Units and Dimensions PDF

1 Units and Dimensions  Introduction To understand any phenomenon in physics we have to perform experiments. Experiments require measurements, and we measure several physical properties like length, mass, time, temperature, pressure etc. Experimental verification of laws & theories also...

1 Units and Dimensions  Introduction To understand any phenomenon in physics we have to perform experiments. Experiments require measurements, and we measure several physical properties like length, mass, time, temperature, pressure etc. Experimental verification of laws & theories also needs measurement of physical properties. Physical Quantity Physical property that can be measured and described by a number is called physical quantity. Types of physical quantities 1. Fundamental quantities (Basic quantities ): The physical quantities which do not depend on any other physical quantities for their measurements are known as fundamental quantities. Examples: Mass Length Time Temperature There are seven base quantities: length, mass, time, current, temperature, amount of substance and luminous intensity. 2. Derived quantities: The physical quantities which depend on one or more fundamental quantities for their measurements are known as derived quantities. Examples: Area Volume Speed Force  Units for measurement The standard used for the measurement of a physical quantity is called a unit. Examples: metre, foot, inch for length kilogram, pound for mass second, minute, hour for time fahrenheit, kelvin for temperature Three Systems of Units (SI) French (international) system [SI]: MKS: meter, Kg, second French system: CGS: centimeter, gram, second British system: FPS: feet, pound, second  Units for measurement Seven fundamental units according the SI system  Rules for writing SI units Full name of unit always starts with small letter ( newton not Newton ) Symbol for unit named after a scientist should be in capital letter. N for newton Symbols for all other units are written in small letters. kg for kilogram One space is left between the last digit of numeral and the symbol of a unit.10 kg not 10kg The units do not have plural forms (6 metre not 6 metres) Full stop should not be used after the units. No space is used between the symbols for units. Nm  Units for measurement Prefixes Prefix Abbreviation Power nano n 10−9 micro  10−6 milli m 10−3 centi c 10−2 deci d 10−1 kilo k 103 mega M 106 giga G 109  Dimensional Analysis Dimension [ ] denotes the physical nature of a quantity. Whether a distance is measured in units of feet or meters, it is still a distance. We say its dimension is length. The symbols of the dimensions of: length, mass, and time are L, M, and T The dimensions of speed v are written distance L [v] = = = [ LT −1 ] time T  Dimensional Analysis Density = Mass / Volume = Mass / (length × width × height) [Density] = [M] / (L × L × L) = [M] / L3 = [ML−3]  ❑ Dimensional Analysis uses Technique to check the correctness of an equation or to assist in deriving an equation Dimensions (length, mass, time, combinations) can be treated as algebraic quantities. (Add, subtract, multiply, divide) Both sides of equation must have the same dimensions. Any relationship can be correct only if the dimensions on both sides of the equation are the same. Cannot give numerical factors: this is its limitation  ❑ Dimensional Analysis uses Technique to check the correctness of an equation or to assist in deriving an equation Dimensions (length, mass, time, combinations) can be treated as algebraic quantities. (Add, subtract, multiply, divide) Both sides of equation must have the same dimensions. Any relationship can be correct only if the dimensions on both sides of the equation are the same. Cannot give numerical factors: this is its limitation  ❑ Dimensional Analysis uses The SI unit of work function of a metal used in photoelectric effect is 1. joule (J) 2. newton (N) 3. pascal (Pa) 4. hertz (Hz) Dimensions of kinetic energy is the same as that of ________ 1. Acceleration 2. Velocity 3. Work 4. Force Farad is the unit of ________ 1. Luminosity 2. Wavelength 3. Permittivity 4. Inertia The surface tension of a liquid is 70 dyne/cm. In MKS system its value is? (a) 70 N/m (b) 7 ✕ 10-2 N/m (c) 7 ✕ 102 N/m (d) 7 ✕ 103 N/m The T= 2π 𝒈 𝑳 expression is dimensionally incorrect (True or False) Is The expression ( u = v−at ) dimensionally correct { where v=final velocity, u = initial velocity, t= time and a is the acceleration} Two different quantities which are to be divided must have the same dimensions. 