AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

5

Der ived Physical Quantities : The physical quantities which can be derived

from fundamental quantities are called derivedquantities.

Example : Area, Volume, Speed, Velocity etc.Unit :

Unit of measurement of a physical quantityis the standard reference of the same physicalquantity which is used for compar ison of thegiven physical quantity.

The meaning of the measurement of a physicalquantity is to find out the number of times its unit iscontained in that physical quantity. Therefore, theprocess of measurement of a physical quantityinvolves:

i) The selection of unit andii) to find out the number of times that unit is

contained in the physical qunatity.For example, if we are asked to measure the

length of a table, the unit to be selected must be thatof length.

Suppose that we use metre as the unit. We placea metre rod successively along the length of the tableand find out the number of times the metre rod iscontained in the length of the table.

Suppose that the length of the table is coveredby the metre rod in three successive placings.

INTRODUCTIONUnder standing of di f f er ent physical

phenomenon requir es the need to measur erelevent physical quantities. Physical quantitiesare those which can be defined and measured.Wi thout measur ement t her e can be noadvancement in physics. Exper iment almeasurements are essential to ver ify theor iticallaws. We use number of physical quantities (indaily life) like length, time, area, volume etc. todescr ibe an event or a phenomenon. Formeasur ing a physical quanti ty, a standar dreference is necessary. This reference is knownas unit1.1 PHYSICAL QUANTITIES

All quantities in terms of which laws ofphysics can be expressed and which can bemeasured directly or indirectly are called physicalquantities

Physical quantities can be classified into two types1) Fundamental Physical Quantities2) Derived Physical Quantities

Fundamental Physical Quantity :A physical quantity which is independent

of any other quantity is called fundamentalphysical quantity.

Ex : Mass, length, time, electric current,thermodynamic temperature, amount of substanceand luminous intensity are taken as fundamentalphysical quantities

UNITS AND DIMENSIONS

h Fundamental and Derived physical quantitiesh SI units

CHAPTER

h Dimensions, Dimentional formulaeh Dimentional analysis

1James Watt was born in Greenock in 1736, the son of a ships chan-dler (trader in canvas, etc). In his late teens he went to London tolearn to be a mathematical and philosophical instrument maker,and when he returned to Glasgow he got a job making instrumentswith Glasgow University, who gave him accomodation and a work-shop. Watts engines were initially used for pumping water fromcornish tin and copper mines. In 1882, 63 years after Watts death, theBritish Association gave his name to the unit of electrical power -and today James Watts name is to be found written on almost everylightbulb in the world. James Watt1736-1819

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

6

Then, 3 is called the numerical value of lengthof the table and metre as the unit of length. We maywrite:

Length of the table = 3 1m =3m

It may be pointed out that it is not sufficient to saythat the length of the table is 3. The unit of the physicalquantity has also to be stated along with the result ofthe measurement. In general,

measure of a physical quantity = numerical valueof the physical qunatity x size of its unit

if u is the size of the unit and n is the numericalvalue of the physical quantity X (the quantity to bemeasured) for the selected unit, then measure of thephysical quantity X= n u

It follows that if the size of the chosen unit issmall, then the numerical value of the quantity willbe large and vice- versa. It is obvious that the measureof the physical quantity is always the same i.e.

n u = constantIf n1 is the numerical value of the physical quantityfor a unit u1 and n2 for a unit u2, then

n1u1 = n2u2

1.2 SYSTEM OF UNITSThere are four main systems of units, namely.

1. CGS (Centimeter, Gramme or Gram, Second)2. FPS (Foot, Pound, Second)3. MKS (Meter, Kilogram, Second)4. SI (System International d' Units)

To measure the fundamental base quantitieslength, mass and time, there are three systemsof units. These systems are called F.P.S. system(British system), C.G.S system (Metric system) andM.K.S. system. The basic units in these systems aregiven in table.

System UnitsLength Mass Time

F.P.S foot pound secondC.G.S centimetre gram secondM.K.S metre kilogram second

An international organization, the ConferenceGeneral des Poids at Measures, or CGPM isinternationally recognised as the authority on thedefinition of units. In English, this body is known as"General Conference on Weights and Measures".The System International de Units, or SI units,was set up in 1960, by the CGPM.Requirements of a Standard Unit :

A standard unit must have following featuresto be accepted world wide.(i) It should be of suitable size i.e. neither too large

nor too small in comparison to the quantity tobe measured.

(ii) It should be very well defined.(iii) It should be independent of time and place.(iv) It should be easily available so that all

laboratories can duplicate and use it as perrequirement.

