CHAPTER TWO
UNITS AND MEASUREMENT
2.1 INTRODUCTION
Measurement of any physical quantity involves comparison
with a certain basic, arbitrarily chosen, internationally
accepted reference standard called unit. The result of a
measurement of a physical quantity is expressed by a
number (or numerical measure) accompanied by a unit.
Although the number of physical quantities appears to be
very large, we need only a limited number of units for
expressing all the physical quantities, since they are inter-
related with one another. The units for the fundamental or
base quantities are called fundamental or base units. The
units of all other physical quantities can be expressed as
combinations of the base units. Such units obtained for the
derived quantities are called derived units. A complete set
of these units, both the base units and derived units, is
known as the system of units.
2.2 THE INTERNATIONAL SYSTEM OF UNITS
In earlier time scientists of different countries were using
different systems of units for measurement. Three such
systems, the CGS, the FPS (or British) system and the MKS
system were in use extensively till recently.
The base units for length, mass and time in these systems
were as follows :
• In CGS system they were centimetre, gram and second
respectively.
• In FPS system they were foot, pound and second
respectively.
• In MKS system they were metre, kilogram and second
respectively.
The system of units which is at present internationally
accepted for measurement is the Système Internationale
d’ Unites (French for International System of Units),
abbreviated as SI. The SI, with standard scheme of symbols,
units and abbreviations, developed by the Bureau
International des Poids et measures (The International
Bureau of Weights and Measures, BIPM) in 1971 were
recently revised by the General Conference on Weights and
Measures in November 2018. The scheme is now for
2.1 Introduction
2.2 The international system of
units
2.3 Measurement of length
2.4 Measurement of mass
2.5 Measurement of time
2.6 Accuracy, precision of
instruments and errors in
measurement
2.7 Significant figures
2.8 Dimensions of physical
quantities
2.9 Dimensional formulae and
dimensional equations
2.10 Dimensional analysis and its
applications
Summary
Exercises
Additional exercises
2020-21
Table 2.1 SI Base Quantities and Units*
international usage in scientific, technical, industrial
and commercial work. Because SI units used decimal
system, conversions within the system are quite simple
and convenient. We shall follow the SI units in this
book.
In SI, there are seven base units as given in Table
2.1. Besides the seven base units, there are two more
units that are defined for (a) plane angle d
θ
as the ratio
of length of arc ds to the radius r and (b) solid angle dΩ
as the ratio of the intercepted area dA of the spherical
surface, described about the apex O as the centre, to
the square of its radius r, as shown in Fig. 2.1(a) and
(b) respectively. The unit for plane angle is radian with
the symbol rad and the unit for the solid angle is
steradian with the symbol sr. Both these are
dimensionless quantities.
UNITS AND MEASUREMENT 17
(a)
(b)
Fig. 2.1 Description of (a) plane angle d
θ
and (b) solid angle
d
Ω
.
Base SI Units
quantity Name Symbol Definition
Length metre m The metre, symbol m, is the SI unit of length. It is defined by taking the
fixed numerical value of the speed of light in vacuum c to be 299792458
when expressed in the unit m s
–1
, where the second is defined in terms of
the caesium frequency
∆
ν
cs.
Mass kilogram kg The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the
fixed numerical value of the Planck constant h to be 6.62607015×10
–34
when
expressed in the unit J s, which is equal to kg m
2
s
–1
, where the metre and
the second are defined in terms of c and
∆
ν
cs.
Time second s The second, symbol s, is the SI unit of time. It is defined by taking the fixed
numerical value of the caesium frequency
∆
ν
cs, the unperturbed ground-
state hyperfine transition frequency of the caesium-133 atom, to be
9192631770 when expressed in the unit Hz, which is equal to s
–1
.
Electric ampere A The ampere, symbol A, is the SI unit of electric current. It is defined by
taking the fixed numerical value of the elementary charge e to be
1.602176634×10
–19
when expressed in the unit C, which is equal to A s,
where the second is defined in terms of
∆
ν
cs.
