CONTENTS
1.1
Physical quantity
1.2
Types of physical quantity
1.3
Fundamental and derived quantities
1.4
Fundamental and derived units
1.5
Prefixes
1.6
Standards of length, mass and time
1.7
Practical Units
1.8
Dimensions of a physical quantity
1.9
Important dimensions of complete physics
1.10
Quantities having same dimensions
1.11
Applications of dimensional analysis
1.12
Limitations of dimensional analysis
1.13
Significant figures
1.14
Rounding off
1.15
Significant figures in calculation
1.16
Order of magnitude
1.17
Errors of measurement
1.18
Propagation of errors
Sample Problems
Practice Problems
Answer Sheet of Practice Problems
Chapter
1
An indication of range over which measurements are made.
Measurements of Time and Length extend over a range of 1040
and measurement of Mass extend over a range 1080.
1050
1040
1030
1020
1010
1 kg
10–10
10–20
10–30
1026
1022
1024
1020
1018
1016
1014
1012
1010
108
106
104
102
1m
10– 2
10– 4
10– 6
10– 8
10–10
10–12
10–14
1015
1013
1017
1011
109
107
105
103
101
10–1
10–3
10–5
10–7
10–9
10– 11
10– 13
10– 15
10– 17
10– 19
10– 21
10– 23
1 sec
–
+
Electron
Proton
DNA
Cell
Fly
Man
Aircraft
Earth
Sun
Galaxy
Universe
+
Diamete
r of
Proton
Diamete
r of
Atom
Bacteria
Length
of Man
Length
of Whale
Diamete
r of
Earth
Diamete
r of Sun
Diamete
r of
Galaxy
Distant
Edge of
Visible
Universe
Age of
Universe
Age of
Earth
Age of
Tajmaha
l
Lecture
Duration
at MMC
Period
of AM
Radio
Period
of
Visible
Light
Period
of
Nuclear
vibratio
n
Mass
Length
Time
50 Units, Dimensions and Measurement
1.1 Physical Quantity.
A quantity which can be measured and by which various physical happenings can be
explained and expressed in form of laws is called a physical quantity. For example length, mass,
time, force etc.
On the other hand various happenings in life e.g., happiness, sorrow etc. are not physical
quantities because these can not be measured.
Measurement is necessary to determine magnitude of a physical quantity, to compare two
similar physical quantities and to prove physical laws or equations.
A physical quantity is represented completely by its magnitude and unit. For example, 10
metre means a length which is ten times the unit of length 1 kg. Here 10 represents the
numerical value of the given quantity and metre represents the unit of quantity under
consideration. Thus in expressing a physical quantity we choose a unit and then find that how
many times that unit is contained in the given physical quantity, i.e.
Physical quantity (Q) = Magnitude × Unit = n × u
Where, n represents the numerical value and u represents the unit. Thus while expressing
definite amount of physical quantity, it is clear that as the unit(u) changes, the magnitude(n)
will also change but product ‘nu’ will remain same.
i.e. n u = constant, or
constant
2211 unun
;
u
n1
i.e. magnitude of a physical quantity and units are inversely proportional to each other
.Larger the unit, smaller will be the magnitude.
1.2 Types of Physical Quantity.
(1) Ratio (numerical value only) : When a physical quantity is a ratio of two similar
quantities, it has no unit.
e.g. Relative density = Density of object/Density of water at 4oC
Refractive index = Velocity of light in air/Velocity of light in medium
Units, Dimensions and Measurement 51
Strain = Change in dimension/Original dimension
Note : Angle is exceptional physical quantity, which though is a ratio of two similar physical
quantities (angle = arc / radius) but still requires a unit (degrees or radians) to specify
it along with its numerical value.
(2) Scalar (Magnitude only) : These quantities do not have any direction e.g. Length, time,
work, energy etc.
Magnitude of a physical quantity can be negative. In that case negative sign indicates that
the numerical value of the quantity under consideration is negative. It does not specify the
direction.
Scalar quantities can be added or subtracted with the help of following ordinary laws of
addition or subtraction.
(3) Vector (magnitude and direction) : e.g. displacement, velocity, acceleration, force etc.
Vector physical quantities can be added or subtracted according to vector laws of addition.
These laws are different from laws of ordinary addition.
Note : There are certain physical quantities which behave neither as scalar nor as vector.
For example, moment of inertia is not a vector as by changing the sense of rotation
its value is not changed. It is also not a scalar as it has different values in different
directions (i.e. about different axes). Such physical quantities are called Tensors.
1.3 Fundamental and Derived Quantities.
(1) Fundamental quantities : Out of large number of physical quantities which exist in
nature, there are only few quantities which are independent of all other quantities and do not
require the help of any other physical quantity for their definition, therefore these are called
absolute quantities. These quantities are also called fundamental or base quantities, as all other
quantities are based upon and can be expressed in terms of these quantities.
(2) Derived quantities : All other physical quantities can be derived by suitable
multiplication or division of different powers of fundamental quantities. These are therefore
called derived quantities.
If length is defined as a fundamental quantity then area and volume are derived from length
and are expressed in term of length with power 2 and 3 over the term of length.
Note : In mechanics Length, Mass and time are arbitrarily chosen as fundamental
quantities. However this set of fundamental quantities is not a unique choice. In fact any three
quantities in mechanics can be termed as fundamental as all other quantities in mechanics can
be expressed in terms of these. e.g. if speed and time are taken as fundamental quantities,
length will become a derived quantity because then length will be expressed as Speed Time.
52 Units, Dimensions and Measurement
and if force and acceleration are taken as fundamental quantities, then mass will be defined as
Force / acceleration and will be termed as a derived quantity.
1.4 Fundamental and Derived Units.
Normally each physical quantity requires a unit or standard for its specification so it
appears that there must be as many units as there are physical quantities. However, it is not so.
It has been found that if in mechanics we choose arbitrarily units of any three physical
quantities we can express the units of all other physical quantities in mechanics in terms of
these. Arbitrarily the physical quantities mass, length and time are choosen for this purpose. So
any unit of mass, length and time in mechanics is called a fundamental, absolute or base unit.
Other units which can be expressed in terms of fundamental units, are called derived units. For
example light year or km is a fundamental units as it is a unit of length while s–1, m2 or kg/m are
derived units as these are derived from units of time, mass and length respectively.
System of units : A complete set of units, both fundamental and derived for all kinds of
physical quantities is called system of units. The common systems are given below –
(1) CGS system : The system is also called Gaussian system of units. In it length, mass and
time have been chosen as the fundamental quantities and corresponding fundamental units are
centimetre (cm), gram (g) and second (s) respectively.
(2) MKS system : The system is also called Giorgi system. In this system also length, mass
and time have been taken as fundamental quantities, and the corresponding fundamental units
are metre, kilogram and second.
(3) FPS system : In this system foot, pound and second are used respectively for
measurements of length, mass and time. In this system force is a derived quantity with unit
poundal.
(4) S. I. system : It is known as International system of units, and is infact extended
system of units applied to whole physics. There are seven fundamental quantities in this
system. These quantities and their units are given in the following table
Quantity
Name of Unit
Symbol
Length
metre
m
Mass
kilogram
kg
Time
second
s
Electric Current
ampere
A
Temperature
Kelvin
K
Amount of Substance
mole
mol
Luminous Intensity
candela
cd
Units, Dimensions and Measurement 53
Besides the above seven fundamental units two supplementary units are also defined –
Radian (rad) for plane angle and Steradian (sr) for solid angle.
Note : Apart from fundamental and derived units we also use very frequently practical
units. These may be fundamental or derived units
e.g., light year is a practical unit (fundamental) of distance while horse power is a
practical unit (derived) of power.
Practical units may or may not belong to a system but can be expressed in any
system of units
e.g., 1 mile = 1.6 km = 1.6 × 103 m.
1.5 S.I. Prefixes.
In physics we have to deal from very small (micro) to very large (macro) magnitudes as
one side we talk about the atom while on the other side of universe, e.g., the mass of an
electron is 9.1 10–31 kg while that of the sun is 2 1030 kg. To express such large or small
magnitudes simultaneously we use the following prefixes :
Power of 10
Prefix
Symbol
1018
exa
E
1015
peta
P
1012
tera
T
109
giga
G
106
mega
M
103
kilo
k
102
hecto
h
101
deca
da
10–1
deci
d
10–1
centi
c
10–3
milli
m
10–6
micro
10–9
nano
n
10–12
pico
p
10–15
femto
f
10–18
atto
a
1.6 Standards of Length, Mass and Time.
54 Units, Dimensions and Measurement
(1) Length : Standard metre is defined in terms of wavelength of light and is called atomic
standard of length.