(v) It should be independent of physical conditionslike temperature, pressure, humidity etc.In SI, there are seven fundamental (base)

physical quantities, length, mass, time, electriccurrent, thermodynamic temperature, luminousintensity and quantity of substance. In addition tothese, two more physical quantities, plane angle andsolid angle are introduced as supplementaryfundamental quantities. Fundamental (base)quantities, their units and dimensional representationare given in table.

Fundamental (base)Quantity

Unit Symbol

Length metre m

Mass kilogramsTime second

Electric current ampere A

Thermodynamictemperature

kelvin K

Luminous Intensity candela

Quantity ofsubstance

mole

kg

cd

mol

Fundamental (base) Quantities in SI

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

7

Plane angle radian rad

Solid angle steradian sr

Supplementary Fundamental (base)Quantities in SI

1.3 DEFINITIONS OF ELEMENTARY(BASIC) UNITS IN SI

1. kilogram(kg) : A kilogram is equal to the massof a cylinder (having diameter = height) ofplatinum irridium alloy (90% platinum and 10%irridium) kept in the International Bureau ofWeights and Measures at Sevres near Paris ofFrance.

2. metre(m) : metre is defined as the length ofpath travelled by light in vacuum in (1/299, 792,458) part of a second.

* Note :In above case, standard of length is definedin terms of unit of time. Hence fundamentalunits are no more independent of each other.

3. second(s) : A second is equal to the duration of9,192,631,770 periods of the radiat ioncorresponding to the unperturbed transitionbetween the two hyperfine levels of groundstate of cesium (133) atom.

4. ampere(A) : An ampere is defined as thatconstant current which when flowing in twostraight parallel conductors of infinite length andof negligible area of cross-section and placedone metre apart in vacuum would producebetween them a force of 2 10-7 newton permetre of length.

5. kelvin(K) : A kelvin is defined as the fraction1

273.16 of the thermodynamic temperature of

the triple point of water. It may be noted that inSI., the triple point of water is taken as afundamental fixed point having a temperatureof 273.16 kelvin by definition.

6. candela(cd) : A candela is as the luminousintensity, in a given direction, of a surface thatemits monochromatic radiation of frequency540 1012Hz and that has the radiant intensityin that direction of 1/683 watt per steradian.

7. mole(mol) : A mole is the amount of substanceof a system, which contains as many elementaryentities as there are atoms in 0.012 kg of carbon-12.The number of atoms in 0.012 kg of C - 12atom is 6.023 1023 and is called the Avogadronumber.

8. radian(rad) : Radian is defined as the anglesubtended at the centre of a circle by an arcwhose length is equal to the radius of the circle.

2 radians = 3600, 1 radian = 3602 = 57

017|45||

9. Steradian(sr) : Steradian is the solid anglesubtended at the centre of a sphere by itssurface, the area of which is equal to the squareof the radius. If d be the solid angle subtendedat the centre of a sphere of radius r by a part its

surface of area S , then 2Sd

r

steradians.

The solid angle subtended at the centre by theentire surface area 2(S 4 R ) of a sphere will

be 2

24 R 4 sr.

R

1.4 FUNDAMENTAL UNITSThe units of the fundamental physical quantities

are called fundamental unitsEx : metre, kilogram, second1.5 DERIVED UNITS

Derived units are the units of the derivedquantities.

newton N=kg m/s2force

joule J=Nm=kg m2/s2Energy, Work, Heat

watt W=J/s=N m/s=kgm2/s3Powerpascal Pa=N/m2=kg/ms2Pressure, Stress

hertz Hz = s-1Frequency

coulomb C = AsElectric chargevolt V=J/Celectric potentialfarad F=coulomb/volt=A2s2/NmCapacitanceohm 2V / A J / A s Resistancehenry H=J/A2=Nm/A2Self inductance

weber WbMagnetic flux

tesla T=Wb/m2

Derived unit Symbol of UnitPhysical Quantity

Magnetic flux

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

8

1.6 MULTIPLES AND SUBMULTIPLES OF SI UNITSTo measure very low or very high values of physical quantities, prefixes are used to represent them.

For example 103m is written as 1 milli metre, 109 second is written as nano second and 106 watt is writtenas 1 Mega watt. The prefixes used to represent multiples and submultiples of SI units are given in table.