Thermo kelvin K The kelvin, symbol K, is the SI unit of thermodynamic temperature.
dynamic It is defined by taking the fixed numerical value of the Boltzmann constant
Temperature k to be 1.380649×10
–23
when expressed in the unit J K
–1
, which is equal to
kg m
2
s
–2
k
–1
, where the kilogram, metre and second are defined in terms of
h, c and
∆
ν
cs.
Amount of mole mol The mole, symbol mol, is the SI unit of amount of substance. One mole
substance contains exactly 6.02214076×10
23
elementary entities. This number is the
fixed numerical value of the Avogadro constant, N
A
, when expressed in the
unit mol
–1
and is called the Avogadro number. The amount of substance,
symbol n, of a system is a measure of the number of specified elementary
entities. An elementary entity may be an atom, a molecule, an ion, an electron,
any other particle or specified group of particles.
Luminous candela cd The candela, symbol cd, is the SI unit of luminous intensity in given direction.
intensity It is defined by taking the fixed numerical value of the luminous efficacy of
monochromatic radiation of frequency 540×10
12
Hz, K
cd
, to be 683 when expressed
in the unit lm W
–1
, which is equal to cd sr W
–1
, or cd sr kg
–1
m
–2
s
3
, where the
kilogram, metre and second are defined in terms of h, c and
∆
ν
cs.
* The values mentioned here need not be remembered or asked in a test. They are given here only to
indicate the extent of accuracy to which they are measured. With progress in technology, the measuring
techniques get improved leading to measurements with greater precision. The definitions of base units
are revised to keep up with this progress.
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PHYSICS18
Table 2.2 Some units retained for general use (Though outside SI)
Note that when mole is used, the elementary
entities must be specified. These entities
may be atoms, molecules, ions, electrons,
other particles or specified groups of such
particles.
We employ units for some physical quantities
that can be derived from the seven base units
(Appendix A 6). Some derived units in terms of
the SI base units are given in (Appendix A 6.1).
Some SI derived units are given special names
(Appendix A 6.2 ) and some derived SI units make
use of these units with special names and the
seven base units (Appendix A 6.3). These are
given in Appendix A 6.2 and A 6.3 for your ready
reference. Other units retained for general use
are given in Table 2.2.
Common SI prefixes and symbols for multiples
and sub-multiples are given in Appendix A2.
General guidelines for using symbols for physical
quantities, chemical elements and nuclides are
given in Appendix A7 and those for SI units and
some other units are given in Appendix A8 for
your guidance and ready reference.
2.3 MEASUREMENT OF LENGTH
You are already familiar with some direct methods
for the measurement of length. For example, a
metre scale is used for lengths from 10
–3
m to 10
2
m. A vernier callipers is used for lengths to an
accuracy of 10
–4
m. A screw gauge and a
spherometer can be used to measure lengths as
less as to 10
–5
m. To measure lengths beyond these
ranges, we make use of some special indirect
methods.
2.3.1 Measurement of Large Distances
Large distances such as the distance of a planet
or a star from the earth cannot be measured
directly with a metre scale. An important method
in such cases is the parallax method.
When you hold a pencil in front of you against
some specific point on the background (a wall)
and look at the pencil first through your left eye
A (closing the right eye) and then look at the
pencil through your right eye B (closing the left
eye), you would notice that the position of the
pencil seems to change with respect to the point
on the wall. This is called parallax. The
distance between the two points of observation
is called the basis. In this example, the basis is
the distance between the eyes.
To measure the distance D of a far away
planet S by the parallax method, we observe it
from two different positions (observatories) A and
B on the Earth, separated by distance AB = b
at the same time as shown in Fig. 2.2. We
measure the angle between the two directions
along which the planet is viewed at these two
points. The ∠ASB in Fig. 2.2 represented by
symbol
θ
is called the parallax angle or
parallactic angle.
As the planet is very far away,
1,
b
D
<<
and
therefore,
θ
is very small. Then we
approximately take AB as an arc of length b of a
circle with centre at S and the distance D as
2020-21
UNITS AND MEASUREMENT 19
t
t
t
t
the radius AS = BS so that AB = b = D
θ
where
θ
is in radians.
