pressure difference between the inside and outside of the bubble. (b) The same soapy water is used to produce a spherical droplet whose radius is one-half that of the bubble, or 0.50 mm. Find the pressure difference between the inside and outside of the droplet. Solution 61 Chapter 3 Surface tension (a) The pressure difference, Pi - Po, between the inside and outside of the soap bubble is given by (b) The pressure difference, Pi - Po, between the inside and outside of the drop is given by: 62 Chapter 3 Surface tension Problems 1. Two soap bubbles ( = 0.025 N/m) have radii of 2.0 and 4.5 mm. For each bubble, determine the difference between the inside and outside pressures. 2. Α small bubble of air in water ( = 0.073 N/m) has a radius of 0.10 mm. Find the difference in pressures between the inside and outside of the bubble. 63 Chapter 4 Viscosity Chapter (4) Viscosity when real fluids flow they have a certain amount of internal friction called viscosity. So, Viscosity is defined as; the resistance of a fluid to flow. It exists in both liquids and gases and is essentially a friction force between different layers of fluid as they move past one another. In liquids the viscosity is due to the cohesive forces between the molecules whilst in gases the viscosity is due to collisions between the molecules. Laminar flow: It sometimes known as streamline flow; occurs when a fluid flows in parallel layers, with no disruption between the layers. Put some transparent glycerol in a tube and then put in some colored glycerol. Let them flow. From the colored glycerol’s shape change, we can see the flow velocity of glycerol is different in different places. 64 Chapter 4 Viscosity  The speed is smaller near the wall of the tube and flow velocity is the biggest in the center of the tube. This means that viscous fluid flows in different layers. We call it Laminar Flow.  When fluid flows in laminar flow, there is internal friction or viscous force between two close layers. The internal friction is caused by the interaction forces between molecules. The internal friction is bigger in liquids than in gases. The basic formula for the frictional force ‘F’ in a liquid was first suggested by NEWTON. He found that the frictional force is:  directly proportional to the area of the surface of liquid (A).  directly proportional to the velocity gradient. If v1, v2 are the velocities of two layers separated 65 Chapter 4 Viscosity by distance h. The velocity gradient between the two layers can be expressed as;( ) in uint (m/s)/m or s-1. So, The constant of proportionality for the fluid is called the coefficient of viscosity η: The greater the coefficient of viscosity η, the greater the force required to move the plate at a velocity v. Definition, Units and Dimensions of coefficient of viscosity The magnitude of η is given by η is defined as the frictional force per unit area of a liquid when it is in a region of unit velocity gradient. 66 Chapter 4 Viscosity Dimensions: The dimensions of a force, F, (=mass x acceleration = mass x velocity/time) are MLT-2. The dimensions of an area, A, are L2. The dimensions of velocity gradient Now Thus η can expressed in units: kgm-1sec-1 Poiseuille's Law (Flow of a fluid through a pipe) In trying to find out what factors control how fast fluids can flow through pipes, the following factors are easy to isolate: 67 Chapter 4 Viscosity - The pressure difference between the ends of the pipe. The bigger the pressure difference, the faster will be the flow. - The length of the pipe. More liquid will flow through a shorter than a longer pipe in the same time. - The radius of the pipe. More liquid will flow through a wide than a narrow pipe in the same time. This dependence is very marked. - The coefficient of viscosity of the liquid. Water flows much more easily than glycerin. Hence the volume of liquid issuing per second (Q) from the pipe depends on: i. The coefficient of viscosity (η). ii. The radius of the pipe (a). iii. The pressure gradient (g). 