The metre is the distance containing 1650763.73 wavelength in vacuum of the radiation
corresponding to orange red light emitted by an atom of krypton-86.
Now a days metre is defined as length of the path travelled by light in vacuum in
1/299,7792, 458 part of a second.
(2) Mass : The mass of a cylinder made of platinum-iridium alloy kept at International
Bureau of Weights and Measures is defined as 1 kg.
On atomic scale, 1 kilogram is equivalent to the mass of 5.0188 1025 atoms of 6C12 (an
isotope of carbon).
(3) Time : 1 second is defined as the time interval of 9192631770 vibrations of radiation in
Cs-133 atom. This radiation corresponds to the transition between two hyperfine level of the
ground state of Cs-133.
1.7 Practical Units.
(1) Length :
(i) 1 fermi = 1 fm = 10–15 m
(ii) 1 X-ray unit = 1XU = 10–13 m
(iii) 1 angstrom = 1Å = 10–10 m = 10–8 cm = 10–7 mm = 0.1
mm
(iv) 1 micron =
m = 10–6 m
(v) 1 astronomical unit = 1 A.U. = 1. 49 1011 m 1.5 1011 m 108 km
(vi) 1 Light year = 1 ly = 9.46 1015 m
(vii) 1 Parsec = 1pc = 3.26 light year
(2) Mass :
(i) Chandra Shekhar unit : 1 CSU = 1.4 times the mass of sun = 2.8 1030 kg
(ii) Metric tonne : 1 Metric tonne = 1000 kg
(iii) Quintal : 1 Quintal = 100 kg
(iv) Atomic mass unit (amu) : amu = 1.67 10–27 kg mass of proton or neutron is of the
order of 1 amu
(3) Time :
(i) Year : It is the time taken by earth to complete 1 revolution around the sun in its orbit.
(ii) Lunar month : It is the time taken by moon to complete 1 revolution around the earth in
its orbit.
1 L.M. = 27.3 days
Units, Dimensions and Measurement 55
(iii) Solar day : It is the time taken by earth to complete one rotation about its axis with
respect to sun. Since this time varies from day to day, average solar day is calculated by taking
average of the duration of all the days in a year and this is called Average Solar day.
1 Solar year = 365.25 average solar day
or average solar day
25.365
1
the part of solar year
(iv) Sedrial day : It is the time taken by earth to complete one rotation about its axis with
respect to a distant star.
1 Solar year = 366.25 Sedrial day = 365.25 average solar day
Thus 1 Sedrial day is less than 1 solar day.
(v) Shake : It is an obsolete and practical unit of time.
1 Shake = 10– 8 sec
1.8 Dimensions of a Physical Quantity.
When a derived quantity is expressed in terms of fundamental quantities, it is written as a
product of different powers of the fundamental quantities. The powers to which fundamental
quantities must be raised in order to express the given physical quantity are called its
dimensions.
To make it more clear, consider the physical quantity force
Force = mass × acceleration
time
velocity mass
time
elength/tim mass
= mass × length × (time)–2
.... (i)
Thus, the dimensions of force are 1 in mass, 1 in length and – 2 in time.
Here the physical quantity that is expressed in terms of the base quantities is enclosed in
square brackets to indicate that the equation is among the dimensions and not among the
magnitudes.
Thus equation (i) can be written as [force] = [MLT–2].
Such an expression for a physical quantity in terms of the fundamental quantities is called
the dimensional equation. If we consider only the R.H.S. of the equation, the expression is
termed as dimensional formula.
Thus, dimensional formula for force is, [MLT – 2].
1.9 Important Dimensions of Complete Physics.
Mechanics
S. N.
Quantity
Unit
Dimension
(1)
Velocity or speed (v)
m/s
[M0L1T –1]
(2)
Acceleration (a)
m/s2
[M0LT –2]
56 Units, Dimensions and Measurement
S. N.
Quantity
Unit
Dimension
(3)
Momentum (P)
kg-m/s
[M1L1T –1]
(4)
Impulse (I)
Newton-sec or kg-m/s
[M1L1T –1]
(5)
Force (F)
Newton
[M1L1T –2]
(6)
Pressure (P)
Pascal
[M1L–1T –2]
(7)
Kinetic energy (EK)
Joule
[M1L2T –2]
(8)
Power (P)
Watt or Joule/s
[M1L2T –3]
(9)
Density (d)
kg/m3
[M1L– 3T 0]
(10)
Angular displacement (
)
Radian (rad.)
[M0L0T 0]
(11)
Angular velocity (
)
Radian/sec
[M0L0T – 1]
(12)
Angular acceleration (
)
Radian/sec2
[M0L0T – 2]
(13)
Moment of inertia (I)
kg-m2
[M1L2T0]
(14)
Torque (
)
Newton-meter
[M1L2T –2]
(15)
Angular momentum (L)
Joule-sec
[M1L2T –1]
(16)
Force constant or spring constant
(k)
Newton/m
[M1L0T –2]
(17)
Gravitational constant (G)
N-m2/kg2
[M–1L3T – 2]
(18)
Intensity of gravitational field (Eg)
N/kg
[M0L1T – 2]
(19)
Gravitational potential (Vg)
Joule/kg
[M0L2T – 2]
(20)
Surface tension (T)
N/m or Joule/m2
[M1L0T – 2]
(21)
Velocity gradient (Vg)
Second–1
[M0L0T – 1]
(22)
Coefficient of viscosity (
)
kg/m-s
[M1L– 1T – 1]
(23)
Stress
N/m2
[M1L– 1T – 2]
(24)
Strain
No unit
[M0L0T 0]
(25)
Modulus of elasticity (E)
N/m2
[M1L– 1T – 2]
(26)
Poisson Ratio (
)
No unit
[M0L0T 0]
(27)
Time period (T)
Second
[M0L0T1]
(28)
Frequency (n)
Hz
[M0L0T –1]
Heat
S. N.
Quantity
Unit
Dimension
(1)
Temperature (T)
Kelvin
[M0L0T0
1]
(2)
Heat (Q)
Joule
[ML2T– 2]
(3)
Specific Heat (c)
Joule/kg-K
[M0L2T– 2
–1]
Units, Dimensions and Measurement 57
S. N.
Quantity
Unit
Dimension
(4)
Thermal capacity
Joule/K
[M1L2T – 2
–1]
(5)
Latent heat (L)
Joule/kg
[M0L2T – 2]
(6)
Gas constant (R)
Joule/mol-K
[M1L2T– 2
– 1]
(7)
Boltzmann constant (k)
Joule/K
[M1L2T– 2
– 1]
(8)
Coefficient of thermal conductivity
(K)
Joule/m-s-K
[M1L1T– 3
– 1]
(9)
Stefan's constant (
)
Watt/m2-K4
[M1L0T– 3
– 4]
(10)
Wien's constant (b)
Meter-K
[M0L1To
1]
(11)
Planck's constant (h)
Joule-s
[M1L2T–1]
(12)
Coefficient of Linear Expansion (
)
Kelvin–1
[M0L0T0
–1]
(13)
Mechanical eq. of Heat (J)
Joule/Calorie
[M0L0T0]
(14)
Vander wall’s constant (a)
Newton-m4
[ML5T– 2]
(15)
Vander wall’s constant (b)
m3
[M0L3T0]
Electricity
S. N.
Quantity
Unit
Dimension
(1)
Electric charge (q)
Coulomb
[M0L0T1A1]
(2)
Electric current (I)
Ampere
[M0L0T0A1]
(3)
Capacitance (C)
Coulomb/volt or Farad
[M–1L– 2T4A2]
(4)
Electric potential (V)
Joule/coulomb
M1L2T–3A–1
(5)
Permittivity of free space (
0)
2
2
meter-Newton
Coulomb
[M–1L–3T4A2]
(6)
Dielectric constant (K)
Unitless
[M0L0T0]
(7)
Resistance (R)
Volt/Ampere or ohm
[M1L2T– 3A– 2]
(8)
Resistivity or Specific resistance
(
)
Ohm-meter
[M1L3T– 3A– 2]
(9)
Coefficient of Self-induction (L)
ampere
secondvolt
or henery or ohm-second
[M1L2T– 2A– 2]
(10)
Magnetic flux (
)
Volt-second or weber
[M1L2T–2A–1]
(11)
Magnetic induction (B)
meterampere
newton
2
meterampere
Joule
2
second
meter
volt
or Tesla
[M1L0T– 2A– 1]
(12)
Magnetic Intensity (H)
Ampere/meter
[M0L– 1T0A1]
(13)
Magnetic Dipole Moment (M)
Ampere-meter2
[M0L2T0A1]