D =
b
θ
(2.1)
Having determined D, we can employ a similar
method to determine the size or angular diameter
of the planet. If d is the diameter of the planet
and
α
the angular size of the planet (the angle
subtended by d at the earth), we have
α
= d/D (2.2)
The angle
α
can be measured from the same
location on the earth. It is the angle between
the two directions when two diametrically
opposite points of the planet are viewed through
the telescope. Since D is known, the diameter d
of the planet can be determined using Eq. (2.2).
Example 2.1 Calculate the angle of
(a) 1
0
(degree) (b) 1′ (minute of arc or arcmin)
and (c) 1″(second of arc or arc second) in
radians. Use 360
0
=2π rad, 1
0
=60′ and
1′ = 60 ″
Answer (a) We have 360
0
= 2π rad
1
0
= (π /180) rad = 1.745×10
–2
rad
(b) 1
0
= 60′ = 1.745×10
–2
rad
1′ = 2.908×10
–4
rad
2.91×10
–4
rad
(c) 1′ = 60″ = 2.908×10
–4
rad
1″ = 4.847×10
–4
rad
4.85×10
–6
rad
t
Example 2.2 A man wishes to estimate
the distance of a nearby tower from him.
He stands at a point A in front of the tower
C and spots a very distant object O in line
with AC. He then walks perpendicular to
AC up to B, a distance of 100 m, and looks
at O and C again. Since O is very distant,
the direction BO is practically the same as
AO; but he finds the line of sight of C shifted
from the original line of sight by an angle
θ
= 40
0
(
θ
is known as ‘parallax’) estimate
the distance of the tower C from his original
position A.
Fig. 2.3
Answer We have, parallax angle
θ
= 40
0
From Fig. 2.3, AB = AC tan
θ
AC = AB/tan
θ
= 100 m/tan 40
0
= 100 m/0.8391 = 119 m
t
Example 2.3 The moon is observed from
two diametrically opposite points A and B
on Earth. The angle
θ
subtended at the
moon by the two directions of observation
is 1
o
54′. Given the diameter of the Earth to
be about 1.276
××
××
×
10
7
m, compute the
distance of the moon from the Earth.
Answer We have
θ
= 1°
54
′
= 114
′
( )
( )
-6
114 60 4.85 10
=
′′
× × ×
rad
=
3.32
×
−
10
2
rad
,
since
6
1 4.85 10 .
=
" rad
−
×
Also
b
=
AB
=
1.276
m
×
10
7
Hence from Eq. (2.1), we have the earth-moon
distance,
D b
=
/
θ
=
1.276 10
3.32
10
7
-2
×
×
8
3.84 10 m
=
×
t
Example 2.4 The Sun’s angular diameter
is measured to be 1920′′. The distance D of
the Sun from the Earth is 1.496 × 10
11
m.
What is the diameter of the Sun ?
Fig. 2.2 Parallax method.
2020-21
PHYSICS20
t
Answer Sun’s angular diameter
α
=
1920"
=
×
×
−
1920
4.85
10
rad
6
=
×
−
9.31
10
rad
3
Sun’s diameter
d
D
=
α
3 11
9.31 10 1.496 10 m
−
= × × ×
9
=1.39 10 m
×
t
2.3.2 Estimation of Very Small Distances:
Size of a Molecule
To measure a very small size, like that of a
molecule (10
–8
m to 10
–10
m), we have to adopt
special methods. We cannot use a screw gauge
or similar instruments. Even a microscope has
certain limitations. An optical microscope uses
visible light to ‘look’ at the system under
investigation. As light has wave like features,
the resolution to which an optical microscope
can be used is the wavelength of light (A detailed
explanation can be found in the Class XII
Physics textbook). For visible light the range of
wavelengths is from about 4000 Å to 7000 Å
(1 angstrom = 1 Å = 10
-10
m). Hence an optical
microscope cannot resolve particles with sizes
smaller than this. Instead of visible light, we can
use an electron beam. Electron beams can be
focussed by properly designed electric and
magnetic fields. The resolution of such an
electron microscope is limited finally by the fact
that electrons can also behave as waves ! (You
will learn more about this in class XII). The
wavelength of an electron can be as small as a
fraction of an angstrom. Such electron
microscopes with a resolution of 0.6 Å have been
built. They can almost resolve atoms and
molecules in a material. In recent times,
tunnelling microscopy has been developed in
which again the limit of resolution is better than
an angstrom. It is possible to estimate the sizes
of molecules.