Multiples and submultiples of SI units

Factor Prefix Symbol Factor Prefix Symbol

1024 Yotta Y 10-1 deci d

1021 Zetta Z 10-2 centi c

1018 Exa E 10-3 milli m

1015 Peta P 10-6 micro m

1012 Tera T 10-9 nano n

109 Giga G 10-12 pico p

106 Mega M 10-15 femto f

103 Kilo k 10-18 atto a

102 Hecto h 10-21 zepto z

101 Deka da 10-24 yocto y

1.7 RULES FOR USING SYMBOLS FOR SI UNITS/PREFIXES1. Full names of units even if they are named after

scientists should not be written with initialcapital letter Ex : newton but not Newton

2. Symbols for a unit named after a scientist shouldhave a capital letter.Ex : N for newton; W for watt; A forampere

3. Punctuation marks should not be used aftersymbol of unit Ex : 100 kg but not 100 kg.100 mm but not 100 mm. (or) 100 m.m.

4. Symbols for units should not take plural formEx : 100 newtons is 100 N but not 100 Ns 50meters is 50 m but not 50ms and but not 50 mts.

1.8 DIMENSIONSDef ini t ion : The demensions of a physicalquantity are the powers (exponents) to which thefundamental (base) quantit ies are raised torepresent that physical quantity.

1.9 DIMENSIONAL FORMULADefinition : The expression showing the powersto which the fundamental quantities must beraised to represent a physical quantity, is calleddimensional formula.

The fundamental physical quantities; namely,mass, length time, temperature, electric current areindicated by the symbols [M], [L], [T], [K] or [ ],[I] or [A] etc. respectively.Examples:5. Force = mass acceleration ; So dimensional

formula of force is [ MLT2 ] =[M1 L1 T2]In the dimensional formula of force, dimensionof mass = 1, dimension of length = 1, dimensionof time = 2.

6. Work =F.S = m a S; So dimensional formulaof work = M LT2 L = [ M1L2T2]In the dimensional formula of work, dimensionof mass = 1, dimension of length = 2, dimensionof time = 2.

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

9

1.10 DIMENSIONAL EQUATIONDefinition : I t is an equation, equating the physical quantity with its dimensional formula.Example : Force [F] = [MLT2] is dimensional equation of force. Dimensional Formulae of Physical QuantitiesSN Physical quantity Formula Dimensional S.I unit Nature scalar

formula (or) vector 1. Volume lengthxbreadthxheight ML3T m3 S

2. Density mass / volume ML3T kg m3 S

3. linear density mass / length ML1T kg m1 S

4. Relative densitydensity of substance

densityof water MLT ---- S

5. a) Energy density Energy / volume ML1T2 N m2

b) Pressure force/Area ML1T2 N m2

c) Strees, youngs, Stress / strain ML1T2 N m2

Rigidity, Bulk

modulie of elasticity

6. a) Velocity displacement / time MLT1 m s1 V

b) Speed distance / time MLT1 m s1 S

c) acceleraion force / mass MLT2 m s2 V

7. Hubbless Constant (H) Velocity / Distance T-1 s-1 -

8. a) Areal velocity Area / time ML2T1 m2 s1 V

b) Coefficient of KinematicCoefficient of viscosity

density ML2T1 S

viscosity

9. a) Momentum (linear) mass velocity ML1T1 kg m s1 V

b) Impulse Force time ML1T1 kg m s1 V

10. a) Moment of force Force displacement ML2T2 N m V

or moment of

couple or torque

b) work Force displacement ML2T2 joule S

c) All energies ML2T2 joule

11. Power Force velocity ML2T3 watt S

12. Moment of Inertia I = MK2 ML2T kg m2 Tensor

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

10

SN Physical quantity FormulaDimensional S.I unit Nature scalarformula (or) vector

13. Angular displacement d / r MLT rad V

14. a) Angular velocity = / t MLT1 rad/s V

b) Velocity gradient dv/dx MLT1 s1 V

c) Frequency = c / MLT1 Hz S

d) decay constant number of fissions/ time MLT1 s1 S

15. Angular acceleration d /dt MLT2 rad s2 V

16. a) Angular I ML2T1 kg m2 s1 VMomentum

b) Planks constant h = Energy

frequency ML2T1 J s S

c) Angularimpulse t ML2T1 J s V

17. Gravitational constant2

2

Fdm

M1L3T2 N m2 kg-2 S

18. Gravitational field intesityFm

0 1 2M L T N kg -1 V

19. a) GravitationalWm

ML2T2 J kg1 S

Potentialb) Latent heat

Qm Q = Heat energy S

c) Calorific valueQm S

20. a) Force constant F/L M T2 N m1 Sb) Spring Constant Sc) Surface Tension S

21. Compressibility1

Bluk modulus M1LT2 Pa1 S

22. Coefficient ofFdvAdx

= ML1T1 Pa s S

Viscosity Pressure Time

23. a) Coefficients of linear,

2 1

1 2 1

l - l=

l t - t MLTK1 K1 S

areal and volume expansion Coefficient of real , apparant expansions

Volume coefficient, Pressure coefficient

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

11

SN Physical quantity Formula Dimensional S.I unit Nature scalar formula (or) vector

