Units_Dim

UNITS AND DIMENSIONS

PHYSICS - I A

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UNITS AND DIMENSIONS

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James Watt was born in Greenock in 1736, the son of a ship’s chandler (trader in canvas, etc). In his late teens he went to London to learn to be a “mathematical and philosophical instrument maker”, and when he returned to Glasgow he got a job making instruments with Glasgow University, who gave him accomodation and a workshop. Watt’s engines were initially used for pumping water from cornish tin and copper mines. In 1882, 63 years after Watt’s death, the British Association gave his name to the unit of electrical power and today James Watt’s name is to be found written on almost every lightbulb in the world.

James Watt 1736-1819

h

Fundamental and Derived physical quantities

h

Dimensions, Dimentional formulae

h

SI units

h

Dimentional analysis

INTRODUCTION

Der ived Physical Quantities :

Under st andi ng of di f f er ent physi cal phenomenon r equir es the need t o measur e r elevent physical quantities. Physical quantities ar e those which can be defined and measur ed. W i t hout measur ement t her e can be no advancement i n physi cs. Exper i ment al measur ements ar e essential to ver ify theor itical laws. We use number of physical quantities (in daily life) like length, time, ar ea, volume etc. to descr i be an event or a phenomenon. For measur ing a physical quantit y, a st andar d reference is necessar y. This reference is known as ‘unit’

The physical quantities which can be derived from fundamental quantities are called derived quantities. Example : Area, Volume, Speed, Velocity etc.

1.1 PHYSI CAL QUANTI TI ES “All quantities in ter ms of which laws of physics can be expr essed and which can be measured directly or indir ectly are called physical quantities” Physical quantities can be classified into two types 1) Fundamental Physical Quantities 2) Derived Physical Quantities Fundamental Physical Quantity : A physical quantity which is independent of any other quantity is called fundamental physical quantity. Ex : Mass, length, time, electric current, thermodynamic temperature, amount of substance and luminous intensity are taken as fundamental physical quantities AKASH MULTIMEDIA

Unit : Unit of measur ement of a physical quantity is the standar d r efer ence of the same physical quantity which is used for compar ison of the given physical quantity. The meaning of the measurement of a physical quantity is to find out the number of times its unit is contained in that physical quantity. Therefore, the process of measurement of a physical quantity involves: i) The selection of unit and ii) 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 that of length. Suppose that we use metre as the unit. We place a metre rod successively along the length of the table and find out the number of times the metre rod is contained in the length of the table. Suppose that the length of the table is covered by the metre rod in three successive placings. 5

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UNITS AND DIMENSIONS

Then, 3 is called the numerical value of length of the table and metre as the unit of length. We may write: Length of the table = 3 × 1m =3m It may be pointed out that it is not sufficient to say that the length of the table is 3. The unit of the physical quantity has also to be stated along with the result of the measurement. In general, measure of a physical quantity = numerical value of the physical qunatity x size of its unit if ‘u’ is the size of the unit and ‘n’ is the numerical value of the physical quantity X (the quantity to be measured) for the selected unit, then measure of the physical quantity X= n u It follows that if the size of the chosen unit is small, then the numerical value of the quantity will be large and vice- versa. It is obvious that the measure of the physical quantity is always the same i.e. n u = constant If n1 is the numerical value of the physical quantity for a unit u1 and n2 for a unit u2, then n1u1 = n2u2 1.2

SYSTEM OF UNI TS

There are four main systems of units, namely. 1. CGS (Centimeter, Gramme or Gram, Second) 2. FPS (Foot, Pound, Second) 3. M K S (Meter, Kilogram, Second) 4. SI (System International d' Units) To measure the fundamental base quantities length, mass and time, there are three systems of units. These systems are called F.P.S. system (British system), C.G.S system (Metric system) and M.K.S. system. The basic units in these systems are given in table. Units

System F.P.S C.G.S M.K.S

Length foot centimetre metre

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Time Mass pound second gram second kilogram second

An international organization, the Confer ence Gener al des Poids at M easur es, or CGPM is internationally recognised as the authority on the definition of units. In English, this body is known as "General Conference on Weights and Measures". The System I nter national de Units, or SI units, was set up in 1960, by the CGPM. Requir ements of a Standar d Unit : A standard unit must have following features to be accepted world wide. (i) It should be of suitable size i.e. neither too large nor too small in comparison to the quantity to be 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 per requirement. (v) It should be independent of physical conditions like temperature, pressure, humidity etc. In SI, there are seven fundamental (base) physical quantities, length, mass, time, electric current, thermodynamic temperature, luminous intensity and quantity of substance. In addition to these, two more physical quantities, plane angle and solid angle are introduced as supplementary fundamental quantities. Fundamental (base) quantities, their units and dimensional representation are given in table. Fundamental (base) Quantities in SI Fundamental (base) Quantity

