1
Chapter
Units arnd Dimensions
REVIEW OF BASIC CONCEPTs
Physical Quantity Name of the Unit Symbol
1.1 THE SI SYSTEM OF UNITS Mass kilogram
Time second
The internationally accepted standard units of the
Electric current ampere
fundamental physical quantities are given in Table 1.1.
Temperature kelvin
Table 1.1 Fundamental SI Units Luminous intensity candela cd
Amount of substance mole nol
Physical Quantity Name of the Unit Symbol
Angle in a plane radian ad
Length metre
Solid angle
steradian
Table 1.2 Dimensional Formulae of some Physical Quantities
Physical Quantity Dimenstonal Formula
Physical Quantity Dimensional Formula
Arca M'LT Heat enerBy ML "T
Volume M'LL'T Entropy ML-TK
Density MLT Specific heat
M°LTK
Velocity M°LT Latent heat M°L'T
Acceleration M'LT Molar specific heat ML T K mol
Momentum MLT Thermal conductivity
MLT K
Angular momentum ML T Wien's constant M°LT'K
Force MLT Stefan's constant MLT'K
Energy, work ML'T Boltzmann's constant ML T K
Power ML T Molar gas constant ML T K mol
Torque, couple T Electric charge TA
Impulse MLT Electric current A
FrequencyY M'LT Electric potential ML T A
Angular frequency M'LT Electric field
MLT A
Angular acceleration M°LT Capacitance M-'LTA
Pressure MLT Inductancee ML TA
Elastic modulii MLT Resistance MLTA
Stress ML T Magnetic flux ML'T A
(Contd.)
, 1.2 Comprehensive Physics- JEE Advanced
Dimens
nsional Formula
Pmysical Quantity
Physical Qhanty Dimensional Formula
or MLT A
Moment of nertia ML'T Magnetic tlux density
Magnetic induction field
Surface tension ML°T Permeability
MLT A
M'L'TA
Viscosity ML T Permittivity
S alional constant M'LT Planck s conistan ML T
Dimensions of at= dimensions of a
PRINCIPLE OF HOMOGENEITY ODE
1.2 DIMENSIONS or [al TLT=[M'LT')
Consider a
simple cquation, Dimensions of brr= dimensions of r
If this is
AtB-C
equation of physics, i.e. if A, B and C are
an
physical quantities, then this equation says that one physl- b Urj-Mur}
cal quantity 4, when added to another
physical quantity B,
gives a third physical quantity C. This
no
equation will have
meaning
of the
in
physics if the nature (i.e. the
dimensions) EXAMPLE 1.2
quantities on the left-hand side of the equation is The pressure P, volume Vand temperature T of a gas
not the same as the nature
side. For
of the quantity on the right-hand are related as
example, if A is a length, B must also be a
length
and the result of addition of A and B
P -b) e7
-
must express a length.
In other words, the
dimensions of both sides of a physical
equation must be identical. This is called the principle of where a, b, and c are constants. Find the dimensions
homogeneity of dimensions f
1.3 USES OF DIMENSIONAL ANALYSIS
Dimensional equations provide a very simple method of sOLUTION
deriving relations between physical quantities involved in
Dimensions of =
dimensions
any physical phenomenon. The analysis of any phenom- of P.
enon carried out by using the method of dimensions1 Dimensions of a =
dimensions of P
called dimensional analysis. This analysIS 1s based on the Also dimensions of b= dimensions of
of
principle homogeneity of dimensions explained above.
There are four important uses of dimensional
Dimensions of =P ]
equationsS:
1. Checking the correctness of an equation.
2. Derivation of the relationship between the physical
=[PV1= [MIL TJx (L')
quantities involved in any phenomenon. =[ML T)
3. Finding the dimensions of constants or variables in NOTE
an equation. 1. Trigonometric function (sin, cos, an, cot are
also
di-
4. Conversion of units from one system to another.
nenstonless. The arguments of these functions anc
dimensionless
EXAMPLE 1.1 Exponential functions are dimensionless. 1neu r
nents are also
The distance x travelled by a body varies with time t as dimensionless
br, where and b are constants
x = at +
Find the dimensions of a and b.
a
EXAMPLE 1.3when a plane wave travels in a medium, the displu
ment y en by
of particle located
a
at x at time f 1s g
SOLUTION y a sin(bt + cx)
the right hand of the
The dimensions of each term on
ofs
a n d c are constants. Find the dimensi
the same as those of the left
given equation must be
hand side. Hence of-
, 1
Units and Dimensions
this cquation must be the same. Equating the powers
sOLUTION of M, Land T, we have
Terms br and ex must be dimensionless. Hence
a-0, bt c=0 and -2c= 1, which give b=and
2
I=km g 2)
Hence
and
tel -['
-[LT - M° LT ")
Note that the dimensions of a are the same as those Thus independent of the mass of the bob
t is
ol y. and is directly proportional to vI and inversely
proportional to Vg
EXAMPLE 14 also be used to convert aa
In the expression The dimensional method can
from one system to another. The method
physical quantity
is based on the fact that the magnitude of a physical
P
X remains the same in every system of its
quantity
Pis pressure, x is a distance and a and b are constants. measurement, 1.e.