24. a) Thermal capacityHeat

Temperature ML2T2K1 J K-1 S

b) EntropyHeat

Temperature ML2T2K1 J K-1 S

(c) Boltzman ConstantPVNT

ML2T2K1 J K-1 S

25. a) Specific heatQ

mL2T2K1 J kg1K-1 S

b) Specific gas constantpr

dT L2T2K1 J kg1K-1 S

26. a) Universal gas constant R = PVnT

ML2T2K1mol1 J mol1 K1 S

b) Molar specific heat CPCV = R ML2T2K1mol1 J mol1 K1 S

27. Stefan's constant 4E

AtMLT3K4 W m2 K4 S

28. Coefficient of Thermal 2 1Qd

A t MLT3K1 W m1K1 S

conductivity

29. a) Thermal resistanceL

KAM-1L2T3K1 W1K S

b) Temperature gradient L L-1K1 K m-1 V

30. Thermal conductanceKAL

ML2T3K1 WK1 S

31. Permeability2

2

Fdm

= 2IF

MLT2I2 H m-1 S

32. Pole strength m = IL IL A m S

33. Magnetic moment M = 2l m I L2 A m2 V

34. Magnetic Induction B = FIL

M T2 I1 tesla V

35. a) Intensity of H = B IL

1 A m1 V

Magnetic fieldb) Intensity of I =

MV

IL1 A m1 VMagnetisation

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

12

SN Physical quantity Formula Dimensional S.I unit Nature scalarformula (or) vector

36. Magnetic flux BA = WI

ML2T2I1 weber S

37. a) Intensity of sound = PowerArea

M T3 watt / meter2 S

b) Intensity of Heatc) Emissive power

38. Spectral emissive power 3Power

LML1T3 watt/meter3 S

39. Pressure gradiantPr essureLength = P/L ML

2T2 Pa m1 V

40. Electric current i MLTI1 ampere S

41. Eelectric charge Q=it IT coulomb S

42. Intensity of electric field E = F/Q MLT3I1 N C-1 V

43. a) Electric PotentialWQ ML

2T3I1 volt S

b) Electro motive force (emf)

44. Electric Capacity C = QV

= WQ2

M1L2T4I2 farad S

45. Electric resistance R = VI

= 2Power

IML2T3I2 ohm S

46. Electric Conductance K = 1R

M1L2T3I2 mho or S

siemen

47. Resistivity = RAL

ML3T3I2 ohm meter S

48. Conductivity = 1 M

1L3T3I2 Sm-1 S

49. Permitivity 0 = 2

2

QFd

M1L3T4I2 farad/metre S

50. a) Self Inductance

b) Mutual inductance = 2energy

current =

/e m f

d i d tM L2T2I2 henry S

51. Linear charge densityql L

-1 T I C m-1 S

52. Surface charge densityqA

L-2 T I C m-2 S

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

13

SN Physical quantity Formula Dimensional S.I unit Nature scalarformula (or) vector

53. Volume charge densityqV

L-3 T I C m-3 S

54. Current densityIA

I L-2 A m-2 V

55. Electro chemical equivalentmit

1 1MI T 1 1kg A s S

56. Decay ConstantdNdt

T-1 s-1 S

57. Mobility dv

EM-1L0T2I A s2 kg-1 S

1.11 DIMENSIONAL CONSTANTS ANDDIMENSIONLESS QUANTITIES.DIMENSIONAL CONSTANT :

Constants having dimensional formula are calleddimensional constants.

Ex : Plancks constant, Universal gravitationalconstant Universal gas constant, Boltz mannsconstant, Stefans constant, Wiens constant, Velocityof light, are Dimensional constants.Dimension less quantities :Physical quant i t ies having no dimensionalformula are called dimension less quantities.

Ex : Angle, Strain, Relative Density,coefficientof friction, coefficient of restitution, poissons ratiohave zero dimensions and are Dimensionlessquantities. Numbers, Ratio of same quantities, ,have no dimensional formula because they will notdepend on fundamental quantities

1.12 PRINCIPLE OF HOMOGENITY OFDIMENSIONS

Statement : Only physical quantities having samedimensions can be added, subtracted or can beequated.