Unit

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

s

Electric current

ampere

A

Thermodynamic temperature

kelvin

K

Luminous Intensity

candela

cd

Quantity of substance

mole

mol

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Supplementary Fundamental (base) Quantities in SI rad

Plane angle

radian

Solid angle

steradian sr

1.3 DEFI NI TI ONS OF EL EM ENTARY (BASI C) UNI TS I N SI 1. kilogram(kg) : A kilogram is equal to the mass of a cylinder (having diameter = height) of platinum irridium alloy (90% platinum and 10% irridium) kept in the International Bureau of Weights and Measures at Sevres near Paris of France. 2. metr e(m) : ‘ metre’ is defined as the length of path travelled by light in vacuum in (1/299, 792, 458) part of a second.

*

3.

4.

5.

Note :

I n above case, standard of length is defined in terms of unit of time. Hence fundamental units are no more independent of each other. second(s) : A second is equal to the duration of 9,192,631,770 periods of t he radiat ion corresponding to the unperturbed transition between the two hyperfine levels of ground state of cesium (133) atom. ampere(A) : An ampere is defined as that constant current which when flowing in two straight parallel conductors of infinite length and of negligible area of cross-section and placed one metre apart in vacuum would produce between them a force of 2 × 10-7 newton per metre of length. kelvin(K ) : A kelvin is defined as the fraction 1 273.16

of the thermodynamic temperature of

the triple point of water. It may be noted that in SI., the triple point of water is taken as a fundamental fixed point having a temperature of 273.16 kelvin by definition. 6.

candela(cd) : A candela is as the luminous intensity, in a given direction, of a surface that emits monochromatic radiation of frequency 540 × 1012Hz and that has the radiant intensity in that direction of 1/683 watt per steradian.

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7.

mole(mol) : A mole is the amount of substance of a system, which contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. The number of atoms in 0.012 kg of C - 12 atom is 6.023 × 1023 and is called the Avogadro number.

8.

r adian(r ad) : Radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle. 360 0 0 | || 2 radians = 360 , 1 radian = 2 = 57 17 45 9. Ster adian(sr ) : Ster adian is the solid angle subtended at the centre of a sphere by its surface, the area of which is equal to the square of the radius. If d be the solid angle subtended at the centre of a sphere of radius r by a part its S surface of area S , then d  2 steradians. r The solid angle subtended at the centre by the entire surface area (S  4R 2 ) of a sphere will 4 R 2  4  sr. be   R2 1.4 FUNDAM ENTAL UNI TS The units of the fundamental physical quantities are called fundamental units Ex : metre, kilogram, second 1.5 DERI VED UNI TS Derived units are the units of the derived quantities. Physical Quantity

Der ived unit

Symbol of Unit

force

newton

N=kg m/s2

Energy, Work, Heat

joule

J=Nm=kg m2/s2

Power

W=J/s=N m/s=kgm2/s3

Pressure, Stress

watt pascal

Pa=N/m2=kg/ms2

Frequency

hertz

Hz = s-1

Electric charge

coulomb

C = As

electric potential

volt

V=J/C

Capacitance

farad

F=coulomb/volt=A2s2/Nm

Resistance

ohm

  V / A  J / A2s

Self inductance

henry

H=J/A2=Nm/A2

Magnetic flux

weber

Magnetic flux

tesla

Wb T=Wb/m2

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1.6

M ULTI PL ES AND SUBM ULTI PL ES OF SI UNI TS To measure very low or very high values of physical quantities, prefixes are used to represent them. For example 10–3m is written as 1 milli metre, 10–9 second is written as nano second and 106 watt is written as 1 Mega watt. The prefixes used to represent multiples and submultiples of SI units are given in table. M ultiples and submultiples of SI units Factor

Pr efix

Symbol

Factor

Pr efix

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 RUL ES FOR USI NG SYM BOL S FOR SI UNI TS/PREFI XES 1.