58 Units, Dimensions and Measurement
S. N.
Quantity
Unit
Dimension
(14)
Permeability of Free Space (
0)
2
ampere
Newton
or
meterampere
Joule
2
or
meterampere
Volt
second
or
meter
ondOhm sec
or
meter
henery
[M1L1T–2A–2]
(15)
Surface charge density (
)
2
metreCoulomb
[M0L–2T1A1]
(16)
Electric dipole moment (p)
meterCoulomb
[M0L1T1A1]
(17)
Conductance (G) (1/R)
1
ohm
[M–1L–2T3A2]
(18)
Conductivity (
) (1/
)
11 meterohm
[M–1L–3T3A2]
(19)
Current density (J)
Ampere/m2
M0L–2T0A1
(20)
Intensity of electric field (E)
Volt/meter, Newton/coulomb
M1L1T –3A–1
(21)
Rydberg constant (R)
m–1
M0L–1T0
1.10 Quantities Having Same Dimensions.
S. N.
Dimension
Quantity
(1)
[M0L0T–1]
Frequency, angular frequency, angular velocity, velocity gradient and decay
constant
(2)
[M1L2T–2]
Work, internal energy, potential energy, kinetic energy, torque, moment of
force
(3)
[M1L–1T–2]
Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy
density
(4)
[M1L1T–1]
Momentum, impulse
(5)
[M0L1T–2]
Acceleration due to gravity, gravitational field intensity
(6)
[M1L1T–2]
Thrust, force, weight, energy gradient
(7)
[M1L2T–1]
Angular momentum and Planck’s constant
(8)
[M1L0T–2]
Surface tension, Surface energy (energy per unit area)
(9)
[M0L0T0]
Strain, refractive index, relative density, angle, solid angle, distance
gradient, relative permittivity (dielectric constant), relative permeability etc.
(10)
[M0L2T–2]
Latent heat and gravitational potential
(11)
[M0L2T–2
–
1]
Thermal capacity, gas constant, Boltzmann constant and entropy
(12)
[M0L0T1]
gRkmgl ,,
, where l = length
g = acceleration due to gravity, m = mass, k = spring constant
(13)
[M0L0T1]
L/R,
LC
, RC where L = inductance, R = resistance, C = capacitance
(14)
[ML2T–2]
2
2
2
2
2, , , , ,, CV
C
q
LIqVVItt
R
V
RtI
where I = current, t = time, q = charge,
L = inductance, C = capacitance, R = resistance
1.11 Application of Dimensional Analysis.
(1) To find the unit of a physical quantity in a given system of units : Writing the
definition or formula for the physical quantity we find its dimensions. Now in the dimensional
Units, Dimensions and Measurement 59
formula replacing M, L and T by the fundamental units of the required system we get the unit of
physical quantity. However, sometimes to this unit we further assign a specific name, e.g.,
Work = Force Displacement
So [W] = [MLT–2] [L] = [ML2T–2]
So its units in C.G.S. system will be g cm2/s2 which is called erg while in M.K.S. system will
be kg m2/s2 which is called joule.
Sample problems based on unit finding
Problem 1. The equation
2
V
a
P
)( bV
= constant. The units of a is [MNR 1995; AFMC 1995]
(a)
5
cmDyne
(b)
4
cmDyne
(c)
3
/cmDyne
(d)
2
/cmDyne
Solution : (b) According to the principle of dimensional homogenity
2
][ V
a
P
][][][][][ 6212 LTMLVPa
][ 25
TML
or unit of a =
gm
×
5
cm
sec–2= Dyne cm4
Problem 2. If
,
2
btatx
where
x
is the distance travelled by the body in kilometre while
t
the time in
seconds, then the units of
b
are [CBSE 1993]
(a) km/s (b) km-s (c) km/s2 (d) km-s2
Solution : (c) From the principle of dimensional homogenity
][][ 2
btx
2
][ t
x
b
Unit of b = km/s2.
Problem 3. The unit of absolute permittivity is [EAMCET (Med.) 1995; Pb. PMT 2001]
(a) Farad - meter (b) Farad / meter (c) Farad/meter 2 (d) Farad
Solution : (b) From the formula
RC 0
4
R
C
4
0
By substituting the unit of capacitance and radius : unit of
0
Farad/ meter.
Problem 4. Unit of Stefan's constant is [MP PMT 1989]
(a)
1
Js
(b)
412 KsJm
(c)
2
Jm
(d)
Js
Solution : (b) Stefan's formula
4
T
At
Q
4
AtT
Q
Unit of
42 sec
Joule
Km
=
412 KsJm
Problem 5. The unit of surface tension in SI system is
[MP PMT 1984; AFMC 1986; CPMT 1985, 87; CBSE 1993; Karnataka CET (Engg/Med.) 1999; DCE 2000, 01]
(a)
2
/cmDyne
(b) Newton/m (c) Dyne/cm (d) Newton/m2
Solution : (b) From the formula of surface tension
l
F
T
By substituting the S.I. units of force and length, we will get the unit of surface tension =
Newton/m
60 Units, Dimensions and Measurement
Problem 6. A suitable unit for gravitational constant is [MNR 1988]
(a)
kg
1
sec
metre
(b)
sec
1
metreNewton
(c)
22
kgmetreNewton
(d)
1
sec
metrekg
Solution : (c) As
221
r
mGm
F
21
2
mm
Fr
G
Substituting the unit of above quantities unit of G =
22
kgmetreNewton
.
Problem 7. The SI unit of universal gas constant (R) is
[MP Board 1988; JIPMER 1993; AFMC 1996; MP PMT 1987, 94; CPMT 1984, 87; UPSEAT 1999]
(a) Watt
11 molK
(b) Newton
11 molK
(c) Joule
11 molK
(d)
11 molKErg
Solution : (c) Ideal gas equation
nRTPV
][][
][][
][
][][
][ 321
Kmole
LTML
nT
VP
R
][][
][ 22
Kmole
TML
So the unit will be
Joule
11 molK
.
(2) To find dimensions of physical constant or coefficients : As dimensions of a physical
quantity are unique, we write any formula or equation incorporating the given constant and
then by substituting the dimensional formulae of all other quantities, we can find the
dimensions of the required constant or coefficient.
(i) Gravitational constant : According to Newton’s law of gravitation
221
r
mm
GF
or
21
2
mm
Fr
G
Substituting the dimensions of all physical quantities
][
]][[
]][[
][ 231
22
TLM
MM
LMLT
G
(ii) Plank constant : According to Planck
hE
or
E
h
Substituting the dimensions of all physical quantities
][
][
][
][ 12
1
22
TML
T
TML
h
(iii) Coefficient of viscosity : According to Poiseuille’s formula
l
pr
dt
dV
8
4
or
)/(8
4
dtdVl
pr
Substituting the dimensions of all physical quantities
][
]/][[
]][[
][ 11
3
421
TML
TLL
LTML
Sample problems based on dimension finding
Problem 8.
2
3YZX
find dimension of
Y
in (MKSA) system, if
X
and
Z
are the dimension of capacity
and magnetic
field respectively [MP PMT 2003]
(a)
1423 ATLM
(b)
2
ML
(c)
4423 ATLM
(d)
4823 ATLM
Units, Dimensions and Measurement 61
Solution : (d)
2
3YZX
][
][
][ 2
Z
X
Y
212
2421
][
][
AMT
ATLM
][ 4823 ATLM
.
Problem 9. Dimensions of
,
1
00
where symbols have their usual meaning, are [AIEEE 2003]
(a)
][ 1
LT
(b)
][ 1TL
(c)
][ 22TL
(d)
][ 22
TL
Solution : (d) We know that velocity of light
00
1
C
2
00
1C
So
21
00
][
1
LT
=
][ 22
TL
.