If x = y + z is dimensionally correct and if xrepresents the physical quantity, the force, then y andz also must represent the same physical quantity i.e.,force. It means that the terms on both sides of adimensional equat ion should have samedimensions. This is called principle of homogeneityof dimensions.

Problem : 1.1Verify the correctness of the equation v=u+at .In the dimensional form

1 1 1 1 1 2 1 1 1L T L T L T T L T The dimension for L on both sides is 1The dimension for T on both sides is 1Hence from the principle of homogenity of dimensionsthe given equation is correct.

1.13 APPLICATIONS OF DIMENSIONALANALYSIS

1) Dimensional formulae can be used toconvert one system of units into anothersystem

2) Dimensional formulae can be used to checkthe correctness of an equation

3) Dimensional formulae can be used toderive relationship among different physicalquantities

4) Dimensional formulae can be used to findunit of a given physical quantity

5) Dimensional formulae can be used todesign our own new system of units

1. To conver t one system of units into anothersystemThis is based on the fact that for a given

physical quantity, numerical value unit = constant. i.e., N1U1= N2U2.

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

14

So when units change, numerical value alsomay change. To convert a physical quantity from onesystem to the other, we write its units in terms ofmass, length and time and then convert into anothersystem. Problem : 1.2

Convert newton into dyneSol. [F] = [MLT2] ; 1 newton = 1 (kg) (m) (s)-2

= 1 (103g) (102cm) (s)-2 = 105 (g) (cm) (s)-2

1newton = 105 dynesAlternate Method : Let the dimensional formula of aphysical quantity be a b c[M L T ].Let n1 be its numerical value in one system in whichunits of fundamental quantities are M1,L1 and T1. Thenthe numerical value 2n in another system in which unitsof fundamental quantities are M2,L2 and T2 is given by

a b c a b c2 2 2 2 1 1 1 1n M L T n M L T

a b c1 1 1

2 12 2 2

M L Tn nM L T

Problem : 1.3Convert the unit of work done from MKS system to CGSsystem

Sol. Dimensional formula of energy is 1 2 2M L T .Let IJ n ergs n J / ergs

1 2 21 1 1

2 12 2 2

M L Tn nM L T

2 21kg 1m 1sn1g 1cm 1s

21

nn

n

2 21000 g 100 cm 1sn1 g 1 cm 1s

2 2n [1000] [100] [1] = 107

71 joule 10 ergs

2. To check the correctness of a given equation :This is based on the principle of homogeneity.

i.e., the dimensions of the terms on both sides of agiven equation must be same.

If the dimensions of each term on both sides aresame, the equations is dimensionally correct,otherwise not correct.

Problem : 1.4

Check the correctness of the formula S = ut + 13 at2

where S is the distance, u is velocity, a is accelerationand t is time.

Sol. Dimensionally,LHS = [L]RHS = [LT-1] [T] + [LT-2] [T]2 = [L] + [L] = [L]

Since, the dimension of each term on both sides are same,the given equation is dimensionally cor rect .

* Note :However, we cannot say anything about the physicalcorrectness of the formula. We know that the correctformula is

S = ut +12

at2

Hence, we conclude that dimensional correctness is noguarantee for physical correctness of the formula.However, dimensional incorrectness guarantees thephysical incorrectness of the formula. Inspite of theabove limitation, the method is still helpful to a greatextent.

Problem : 1.5

Consider the equation T 2g

and check whetherit is cor rect or not.

we know, dimensional formulae of T, and g are [T],[L], [LT2] respectively.

Dimensional formula of 22L T T

g LT

Dimensional formula of left side is also [T]. As 2 is adimensionless quantity, the dimensional formula of lefthand side of the equation is equal to dimensional formulaof right hand side. Hence we conclude that the givenequation is a correct one.

3. To deduce relation between the physicalquantities :This is also based on the principle of

homogeneity. If one knows the quantities on whicha particular physical quantity depends and if oneguesses that this dependence is of product type, themethod of dimensional analysis may be helpful inthe derivation of the relation.

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

15

Problem : 1.6

Derive an expression for the time period (T) of a simplependulum which may depend upon the mass(m) of thebob, length (l) of the pendulum and acceleration due togravity (g).

Sol. Let T = kma lb gc where k is a dimensionless constant.Writing the equation in dimensional form, we have

[M0 L0 T1] = [M]a [L]b [LT-2]c = [Ma Lb+c T-2c]Equating exponents of M, L and T on both sides, we get

a = 0, b + c = 0, 2c = 1.Solving the eq., we get a = 0, b = 1/2, c = -1/2

Hence, T kg

, where k is constant.