Full names of units even if they are named after scientists should not be written with initial capital letter Ex : newton but not Newton 2. Symbols for a unit named after a scientist should have a capital letter. Ex : ‘N’ for newton; ‘W’ for watt; ‘A’ for ampere 3. Punctuation marks should not be used after symbol 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 form Ex : 100 newtons is 100 N but not 100 Ns 50 meters is 50 m but not 50ms and but not 50 mts. 1.8 DI M ENSI ONS Def init ion : The demensions of a physical quantity ar e the powers (exponents) to which the fundamental (base) quantit ies ar e r aised to represent that physical quantity. AKASH MULTIMEDIA

1.9 DI M ENSI ONAL FORM UL A Definition : The expr ession showing the power s to which the fundamental quantities must be r aised to represent a physical quantity, is called dimensional for mula. The fundamental physical quantities; namely, mass, length time, temperature, electric current are indicated by the symbols [M], [L], [T], [K] or [  ], [I] or [A] etc. respectively. Examples: 5. Force = mass × acceleration ; So dimensional formula of force is [ MLT–2 ] =[M1 L1 T–2] In the dimensional formula of force, dimension of mass = 1, dimension of length = 1, dimension of time = –2. 6. Work =F.S = m a S; So dimensional formula of work = M LT–2 L = [ M1L2T–2] In the dimensional formula of work, dimension of mass = 1, dimension of length = 2, dimension of time = –2. 8

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1.10 DI M ENSI ONAL EQUATI ON Definition : I t is an equation, equating the physical quantity with its dimensional for mula. Example : Force [F] = [MLT–2] is dimensional equation of force. Dimensional For mulae of Physical Quantities SN Physical quantity

For mula

Dimensional for mula

S.I unit

Natur e scalar (or ) vector

1. Volume

lengthxbreadthxheight

MºL3 Tº

m3

S

2. Density

mass / volume

ML–3 Tº

kg m–3

S

3. linear density

mass / length

ML–1 Tº

kg m–1

S

4. Relative density

density of substance densityof water

MºLºTº

----

S

5. a)

Energy density

Energy / volume

ML–1T–2

N m–2

–

b)

Pressure

force/Area

ML–1 T–2

N m–2

–

c)

Strees, young’s,

Stress / strain

ML–1 T–2

N m–2

–

Velocity

displacement / time

MºLT –1

m s–1

V

b)

Speed

distance / time

MºLT –1

m s–1

S

c)

acceleraion

force / mass

MºLT –2

m s–2

V

7. Hubbles’s Constant (H)

Velocity / Distance

T-1

s-1

-

8. a) Areal velocity

Area / time

MºL2T–1

m2 s–1

V

Coefficient of vis cosity density

MºL2T–1

Momentum (linear)

mass  velocity

ML1T–1

kg m s–1

V

Impulse

Force  time

ML1T–1

kg m s–1

V

Moment of force

Force  displacement

ML2T–2

Nm

V

Force  displacement

ML2T–2

joule

S

ML2T–2

joule

Rigidity, Bulk modulie of elasticity 6. a)

b) Coefficient of Kinematic

S

viscosity 9. a) b) 10. a)

or moment of couple or torque b)

work

c)

All energies

11. Power

Force  velocity

ML2T–3

watt

S

12. Moment of Inertia

I = MK2

ML2Tº

kg m2

Tensor

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SN

Physical quantity

For mula

Dimensional for mula

S.I unit

Natur e scalar (or ) vector

13. Angular displacement

  d/r

MºLºTº

rad

V

14. a)

Angular velocity

 = / t

MºLºT –1

rad/s

V

b)

Velocity gradient

dv/dx

MºLºT –1

s–1

V

c)

Frequency

 = c/

MºLºT –1

Hz

S

d)

decay constant

number of fissions/ time

MºLºT–1

s–1

S

15. Angular acceleration

d  /dt

MºLºT –2

rad s–2

V

16. a)

I

ML2T–1

kg m2 s–1

V

Angular Momentum

b)

Planks constant

Energy h = frequency

ML2T–1

Js

S

c)

Angular impulse

 t

ML2T–1

Js

V

17. Gravitational constant

Fd 2 m2

M–1 L3 T–2

N m2 kg-2

S

18. Gravitational field intesity

F m

M 0 L1 T 2

N kg

19. a)

W m

MºL2T–2

J kg–1

Gravitational

b)

Potential Latent heat

c)

Calorific value

Q m Q m

-1

Q = Heat energy

V

S S S

F/L

M T–2

N m–1

S S S

21. Compressibility

1 Bluk mod ulus

M–1 LT 2

Pa–1

S

22. Coefficient of

   F     A dv  =  dx  Pressure  Time

ML–1 T–1

Pa s

S

MºLºTºK–1

K–1

S

20. a) b) c)