Problem 10. If L, C and R denote the inductance, capacitance and resistance respectively, the
dimensional formula for
LRC2
is [UPSEAT 2002]
(a)
][ 012 ITML
(b)
][ 0300 ITLM
(c)
][ 2621 ITLM
(d)
][ 0200 ITLM
Solution : (b)
][ 2LRC
=
L
R
LC 22
=
L
R
LC 2
)(
and we know that frequency of LC circuits is given by
LC
f1
2
1
i.e., the dimension of LC
is equal to
][ 2
T
and
R
L
gives the time constant of
RL
circuit so the dimension of
R
L
is equal to [T].
By substituting the above dimensions in the given formula
][][][)( 31222 TTT
L
R
LC
.
Problem 11. A force F is given by
,
2
btatF
where
t
is time. What are the dimensions of a and b
[BHU 1998; AFMC 2001]
(a)
3
MLT
and
42
TML
(b)
3
MLT
and
4
MLT
(c)
1
MLT
and
0
MLT
(d) and
Solution : (b) From the principle of dimensional homogenity
][][ atF
T
MLT
t
F
a2
][
][ 3
MLT
Similarly
][][ 2
btF
2
2
2
][ T
MLT
t
F
b
][ 4
MLT
.
Problem 12. The position of a particle at time t is given by the relation
),1()( 0t
c
v
tx
where
0
v
is a
constant and
0
. The dimensions of
0
v
and
are respectively [CBSE 1995]
(a)
110
TLM
and
1
T
(b)
010 TLM
and
1
T
(c)
110
TLM
and
2
LT
(d)
110
TLM
and
T
Solution : (a) From the principle of dimensional homogeneity
][ t
= dimensionless
][
1
][ 1
T
t
Similarly
][
]
][ 0
v
x
][][][]][[][ 11
0 LTTLxv
.
62 Units, Dimensions and Measurement
Problem 13. The dimensions of physical quantity X in the equation Force
Density
X
is given by [DCE 1993]
(a)
241
TLM
(b)
122 TLM
(c)
222 TLM
(d)
121 TLM
Solution : (c) [X] = [Force] × [Density] =
][][ 32 MLMLT
=
][ 222 TLM
.
Problem 14. Number of particles is given by
12
12 xx
nn
Dn
crossing a unit area perpendicular to X- axis
in unit time, where
1
n
and
2
n
are number of particles per unit volume for the value of x
meant to
2
x
and
.
1
x
Find dimensions of D called as diffusion constant [CPMT 1979]
(a)
20 LTM
(b)
420
TLM
(c)
30
LTM
(d)
120
TLM
Solution : (d) (n) = Number of particle passing from unit area in unit time =
tA
eof particl No.
][][
][
2
000
TL
TLM
=
][ 12 TL
][][ 21 nn
No. of particle in unit volume =
][ 3
L
Now from the given formula
][
]][[
][
12
12 nn
xxn
D
][
][][
3
12
L
LTL
][ 12
TL
.
Problem 15. E, m, l and G denote energy, mass, angular momentum and gravitational constant
respectively, then the dimension of
25
2
Gm
El
are [AIIMS 1985]
(a) Angle (b) Length (c) Mass (d) Time
Solution : (a)
][E
= energy =
][ 22
TML
, [m] = mass = [M], [l] = Angular momentum =
][ 12
TML
[G] = Gravitational constant =
][ 231 TLM
Now substituting dimensions of above quantities in
25
2
Gm
El
=
22315
21222
][][
][][
TLMM
TMLTML
=
][ 000 TLM
i.e., the quantity should be angle.
Problem 16. The equation of a wave is given by
AY
sin
k
v
x
where
is the angular velocity and
v
is the linear velocity. The dimension of k is [MP PMT 1993]
(a)
LT
(b)
T
(c)
1
T
(d)
2
T
Solution : (b) According to principle of dimensional homogeneity
v
x
k][
=
][
1T
LT
L
.
Problem 17. The potential energy of a particle varies with distance x from a fixed origin as
,
2Bx
xA
U
where A and B are dimensional constants then dimensional formula for AB is
(a) ML7/2T
2
(b)
22/11
TML
(c)
22/92
TLM
(d)
32/13
TML
Solution : (b) From the dimensional homogeneity
][][ 2Bx
[B] = [L2]
As well as
][][
][][
2
2/1
Bx
xA
U
][
][][
][ 2
2/1
22
L
LA
TML
][][ 22/7
TMLA
Units, Dimensions and Measurement 63
Now
][][][ 222/7 LTMLAB
][ 22/11
TML
Problem 18. The dimensions of
2
1
2
0E
(
0
= permittivity of free space ; E = electric field ) is [IIT-JEE 1999]
(a)
1
MLT
(b) ML
2
T
2
(c) ML
1
T
2
(d)
12
TML
Solution : (c) Energy density =
Volume
Energy
2
12
0E
][ 21
3
22
TML
L
TML
Problem 19. You may not know integration. But using dimensional analysis you can check on some
results. In the integral
1sin
)2(
1
2/12 a
x
a
xax
dx n
the value of n is
(a) 1 (b) – 1 (c) 0 (d)
2
1
Solution : (c) Let x = length
][][ LX
and
][][ Ldx
By principle of dimensional homogeneity
a
x
dimensionless
][][][ Lxa
By substituting dimension of each quantity in both sides:
][
][
][
2/122 n
L
LL
L
0 n
Problem 20. A physical quantity
m
lB
P22
where B= magnetic induction, l= length and m = mass. The
dimension of P is
(a)
3
MLT
(b)
42
TML
I–2 (c)
ITLM 422
(d)
22 IMLT
Solution : (b) F = BIL
][][
][
][][
][
][ofDimension 2
LI
MLT
LI
F
B
=
][ 12 IMT
Now dimension of
][
][][
][ 221222
M
LIMT
m
lB
P
][ 242
ITML
Problem 21. The equation of the stationary wave is y=
xct
a2
cos
2
sin2
, which of the following
statements is wrong
(a) The unit of
ct
is same as that of
(b) The unit of x is same as that of
(c) The unit of
c
2
/
is same as that of
x
2
/
t (d) The unit of c/
is same
as that of
/x
Solution : (d) Here,
ct2
as well as
x2
are dimensionless (angle) i.e.
000
22 TLM
xct
So (i) unit of c t is same as that of
(ii) unit of x is same as that of
(iii)
t
xc
22
and (iv)
x
is unit less. It is not the case with
.
c
(3) To convert a physical quantity from one system to the other : The measure of a
physical quantity is nu = constant
64 Units, Dimensions and Measurement
If a physical quantity X has dimensional formula [MaLbTc] and if (derived) units of that
physical quantity in two systems are
][ 111 cba TLM
and
][ 222 cba TLM
respectively and n1 and n2 be the
numerical values in the two systems respectively, then
][][ 2211 unun
][][ 22221111 cbacba TLMnTLMn
cba
T
T
L
L
M
M
nn
2
1
2
1
2
1
12
where M1, L1 and T1 are fundamental units of mass, length and time in the first (known)
system and M2, L2 and T2 are fundamental units of mass, length and time in the second
(unknown) system. Thus knowing the values of fundamental units in two systems and
numerical value in one system, the numerical value in other system may be evaluated.
Example : (1) conversion of Newton into Dyne.
The Newton is the S.I. unit of force and has dimensional formula [MLT–2].
So 1 N = 1 kg-m/ sec2
By using
cba
T
T
L
L
M
M
nn
2
1
2
1
2
1
12
21
1
1
sec
sec
cm
m
gm
kg
2
1
2
1
31010
1
sec
sec
cm
cm
gm
gm
5
10
1 N = 105 Dyne
(2) Conversion of gravitational constant (G) from C.G.S. to M.K.S. system
The value of G in C.G.S. system is 6.67 10–8 C.G.S. units while its dimensional formula is
[M–1L3T–2]
So G = 6.67 10–8 cm3/g s2
By using
cba
T
T
L
L
M
M
nn
2
1
2
1
2
1
12
23
1
8
1067.6
sec
sec
m
cm
kg
gm
11
23
2
1
3
81067.6
1010
1067.6
sec
sec
cm
cm
gm
gm
G = 6.67 10–11 M.K.S. units
Sample problems based on conversion
Problem 22. A physical quantity is measured and its value is found to be
nu
where
n
numerical value
and
u
unit.
Then which of the following relations is true [RPET 2003]
(a)
2
un
(b)
un
(c)
un
(d)
u
n1
Solution : (d) We know
nuP
constant
2211 unun
or
u
n1
.