*Problem : 1.7Derive an expression for the velocity of sound (V) whichmay depend upon the modulus of elasticity (E) of themedium, and density (d) of the medium.

Sol. Let V= a bkE d where k is the proportionality constant.

Dimensional formula of v, E and d are0 1 1 2 3 0V M LT ; E ML T ;d ML T

As k has no dimensionsa b0 1 1 2 3 0M LT ML T ML T

a b a 3b 2a0 1M LT M L T Equating dimensions of M, L and T on either side of theequationWe get, a +b = 0; a3b = 1 and 2a =12a = 1 a = 1/2; a+b = 0 b = 1/2

V = k E1/2 d1/2; V = k EdExperimentally the value of k is found to be one.

EVd

*Problem : 1.8

I f E, M,L and G denote energy, mass, angular momen-tum and universal Gravitational constant respectively,

prove that 2

5 2EL

M G is a dimensionless quantity.

Sol. Taking dimensional formulaeenergy (E) = ML2T2

mass (M) = ML0T0

Angular Momentum (L) = ML2T-1Universal Gravitational constnat

(G) = M1L3T2

Substituting in 2

5 2EL

M G, we get

22 2 2 1 1 2 2 4 2 2

5 2 5 2 0 6 0 40 0 1 3 2

ML T ML T M L TM L TML T M L T

= A dimensionless quantity.

Problem : 1.9I f the equation of state of a gas is expressed as

2

aP V b RT

V where P is the pressure, V is

the volume and T the absolute temperature and a, b, Rare constnats, then find the dimensions of a and b.

Sol. By principle of homogentiy of dimensions P can addedto P only. It means

2a

V also gives pressure.

Dimension formulae for pressure (P) = M1L1T2 and Vol-ume (V) = M0L3T0.

Since 2a

V = pressure

1 1 2

20 3 0

a M L TM L T

1 1 20 6 0a M L T

M L T

a = M1L5T2

Similarly, b will have same dimensions as volumeV b = volume b = M0L3T0

*Problem : 1.10The dimensional formula of the product of two physi-cal quantities P and Q is ML2T2. The dimensional for-mula of P/Q is ML0T2. Then which physical quantitiesare represented by P and Q ?

Sol. Given P Q has dimensions ML2T2

P/Q has dimensions ML0T2

Hence, (PQ) 2 2 0 2P ML T ML TQ

P2 = M2L2T4

P = ML1T2 = Force ......(1)

given PQ

= M1L0T2 ; 1 2

0 2ML T ML TQ

1 2

0 2ML TQML T

= L = Displacement ...............(2)

P and Q are respectively the force and the displacement

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

16

*Problem : 1.11I f pressure P, velocity v and time T are taken as thefundamental (base) quantities, find the dimensional for-mula of force.

Sol. Let force = F = PaVbTc

(F) = M1L1T2, (P) = M1L1T2, (V) = M0L1T1,(T) = M0L0T1

MLTT2 = (ML1T2)a (M0L1T1)b(M0L0T1)c

MLT2 = MaLa+bT2a-b+c Comparing powers of M, L, T on both sidesa = 1, a + b = 1 b = 2 and 2a b + c = 2; 2 2 + c = 2; c = 2 Force (F) = PV2T2.

*Problem : 1.12 A gas bubble from an explosion under water, oscillateswith a period proportional to PabEc where p is thestatic pressure, is the density of water and E is thetotal energy of the explosion.Find the values of a, b,and c.

Sol. Let, T PabEc

Dimensional formula of-1 -2 -3 2 2p ML T ; ML ; E ML T

1 1 2 a 3 b 2 2 c T (ML T ) (ML ) (ML T ) Comparing powers of M, L and T on both sidesa + b + c = 0 -----(1)-a - 3b + 2c = 0 -----(2)

-2a - 2c = 1 ----- (3)

From (3) and (1) : a + c = 1 2

and 1b = 21-a +2c = 3 1 2a +c = - ;

2 3c = 11 c = 3

and 1 1 5a = - - = - 3 2 6

The values of a, b, c are respectively,,5 1 1

, , .6 2 3

*Problem : 1.13The period T0 of a planet above the sun of mass M ina circular orbit of radius R depends on M, R and Gwhere G is the gravitational constant. Find expressionfor time period by dimensional methods.(Hint : The solution is similar to above problem)

Ans : 3

0R

T = K GM

*Problem : 1.14The freguency n of a vibrating string depends up onits length l , linear density m (mass per unit length)and tension T in the string. Derive an expression for thefrequency of the string. (Solution is simlar to 1.12prob-lem)Ans.

k Tn = .l m

*Problem : 1.15

Derive an expression for the rate of flow of a liquidthrough a capillary tube. Assume that the rate of flow

depends on (i) pressure gradient Pl , (ii) The radius,

r and (iii) the coefficient of viscosity, . The value of

the proportionatity constant k = .8

(Solution is simlarto 1.12problem)

Ans. 4 4V p r pr = k k

t 8 l 8

l

4. To find unit of a given physical quantity:By using the formula of a physical quantity in

terms of basic quantities, we can find its dimensionalformula. So, in the dimensional formula, by replacingM, L and T by the fundamental units, we get theunit of physical quantity.

