Dimension Analysis:-

The fundamental quantities, irrespective of the system of units used for their measurement and their corresponding quantitative values, characterize different independent groups or classes of basic physical quantities. In other words, a fundamental quantity belongs to a class of physical quantities of its own kind only, each of them having no dependence with other fundamental quantities. The attribute that is common to a class of physical quantities is what is called their dimensionality.

Most physical quantities can be expressed in terms of combinations of five basic dimensions. These aremass (M), length (L), time (T), electrical current (I), and temperature, represented by the Greek letter theta ().These five dimensions have been chosen as being basic because they are easy to measure in experiments. Dimensions aren't the same as units. For example, the physical quantity, speed, may be measured in units of meters per second, miles per hour etc.; but regardless of the units used, speed is always a length divided a time, so we say that the dimensions of speed are length divided by time, or simply L/T. Similarly, the dimensions of area are L2since area can always be calculated as a length times a length. For example, although the area of a circle is conventionally written asr2, we could write it asr (which is a length) r (another length).Dimensions of a physical quantity are the powers to which the fundamental units be raised in order to represent that quantity.A dimension deal with qualitative part of measurement.By international agreement a small number of physical quantities such as length, time etc. are chosen and assigned standards. These quantities are called base quantities and their units as base units. All other physical quantities are expressed in terms of these base quantities. The units of these dependent quantities are called derived units.

The standard for a unit should have the following characteristics.(a) It should be well defined.(b) It should be invariable (should not change with time)(c) It should be convenient to use(d) It should be easily accessibleThe 14th general conference on weights and measures (in France) picked seven quantities as base quantities, thereby forming theInternational System of Unitsabbreviated as SI (System de International) system.Dimensional Formula and Dimensional Equation:-Dimensional formula of a physical quantity is the formula which tells us how and which of the fundamental units have been used for the measurement of that quantity.An equation written in the following manner is called dimensional equation.Area = [M0L2T0]How to Write Dimensions of Physical Quantities:-Dimensions of a physical quantity can be determined as follows:(a) Write the formula for that quantity, with the quantity on L.H.S. of the equation.(b) Convert all the quantities on R.H.S. into the fundamental quantities mass, length and time.(c) Substitute M,L, and T for mass, length and time respectively.(d) Collect terms of M,L and T and find their resultant powers (a,b,c) which give the dimensions of the quantity in mass, length and time respectively.Base quantities and their units:-The seven base quantities and their units are,

Base quantityUnitSymbol

LengthMeterM

MassKilogramKg

TimeSecondSec

Electric currentAmpereA

TemperatureKelvinK

Luminous intensityCandelaCd

Amount of substanceMoleMole

Derived units:-We can define all the derived units in terms of base units. For example, speed is defined to be the ratio of distance to time.Unit of Speed = (unit of distance (length))/(unit of time)= m/s = ms-1(Read as meter per sec.)

SOME DERIVED SI UNITS AND THEIR SYMBOLS:-QuantityUnitSymbolExpress in base units

ForcenewtonNKg-m/sec2

WorkjoulesJKg-m2/sec2

PowerwattWKg-m2/sec3

PressurePascalPaKg m-1/S2

Note:The following conventions are adopted while writing a unit. Even if a unit is named after a person the unit is not written capital letters. i.e. we write joules not Joules. For a unit named after a person the symbol is a capital letter e.g. for joules we write J and the rest of them are in lowercase letters e.g. seconds is written as s. The symbols of units do not have plural form i.e. 70 m not 70 ms or 10 N not 10Ns. Not more than one solids is used i.e. all units of numerator written together before the / sign and all in the denominator written after that. i.e. It is 1 ms-2or 1 m/s-2not 1m/s/s. Punctuation marks are not written after the unit e.g. 1 litre = 1000 cc not 1000 c.c.It has to be borne in mind that SI system of units is not the only system of units that is followed all over the world. There are some countries (though they are very few in number) which use different system of units. For example: the FPS (Foot Pound Second) system or the CGS (Centimeter Gram Second) system.