Units, Dimensions and Measurement 65
Problem 23. In C.G.S. system the magnitude of the force is 100 dynes. In another system where the
fundamental physical quantities are kilogram, metre and minute, the magnitude of the force
is [EAMCET 2001]
(a) 0.036 (b) 0.36 (c) 3.6 (d) 36
Solution : (c)
100
1n
,
gM
1
,
cmL
1
,
sec
1T
and
kgM
2
,
meterL
2
,
inute
2mT
,
1x
,
1y
,
2z
By substituting these values in the following conversion formula
2
2
1
2
1
2
1
12
T
T
L
L
M
M
nn
yx
21
1
2minute
sec
100
meter
cm
kg
gm
n
21
2
1
3
2sec60
sec
1010
100
cm
cm
gm
gm
n
6.3
Problem 24. The temperature of a body on Kelvin scale is found to be X K. When it is measured by a
Fahrenheit thermometer, it is found to be X F. Then X is [UPSEAT 200]
(a) 301.25 (b) 574.25 (c) 313 (d) 40
Solution : (c) Relation between centigrade and Fahrenheit
9
32
5
273
FK
According to problem
9
32
5
273
XX
313X
.
Problem 25. Which relation is wrong [RPMT 1997]
(a) 1 Calorie = 4.18 Joules (b) 1Å =10–10 m
(c) 1 MeV = 1.6 × 10–13 Joules (d) 1 Newton =10–5 Dynes
Solution : (d) Because 1 Newton =
5
10
Dyne.
Problem 26. To determine the Young's modulus of a wire, the formula is
;. l
L
A
F
Y
where L= length, A=
area of cross- section of the wire,
L
Change in length of the wire when stretched with a
force F. The conversion factor to change it from CGS to MKS system is [MP PET 1983]
(a) 1 (b) 10 (c) 0.1 (d) 0.01
Solution : (c) We know that the dimension of young's modulus is
][ 21 TML
C.G.S. unit : gm
21 sec
cm
and M.K.S. unit : kg. m–1 sec–2 .
By using the conversion formula:
2
2
1
1
2
1
1
2
1
12
T
T
L
L
M
M
nn
21
1
sec
sec
meter
cm
kg
gm
Conversion factor
2
1
2
1
3
1
2sec
sec
1010
cm
cm
gm
gm
n
n
1.0
10
1
Problem 27. Conversion of 1 MW power on a new system having basic units of mass, length and time
as 10kg, 1dm and 1 minute respectively is
(a)
unit
12
1016.2
(b)
unit
12
1026.1
(c)
unit
10
1016.2
(d)
66 Units, Dimensions and Measurement
Solution : (a)
][][ 32
TMLP
Using the relation
zyx
T
T
L
L
M
M
nn
2
1
2
1
2
1
12
3
21
6min1
1
1
1
10
1
101
s
dm
m
kg
kg
[As
WMW 6
101
]
3
2
660
1
1
10
10
1
10
sec
sec
dm
dm
kg
kg
12
1016.2
unit
Problem 28. In two systems of relations among velocity, acceleration and force are respectively
,
1
2
2vv
12 aa
and
.
1
2
F
F
If
and
are constants then relations among mass, length
and time in two systems are
(a)
1
3
21
2
2
212 ,, T
TLLMM
(b)
2
121
3
3
21
22
2,,
1
TTLLMM
(c)
121
2
2
21
3
3
2,, TTLLMM
(d)
1
3
3
21
2
21
2
2
2,, TTLLMM
Solution : (b)
2
12 vv
2
1
11
1
22 ][][ TLTL
......(i)
12 aa
][][ 2
11
2
22 TLTL
......(ii)
and
1
2F
F
1
][][ 2
111
2
222 TLMTLM
......(iii)
Dividing equation (iii) by equation (ii) we get
)( 1
2M
M
22 1
B
M
Squaring equation (i) and dividing by equation (ii) we get
3
3
12
LL
Dividing equation (i) by equation (ii) we get
2
12
TT
Problem 29. If the present units of length, time and mass (m, s, kg) are changed to 100m, 100s, and
10
1
kg then
(a) The new unit of velocity is increased 10 times (b) The new unit of
force is decreased
1000
1
times
(c) The new unit of energy is increased 10 times (d) The new unit of
pressure is increased 1000 times
Solution : (b) Unit of velocity = m/sec ; in new system =
secsec100
100 mm
(same)
Unit of force
2
sec
mkg
; in new system
sec100sec100
100
10
1
m
kg
2
sec
1000
1mkg
Units, Dimensions and Measurement 67
Unit of energy
2
2
sec
mkg
; in new system
sec100sec100
100100
10
1
mm
kg
2
2
10 sec
mkg
Unit of pressure
2
sec
m
kg
; in new system
secsec
mkg 100100
1
100
1
10
1
2
7
10 secm
kg
Problem 30. Suppose we employ a system in which the unit of mass equals 100 kg, the unit of length
equals 1 km and the unit of time 100 s and call the unit of energy eluoj (joule written in
reverse order), then
(a) 1 eluoj = 104 joule (b) 1 eluoj = 10-3 joule (c) 1 eluoj = 10-4 joule (d) 1 joule = 103 eluoj
Solution : (a)
][][ 22
TMLE
1 eluoj
22 sec]100[]1[]100[
kmkg
2426 sec1010100
mkg
224 sec10
mkg
Joule
4
10
Problem 31. If 1gm cms–1 = x Ns, then number x is equivalent to
(a)
1
101
(b)
2
103
(c)
4
106
(d)
5
101
Solution : (d)
1
-
scmgm
123 1010 smkg
15
10 smkg
= 10–5 Ns
(4) To check the dimensional correctness of a given physical relation : This is based on
the ‘principle of homogeneity’. According to this principle the dimensions of each term on both
sides of an equation must be the same.
If
DEFBCAX 2
)(
,
then according to principle of homogeneity [X] = [A] = [(BC)2]
][ DEF
If the dimensions of each term on both sides are same, the equation is dimensionally
correct, otherwise not. A dimensionally correct equation may or may not be physically correct.
Example : (1)
22 /rmvF
By substituting dimension of the physical quantities in the above relation –
2212 ]/[]][[][ LLTMMLT
i.e.
][][ 22 MTMLT
As in the above equation dimensions of both sides are not same; this formula is not correct
dimensionally, so can never be physically.
(2)
2
)2/1( atuts
By substituting dimension of the physical quantities in the above relation –
[L] = [LT–1][T] – [LT–2][T2]
i.e. [L] = [L] – [L]
As in the above equation dimensions of each term on both sides are same, so this equation
is dimensionally correct. However, from equations of motion we know that
2
)2/1( atuts
Sample problems based on formulae checking
68 Units, Dimensions and Measurement
Problem 32. From the dimensional consideration, which of the following equation is correct [CPMT 1983]
(a)
GM
R
T3
2
(b)
3
2R
GM
T
(c)
2
2R
GM
T
(d)
GM
R
T2
2
Solution : (a)
GM
R
T3
2
2
3
2gR
R
g
R
2
[As GM = gR2]
Now by substituting the dimension of each quantity in both sides.
2/1
2
][
LT
L
T
][T
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 33. A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another
cubical block B of the same dimensions and of low modulus of rigidity
such that the
lower face of A completely covers the upper face of B. The lower face of B is rigidly held on
a horizontal surface. A small force F is applied perpendicular to one of the side faces of A.
After the force is withdrawn block A executes small oscillations. The time period of which
is given by [IIT-JEE 1992]
(a)
L
M
2
(b)
M
L
2
(c)
ML
2
(d)
L
M
2
Solution : (d) Given m = mass = [M],
= coefficient of rigidity =
][ 21 TML
, L = length = [L]
By substituting the dimension of these quantity we can check the accuracy of the given
formulae
2/1
][][
][
2][
L
M
T
=
2/1
21
LTML
M
= [T].
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 34. A small steel ball of radius r is allowed to fall under gravity through a column of a viscous
liquid of coefficient of viscosity. After some time the velocity of the ball attains a constant
value known as terminal velocity
.