Example: Force = mass x acceleration[F] = [M] [LT2] = [MLT2]Hence, units of force will be kg ms2 which is

also called newton(N)

5. To design our own new system of dimensions:In place of M, L and T, we can take any other

three physical quantities as the fundamental physicalquantities with the condition that they should beindependent of each other and not derivable mutually.Suppose in a new system, force(F), length (L), andtime (T) are taken as fundamental physical quantitiesthen dimensional formulae of other quantities likemass, velocity, acceleration etc., can be derived interms of F, L and T.

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

17

*Problem : 1.16I f energy E, velocity v and time T are chosen as thefundamental quantities. Find dimensional formula ofsurface tension ?

Sol. S. T = MT-2, E = ML2T2, v = LT1, t = T1a b c S. T (E) (v) (t) and MT-2 =

(ML2T-2)a(LT-1)b(T)c

Comparing power of M, L and T on either sidea = 1, 2a + b = 0 b = - 2 -2a - b + c = - 2 c = - 2

1 2 2 S. T = E (V) (T)

Problem : 1.17Suppose you design your system of dimensions andtake veloci ty (V), Planck constant (h) andgravitational constant (G) as the fundamentalquantities. What will be the dimensions of length (L)in this system ?

Sol. We know,[V] = [LT-1], [h] = [ML2T-1],[G] = [M-1L3T-2] Let L = VahbGc

M0L1T0 = (LT1)a(ML2T1)b(M1L3T2)c

comparing the powers of M, L, T both sidesb c = 0 .......... (1) a + 2b + 3c = 1 ........... (2) a b 2c = 0 ............. (3)on solving (1), (2), (3) we get a = 3/2, b = 1/2, c = 1/2L = V3/2 h1/2G1/2

*Problem : 1.18I f unit of mass is taken as 1 kg, of time as 1 minute andthat acceleration due to gravity is taken as9.81 ms-2,what is the unit of energy ?

Sol. New unit of energy = E1New unit of mass M1 = 1kgNew unit of time T1 = 1 minute = 60 secNew unit of length = L1E1 = [M1 L1T1

-2]New unit of acceleration due gravity,g1 = 9.81 ms

2

g1 = L1T1-2; L1 = g1T1

-2 = 9.81 (60)2 meter

2 2 2 -2 -2

1 E = 1 kg (9.81 60 ) m (60) S

2 -21

1 9.81 60 60 9.81 60 60 E = kg m s60 60

5 2 -21 E = 3.464 10 kg m sec The new unit of energy 51E = 3.464 10 joule.

1.14 LIMITATIONS ON THE USES OFDIMENSIONAL ANALYSIS

There is no limitation as far as the first use(conversion of units) is concerned. As for the seconduse (checking the dimensional accuracy of the givenrelationship) is concerned, if both the LHS and RHSare dimensionless, we cannot know whether thequantities in the numerator and denominator are incorrect or reverse order. The rest of the uses havefollowing limitations.

(i) Proportionality constants cannot bedetermined by dimensional analysis.

(ii)Formulae containing non-algebraic functionsi.e., sin, cos, log, exponential etc. cannot be derived.

(iii)Formulae which contain two or more termsin the RHS cannot be derived.

Example : S = ut + 12

at2.

(iv)Formulae containg dimensional constantscannot be derived.

Example : F = 1 22Gm m

d(v) Dimensional analysis does not differentiate

between a scalar and a vector quantity.(vi) If we do not know the physical quantities

on which a physical quantitiy depends, we cannoteven proceed to derive a formula.

(vii) It is difficult to apply dimensional methodsif the physical quantity depends upon more thanthree fundamental quantities the length, mass andtime.

Short Answer Questions1. What are fundamantal quantities ? State their

units in SI.

2. Explain the uses of dimensional methods withexamples.

3. Is there any limit for the number of fundemental(base) quantities ? What are supplementaryfundamendal quantities? Define their units.