Force constant Spring Constant Surface Tension

Viscosity 23. a)

Coefficients of linear,

=

 l 2 - l1  l 1  t 2 - t1 

areal and volume expansion Coefficient of real , apparant expansions Volume coefficient, Pressure coefficient

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For mula

SN Physical quantity 24. a) b)

b)

Natur e scalar (or ) vector

Heat Temperature

ML2T –2 K–1

J K-1

S

Entropy

Heat Temperature

ML2T –2 K–1

J K-1

S

PV NT

ML2T –2 K–1

J K-1

S

Q m

L2 T–2 K–1

J kg–1 K-1

S

L2 T–2 K–1

J kg–1K-1

Specific heat

b) Specific gas constant

26. a)

S.I unit

Thermal capacity

(c) Boltzman Constant 25. a)

Dimensional for mula

Universal gas constant Molar specific heat

r

p dT

PV nT CP–CV = R R=

S

ML2T–2K–1mol–1

J mol–1 K–1

S

ML2T–2K–1mol–1

J mol–1 K–1

S

27. Stefan's constant

E At4

MLºT –3 K–4

W m–2 K–4

S

28. Coefficient of Thermal

Qd A  2  1  t

MLT–3 K–1

W m–1 K–1

S

L KA

M-1 L–2 T 3 K1

W–1 K

S

 L

L-1 K 1

K m-1

V

30. Thermal conductance

KA L

ML2T –3 K–1

WK–1

S

31. Permeability



MLT–2 I–2

H m-1

S

32. Pole strength

m = IL

IL

Am

S

33. Magnetic moment

M = 2l  m

I L2

A m2

V

34. Magnetic Induction

B=

M T–2 I–1

tesla

V

IL–1

A m–1

V

IL–1

A m–1

V

conductivity 29. a) Thermal resistance b) Temperature gradient

35. a)

Intensity of

F IL B H= 

Magnetic field b)

Intensity of Magnetisation

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Fd2 F = m2 I2

I=

M V

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SN Physical quantity

For mula

36. Magnetic flux

  BA =

37. a)

=

b) c)

Intensity of sound

Dimensional for mula

W I

Power Area

S.I unit

Natur e scalar (or ) vector

ML2 T–2 I–1

weber

S

M T–3

watt / meter2

S

Intensity of Heat Emissive power

38. Spectral emissive power

Power L3

ML–1 T–3

watt/meter 3

S

39. Pressure gradiant

Pr essure Length = P/L

ML–2T–2

Pa m–1

V

40. Electric current

i

MºLºTºI1

ampere

S

41. Eelectric charge

Q=it

IT

coulomb

S

42. Intensity of electric field

E = F/Q

MLT–3 I–1

N C-1

V

43. a) Electric Potential

W Q

ML2 T–3 I–1

volt

S

M–1 L–2 T 4I2

farad

S

ML2 T–3 I–2

ohm

S

M–1 L–2 T 3I2

mho or

S

b) Electro motive force (emf) 44. Electric Capacity

C=

45. Electric resistance

R=

46. Electric Conductance

Q Q2 = V W

V Power = 2 I I 1 K= R

siemen 47. Resistivity

 =

RA L

ML3 T–3 I–2

ohm meter

S

48. Conductivity

 =

1 

M–1 L–3 T 3I2

Sm-1

S

49. Permitivity

0 =

Q2 Fd 2

M–1 L–3 T 4I2

farad/metre

S

M L2T–2I–2

henry

S

50. a) b)

Self Inductance Mutual inductance =

e nerg y cu rren t 2

=

em f di / dt

51. Linear charge density

q l

L-1 T I

C m-1

S

52. Surface charge density

q A

L-2 T I

C m-2

S

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SN

Physical quantity

53. Volume charge density 54. Current density

For mula

q V I A

Dimensional for mula

S.I unit

Natur e scalar (or ) vector

L-3 T I

C m-3

S

I L-2

A m-2

V

55. Electro chemical equivalent

m it

MI 1 T 1

kg A1 s 1

S

56. Decay Constant

dN dt

T-1

s-1

S

M-1L0T2I

A s2 kg-1

S

57. Mobility

vd E

1.11 DI M ENSI ONAL CONSTANTS AND DI M ENSI ONL ESS QUANTI TI ES. DI M ENSI ONAL CONSTANT : Constants having dimensional formula ar e called dimensional constants. Ex : Planck’s constant, Universal gravitational constant Universal gas constant, Boltz mann’s constant, Stefan’s constant, Wien’s constant, Velocity of light, are Dimensional constants. Dimension less quantities : Physical quant it ies having no dimensional for mula are called dimension less quantities. Ex : Angle, Strain, Relative Density,coefficient of friction, coefficient of restitution, poisson’s ratio have zero dimensions and are Dimensionless quantities. Numbers, Ratio of same quantities,  ,