A simple method for estimating the molecular
size of oleic acid is given below. Oleic acid is a
soapy liquid with large molecular size of the
order of 10
–9
m.
The idea is to first form mono-molecular layer
of oleic acid on water surface.
We dissolve 1 cm
3
of oleic acid in alcohol to
make a solution of 20 cm
3
. Then we take 1 cm
3
of this solution and dilute it to 20 cm
3
, using
alcohol. So, the concentration of the solution is
equal to
1
20 20
3
×
cm
of oleic acid/cm
3
of
solution. Next we lightly sprinkle some
lycopodium powder on the surface of water in a
large trough and we put one drop of this solution
in the water. The oleic acid drop spreads into a
thin, large and roughly circular film of molecular
thickness on water surface. Then, we quickly
measure the diameter of the thin film to get its
area A. Suppose we have dropped n drops in
the water. Initially, we determine the
approximate volume of each drop (V cm
3
).
Volume of n drops of solution
= nV cm
3
Amount of oleic acid in this solution
=
nV
1
20 20×
cm
3
This solution of oleic acid spreads very fast
on the surface of water and forms a very thin
layer of thickness t. If this spreads to form a
film of area A cm
2
, then the thickness of the
film
t =
Volume of
the film
Area of th
e film
or,
cm
20 20
nV
t
A
=
×
(2.3)
If we assume that the film has mono-molecular
thickness, then this becomes the size or diameter
of a molecule of oleic acid. The value of this
thickness comes out to be of the order of 10
–9
m.
Example 2.5 If the size of a nucleus (in
the range of 10
–15
to 10
–14
m) is scaled up
to the tip of a sharp pin, what roughly is
the size of an atom ? Assume tip of the pin
to be in the range 10
–5
m to 10
–4
m.
Answer The size of a nucleus is in the range of
10
–15
m and 10
–14
m. The tip of a sharp pin is
taken to be in the range of 10
–5
m and 10
–4
m.
Thus we are scaling up by a factor of 10
10
. An
atom roughly of size 10
–10
m will be scaled up to a
size of 1 m. Thus a nucleus in an atom is as small
in size as the tip of a sharp pin placed at the centre
of a sphere of radius about a metre long.
t
2020-21
UNITS AND MEASUREMENT 21
2.3.3 Range of Lengths
The sizes of the objects we come across in the
universe vary over a very wide range. These may
vary from the size of the order of 10
–14
m of the
tiny nucleus of an atom to the size of the order
of 10
26
m of the extent of the observable universe.
Table 2.3 gives the range and order of lengths
and sizes of some of these objects.
We also use certain special length units for
short and large lengths. These are
1 fermi = 1 f = 10
–15
m
1 angstrom = 1 Å = 10
–10
m
1 astronomical unit = 1 AU (average distance
of the Sun from the Earth)
= 1.496 × 10
11
m
1 light year = 1 ly= 9.46 × 10
15
m (distance
that light travels with velocity of
3 × 10
8
m s
–1
in 1 year)
1 parsec = 3.08 × 10
16
m (Parsec is the
distance at which average radius of earth’s orbit
subtends an angle of 1 arc second)
2.4 MEASUREMENT OF MASS
Mass is a basic property of matter. It does not
depend on the temperature, pressure or location
of the object in space. The SI unit of mass is
kilogram (kg). It is defined by taking the fixed
numerical value of the Plank Constant h to be
6.62607015×10
–34
when expressed in the unit of
Js which is equal to kg m
2
s
–1
, where the
metre and the second are defined is terms of C
and
∆
ν
cs.
While dealing with atoms and molecules, the
kilogram is an inconvenient unit. In this case,
there is an important standard unit of mass,
called the unified atomic mass unit (u), which
has been established for expressing the mass
of atoms as
1 unified atomic mass unit = 1u
= (1/12) of the mass of an atom of carbon-12
isotope
(
)
6
12
C
including the mass of electrons
= 1.66 × 10
–27
kg
Mass of commonly available objects can be
determined by a common balance like the one
used in a grocery shop. Large masses in the
universe like planets, stars, etc., based on
Newton’s law of gravitation can be measured by
using gravitational method (See Chapter 8). For
measurement of small masses of atomic/sub-
atomic particles etc., we make use of mass
spectrograph in which radius of the trajectory
is proportional to the mass of a charged particle
moving in uniform electric and magnetic field.