T
v
The terminal velocity depends on (i) the mass of the
ball. (ii)
(iii) r and (iv) acceleration due to gravity g. which of the following relations is
dimensionally correct [CPMT 1992; CBSE 1992; NCERT 1983; MP PMT 2001]
(a)
r
mg
vT
(b)
mg
r
vT
(c)
rmgvT
(d)
Solution : (a) Given
T
v
= terminal velocity =
][ 1
LT
, m = Mass = [M], g = Acceleration due to gravity =
][ 2
LT
r = Radius = [L],
= Coefficient of viscosity =
][
By substituting the dimension of each quantity we can check the accuracy of given formula
r
mg
vT
Units, Dimensions and Measurement 69
][][
][][
][ 11
2
1
LTML
LTM
LT
=
][ 1
LT
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 35. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity
flowing per second through a tube of radius
r
and length l and having a pressure difference
p across its end, is
(a)
l
pr
V
8
4
(b)
4
8pr
l
V
(c)
4
8
r
lp
V
(d)
4
8lr
p
V
Solution : (a) Given V = Rate of flow =
][
sec
Volume 13
TL
, P = Pressure =
][ 21 TML
, r = Radius = [L]
= Coefficient of viscosity =
][ 11 TML
, l = Length = [L]
By substituting the dimension of each quantity we can check the accuracy of the formula
l
rP
V
8
4
][][
][][
][ 11
421
13
LTML
LTML
TL
=
][ 13
TL
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 36. With the usual notations, the following equation
)12(
2
1 tauSt
is
(a) Only numerically correct (b) Only dimensionally
correct
(c) Both numerically and dimensionally correct (d) Neither numerically
nor dimensionally correct
Solution : (c) Given
t
S
= distance travelled by the body in tth sec.=
][ 1
LT
, a = Acceleration =
][ 2
LT
,
v = velocity =
][ 1
LT
, t = time = [T]
By substituting the dimension of each quantity we can check the accuracy of the formula
)12(
2
1 tauSt
][][][][ 211 TLTLTLT
][][][ 111 LTLTLT
Since the dimension of each terms are equal therefore this equation is dimensionally
correct. And after deriving this equation from Kinematics we can also proof that this
equation is correct numerically also.
Problem 37. If velocity
,v
acceleration A and force F are chosen as fundamental quantities, then the
dimensional formula of angular momentum in terms of
Av,
and
F
would be
(a)
1
FA
v (b)
23
AFv
(c)
12
AFv
(d)
122
AvF
Solution : (b) Given, v = velocity =
][ 1
LT
, A = Acceleration =
][ 2
LT
, F = force =
][ 2
MLT
By substituting, the dimension of each quantity we can check the accuracy of the formula
70 Units, Dimensions and Measurement
[Angular momentum] =
23
AFv
][ 12
TML
22312 ][][][
LTLTMLT
=
][ 12
TML
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 38. The largest mass (m) that can be moved by a flowing river depends on velocity (v), density
(
) of river water and acceleration due to gravity (g). The correct relation is
(a)
2
42
g
v
m
(b)
2
6
g
v
m
(c)
3
4
g
v
m
(d)
3
6
g
v
m
Solution : (d) Given, m = mass = [M], v = velocity =
][ 1
LT
,
= density =
][ 3
ML
, g = acceleration due to
gravity = [LT–2]
By substituting, the dimension of each quantity we can check the accuracy of the formula
3
6
g
v
Km
32
613
][
]][[
][
LT
LTML
M
= [M]
L.H.S. = R.H.S. i.e., the above formula is Correct.
(5) As a research tool to derive new relations : If one knows the dependency of a physical
quantity on other quantities and if the dependency is of the product type, then using the method
of dimensional analysis, relation between the quantities can be derived.
Example : (i) Time period of a simple pendulum.
Let time period of a simple pendulum is a function of mass of the bob (m), effective length
(l), acceleration due to gravity (g) then assuming the function to be product of power function
of m, l and g
i.e.,
zyx glKmT
; where K = dimensionless constant
If the above relation is dimensionally correct then by substituting the dimensions of
quantities –
[T] = [M]x [L]y [LT–2]z
or [M0L0T1] = [MxLy+zT–2z]
Equating the exponents of similar quantities x = 0, y = 1/2 and z = – 1/2
So the required physical relation becomes
g
l
KT
The value of dimensionless constant is found (2
) through experiments so
g
l
T
2
Units, Dimensions and Measurement 71
(ii) Stoke’s law : When a small sphere moves at low speed through a fluid, the viscous
force F, opposing the motion, is found experimentally to depend on the radius r, the velocity of
the sphere v and the viscosity
of the fluid.
So F = f (
, r, v)
If the function is product of power functions of
, r and v,
zyx vrKF
; where K is
dimensionless constant.
If the above relation is dimensionally correct
zyx LTLTMLMLT ][][][][ 1112
or
][][ 2zxzyxx TLMMLT
Equating the exponents of similar quantities x = 1; – x + y + z = 1 and – x – z = – 2
Solving these for x, y and z, we get x = y = z = 1
So eqn (i) becomes F = K
rv
On experimental grounds, K = 6
; so F = 6
rv
This is the famous Stoke’s law.
Sample problem based on formulae derivation
Problem 39. If the velocity of light (c), gravitational constant (G) and Planck's constant (h) are chosen
as fundamental units, then the dimensions of mass in new system is [UPSEAT 2002]
(a)
2/12/12/1 hGc
(b)
2/12/12/1
hGc
(c)
2/12/12/1 hGc
(d)
2/12/12/1 hGc
Solution : (c) Let
zyx hGcm
or
zyx hGcKm
By substituting the dimension of each quantity in both sides
zyx TMLTLMLTKTLM ][][][][ 122311001
][ 223 zyxzyxzy TLM
By equating the power of M, L and T in both sides :
1 zy
,
023 zyx
,
02 zyx
By solving above three equations
2/1x
,
2/1y
and
2/1z
.
2/12/12/1 hGcm
Problem 40. If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius
(r) of the drop and density
)(
of the liquid, then the expression of T is [AMU (Med.) 2000]
(a)
SrKT /
3
(b)
SrKT /
32/1
(c)
2/13 /SrKT
(d) None of these
Solution : (a) Let
zyxrST
or T =
zyx rSK
By substituting the dimension of each quantity in both sides
zyx MLLMTKTLM ][][][][ 32100
][ 23 xzyzx TLM
By equating the power of M, L and T in both sides
0 zx
,
03 zy
,
12 x
By solving above three equations
2/1x
,
2/3y
,
2/1z
So the time period can be given as,
S
r
KrSKT 3
2/12/32/1
.
72 Units, Dimensions and Measurement
Problem 41. If P represents radiation pressure, C represents speed of light and Q represents radiation
energy striking a unit area per second, then non-zero integers x, y and z such that
zyx CQP
is dimensionless, are
[AFMC 1991; CBSE 1992; CPMT 1981, 92; MP PMT 1992]
(a)
1,1,1 zyx
(b)
1,1,1 zyx
(c)
1,1,1 zyx
(d)
1,1,1 zyx
Solution : (b)
000
][ TLMCQP zyx
By substituting the dimension of each quantity in the given expression
zyx LTMTTML ][][][ 1321
00032 ][ TLMTLM zyxzxyx
by equating the power of M, L and T in both sides:
0 yx
,
0 zx
and
032 zyx
by solving we get
1,1,1 zyx
.
Problem 42. The volume V of water passing through a point of a uniform tube during t seconds is related
to the cross-sectional area A of the tube and velocity u of water by the relation
tuAV
,
which one of the following will be true
(a)
(b)
(c)
(d)
Solution : (b) Writing dimensions of both sides
][][][][ 123 TLTLL
][][ 203
TLTL
By comparing powers of both sides
32
and
0
Which give
and
)3(
2
1
i.e.
.
Problem 43. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional
formula for mass will be
(a)
EFV 02
(b)
20FEV
(c)
02EVF
(d)
EFV 02
Solution : (d) Let
cba EFVM
Putting dimensions of each quantities in both side
cba TMLMLTLTM][][][][ 2221
Equating powers of dimensions. We have
,1 cb
02 cba
and
022 cba
Solving these equations,
,2a
b = 0 and c = 1
So
][ 02 EFVM
Problem 44. Given that the amplitude A of scattered light is :
(i) Directly proportional to the amplitude (A0) of incident light.