68 Chapter 4 Viscosity The pressure gradient = P/L, where P is the pressure difference between the ends of the pipe and L is its length. Thus x, y, z being indices which require to be found, suppose: η  Now the dimensions of Q (volume per second) are: L3 T-1  the dimensions of η are ML-1T-1  the dimensions of g are By equating the dimensions on both sides, 69 Chapter 4 Viscosity ( ) ( ) Equating the respective indices of M, L, and T on both sides, we have x+z=0 -x + y -2z = 3 X + 2z = 1 Solving we obtain x = -1, z = 1, y = 4. This lead to: The constant k was found experimentally to equal π/8 ( ) Stoke’s law and terminal velocity When a small object, such as steel ball-bearing, is dropped into a viscous liquid like glycerin it accelerates at first, but its velocity soon reaches a steady value known as the terminal velocity. 70 Chapter 4 Viscosity Suppose a sphere of radius ‘r’ is dropped into a viscous liquid of coefficient of viscosity ‘η’ and its velocity at an instant is ‘v’. The frictional force ‘F’ can be partly found by the method of dimensions. Thus, suppose F=k η , where k is constant. The dimensions of η are M L-1 T-1 and the dimensions of v are L T-1. ( ) ( ) Equating the respective indices of M, L, and T on both sides, we have 71 Chapter 4 Viscosity y = 1, x – y + z = 1, -y – z = -2. Hence z = 1, x = 1, y = 1, consequently Stokes showed mathematically that the constant k equal 6π, and he arrived the formula This equation called Stoke’s law. Determination of viscosity coefficient using Stoke’s law Suppose a tall glass vessel is filled with the liquid, and a small ball-bearing is dropped gently into the liquid so that it falls along the axis of the vessel as shown in the figure. Toward the middle of the liquid the ball reaches its terminal velocity v. In this case: 72 Chapter 4 Viscosity  The viscous force acting upwards ↑↑↑ ( )  The up thrust due to the liquid on the ball acting upwards ↑↑↑ ( )  The weight of the ball acting downwards ↓↓↓ ( ) where, ρL and ρs are the density of liquid and ball, respectively. 73 Chapter 4 Viscosity At terminal velocity, the ball move with constant velocity so the two forces acting upwards equal the downward force (weight of the ball) ( ) ( ) ( ) When this process is repeated with a liquid of coefficient of viscosity η1 and density ρL1, using the same ball- bearing, then ( ) where v1, is the new terminal velocity ( ) ( ) Thus knowing v, v1, ρs, ρL and ρL1, the coefficient of viscosity can be compared. In very accurate work a correction is required for the effect of the walls of the vessel containing the liquid. 74 Chapter 4 Viscosity EQUATION OF CONTINUITY Consider an ideal fluid flowing through a pipe of varying cross sectional area A. The volume V1 of fluid and mass m1 flowing past (1) in a very small time interval Δt is V1 = A1 v1 Δt m1 = ρ1 A1 v1 Δt Similarly the volume and mass of fluid flowing past (2) in time Δt is V2 = A2 v2 Δt m2 = ρ2 A2 v2 Δt 75 Chapter 4 Viscosity When the flow is steady all the material which goes past (1) must go past (2) in the same time (or else it will be continually piling up past somewhere) and since the fluid is incompressible its density does not change Ρ1= ρ2 = ρ Therefore we must have m1 = m2 ρ A1 v1 Δt = ρ A2 v2 Δt A1 v1 = A2 v2 If the fluid is approximately incompressible, i.e. if its density never changes by very much, then the equation of continuity, as we quoted it, is approximately true. The quantity A v which measures the volume of the fluid that flows past any point of the tube divided by time is called the volume flow rate Q = dV/dt The equation of continuity is often expressed as Q = A v = constant ⇒ If A decreases then v increases 76 Chapter 4 Viscosity ⇒ If A increases then v decreases In complicated patterns of streamline flow, the streamlines effectively define flow tubes. So the equation of continuity says that where streamlines crowd together the flow speed must increase. Bernoulli's Principle: Bernoulli obtained a relation between the pressure and velocity at different parts of a moving incompressible fluid. If the viscosity is negligibly small, there are no frictional forces to overcome. In this case the work done by pressure forces equal the sum of both kinetic energy and the potential energy. 77 Chapter 4 Viscosity Derivation of Bernoulli's equation Mass element m moves from (1) to (2) as shown in the figure: 𝑚= 1∆𝑋1= 2∆𝑋2= ∆𝑉 where ∆𝑉= 1∆𝑋1= 2∆𝑋2 Equation of continuity A V = constant Since v1 r da v when T = 35°C? 151