Problem : 1.1 Verify the correctnessof the equation v=u+at . In the dimensional form  L1T 1   L1T 1    L1T 2   T1    L1 T 1           The dimension for L on both sides is 1

The dimension for T on both sides is –1 Hence from the principle of homogenity of dimensions the given equation is correct.

1.13 APPL I CATI ONS OF DI M ENSI ONAL ANALYSI S 1) Dimensional formulae can be used to convert one system of units into another system 2)

have no dimensional formula because they will not depend on fundamental quantities

Dimensional formulae can be used to check the correctness of an equation

3)

1.12 PRI NCI PL E OF HOM OGENI TY OF DI M ENSI ONS

Dimensional formulae can be used to derive relationship among different physical quantities

4)

Dimensional formulae can be used to find unit of a given physical quantity

5)

Dimensional formulae can be used to design our own new system of units

Statement : Only physical quantities having same dimensions can be added, subtr acted or can be equated.

If x = y + z is dimensionally correct and if x represents the physical quantity, the force, then y and z also must represent the same physical quantity i.e., force. It means that the ter ms on both sides of a di mensi onal equat i on shoul d have same dimensions. This is called principle of homogeneity of dimensions. AKASH MULTIMEDIA

1. To conver t one system of units into another system This is based on the fact that for a given physical quantity, numerical value × unit = constant. i.e., N1U1= N2U2. 13

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So when units change, numerical value also may change. To convert a physical quantity from one system to the other, we write its units in terms of mass, length and time and then convert into another system.

Problem : 1.4 1 2 at 3 where S is the distance, u is velocity, a is acceleration and t is time.

Check the correctness of the formula S = ut +

Sol. Dimensionally,

Problem : 1.2 Convert newton into dyne

LHS = [L] RHS = [LT-1] [T] + [LT-2] [T]2 = [L] + [L] = [L]

Sol. [F] = [MLT–2] ; 1 newton = 1 (kg) (m) (s)-2

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

= 1 ´ (103g) (102cm) (s)-2 = 105 (g) (cm) (s)-2 1newton = 105 dynes Alter nate M ethod : Let the dimensional formula of a physical quantity be [M a Lb T c ]. Let n1 be its numerical value in one system in which units of fundamental quantities are M1,L1 and T1. Then the numerical value n 2 in another system in which units of fundamental quantities are M2,L2 and T2 is given by

*

However, we cannot say anything about the physical correctness of the formula. We know that the correct formula is 1 2 at 2 Hence, we conclude that dimensional correctness is no guarantee for physical correctness of the formula. However, dimensional incorrectness guarantees the physical incorrectness of the formula. Inspite of the above limitation, the method is still helpful to a great extent.

S = ut +

n 2  M2a Lb2 T2c   n1  M1a Lb1 T1c  a

b

M  L  T   n 2  n1  1   1   1   M2   L 2   T2 

c

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

Note :

Problem : 1.5 Consider the equation T  2

Sol. Dimensional formula of energy is M L T . 1 2 2

it is cor rect or not.

 and check whether g

Let IJ  n ergs  n  J / ergs 1

2

M  L  T   n 2  n1  1   1   1   M2   L 2   T2  2

1 kg   1m  1s   n      1g   1cm  1s 

2

we know, dimensional formulae of T,  and g are [T],

2

[L], [LT–2] respectively.

 Dimensional formula of

 n2   n  n 1

2

 1000 g   100 cm  1s  n      1 g   1 cm  1s 

 L   T2   T  g LT 2

Dimensional formula of left side is also [T]. As 2 is a dimensionless quantity, the dimensional formula of left hand side of the equation is equal to dimensional formula of right hand side. Hence we conclude that the given

2

 n  [1000]  [100]2  [1]2 = 107

equation is a correct one.

7

 1 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 a given equation must be same. If the dimensions of each term on both sides are same, the equations is dimensionally correct, otherwise not correct. AKASH MULTIMEDIA

3.

