KEY CONCEPT
PHYSICAL QUANTITIES AND UNITS
Physical quantities :
All quantities that can be measured are called physical quantities. eg. time, length, mass, force, work
done, etc. In physics we study about physical quantities and their inter relationship.
Measurement :
Measurement is the comparison of a quantity with a standard of the same physical quantity.
Classification :
Physical quantities can be classified on the following bases :
I. Based on their directional properties
1. Scalars:The physical quantities which have only magnitude but no direction are called scalar quantities.
Ex. mass, density, volume, time, etc.
2. Vectors : The physical quantities which have both magnitude and direction and obey laws of vector
algebra are called vector quantities. Ex. displacement, force, velocity, etc.
II. Based on their dependency
1. Fundamental or base quantities : A set of physical quantities which are completely independent of
each other and all other physical quantities can be expressed in terms of these physical quantities is
called Set of Fundamental Quantities.
2. Derived quantities : The quantities which can be expressed in terms of the fundamental quantities
are known as derived quantities. Ex. Speed (= distance/time), volume, acceleration, force, pressure,
etc.
Physical quantities can also be classified as dimensional and dimensionless quantities or constants
and variables.
Ex. Classify the quantities displacement, mass, force, time, speed, velocity, acceleration, moment of inertia,
pressure and work under the following categories :
(a) base and scalar (b) base and vector (c) derived and scalar (d) derived and vector
Ans. (a) mass, time (b) displacement (c) speed, pressure, work
(d) force, velocity, acceleration
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Units of Physical Quantities
The chosen reference standard of measurement in multiples of which, a physical quantity is expressed
is called the unit of that quantity. Four basic properties of units are :
1. They must be well defined.
2. They should be easily available and reproducible.
3. They should be invariable e.g. step as a unit of length is not invariable.
4. They should be accepted to all.
System of Units :
1. FPS or British Engineering system :
In this system length, mass and time are taken as fundamental quantities and their base units are foot
(ft), pound (lb) and second (s) respectively.
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,2 JEE-Physics ALLEN
2. CGS or Gaussian system :
In this system the fundamental quantities are length, mass and time and their respective units are
centimetre (cm), gram (g) and second (s).
3. MKS system :
In this system also the fundamental quantities are length, mass and time but their fundamental units
are metre (m), kilogram (kg) and second (s) respectively.
4. International system (SI) of units :
This system is modification over the MKS system and so it is also known as Rationalised MKS
system. Besides the three base units of MKS system four fundamental and two supplementary units
are also included in this system.
Classification of Units : The units of physical quantities can be classified as follows :
1. Fundamental or base units :
The units of fundamental quantities are called base units. In SI there are seven base units.
SI BASE QUANTITIES AND THEIR UNITS
S.No. Physical quantity SI unit Symbol
1. Length metre m
2. Mass kilogram kg
3. Time second s
4. Temperature Kelvin K
5. Electric current ampere A
6. Luminous intensity candela cd
7. Amount of substance mole mol
2. Derived units :
The units of derived quantities or the units that can be expressed in terms of the base units are called
derived units
unit of distance metre
Ex. Unit of speed = = = ms-1
unit of time second
Some derived units are named in honour of great scientists.
• Unit of force – newton (N) • Unit of frequency – hertz (Hz) etc.
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UNITS OF SOME PHYSICAL QUANTITIES IN DIFFERENT SYSTEMS
Type of Physical CGS MKS FPS
Physical Quantity (Originated in (Originated in (Originated in
Quantity France) France) Britain)
Length cm m ft
Fundamental Mass g kg lb
Time s s s
Force dyne newton(N) poundal
Work or
Derived erg joule(J) ft-poundal
Energy
Power erg/s watt(W) ft-poundal/s
E
, ALLEN Unit & Dimension, Basic Maths and Vector 3
Dimensions :
Dimensions of a physical quantity are the powers (or exponents) to which the base quantities are
raised to represent that quantity. To make it clear, consider the physical quantity force. As we shall
learn later, force is equal to mass times acceleration. Acceleration is change in velocity divided by
time interval. Velocity is length divided by time interval. Thus,
force = mass × acceleration
velocity length / time
= mass × = mass × = mass × length × (time)–2
time time
Thus, the dimensions of force are 1 in mass, 1 in length and –2 in time. The dimensions in all other
base quantities are zero.
1. Dimensional formula :
The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to
remind that the equation is among the dimensions and not among the magnitudes. Thus above equation
may be written as [force]= MLT –2.