2.4.1 Range of Masses
The masses of the objects, we come across in
the universe, vary over a very wide range. These
may vary from tiny mass of the order of 10
-30
kg
of an electron to the huge mass of about 10
55
kg
of the known universe. Table 2.4 gives the range
and order of the typical masses of various
objects.
Table 2.3 Range and order of lengths
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PHYSICS22
Table 2.4 Range and order of masses
2.5 MEASUREMENT OF TIME
To measure any time interval we need a clock.
We now use an atomic standard of time, which
is based on the periodic vibrations produced in
a cesium atom. This is the basis of the caesium
clock, sometimes called atomic clock, used in
the national standards. Such standards are
available in many laboratories. In the caesium
atomic clock, the second is taken as the time
needed for 9,192,631,770 vibrations of the
radiation corresponding to the transition
between the two hyperfine levels of the ground
state of caesium-133 atom. The vibrations of the
caesium atom regulate the rate of this caesium
atomic clock just as the vibrations of a balance
wheel regulate an ordinary wristwatch or the
vibrations of a small quartz crystal regulate a
quartz wristwatch.
The caesium atomic clocks are very accurate.
In principle they provide portable standard. The
national standard of time interval ‘second’ as
well as the frequency is maintained through four
cesium atomic clocks. A caesium atomic clock
is used at the National Physical Laboratory
(NPL), New Delhi to maintain the Indian
standard of time.
In our country, the NPL has the responsibility
of maintenance and improvement of physical
standards, including that of time, frequency, etc.
Note that the Indian Standard Time (IST) is
linked to this set of atomic clocks. The efficient
caesium atomic clocks are so accurate that they
impart the uncertainty in time realisation as
± 1 × 10
–15
, i.e. 1 part in 10
15
. This implies that
the uncertainty gained over time by such a
device is less than 1 part in 10
15
; they lose or
gain no more than 32 µs in one year. In view of
the tremendous accuracy in time measurement,
the SI unit of length has been expressed in terms
the path length light travels in certain interval
of time (1/299, 792, 458 of a second) (Table 2.1).
The time interval of events that we come
across in the universe vary over a very wide
range. Table 2.5 gives the range and order of
some typical time intervals.
You may notice that there is an interesting
coincidence between the numbers appearing
in Tables 2.3 and 2.5. Note that the ratio of the
longest and shortest lengths of objects in our
universe is about 10
41
. Interestingly enough,
the ratio of the longest and shortest time
intervals associated with the events and objects
in our universe is also about 10
41
. This number,
10
41
comes up again in Table 2.4, which lists
typical masses of objects. The ratio of the
largest and smallest masses of the objects in
our universe is about (10
41
)
2
. Is this a curious
coincidence between these large numbers
purely accidental ?
2.6 ACCURACY, PRECISION OF INSTRUMENTS
AND ERRORS IN MEASUREMENT
Measurement is the foundation of all
experimental science and technology. The result
of every measurement by any measuring
instrument contains some uncertainty. This
uncertainty is called error. Every calculated
quantity which is based on measured values,
also has an error. We shall distinguish between
two terms: accuracy and precision. The
accuracy of a measurement is a measure of how
close the measured value is to the true value of
the quantity. Precision tells us to what resolution
or limit the quantity is measured.
The accuracy in measurement may depend on
several factors, including the limit or the resolution
of the measuring instrument. For example, suppose
the true value of a certain length is near 3.678 cm.
In one experiment, using a measuring instrument
of resolution 0.1 cm, the measured value is found to
be 3.5 cm, while in another experiment using a
measuring device of greater resolution, say 0.01 cm,
the length is determined to be 3.38 cm. The first
measurement has more accuracy (because it is
2020-21
UNITS AND MEASUREMENT 23
closer to the true value) but less precision (its
resolution is only 0.1 cm), while the
second measurement is less accurate but
more precise. Thus every measurement is
approximate due to errors in measurement. In
general, the errors in measurement can be
broadly classified as (a) systematic errors and
(b) random errors.