(ii) Directly proportional to the volume (V) of the scattering particle
(iii) Inversely proportional to the distance (r) from the scattered particle
(iv) Depend upon the wavelength (
) of the scattered light. then:
(a)
1
A
(b)
2
1
A
(c)
3
1
A
(d)
4
1
A
Solution : (b) Let
r
VKA
Ax
0
By substituting the dimension of each quantity in both sides
Units, Dimensions and Measurement 73
][
][].[][
][ 3
L
LLL
Lx
][][ 3x
LL
;
13 x
or
2x
2
A
1.12 Limitations of Dimensional Analysis.
Although dimensional analysis is very useful it cannot lead us too far as,
(1) If dimensions are given, physical quantity may not be unique as many physical
quantities have same dimensions. For example if the dimensional formula of a physical quantity
is
][ 22
TML
it may be work or energy or torque.
(2) Numerical constant having no dimensions [K] such as (1/2), 1 or 2
etc. cannot be
deduced by the methods of dimensions.
(3) The method of dimensions can not be used to derive relations other than product of
power functions. For example,
2
)2/1( tatus
or
tay
sin
cannot be derived by using this theory (try if you can). However, the dimensional
correctness of these can be checked.
(4) The method of dimensions cannot be applied to derive formula if in mechanics a
physical quantity depends on more than 3 physical quantities as then there will be less number
(= 3) of equations than the unknowns (>3). However still we can check correctness of the given
equation dimensionally. For example
mglT 12
can not be derived by theory of dimensions
but its dimensional correctness can be checked.
(5) Even if a physical quantity depends on 3 physical quantities, out of which two have
same dimensions, the formula cannot be derived by theory of dimensions, e.g., formula for the
frequency of a tuning fork
vLdf )/( 2
cannot be derived by theory of dimensions but can be
checked.
1.13 Significant Figures.
Significant figures in the measured value of a physical quantity tell the number of digits in
which we have confidence. Larger the number of significant figures obtained in a measurement,
greater is the accuracy of the measurement. The reverse is also true.
The following rules are observed in counting the number of significant figures in a given
measured quantity.
(1) All non-zero digits are significant.
Example : 42.3 has three significant figures.
243.4 has four significant figures.
74 Units, Dimensions and Measurement
24.123 has five significant figures.
(2) A zero becomes significant figure if it appears between to non-zero digits.
Example : 5.03 has three significant figures.
5.604 has four significant figures.
4.004 has four significant figures.
(3) Leading zeros or the zeros placed to the left of the number are never significant.
Example : 0.543 has three significant figures.
0.045 has two significant figures.
0.006 has one significant figures.
(4) Trailing zeros or the zeros placed to the right of the number are significant.
Example : 4.330 has four significant figures.
433.00 has five significant figures.
343.000 has six significant figures.
(5) In exponential notation, the numerical portion gives the number of significant figures.
Example : 1.32 10–2 has three significant figures.
1.32 104 has three significant figures.
1.14 Rounding Off.
While rounding off measurements, we use the following rules by convention:
(1) If the digit to be dropped is less than 5, then the preceding digit is left unchanged.
Example :
82.7x
is rounded off to 7.8, again
94.3x
is rounded off to 3.9.
(2) If the digit to be dropped is more than 5, then the preceding digit is raised by one.
Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8.
(3) If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is
raised by one.
Example : x = 16.351 is rounded off to 16.4, again x = 6.758 is rounded off to 6.8.
(4) If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if
it is even.
Example : x = 3.250 becomes 3.2 on rounding off, again x = 12.650 becomes 12.6 on
rounding off.
(5) If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one,
if it is odd.
Example : x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off to 16.2.
1.15 Significant Figures in Calculation.
Units, Dimensions and Measurement 75
In most of the experiments, the observations of various measurements are to be combined
mathematically, i.e., added, subtracted, multiplied or divided as to achieve the final result.
Since, all the observations in measurements do not have the same precision, it is natural that
the final result cannot be more precise than the least precise measurement. The following two
rules should be followed to obtain the proper number of significant figures in any calculation.
(1) The result of an addition or subtraction in the number having different precisions
should be reported to the same number of decimal places as are present in the number having
the least number of decimal places. The rule is illustrated by the following examples :
(i) 33.3 (has only one decimal place)
3.11
+ 0.313
36.723 (answer should be reported to one decimal place)
Answer = 36.7
(ii) 3.1421
0.241
+ 0.09 (has 2 decimal places)
3.4731 (answer should be reported to 2 decimal places)
Answer = 3.47
(iii) 62.831 (has 3 decimal places)
– 24.5492
38.2818 (answer should be reported to 3 decimal places after
rounding off)
Answer = 38.282
(2) The answer to a multiplication or division is rounded off to the same number of
significant figures as is possessed by the least precise term used in the calculation. The rule is
illustrated by the following examples :
(i) 142.06
0.23 (two significant figures)
32.6738 (answer should have two significant figures)
Answer = 33
(ii) 51.028
1.31 (three significant figures)
66.84668
Answer = 66.8
76 Units, Dimensions and Measurement
(iii)
2112676.0
26.4
90.0
Answer = 0.21
1.16 Order of Magnitude.
In scientific notation the numbers are expressed as, Number
x
M10
. Where M is a
number lies between 1 and 10 and x is integer. Order of magnitude of quantity is the power of
10 required to represent the quantity. For determining this power, the value of the quantity has
to be rounded off. While rounding off, we ignore the last digit which is less than 5. If the last
digit is 5 or more than five, the preceding digit is increased by one. For example,
(1) Speed of light in vacuum
smms /10103818
(ignoring 3 < 5)
(2) Mass of electron
kgkg 3031 10101.9
(as 9.1 > 5).
Sample problems based on significant figures
Problem 45. Each side a cube is measured to be 7.203 m. The volume of the cube up to appropriate
significant figures is
(a) 373.714 (b) 373.71 (c) 373.7 (d) 373
Solution : (c) Volume
33 )023.7( a
3
715.373 m
In significant figures volume of cube will be
3
7.373 m
because its side has four significant
figures.
Problem 46. The number of significant figures in 0.007
2
m
is
(a) 1 (b) 2 (c) 3 (d) 4
Solution : (a)
Problem 47. The length, breadth and thickness of a block are measured as 125.5 cm, 5.0 cm and 0.32 cm
respectively. Which one of the following measurements is most accurate
(a) Length (b) Breadth (c) Thickness (d) Height
Solution : (a) Relative error in measurement of length is minimum, so this measurement is most
accurate.
Problem 48. The mass of a box is 2.3 kg. Two marbles of masses 2.15 g and 12.39 g are added to it. The
total mass of the box to the correct number of significant figures is
(a) 2.340 kg (b) 2.3145 kg. (c) 2.3 kg (d) 2.31 kg
Solution : (c) Total mass
kg31.201239.000215.03.2
Total mass in appropriate significant figures be 2.3 kg.
Problem 49. The length of a rectangular sheet is 1.5 cm and breadth is 1.203 cm. The area of the face of
rectangular sheet to the correct no. of significant figures is :
(a) 1.8045
2
cm
(b) 1.804
2
cm
(c) 1.805
2
cm
(d) 1.8
2
cm
Solution : (d) Area
2
8045.1203.15.1 cm
2
8.1 cm
(Upto correct number of significant figure).
Units, Dimensions and Measurement 77
Problem 50. Each side of a cube is measured to be 5.402 cm. The total surface area and the volume of
the cube in appropriate significant figures are :
(a) 175.1
2
cm
, 157
2
cm
(b) 175.1
2
cm
, 157.6
3
cm
(c) 175
2
cm
, 157
2
cm
(d) 175.08
2
cm
, 157.639
3
cm
Solution : (b) Total surface area =
22 09.175)402.5(6 cm
2
1.175 cm
(Upto correct number of significant
figure)
Total volume
33 64.175)402.5( cm
3
6.175 cm
(Upto correct number of significant figure).
Problem 51. Taking into account the significant figures, what is the value of 9.99 m + 0.0099 m
(a) 10.00 m (b) 10 m (c) 9.9999 m (d) 10.0 m
Solution : (a)
mmm 999.90099.099.9
m00.10
(In proper significant figures).
Problem 52. The value of the multiplication 3.124
4.576 correct to three significant figures is
(a) 14.295 (b) 14.3 (c) 14.295424 (d) 14.305
Solution : (b)
295.14576.4124.3
=14.3 (Correct to three significant figures).
Problem 53. The number of the significant figures in 11.118
10
6
V is
(a) 3 (b) 4 (c) 5 (d) 6
Solution : (c) The number of significant figure is 5 as
6
10
does not affect this number.