To deduce r elation between the physical quantities :

This is also based on t he principle of homogeneity. If one knows the quantities on which a particular physical quantity depends and if one guesses that this dependence is of product type, the method of dimensional analysis may be helpful in the derivation of the relation. 14

UNITS AND DIMENSIONS

PHYSICS - I A Problem : 1.6 Derive an expression for the time period (T) of a simple pendulum which may depend upon the mass(m) of the bob, length (l) of the pendulum and acceleration due to gravity (g). Sol. Let T = kma l b 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]

Angular Momentum (L) = ML2T-1 Universal Gravitational constnat (G) = M–1L3T–2 EL2 Substituting in , we get M 5G 2 1 2

 ML T  ML T   ML T   M L T  2

0

2

2

0 5

1 3

2 2



M1 2 L2  4 T 2  2 M5 2 L0  6 T 0  4

= A dimensionless quantity.

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  k

 , where k is constant. g

* Problem : 1.7 Derive an expression for the velocity of sound (V) which may depend upon the modulus of elasticity (E) of the medium, and density (d) of the medium. Sol. Let V= kEa d b where k is the proportionality constant.

Dimensional formula of v, E and d are V   M 0 LT 1  ; E   ML1 T 2  ; d   ML3 T 0 

As k has no dimensions 1 2  a

0 b

 M0 LT 1    ML T  3      ML T  a b

  M 0 LT 1    M 

 L a 3b  T 2a

Equating dimensions of M, L and T on either side of the equation We get, a +b = 0; –a–3b = 1 and –2a =–1 –2a = –1  a = 1/2; a+b = 0  b = –1/2 E d Experimentally the value of k is found to be one.

 V = k E1/2 d–1/2;

V 

V=k

E d

* Problem : 1.8 I f E, M,L and G denote energy, mass, angular momentum and universal Gravitational constant respectively, EL2 prove that is a dimensionless quantity. M 5 G2 Sol. Taking dimensional formulae energy (E) = ML2T–2 mass (M) = ML0T0

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Problem : 1.9 I f the equati on of state of a gas is expressed as a   P  2   V  b  RT where P is the pressure, V is V the volume and T the absolute temperature and a, b, R are constnats, then find the dimensions of ‘a’and ‘b’.

Sol. By principle of homogentiy of dimensions P can added a to P only. It means 2 also gives pressure. V Dimension formulae for pressure (P) = M1L–1T–2 and Volume (V) = M0L3T0. a Since 2 = pressure V a a  0 6 0  M1L1T 2   M1L1 T2 2 M LT M0 L3 T 0







a = M1L5T–2

Similarly, b will have same dimensions as volume  b = M0L3T0 V – b = volume * Problem : 1.10 The dimensional formula of the product of two physical quantities P and Q is ML 2T–2. The dimensional formula of P/Q is ML 0T–2. Then which physical quantities are represented by P and Q ? Sol. Given P × Q has dimensions ML2T–2

P/Q has dimensions ML0T–2  P Hence, (PQ)    ML2 T 2 ML0 T 2  Q







P2 = M2L2T–4  P = ML1T–2 = Force ......(1)

given Q

P ML1T 2 = M1L0T–2 ;  ML0 T 2 Q Q

ML1T 2 ML0 T 2

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

 P and Q are respectively the force and the displacement

15

PHYSICS - I A

UNITS AND DIMENSIONS * Problem : 1.11

* Problem : 1.14

I f pressure P, velocity v and time T are taken as the fundamental (base) quantities, find the dimensional for-

The freguency ‘n’ of a vibrating string depends up on its length ‘l ’, linear density m (mass per unit length) and tension T in the string. Derive an expression for the frequency of the string. (Solution is simlar to 1.12problem) k T . Ans. n = l m

mula of force. Sol. Let force = F = PaVbTc

(F) = M1 L 1T –2 , (P) = M1 L–1 T –2 , (V) = M0 L1 T –1 , (T) = M0L0T1  MLT T–2 = (ML–1T–2)a (M0L1T–1)b(M0L0T1)c

MLT–2 = MaL–a+bT–2a-b+c  Comparing powers of M, L, T on both sides a = 1, –a + b = 1  b = 2 and – 2a – b + c = – 2; – 2 – 2 + c = – 2; c = 2

* Problem : 1.15 Derive an expression for the rate of flow of a liquid through a capillary tube. Assume that the rate of flow  P depends on (i) pressure gradient  l  , (ii) The radius, r and (iii) the coefficient of viscosity,  . The value of

 Force (F) = PV2T2.