Find the dimensional formula for b. ()
where u and u are the two units of measurement of
sOLUTION
quantity X and ni and n, are their respective numerical
values. Suppose M, Li and T are the fundamental units
- of mass, length and time in one system of measurement
and M2, La and T, in the second system of measurement,
Also ax is dimensionless. Hence [a] =[L 1 Let a, b andc be the dimensions of mass, length and time
of quantity A, the units of measurement u and u, Will be
[P [MLT] M LT
[MLT and
also
principle of homogeneity of dimensions
can
The
be used to find the dependence of a physical quantity
Using these in Eq. (1), we have
on other physical quantities. nM LT]-n:[M; L, TE]
EXAMPLE 1.5 (2)
The time period () of a simple pendulum may depend
the mass of the bob, / the length of the string
upon m
and g the acceleration due to gravity. Find the depen- Knowing (M,, Li, T), (M2, L2, T2), (a, b and c) and the
in
value of n in system 1, we can calculate the value of n,
dence of r on m, I
andg system 2 from Eq. (2).
SOLUTION
Let toc mlg EXAMPLE 1.6
Dyne is the unit of force in the c.g.s. system and new
or t =k mW'g ton is the unit of force in the SI system. Convert
where k is a dimensionless constant. dyne into newton.
the dimensions of each quantity, we have
Writing
SOLUTION
IT]-[M" LP [LT T The dimensional formula of force is
or [M LT] = [M°1°*T] F] =[M L'T]
of homogeneity of dimen- Therefore, a= 1, b= 1 and c=-2
According to the principle
either side of
sions, the dimensions of all the terms on
Chapter
Units arnd Dimensions
REVIEW OF BASIC CONCEPTs
Physical Quantity Name of the Unit Symbol
1.1 THE SI SYSTEM OF UNITS Mass kilogram
Time second
The internationally accepted standard units of the
Electric current ampere
fundamental physical quantities are given in Table 1.1.
Temperature kelvin
Table 1.1 Fundamental SI Units Luminous intensity candela cd
Amount of substance mole nol
Physical Quantity Name of the Unit Symbol
Angle in a plane radian ad
Length metre
Solid angle
steradian
Table 1.2 Dimensional Formulae of some Physical Quantities
Physical Quantity Dimenstonal Formula
Physical Quantity Dimensional Formula
Arca M'LT Heat enerBy ML "T
Volume M'LL'T Entropy ML-TK
Density MLT Specific heat
M°LTK
Velocity M°LT Latent heat M°L'T
Acceleration M'LT Molar specific heat ML T K mol
Momentum MLT Thermal conductivity
MLT K
Angular momentum ML T Wien's constant M°LT'K
Force MLT Stefan's constant MLT'K
Energy, work ML'T Boltzmann's constant ML T K
Power ML T Molar gas constant ML T K mol
Torque, couple T Electric charge TA
Impulse MLT Electric current A
FrequencyY M'LT Electric potential ML T A
Angular frequency M'LT Electric field
MLT A
Angular acceleration M°LT Capacitance M-'LTA
Pressure MLT Inductancee ML TA
Elastic modulii MLT Resistance MLTA
Stress ML T Magnetic flux ML'T A
(Contd.)
, 1.2 Comprehensive Physics- JEE Advanced
Dimens
nsional Formula
Pmysical Quantity
Physical Qhanty Dimensional Formula
or MLT A
Moment of nertia ML'T Magnetic tlux density
Magnetic induction field
Surface tension ML°T Permeability
MLT A
M'L'TA
Viscosity ML T Permittivity
S alional constant M'LT Planck s conistan ML T
Dimensions of at= dimensions of a
PRINCIPLE OF HOMOGENEITY ODE
1.2 DIMENSIONS or [al TLT=[M'LT')
Consider a
simple cquation, Dimensions of brr= dimensions of r
If this is
AtB-C
equation of physics, i.e. if A, B and C are
an
physical quantities, then this equation says that one physl- b Urj-Mur}
cal quantity 4, when added to another
physical quantity B,
gives a third physical quantity C. This
no
equation will have
meaning
of the
in
physics if the nature (i.e. the
dimensions) EXAMPLE 1.2
quantities on the left-hand side of the equation is The pressure P, volume Vand temperature T of a gas
not the same as the nature
side. For
of the quantity on the right-hand are related as
example, if A is a length, B must also be a
length
and the result of addition of A and B
P -b) e7
-
must express a length.