Systematic errors
The systematic errors are those errors that
tend to be in one direction, either positive or
negative. Some of the sources of systematic
errors are :
(a) Instrumental errors that arise from the
errors due to imperfect design or calibration
of the measuring instrument, zero error in
the instrument, etc. For example, the
temperature graduations of a thermometer
may be inadequately calibrated (it may read
104 °C at the boiling point of water at STP
whereas it should read 100 °C); in a vernier
callipers the zero mark of vernier scale may
not coincide with the zero mark of the main
scale, or simply an ordinary metre scale may
be worn off at one end.
(b) Imperfection in experimental technique
or procedure To determine the temperature
of a human body, a thermometer placed
under the armpit will always give a
temperature lower than the actual value of
the body temperature. Other external
conditions (such as changes in temperature,
humidity, wind velocity, etc.) during the
experiment may systematically affect the
measurement.
(c) Personal errors that arise due to an
individual’s bias, lack of proper setting of
the apparatus or individual’s carelessness
in taking observations without observing
proper precautions, etc. For example, if you,
by habit, always hold your head a bit too far
to the right while reading the position of a
needle on the scale, you will introduce an
error due to parallax.
Systematic errors can be minimised by
improving experimental techniques, selecting
better instruments and removing personal bias
as far as possible. For a given set-up, these
errors may be estimated to a certain extent and
the necessary corrections may be applied to the
readings.
Random errors
The random errors are those errors, which occur
irregularly and hence are random with respect
Table 2.5 Range and order of time intervals
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PHYSICS24
to sign and size. These can arise due to random
and unpredictable fluctuations in experimental
conditions (e.g. unpredictable fluctuations in
temperature, voltage supply, mechanical
vibrations of experimental set-ups, etc), personal
(unbiased) errors by the observer taking
readings, etc. For example, when the same
person repeats the same observation, it is very
likely that he may get different readings
everytime.
Least count error
The smallest value that can be measured by the
measuring instrument is called its least count.
All the readings or measured values are good only
up to this value.
The least count error is the error
associated with the resolution of the instrument.
For example, a vernier callipers has the least
count as 0.01cm; a spherometer may have a
least count of 0.001 cm. Least count error
belongs to the category of random errors but
within a limited size; it occurs with both
systematic and random errors. If we use a metre
scale for measurement of length, it may have
graduations at 1 mm division scale spacing or
interval.
Using instruments of higher precision,
improving experimental techniques, etc., we can
reduce the least count error. Repeating the
observations several times and taking the
arithmetic mean of all the observations, the
mean value would be very close to the true value
of the measured quantity.
2.6.1 Absolute Error, Relative Error and
Percentage Error
(a) Suppose the values obtained in several
measurements are a
1
, a
2
, a
3
...., a
n
. The
arithmetic mean of these values is taken as
the best possible value of the quantity under
the given conditions of measurement as :
a
mean
= (a
1
+a
2
+a
3
+...+a
n
) / n (2.4)
or,
a a / n
mean i
i 1
n
=
=
∑
(2.5)
This is because, as explained earlier, it is
reasonable to suppose that individual
measurements are as likely to overestimate
as to underestimate the true value of the
quantity.
The magnitude of the difference
between the individual measurement and
the true value of the quantity is called the
absolute error of the measurement. This
is denoted by |∆a
|. In absence of any other
method of knowing true value, we considered
arithmatic mean as the true value. Then the
errors in the individual measurement values
from the true value, are
∆a
1
= a
1
– a
mean
,
∆a
2
= a
2
– a
mean
,
.... .... ....
.... .... ....
∆a
n
= a
n
– a
mean
The ∆a
calculated above may be positive in
certain cases and negative in some other
cases. But absolute error |∆a| will always
be positive.
(b) The arithmetic mean of all the absolute errors
is taken as the final or mean absolute error
of the value of the physical quantity a. It is
represented by ∆a
mean
.