* Problem : 1.12 A gas bubble from an explosion under water, oscillates with a period proportional to Pa  bEc where p is the static pressure, ρ is the density of water and E is the total energy of the explosion.Find the values of a, b, and c. Sol. Let, T  Pa  bEc

Dimensional formula of p  ML-1 T-2 ;   ML-3 ; E  ML2 T2  T1  (ML1 T 2 )a (ML3 ) b (ML2 T 2 )c

Comparing powers of M, L and T on both sides -----(1) a+b+c=0 -----(2) -a - 3b + 2c = 0 ----- (3) 1 1 From (3) and (1) : a + c =  and b = 2 2 1 -a +2c = 3× 1 2 a +c = - ; 2 3c = 1 1 1 1 5  c = and a = - - = 3 3 2 6  The values of a, b, c are respectively,, 5 1 1  , , . 6 2 3 -2a - 2c = 1

* Problem : 1.13 The period T0 of a planet above the sun of mass ‘M’in a circular orbit of radius R depends on M, R and G where G is the gravitational constant. Find expression for time period by dimensional methods. (Hint : The solution is similar to above problem) Ans : T0 = K

R3 GM

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the proportionatity constant k = to 1.12problem) Ans.

4.

 p r 4    pr 4  V =k     t  l   8  l 

 . (Solution is simlar 8    k  8   

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 dimensional formula. So, in the dimensional formula, by replacing M, L and T by the fundamental units, we get the unit of physical quantity. Example: Force = mass x acceleration [F] = [M] [LT–2] = [MLT–2] Hence, units of force will be kg ms–2 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 physical quantities with the condition that they should be independent of each other and not derivable mutually. Suppose in a new system, force(F), length (L), and time (T) are taken as fundamental physical quantities then dimensional formulae of other quantities like mass, velocity, acceleration etc., can be derived in terms of F, L and T. 16

UNITS AND DIMENSIONS

PHYSICS - I A * Problem : 1.16 I f energy E, velocity v and time T are chosen as the fundamental quantities. Find dimensional formula of surface tension ? Sol. S. T = MT-2, E = ML2T2, v = LT–1, t = T1  S. T  (E)a (v) b (t)c and MT -2 =

(ML2 T -2 )a (LT-1)b(T)c Comparing power of M, L and T on either side a = 1, 2a + b = 0  b = - 2 -2a - b + c = - 2 1

 S. T = E (V)

 c=-2 2

(T) 2

Problem : 1.17 Suppose you design your system of dimensions and take vel oci ty (V), Pl anck constant (h) and gravitati onal constant (G) as the f undamental quantities. 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 = (LT–1)a(ML2T–1)b(M–1L3T–2)c comparing the powers of M, L, T both sides b – 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 = 1/2 L = V–3/2 h1/2G1/2 * Problem : 1.18 I f unit of mass is taken as 1 kg, of time as 1 minute and that acceleration due to gravity is taken as9.81 ms-2, what is the unit of energy ? Sol. New unit of energy = E1 New unit of mass M1 = 1kg New unit of time T1 = 1 minute = 60 sec New unit of length = L1 E1 = [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 E 1 = 1 kg × (9.81 × 60 2 ) 2 m 2 (60) -2 S -2    E1 =

1 × 9.81 × 60 × 60 × 9.81 × 60 × 60 kg m2 s-2 60 × 60

E1 = 3.464 × 105 kg m2 sec-2  The new unit of energy E1 = 3.464 × 105 joule.

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1.14 L I M I TATI ONS ON THE USES OF DI M ENSI ONAL ANALYSI S There is no limitation as far as the first use (conversion of units) is concerned. As for the second use (checking the dimensional accuracy of the given relationship) is concerned, if both the LHS and RHS are dimensionless, we cannot know whether the quantities in the numerator and denominator are in correct or reverse order. The rest of the uses have following limitations. (i) Proportionalit y co nst ants cannot be determined by dimensional analysis. (ii)Formulae containing non-algebraic functions i.e., sin, cos, log, exponential etc. cannot be derived. (iii)Formulae which contain two or more terms in the RHS cannot be derived. 1 Example : S = ut + at2. 2 (iv)Formulae containg dimensional constants cannot be derived.