In other words, the
dimensions of both sides of a physical
equation must be identical. This is called the principle of where a, b, and c are constants. Find the dimensions
homogeneity of dimensions f
1.3 USES OF DIMENSIONAL ANALYSIS
Dimensional equations provide a very simple method of sOLUTION
deriving relations between physical quantities involved in
Dimensions of =
dimensions
any physical phenomenon. The analysis of any phenom- of P.
enon carried out by using the method of dimensions1 Dimensions of a =
dimensions of P
called dimensional analysis. This analysIS 1s based on the Also dimensions of b= dimensions of
of
principle homogeneity of dimensions explained above.
There are four important uses of dimensional
Dimensions of =P ]
equationsS:
1. Checking the correctness of an equation.
2. Derivation of the relationship between the physical
=[PV1= [MIL TJx (L')
quantities involved in any phenomenon. =[ML T)
3. Finding the dimensions of constants or variables in NOTE
an equation. 1. Trigonometric function (sin, cos, an, cot are
also
di-
4. Conversion of units from one system to another.
nenstonless. The arguments of these functions anc
dimensionless
EXAMPLE 1.1 Exponential functions are dimensionless. 1neu r
nents are also
The distance x travelled by a body varies with time t as dimensionless
br, where and b are constants
x = at +
Find the dimensions of a and b.
a
EXAMPLE 1.3when a plane wave travels in a medium, the displu
ment y en by
of particle located
a
at x at time f 1s g
SOLUTION y a sin(bt + cx)
the right hand of the
The dimensions of each term on
ofs
a n d c are constants. Find the dimensi
the same as those of the left
given equation must be
hand side. Hence of-
, 1
Units and Dimensions
this cquation must be the same. Equating the powers
sOLUTION of M, Land T, we have
Terms br and ex must be dimensionless. Hence
a-0, bt c=0 and -2c= 1, which give b=and
2
I=km g 2)
Hence
and
tel -['
-[LT - M° LT ")
Note that the dimensions of a are the same as those Thus independent of the mass of the bob
t is
ol y. and is directly proportional to vI and inversely
proportional to Vg
EXAMPLE 14 also be used to convert aa
In the expression The dimensional method can
from one system to another. The method
physical quantity
is based on the fact that the magnitude of a physical
P
X remains the same in every system of its
quantity
Pis pressure, x is a distance and a and b are constants. measurement, 1.e.
Find the dimensional formula for b. ()
where u and u are the two units of measurement of
sOLUTION
quantity X and ni and n, are their respective numerical
values. Suppose M, Li and T are the fundamental units
- of mass, length and time in one system of measurement
and M2, La and T, in the second system of measurement,
Also ax is dimensionless. Hence [a] =[L 1 Let a, b andc be the dimensions of mass, length and time
of quantity A, the units of measurement u and u, Will be
[P [MLT] M LT
[MLT and
also
principle of homogeneity of dimensions
can
The
be used to find the dependence of a physical quantity
Using these in Eq. (1), we have
on other physical quantities. nM LT]-n:[M; L, TE]
EXAMPLE 1.5 (2)
The time period () of a simple pendulum may depend
the mass of the bob, / the length of the string
upon m
and g the acceleration due to gravity. Find the depen- Knowing (M,, Li, T), (M2, L2, T2), (a, b and c) and the
in
value of n in system 1, we can calculate the value of n,
dence of r on m, I
andg system 2 from Eq. (2).
SOLUTION
Let toc mlg EXAMPLE 1.6
Dyne is the unit of force in the c.g.s. system and new
or t =k mW'g ton is the unit of force in the SI system. Convert
where k is a dimensionless constant. dyne into newton.
the dimensions of each quantity, we have
Writing
SOLUTION
IT]-[M" LP [LT T The dimensional formula of force is
or [M LT] = [M°1°*T] F] =[M L'T]
of homogeneity of dimen- Therefore, a= 1, b= 1 and c=-2
According to the principle
either side of
sions, the dimensions of all the terms on