Problem 54. If the value of resistance is 10.845 ohms and the value of current is 3.23 amperes, the
potential difference is 35.02935 volts. Its value in significant number would be [CPMT 1979]
(a) 35 V (b) 35.0 V (c) 35.03 V (d) 35.025 V
Solution : (b) Value of current (3.23 A) has minimum significant figure (3) so the value of potential
difference
)( IRV
have only 3 significant figure. Hence its value be 35.0 V.
1.17 Errors of Measurement.
The measuring process is essentially a process of comparison. Inspite of our best efforts,
the measured value of a quantity is always somewhat different from its actual value, or true
value. This difference in the true value of a quantity is called error of measurement.
(1) Absolute error : Absolute error in the measurement of a physical quantity is the
magnitude of the difference between the true value and the measured value of the quantity.
Let a physical quantity be measured n times. Let the measured value be a1, a2, a3, ….. an. The
arithmetic mean of these value is
n
aaa
an
m....
21
Usually, am is taken as the true value of the quantity, if the same is unknown otherwise.
By definition, absolute errors in the measured values of the quantity are
11 aaa m
22 aaa m
………….
78 Units, Dimensions and Measurement
nmn aaa
The absolute errors may be positive in certain cases and negative in certain other cases.
(2) Mean absolute error : It is the arithmetic mean of the magnitudes of absolute errors in
all the measurements of the quantity. It is represented by
.a
Thus
n
aaa
an||.....|||| 21
Hence the final result of measurement may be written as
aaa m
This implies that any measurement of the quantity is likely to lie between
)( aam
and
).(aam
(3) Relative error or Fractional error : The relative error or fractional error of
measurement is defined as the ratio of mean absolute error to the mean value of the quantity
measured. Thus
Relative error or Fractional error
m
a
a
valuemean
error absolute mean
(4) Percentage error : When the relative/fractional error is expressed in percentage, we
call it percentage error. Thus
Percentage error
%100
m
a
a
1.18 Propagation of Errors.
(1) Error in sum of the quantities : Suppose x = a + b
Let a = absolute error in measurement of a
b = absolute error in measurement of b
x = absolute error in calculation of x i.e. sum of a and b.
The maximum absolute error in x is
)( bax
Percentage error in the value of
%100
)(
ba
ba
x
(2) Error in difference of the quantities : Suppose x = a – b
Let a = absolute error in measurement of a,
b = absolute error in measurement of b
x = absolute error in calculation of x i.e. difference of a and b.
The maximum absolute error in x is
)( bax
Percentage error in the value of
%100
)(
ba
ba
x
(3) Error in product of quantities : Suppose x = a b
Let a = absolute error in measurement of a,
Units, Dimensions and Measurement 79
b = absolute error in measurement of b
x = absolute error in calculation of x i.e. product of a and b.
The maximum fractional error in x is
b
b
a
a
x
x
Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in
value of b)
(4) Error in division of quantities : Suppose
b
a
x
Let a = absolute error in measurement of a,
b = absolute error in measurement of b
x = absolute error in calculation of x i.e. division of a and b.
The maximum fractional error in x is
b
b
a
a
x
x
Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in
value of b)
(5) Error in quantity raised to some power : Suppose
m
n
b
a
x
Let a = absolute error in measurement of a,
b = absolute error in measurement of b
x = absolute error in calculation of x
The maximum fractional error in x is
b
b
m
a
a
n
x
x
Percentage error in the value of x = n (Percentage error in value of a) + m (Percentage
error in value of b)
Note : The quantity which have maximum power must be measured carefully because it's
contribution to error is maximum.
Sample problems based on errors of measurement
Problem 55. A physical parameter a can be determined by measuring the parameters b, c, d and e
using the relation a =
edcb /
. If the maximum errors in the measurement of b, c, d
and e are
1
b
%,
1
c
%,
1
d
% and
1
e
%, then the maximum error in the value of a
determined by the experiment is [CPMT 1981]
(a) (
1111 edcb
)% (b) (
1111 edcb
)%
(c) (
1111 edcb
)% (d) (
1111 edcb
)%
Solution : (d)
edcba /
So maximum error in a is given by
80 Units, Dimensions and Measurement
100.100.100.100.100
max
e
e
d
d
c
c
b
b
a
a
%
1111 edcb
Problem 56. The pressure on a square plate is measured by measuring the force on the plate and the
length of the sides of the plate. If the maximum error in the measurement of force and
length are respectively 4% and 2%, The maximum error in the measurement of pressure is [CPMT 1993]
(a) 1% (b) 2% (c) 6% (d) 8%
Solution : (d)
2
l
F
A
F
P
, so maximum error in pressure
)(P
1002100100
max
l
l
F
F
P
P
= 4% + 2 × 2% = 8%
Problem 57. The relative density of material of a body is found by weighing it first in air and then in
water. If the weight in air is (5.00
05.0
) Newton and weight in water is (4.00
0.05)
Newton. Then the relative density along with the maximum permissible percentage error is
(a) 5.0
11% (b) 5.0
1% (c) 5.0
6% (d) 1.25
5%
Solution : (a) Weight in air
N)05.000.5(
Weight in water
N)05.000.4(
Loss of weight in water
N)1.000.1(
Now relative density
waterin loss weight
air inweight
i.e. R . D
1.000.1
05.000.5
Now relative density with max permissible error
100
00.1
1.0
00.5
05.0
00.1
00.5
)%101(0.5
%110.5
Problem 58. The resistance R =
i
V
where V= 100
5 volts and i = 10
0.2 amperes. What is the total
error in R
(a) 5% (b) 7% (c) 5.2% (d)
2
5
%
Solution : (b)
I
V
R
100100100
max
I
I
V
V
R
R
100
10
2.0
100
100
5
)%25(
= 7%
Problem 59. The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56
s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is
(a) 0.1 s (b) 0.11 s (c) 0.01 s (d) 1.0 s
Solution : (b) Average value
5
80.271.242.256.263.2
sec62.2
Now
01.062.263.2|| 1T
06.056.262.2|| 2T
Units, Dimensions and Measurement 81
20.042.262.2|| 3T
09.062.271.2|| 4T
18.062.280.2|| 5T
Mean absolute error
5
|||||||||| 54321 TTTTT
T
sec11.0108.0
5
54.0
Problem 60. The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its
diameter is measured with venier calipers having least count 0.01 cm. Given that length is
5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume
will be
(a) 1% (b) 2% (c) 3% (d) 4%
Solution : (c) Volume of cylinder
lrV 2
Percentage error in volume
100100
2
100
l
l
r
r
V
V
100
0.5
1.0
100
0.2
01.0
2
)%21(
=
%3
Problem 61. In an experiment, the following observation's were recorded : L = 2.820 m, M = 3.00 kg, l =
0.087 cm, Diameter D = 0.041 cm Taking g = 9.81
2
/sm
using the formula , Y=
lD
Mg
2
4
, the
maximum permissible error in Y is
(a) 7.96% (b) 4.56% (c) 6.50% (d) 8.42%
Solution : (c)
lD
MgL
Y2
4
so maximum permissible error in Y =
100
2
100
l
l
D
D
L
L
g
g
M
M
Y
Y
100
87
1
41
1
2
9820
1
81.9
1
300
1
%5.6100065.0
Problem 62. According to Joule's law of heating, heat produced
2
IH
Rt, where I is current, R is
resistance and t is time. If the errors in the measurement of I, R and t are 3%, 4% and 6%
respectively then error in the measurement of H is
(a)
17% (b)
16% (c)
19% (d)
25%
Solution : (b)
tRIH 2
100
2
100
t
t
R
R
I
I
H
H
)%6432(
%16
Problem 63. If there is a positive error of 50% in the measurement of velocity of a body, then the error
in the measurement of kinetic energy is
(a) 25% (b) 50% (c) 100% (d) 125%
82 Units, Dimensions and Measurement
Solution : (c) Kinetic energy
2
2
1mvE
100
2
100
v
v
m
m
E
E
Here
0m
and
%50100
v
v
%100502100
E
E
Problem 64. A physical quantity P is given by P=
2
3
4
2
1
3
DC
BA
. The quantity which brings in the maximum
percentage error in P is
(a) A (b) B (c) C (d) D
Solution : (c) Quantity C has maximum power. So it brings maximum error in P.