Such an expression for a physical quantity in terms of the base quantities is called the dimensional
formula. Thus, the dimensional formula of force is MLT–2. The two versions given below are equivalent
and are used interchangeably.
(a) The dimensional formula of force is MLT–2.
(b) The dimensions of force are 1 in mass, 1 in length and –2 in time.
The dimensional formula of any physical quantity is that expression which represents how and which
of the base quantities are included in that quantity.
Ex. Dimensional formula of mass is [M1L0 T0] and that of speed (= distance/time) is [M0L1T–1]
2. Applications of dimensional analysis :
(i) To convert a physical quantity from one system of units to the other :
This is based on a fact that magnitude of a physical quantity
remains same whatever system is used for measurement n2 = numerical value in II system
i.e. magnitude = numeric value (n) × n 1 = numerical value in I system
M1 = unit of mass in I system
unit (u) = constant or n1u1 = n2u2
M2 = unit of mass in II system
So if a quantity is represented by [MaLbTc] L1 = unit of length in I system
L2 = unit of length in II system
a b c
é u1 ù é M1 ù é L1 ù é T1 ù T1 = unit of time in I system
Then n 2 = n1 ê u ú = n1 ê M ú ê L ú ê T ú T2 = unit of time in II system
ë 2û ë 2û ë 2û ë 2û
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Ex. 1m = 100 cm= 3.28 ft = 39.4 inch
(SI) (CGS) (FPS)
Ex. The acceleration due to gravity is 9.8 m s–2. Give its value in ft s–2
Sol. As 1m = 3.2 ft \ 9.8 m/s2 = 9.8 × 3.28 ft/s2 = 32.14 ft/s2 » 32 ft/s2
Ex. Convert 1 newton (SI unit of force) into dyne (CGS unit of force)
Sol. The dimensional equation of force is [F] = [M1 L1 T–2]
Therefore if n1, u1, and n2, u2 corresponds to SI & CGS units respectively, then
1 1 -2 -2
é M1 ù é L1 ù é T1 ù é kg ù é m ù é s ù
n2 = n1 ê ú ê ú ê ú =1 ê úê ú ê ú = 1 × 1000 × 100 × 1= 105
M L
ë 2û ë 2û ë 2û T ë g ûë cm ûë s û
\ 1 newton = 105 dyne.
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, 4 JEE-Physics ALLEN
Q. The value of Gravitational constant G in MKS system is 6.67 × 10–11 N–m2/kg2 .
What will be its value in CGS system ?
Ans. 6.67 × 10–8 cm3/g s2
(ii) To check the dimensional correctness of a given physical relation :
If in a given relation, the terms on both the sides have the same dimensions, then the relation is
dimensionally correct. This is known as the principle of homogeneity of dimensions.
L
Ex. Check the accuracy of the relation T = 2 p for a simple pendulum using dimensional analysis.
g
Sol. The dimensions of LHS = the dimension of T = [M0 L0 T1]
12
æ dimensions of length ö
The dimensions of RHS = ç ÷ (Q 2p is a dimensionless constant)
è dimensions of acceleration ø
12
æ L ö
=ç -2 ÷ = (T2)1/2 = [T] = [ M0 L0 T1]
è LT ø
Since the dimensions are same on both the sides, the relation is correct.
(iii) To derive relationship between different physical quantities :
Using the same principle of homogeneity of dimensions new relations among physical quantities can
be derived if the dependent quantities are known.
Ex. It is known that the time of revolution T of a satellite around the earth depends on the universal
gravitational constant G, the mass of the earth M, and the radius of the circular orbit R. Obtain an
expression for T using dimensional analysis.
Sol. We have [T] = [G]a [M]b [R]c
[M]0 [L]0 [T]1 = [M]–a [L]3a [T]–2a × [M]b × [L]c = [M]b–a [L]c+3a [T]–2a
Comparing the exponents
1
For [T] : 1 = –2a Þ a = –
2
1
For [M] : 0 = b – a Þ b = a = –
2
3
For [L] : 0 = c + 3a Þ c = –3a =
2
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R3
Putting the values we get T = G–1/2 M–1/2 R3/2 =
GM
R3
So the actual expression is T = 2 p
GM
Limitations of this method :
(i) In Mechanics the formula for a physical quantity depending on more than three physical quantities
cannot be derived. It can only be checked.
(ii) This method can be used only if the dependency is of multiplication type. The formulae containing
exponential, trigonometrical and logarithmic functions can't be derived using this method. Formulae
containing more than one term which are added or subtracted like s = ut +at2 /2 also can't be derived.
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