Dimensions:-The unit of any derived quantity depends upon one or more fundamental units. This dependence can be expressed with the help of dimensions of that derived quantity. In other words, the dimensions of a physical quantity show how its unit is related to the fundamental units.To express dimensions, each fundamental unit is represented by a capital letter. Thus the unit of length is denoted by L, unit of mass by M. Unit of time by T, unit of electric current by I, unit of temperature by K and unit of luminous intensity by C.Remember that speed will always remain distance covered per unit of time, whatever is the system of units, so the complex quantity speed can be expressed in terms of length L and time T. Now, we say that dimensional formula of speed is LT-2. We can relate the physical quantities to each other (usually we express complex quantities in terms of base quantities) by a system of dimensions.Dimension of a physical quantity are the powers to which the fundamental quantities must be raised to represent the given physical quantity.Example:-Density of a substance is defined to be the mass contained in unit volume of the substance.Hence, [density] = ([mass])/([volume]) = M/L3= ML-3So, the dimensions of density are 1 in mass, -3 in length and 0 in time.Hence the dimensional formula of density is written as[]= ML-3T0It is to be noted that constants such as , or trigonometric functions such as sin wt have no units or dimensions because they are numbers, ratios which are also numbers. While stating the dimensions of a body, only the powers of fundamental units should be mentioned. While starting the dimensional formula the fundamental units have to be raised through certain powers. The dimensional formula of a quantity represents its qualitative nature only. Therefore, it must be enclosed inside brackets [MaLbTc]. Dimensions of a physical quantity are independent of the system of units. Quantities having similar dimensions can be added to or subtracted from each other. Multiplication/division of dimensions of two physical quantities (may be same or different) results in production of dimensions of a third quantity. Dimensions of a physical quantity can be obtained from its units and vice-versa. Two different physical quantities may have same dimensions. Dimensional analysis permits conversion of quantities from one system to another only in their fundamental values. Therefore, values of n1must be noted when the unit is fundamental. In case a relation contains more than one terms on either side of equation, the relation will be correct only if all the terms involved in that relation have same dimensions.

Uses of Dimensional Analysis:-Dimensional analysis has been put to following three uses:-

Limitations of Dimensional Analysis:-Owing to the process of its development, the process of dimensional analysis is subjected to the following limitations:(a) It gives no information regarding the constant of proportionality involved in the equation.(b) It cannot be used to derive an expression involving trigonometric functions.(c)It cannot be used to derive an expression involving exponential functions.(e)It cannot be used to derive an expression for a physical quantity which depends upon factors more than three.

Question 1:-Which of the following is the dimensionless quantity:(a) force of surface tension (b) strain(c) stress (d) co-efficient of viscosityQuestion 2:-Dimensions of velocity gradient are same as that of(a) time period (b) frequency(c) angular acceleration (d) accelerationQuestion 3:-Out of the following pairs, only one pair does not have identical dimensions. It is(a) angular momentum and Plancks Constant(b) moment of Inertia and moment of force(c) work and torque(d) impulse and momentumQuestion 4:-The dimension of Plancks constant is same as that of:(a) angular momentum (b) linear momentum(c) work (d) coefficient of viscosityQuestion 5:-If K represents kinetic energy, V velocity and T time, and these are chosen as the fundamental units then, the units of surface tension will be:(a) KV-2T-2 (b) K-1V-2T-1 (c) KV-1T-1 (d) KV-2T2

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Applications of DimensionsThe physical quantities those can be expressed in terms of fundamental physical quantities are called derived physical quantities. All the physical quantities of interest can be derived from the base quantities. The power (exponent) of base quantity that enters into the expression of a physical quantity, is called the dimension of the quantity in that base.Broadly speaking, dimension is the nature of a Physical quantity. Understanding of this nature helps us in many ways.Following are some of the applications of the theory of dimensional analysis in Physics:

(a) To find the unit of a given physical quantity in a given system of units:-By expressing a physical quantity in terms of basic quantity we find its dimensions. In the dimensional formula replacingM, L, Tby the fundamental units of the required system, we get the unit of physical quantity. However, sometimes we assign a specific name to this unit.Problem 1:-Find the dimension of Force.Solution:-Force is numerically equal to the product of mass and accelerationForce = mass x accelerationor [F] = mass x velocity/time= mass x (displacement/(time)2) = mass x (length/(time)2)= [M] x [LT-2] = [MLT-2]Its unit in SI system will be Kgms-2which is given a specific name newton (N).Similarly, its unit in CGS system will be gmcms-2which is called dyne.