1. Which of the following is a dimensionless quantity? A) force B) mass C) preasure D) none of above  2 Deformation  Introduction A change in shape due to the application of force is known as deformation. Objects undergo deformation which may be like squashing, twisting, ripping, or pulling apart the object. Permanent Deformation – Also known as plastic deformation, it is irreversible. It is a type of deformation that stays even after the removal of applied forces. Temporary Deformation – Also known as elastic deformation, it is reversible. It is a type of deformation that disappears after the removal of applied forces. (Elasticity) Property by which a deformed body tends to return to its original size and shape when the deforming stress is removed Force of Recovery The force which a deformed object exerts to return to its original state is a measure of the elasticity of the substance. (-ve sign) Elastic body: is the one that returns to its original shape after deformation. Inelastic body. is the one that does not return to its original shape after deformation. (Dough, Clay, Wood, Hook’s law Beyond the elastic limit, Elongation is proportional to Tension Hooke’s law is formulated as F = −kx. k is called the stiffness constant (Hook’s constant) 1. In this expression F no longer means the applied force but rather means the equal and oppositely directed restoring force that causes elastic materials to return to their original dimensions. 2. may also be expressed in terms of stress and strain. Hooke’s law can be explained by the fact that small displacements of their constituent molecules, atoms, or ions from normal positions is also proportional to the force that causes the displacement. From the origin till the proportional limit nearing yield strength, the straight line implies that the material follows Hooke’s law. Beyond the elastic limit between proportional limit and yield strength, the material loses its elasticity and exhibits plasticity. The area under the curve from origin to the proportional limit falls under the elastic range. The area under the curve from a proportional limit to the rupture/fracture point falls under the plastic range. The material’s ultimate strength is defined based on the maximum ordinate value given by the stress-strain curve (from origin to rupture). The value provides the rupture with strength at a point of rupture.  STRESS The quantity that describes the magnitude of forces that cause deformation is known as stress. It is defined as force per unit area. STRAIN The quantity that describes the deformation undergone is known as strain. The strain is given as a fractional change in either the length, geometry or volume. It is a dimensionless number. Stress = Elastic Modulus × Strain When an object has a large value of elastic modulus, the effect of stress is small. A small elastic modulus means that stress produces noticeable deformation. For example: ❑ stress on a rubber band produces a larger strain as compared to a steel band of the same size. ❑ The elastic modulus of a rubber band is smaller than the elastic modulus of the steel band. ❑ Elastic moduli of different materials are measured based on physical conditions, such as varying temperature, and collected in engineering data tables for reference. These tables are valuable references for industry and for anyone involved in engineering or construction.  Hook’s law ❖ Hooke’s law applies to a perfectly elastic material and does not apply beyond the elastic limit of any material. ❖ Hooke’s Law is considered linear as the law states that the restoring (recovery) force is proportional to the displacement. ❖ Hooke’s Law does not apply to all materials as it helps us understand how a stretchy object will behave when stretched or compacted. Types of stress Tensile Stress – When forces pull an object and cause its elongation such as the stretching of a rubber band. Compressive Stress – When the forces result in the compression of an object. Bulk Stress – When an object is squeezed from all sides like the submarine in the depth of the ocean. Shear Stress – A type of stress where the deforming stress acts tangentially to the object’s surface. Types of strain Longitudinal Strain Volumetric Strain Shear Strain To describe the effects of these stresses, three coefficients of elasticity called modules are necessary. They are: 1. The modulus of stretch or Young’s modulus; 2. The volume or bulk modulus; 3. The modulus of rigidity or shear. Modulus of stretch: Young’s modulus Consider an elastic long bar of length ’L’ and cross-sectional area ‘A’, as shown in Fig.4. If the bar is clamped at one end and an external force ‘F’ is applied longitudinally along the bar perpendicular to its cross section A internal forces in the bar resist stretching, but the bar attains an equilibrium in which its length is greater, and F is exactly balanced by internal forces. The ratio of the internal force (equal to F) to the cross-sectional area A is the tensile stress: The units of stress is N/m2 = Pa, other units are dyne/cm2 (1 newton = 105 dyne) The ratio (ΔL/L) is the tensile strain, where ΔL is the elongation of the elastic bar. If the solid obeys Hook’s law, we then use eqn. (2) to define Young’s modulus, Y, as: Y= Tensile stress / tensile strain Tensile stress =F/A  Stress-versus-strain curve for an elastic solid 1. Stress-versus-strain curve for an elastic solid. 2. It’s possible to exceed the elastic limit of a substance by applying a sufficiently great stress 3. At the elastic limit, the stress-strain curve departs from a straight line. 4. A material subjected to a stress beyond this limit ordinarily doesn’t return to its original length when the external force is removed. 5. As the stress is increased further, it surpasses the ultimate strength: the greatest stress the substance can withstand without breaking. 