Dimensional analysis

UNITS AND MEASURMENTSPhysical quantitiesThe quantities that can be measured directly or indirectly and it terms of which laws of physics can be expressed are called as physical quantities.Fundamental quantitiesThe physical quantities which are independent of other physical quantities are called as fundamental quantities. Ex. Mass, length, time, temperature etcDerived quantitiesThe physical quantities which depend upon the physical quantities and can be mathematically derived using them are called as derived quantities. Ex. Area, volume density, speed, acceleration, force, work, power etcPhysical UnitThe standard amount of a physical quantity chosen to measure the physical quantity of the same kind is called a physical unit.Characteristics of an ideal unitIt should be well defined.It should be of convenient size.It should not change with time.It should not change with physical conditions such as temperature, pressure etc.It should be easily reproducible.It should be easily accessible.It should be universally accepted.It should be indestructible.Fundamental unitsThe physical units which can neither be derived from on e another nor they can be resolves any further into simpler units are called as fundamental units.Ex. meter, kilogram, second, Kelvin etcDerived unitsThe physical units which can be expressed in terms of the fundamental units are called as derived units. Ex. m/s, newton, joule, kg/m3 etc.Coherent system of unitsA system of units based on a set of fundamental quantities in which all derived units can be derived by simple multiplication or division of fundamental units without introducing any numerical factor is called a coherent system of units.Advantages of SI It is a coherent system of units.It is a rational system of units as it uses only one unit for a given physical quantity.It is a metric system of units as the multiples and submultiples of SI units can be expressed as powers of 10.It is an internationally accepted system.Fundamental units in SI; The international system of unitsS. No.Fundamental physical quantityFundamental unitSymbolDimensional symbol1LengthmetremL2MasskilogramkgM3TimesecondsT4TemperatureKelvinKK or θ5Electric currentAmpereAA6Luminous intensitycandelacdcd7Amount of substancemolemolmolSupplementary units in SIS. No.Supplementary physical quantitySupplementary unitSymbol1Plane angleradianrad2Solid anglesteradiansrDimensions of a physical quantityThe powers to which the fundamental quantity are raised to represent that quantity completely are called as the dimensions of that physical quantity.ExampleName two physical quantities which have units but no dimensions.Plane angle and solid angle.Applications of Dimensional analysisTo convert a physical quantity from one system of units to other.To check the correctness of a given physical relation.To derive a relationship between different physical quantities.To convert a physical quantity from one system of units to otherEx. Convert 1 joule into erg.The dimensional formula of energy is [ML2T-2]. Thus its dimensions in mass, length and time are a= 1, b= 2 and c=-2 respectively.SIn1 = 1M1 = 1 kg = 1000 gL1 = 2 m = 200 cmT1 = 1 scgsn2 = ?M2 = 1 gL2 = 1 cmT2 = 1 sSolutionNow,1 Joule = 107 ergThus, To check the correctness of a given physical equationPrinciple of homogeneityA physically correct equation is always dimensionally correct i.e. the dimensions of all the terms on both sides of the equation are the same.Ex. Check whether the following equation is dimensionally correct.S = ut + ½ at2SolutionThe dimensions of s = [M0LT0]The dimensions of ut = [M0LT-1] [M0L0T] = [M0LT0]The dimensions of at2 = [M0LT-2] [M0L0T] 2 = [M0LT-2] [M0L0T2] = [M0LT0]Since the dimensions of all the terms on both sides of the given equation are same hence it is dimensionally correct. To derive a relationship between different physical quantitiesEx. The time period ‘T’ of oscillation of a simple pendulum depends upon the mass of the bob ‘m’, length of the pendulum ‘l’ and acceleration due to gravity at the place ‘g’. Derive an expression for its time period using dimensional analysis.Solution Let ………………… Where K is a dimensionless constant.Writing the dimensional formulae on both sides we get,Comparing the dimensions both sides we get,a = 0, -2c = 1 or c = -½b +c = 0 or b + (-½) = 0 or b = ½a = 0, b+c = 0 and -2c = 1Solving the equations we geta = 0, b = ½ and c = -½Putting the values of a, b and c in eq 1 we get, This is the required relation.