(b) To find dimensions of physical constants or coefficients:-The dimension of a physical quantity is unique because it is the nature of the physical quantity and the nature does not change. If we write any formula or equation incorporating the given physical constant, we can find the dimensions of the required constant or co-efficient.Problem 2:-Find the dimension of gravitational constantG.Solution:-From Newtons law of Gravitation, the exerted by one mass upon another isF=G (m1m2)/r2or G=(Fr2)/(m1m1)or [G] = [MLT-2][L2] / ([M][M]) = [M-1L3T-2]We can find its SI unit which is m3/Kgs2.(c) To convert a physical quantity from one system of units to another:-This is based on the fact that for a given physical quantity, magnitude x unit = constantSo, when unit changes, magnitude will also change.Problem 3:-Convert one Newton into dyne.Solution:-Dimensional formula for Newton = [MLT-2]Or 1 N = 1 Kg m/s2; But 1 kg = 103g and 1 m = 102cmTherefore 1 N = ((103g)(102cm))/s2= 105g cm/s2= 105dyne(d) To check the dimensional correctness of a given physical relation:-This is based on the principle that the dimensions of the terms on both sides on an equation must be same. This is known as theprinciple of homogeneity. If the dimensions of the terms on both sides are same, the equation is dimensionally correct, otherwise not.Caution:It is not necessary that a dimensionally correct equation is also physically correct but a physically correct equation has to be dimensionally correct.Problem 4:-Consider the formula, T=2(l/g)Where T is the time period of oscillation of a simple pendulum in a simple harmonic motion, l and g are the length of the pendulum and gravitational constants respectively. Check this formula, whether it is correct or not, using the concept of dimension.Solution:-As we know [g] = [LT-2]Therefore [T] = (([L])/([LT-2])) = [T] sThus the above equation is dimensionally correct (homogenous) and later you will come to know that it is physically also correct.____________________________________________________________________________________________________Problem 5:-Consider the formula s=ut -1/3 at2. Check this formula whether it is correct or not, using the concept of dimension.Solution:-Dimensionally[L] = [LT-1] [T] [LT-2] [T2]=>[L] = [L] [L]In this case also the formula is dimensionally correct but, you know that it is physically incorrect as the correct formula is given byS = ut + at2(e) As a research tool to derive new relations:-One of the aims of scientific research is to discover new laws relating different physical quantities. The theory of dimensions (in the light of principal of homogeneity) provides us with a powerful tool of research in the preliminary stages of investigation [It must be again emphasized that mere dimensional correctness of an equation does not ensure its physical correctness

Functions of dimensionless variables are dimensionless. Dimensionless functions must have dimensionless arguments. Using dimensional analysis we cannot find the value of dimensionless constant. We cannot derive the relation containing exponential and trigonometricfunctions. It cannot inform that whether a quantity is scalar or vector. It cannot find the exact nature of plus or minus, connecting two or more terms informula. The relation containing more than three physical quantities cannot be derivedusing dimensional analysis.

Limitations of the theory of dimensions:-The limitations are as follows:-(i)If dimensions are given, physical quantity may not be unique as many physical quantities have the same dimension. For example, if the dimensional formula of a physical quantity is [ML2T-2] it may be work or energy or even moment of force.(ii)Numerical constants, having no dimensions, cannot be deduced by using the concepts of dimensions.(iii)The method of dimensions cannot be used to derive relations other than product of power functions. Again, expressions containing trigonometric or logarithmic functions also cannot be derived using dimensional analysis, e.g.s = ut + 1/3 at2 or y = a sincot or P= P0exp[(Mgh)/RT]cannot be derived. However, their dimensional correctness can be verified.(iv)If a physical quantity depends on more than three physical quantities, method of dimensions cannot be used to derive its formula. For such equations, only the dimensional correctness can be checked. For example, the time period of a physical pendulum of moment of inertia I, mass m and length l is given by the following equation.T = 2(I/mgl) (I is known as the moment of Inertia with dimensions of [ML2] through dimensional analysis), though we can still check the dimensional correctness of the equation (Try to check it as an exercise).(v)Even if a physical quantity depends on three Physical quantities, out of which two have the same dimensions, the formula cannot be derived by theory of dimensions, and only its correctness can be checked e.g. we cannot derive the equation.

Question 1:-If force, time and velocity are treated as fundamental quantities then dimensional formula of energy will be,(a) [FTV] (b) [FT2V] (c) [FTV2] (d) [FT2V2]Question 2:-Which one of the following physical quantities do not have the same dimensions.(a) Pressure, Youngs Modulus, Stress (b) Electromotive Force, Voltage, Potential(c) Heat, Work, Energy (d) Electric Dipole, Electric Field, FluxQuestion 3:-The pairs having same dimensional formula-(a) Angular Momentum, Torque(b) Torque, Work(c) Planks Constant, Boltzmanns Constant(d) Gas Constant, PressureQuestion 4:-If F = ax + bt2+ c where F is force, x is distance and t is time. Then what is dimension of axc/bt2?(a) [ML2T-2] (b) [MLT-2] (c) [M0L0T0] (d) [MLT-1]Question 5:-The dimensional formula for angular momentum is,(a) [ML2T-2] (b) [ML2T-1] (c) [MLT-1] (d) [M0L2T-2]

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