6. The breaking point for brittle materials is just beyond the ultimate strength. For ductile metals like copper and gold, after passing the point of ultimate strength, the metal thins and stretches at a lower stress level before breaking. 7. Ductile materials are those materials which show large plastic range beyond elastic limit. eg:- copper. 8. Brittle materials are those materials which show very small plastic range beyond elastic limit. eg:- Glass. The Young’s modulus (E) is a property of the material that tells us how easily it can stretch and deform and is defined as the ratio of tensile stress (σ) to tensile strain (ε). For different types of materials, the stress-strain plots can look very different. Brittle materials tend to be very strong because they can withstand a lot of stress, they don’t stretch very much and will break suddenly. Ductile materials have a larger elastic region where the stress-strain relationship is linear, but at the first turnover (the elastic limit), the linearity breaks down and the material can no longer return to its original form. The second peak is the ultimate tensile strength and it tells us the maximum stress a material can withstand before breaking. Plastic materials are not very strong but can withstand a lot of strain. Young's modulus is given by the gradient of the line in a stress-strain plot.  Modulus of stretch: Young’s modulus Strain is a dimensionless quantity. Then, Y has units of stress, i.e. N/m2. Young’s modulus is typically used to characterize rod or wire stressed either under tension or compression. When a material elongated (or compressed) under tensile stress, the dimension perpendicular to the direction of stress becomes shorter (or longer), as shown in Fig. 4, by an amount proportional to the change in length. In addition, the longitudinal strain (ΔL/L) is responsible for the decrease in cross- section of the rod or wire and is represented by (Δr/r). The ratio of lateral to longitudinal strain, is called Poisson’s ratio. It happens that σ is nearly independent of the magnitude of the strain and is related to other elastic constants. / 𝜎= −(Δ𝑟/𝑟) (ΔL/L) Bulk modulus The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume. Its unit is the Pascal. Its the ratio of the stress on the body to the body's fractional decrease in volume. Bulk modulus The bulk modulus B can be formally defined by the equation: B= -V ∂P/∂V where P is pressure, V is volume, and ∂V is the decrease in the volume. The inverse of the bulk modulus gives a substance's compressibility. K= (1/B)= - ∂V/( V ∂P) Shear Stress and Strain shear stress = F7/A shear strain = x/h = tanf shear modulus = S Shear Stress and Strain stress = S  strain F| | x F| | =S or = S tan f  Sf A h A  Force in bar due to contraction or expansion: When a bar is heated, and then prevented from contracting as it cools, a force is exerted at the ends of the bar. If the change in length is DL for a change in temperature DT, then DL =  L DT Y : Young’s modulus a : cross-sectional area, α: linear coefficient of expansion L: original length of the bar F FL since Y = A = DL ADL L  YA   YA  F =  DL =  (LDT ) = Y A  DT  L   L   Energy stored in a wire: Consider a wire with original length L and stretched by a length ΔL when a force F is applied at one end. Within the elastic range, the extension is directly proportional to force F We can prove that the work done to stretch the wire is 1 W= F DL (Joule OR erg) 2 This is the amount of energy stored in the By dividing by the volume of the wire wire. (AL) W 1  F  DL  =    AL 2  A  L  1 = stress x strain 2 1 Energy per c.c. = stress x strain 2 Example: A load 250 kg hanging form a metal wire of length 6 m and cross section 0.2 cm2. The length of the wire is increased by 0.4 cm. Calculate the value of Young’s modulus for the metal of which the wire is composed. F 250 x9.8 stress = = −4 = 1.225x108 N/m 2 A 0.2 x10 DL 0.4 x10 −2 strain = = = 0.666x10−3 L 6 stress 1.225 x108 Y= = = 1.84x1011 N/m 2 strain 0.666 x10 −3 Exerccise: A load hanging from a steel wire of length 3 m and cross section 2x10-3 m2 was found to stretch the wire 0.4 cm above its no-load length. If Y for the steel is 1.48x1011 N/m2, calculate the value of the load. (d) 900 kg (c) 50 kg (a) 402 kg (b) 4000 kg Exerccise: A wire of length 4 m and 2 mm in diameter stretches by 0.5 mm when 8 kg- wt is hung on it. Find Young’s modulus for the material of the wire. (a) 9.98x1010 N/m2 (b) 1.2x1010 N/m2 (c) 1.2x1011 N/m2 Example A solid lead sphere of volume 0.5 m3 is dropped in the ocean to a depth where the water pressure is 2x107 N/m2. Lead has a bulk modulus of 7.7x109 N/m2. What is the change in volume of the sphere? DP DP Bulk modulus B = − DV = −V DV B V0 2x107 −3 ΔV = −0.5x = −1.3x10 m -3 7.7x109 Example of a wire of length 3 m Compute the work done in the elongation by 2 mm and the diameter 1mm, if the Young’s modulus is Y= 2x1012 dyne/cm2. (Calculate the restored energy per unit volume) 1 Energy per c.c. = stress x strain 2 stress Y=  stress = Y x strain strain  DL  2 1 1 Energy per c.c. = Y x strain = Y x  2  2 2  L  2 1  0.2  = 2x1012 x   = 4.44x10 erg 5 2  300  MCQs MCQs