Gm1m 2 d2 (v) Dimensional analysis does not differentiate between a scalar and a vector quantity. Example : F =

(vi) If we do not know the physical quantities on which a physical quantitiy depends, we cannot even proceed to derive a formula. (vii) It is difficult to apply dimensional methods if the physical quantity depends upon more than three fundamental quantities the length, mass and time.  Shor t Answer Questions 1.

What are fundamantal quantities ? State their units in SI.

2.

Explain the uses of dimensional methods with examples.

3.

Is there any limit for the number of fundemental (base) quantities ? What are supplementary fundamendal quantities? Define their units. 17

PHYSICS - I A

UNITS AND DIMENSIONS

4.

What are limitations of dimensional methods?

5.

Using dimensio nal methods,verify t he correctness of the following relations 2 1) a=v / r

2) F= mr  2

3) S.T = rhdg/2. In the above formulae ‘r’ stands for radius, ‘m’ for mass, ‘  ’ for angular velocity, ‘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 in S.I . find the units of constants A, B and C.

 1. 2.

Ver y Shor t Answer Questions

What are fundamental and derived physical quantities ? Give examples.

12. The air pressure in a tyre is 10Nm–2, Express it in dynes/sq.cm 13. Which physical quantity has same dimensional formula as surface tension? 14. Find out the quantity gl using dimensional formula where g is acceleration due to gravity and l is wavelength of water waves. 15. Which physical quantities have units Jm–1 and Jm–2 16. State two physical quantities having their units as Pa. 17. If energy E and Volume V are the fundamental quatnit ies which physical quant ity has dimesnional formula EV–1 18. S.T for water is 70dyne/cm. Express it in SI.

What is meant by ‘dimension’of a physical quantity ? What is meant by dimensional formula ?

19. What is the convenience in using SI.

3.

Which physical quantit y has negative dimensions in mass ?

1.

4.

State two constants, which have dimensions.

5.

What is the physical quantity represented by gR

6. 7.

Assess Your self

Then the unit was defined in October 1983 in terms of wavelength of Kr–86 radiation.

What is the physical quantity represented by PV?

Now the present unit is defined in terms of distance travelled by light.

If density of wood is 0.8gm/cc. Find its value in SI

8.

If units of length and force are increased by four times. How much increase will be there in unit of pressure.

9.

If units of mass, length and time are double what happends to unit of energy.

10. If force F, length L and time T are fundamental quantities. Find the dimensional formula of mass. 11. Prove that energy per unit volume is pressure. AKASH MULTIMEDIA

The unit of length, metre was originally defined as the distance between two fine lines engraved on gold plugs near the ends of a bar of platinum–iridium alloy that is kept at 00C.

In your opinion, which of the two essential requirements for a standard unit–(1) availability invariability played the key role in this modification ? Ans.Invariability 2.

Originally the ‘foot’ of human being is taken as a unit of length. What condition prevents the use of human foot as a standard scientific fundamental (base) unit ?

Ans.Invariability 18

UNITS AND DIMENSIONS

PHYSICS - I A

3.

As far as mechanics is concerned all systems of units are having the same dimensions for a given physical quantity. But for thermodynamics, electricity, magnetism and electromagnetics this is not the case. Why?

Ans.This is because in SI we have got dimensions for temperaturre and current in addition to length, mass and time. 4.

Two physical quantities are having the same dimensions. 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 same dimensions. Should their units be necessarily the same ? Give an example supporting your answer ?

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 quantity as per the definition. 7.

A dimensionally wrong equation in which principle of homogeneity of dimensions is violated must be wrong. Can you say that, a dimensionally correct equation should always be right ?

Ans. Need not be 8.

State about the correctness or otherwise of the following two statements.

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i)

A physical quantity can have dimensions but no units.

ii)

A physical quantity can have units but no dimensions.

Ans.(i) Not correct 9.

(ii) Correct

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

Ans.No. On left hand side there is only one term, but on the right hand side there are more than one terms. 10. Coulomb’s law is F  k. q1q 2 . In C.G.S. r2 system, unit of charge is so defined that the force in air or vacuum will be 1 dyne when two identical charges of unit magnitude are kept at a distance of r = 1 cm. This makes K=1 in CGS system. But in SI the charges 1 C is defined as the amount of charge flowing when 1A of current passes for 1s. That is 1C=1A × 1s. The unit of current, ampere is defined through electromagnetics (refer to unit of current in SI). Consequently the value of F in air or vacuum comes out to be 9 × 109N when two identical charges of 1C magnitude are placed in air or vacuum at 1 m apart. What are the consequences of these results? Ans.The constant of proportionality K has got units and dimensions.

19