UNITS AND DIMENSIONS

AKASH MULTIMEDIA

PHYSICS - I A

18

4. What are limitations of dimensional methods?

5. Using dimensional methods,verify thecorrectness of the following relations

1) 2a=v / r 2) F= mr 2

3) S.T = rhdg/2. In the above formulae r standsfor radius, m for mass, for angularvelocity, v for linear velocity, h for height,d for density and S.T for sur face tension.

6. The velocity of an object varies with time as v=At2 + Bt + C. If units of v and t are expressed inS.I . find the units of constants A, B and C.

Very Short Answer Questions1. What are fundamental and derived physical

quantities ? Give examples.

2. What is meant by dimensionof a physicalquantity ? What is meant by dimensionalformula ?

3. Which physical quantity has negativedimensions in mass ?

4. State two constants, which have dimensions.

5. What is the physical quantity representedby gR

6. What is the physical quantity represented byPV?

7. If density of wood is 0.8gm/cc. Find its valuein SI

8. If units of length and force are increased by fourtimes. How much increase will be there in unitof pressure.

9. If units of mass, length and time are double whathappends to unit of energy.

10. If force F, length L and time T are fundamentalquantities. Find the dimensional formula ofmass.

11. Prove that energy per unit volume is pressure.

12. The air pressure in a tyre is 10Nm2, Express itin dynes/sq.cm

13. Which physical quantity has same dimensionalformula as surface tension?

14. Find out the quantity gl using dimensionalformula where g is acceleration due to gravityand l is wavelength of water waves.

15. Which physical quantities have units Jm1 andJm2

16. State two physical quantities having their unitsas Pa.

17. If energy E and Volume V are the fundamentalquatnit ies which physical quant ity hasdimesnional formula EV1

18. S.T for water is 70dyne/cm. Express it in SI.

19. What is the convenience in using SI.

Assess Yourself

1. The unit of length, metre was originally definedas the distance between two fine lines engravedon gold plugs near the ends of a bar of plati-numiridium alloy that is kept at 00C.

Then the unit was defined in October 1983 interms of wavelength of Kr86 radiation.

Now the present unit is defined in terms of dis-tance travelled by light.

In your opinion, which of the two essentialrequirements for a standard unit(1) availabilityinvariability played the key role in thismodification ?

Ans.Invariability

2. Originally the foot of human being is taken asa unit of length. What condition prevents theuse of human foot as a standard scientificfundamental (base) unit ?

Ans.Invariability

AKASH MULTIMEDIA

PHYSICS - I A UNITS AND DIMENSIONS

19

3. As far as mechanics is concerned all systems ofunits are having the same dimensions for a givenphysical quantity. But for thermodynamics, elec-tricity, magnetism and electromagnetics this isnot the case. Why?

Ans.This is because in SI we have got dimensionsfor temperaturre and current in addition to length,mass and time.

4. Two physical quantities are having the samedimensions. Does it mean that they are the same?Give an example supporting your answer.

Ans.No. Torque and work

5. Two physical quantities are having the samedimensions. Should their units be necessarilythe same ? Give an example supporting youranswer ?

Ans.No. Torque (Nm) and work (J)

6. Why is the dimension of one fundamental (base)quantity interms of any other fundamental (base)quantity is always zero ?

Ans.Other wise, it will not be fundamental quantityas per the definition.

7. A dimensionally wrong equation in which prin-ciple of homogeneity of dimensions is violatedmust be wrong. Can you say that, a dimension-ally correct equation should always be right ?

Ans. Need not be

8. State about the correctness or otherwise of thefollowing two statements.

i) A physical quantity can have dimensions butno units.

ii) A physical quantity can have units but nodimensions.

Ans.(i) Not correct (ii) Correct

9. Can we derive the kinematic relations v=u+at,s = ut + 1/2at2 and v2 = u2 + 2as. Explain thereason for your answer.

Ans.No. On left hand side there is only one term,but on the right hand side there are more thanone terms.

10. Coulombs law is 1 22

q qF k.r

. In C.G.S.

system, unit of charge is so defined that the forcein air or vacuum will be 1 dyne when twoidentical charges of unit magnitude are kept ata distance of r = 1 cm. This makes K=1 inCGS system.

But in SI the charges 1 C is defined as the amountof charge flowing when 1A of current passesfor 1s. That is 1C=1A 1s. The unit of current,ampere is defined through electromagnetics(refer to unit of current in SI). Consequentlythe value of F in air or vacuum comes out to be9 109N when two identical charges of 1Cmagnitude are placed in air or vacuum at 1 mapart. What are the consequences of theseresults?

Ans.The constant of proportionality K has got unitsand dimensions.