Thus,
∆a
mean
= (|∆a
1
|+|∆a
2
|+|∆a
3
|+...+ |∆a
n
|)/n
(2.6)
=
=
∑
i 1
n
|∆a
i
|/n (2.7)
If we do a single measurement, the value we
get may be in the range a
mean
± ∆a
mean
i.e. a = a
mean
± ∆a
mean
or,
a
mean
– ∆a
mean
≤ a ≤ a
mean
+ ∆a
mean
(2.8)
This implies that any measurement of the
physical quantity a is likely to lie between
(a
mean
+ ∆a
mean
) and (a
mean
− ∆a
mean
).
(c) Instead of the absolute error, we often use
the relative error or the percentage error
(δa). The relative error is the ratio of the
mean absolute error
∆∆
∆∆
∆
a
mean
to the mean
value a
mean
of the quantity measured.
2020-21
UNITS AND MEASUREMENT 25
t
t
Relative error = ∆a
mean
/a
mean
(2.9)
When the relative error is expressed in per
cent, it is called the percentage error (δa).
Thus, Percentage error
δa = (∆a
mean
/a
mean
) × 100% (2.10)
Let us now consider an example.
Example 2.6 Two clocks are being tested
against a standard clock located in a
national laboratory. At 12:00:00 noon by
the standard clock, the readings of the two
clocks are :
Clock 1 Clock 2
Monday 12:00:05 10:15:06
Tuesday 12:01:15 10:14:59
Wednesday 11:59:08
10:15:18
Thursday 12:01:50 10:15:07
Friday 11:59:15 10:14:53
Saturday 12:01:30 10:15:24
Sunday 12:01:19 10:15:11
If you are doing an experiment that requires
precision time interval measurements, which
of the two clocks will you prefer ?
Answer The range of variation over the seven
days of observations is 162 s for clock 1, and
31 s for clock 2. The average reading of clock 1
is much closer to the standard time than the
average reading of clock 2. The important point
is that a clock’s zero error is not as significant
for precision work as its variation, because a
‘zero-error’ can always be easily corrected.
Hence clock 2 is to be preferred to clock 1. t
Example 2.7 We measure the period of
oscillation of a simple pendulum. In
successive measurements, the readings
turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s
and 2.80 s. Calculate the absolute errors,
relative error or percentage error.
Answer The mean period of oscillation of the
pendulum
(
)
T =
+ + + +2.63 2.56 2.42 2.71 2.80 s
5
=
13.12
5
s
= 2.624 s
= 2.62 s
As the periods are measured to a resolution
of 0.01 s, all times are to the second decimal; it
is proper to put this mean period also to the
second decimal.
The errors in the measurements are
2.63 s – 2.62 s = 0.01 s
2.56 s – 2.62 s = – 0.06 s
2.42 s – 2.62 s = – 0.20 s
2.71 s – 2.62 s = 0.09 s
2.80 s – 2.62 s = 0.18 s
Note that the errors have the same units as the
quantity to be measured.
The arithmetic mean of all the absolute errors
(for arithmetic mean, we take only the
magnitudes) is
∆
Τ
mean
= [(0.01+ 0.06+0.20+0.09+0.18)s]/5
= 0.54 s/5
= 0.11 s
That means, the period of oscillation of the
simple pendulum is (2.62 ± 0.11) s i.e. it lies
between (2.62 + 0.11) s and (2.62 – 0.11) s or
between 2.73 s and 2.51 s. As the arithmetic
mean of all the absolute errors is 0.11 s, there
is already an error in the tenth of a second.
Hence there is no point in giving the period to a
hundredth. A more correct way will be to write
T = 2.6 ± 0.1 s
Note that the last numeral 6 is unreliable, since
it may be anything between 5 and 7. We indicate
this by saying that the measurement has two
significant figures. In this case, the two
significant figures are 2, which is reliable and
6, which has an error associated with it. You
will learn more about the significant figures in
section 2.7.
For this example, the relative error or the
percentage error is
δa = × =
01
100 4
.
2.6
% t
2.6.2 Combination of Errors
If we do an experiment involving several
measurements, we must know how the errors
in all the measurements combine. For example,
2020-21
PHYSICS26
t
mass density is obtained by deviding mass by
the volume of the substance. If we have errors
in the measurement of mass and of the sizes or
dimensions, we must know what the error will
be in the density of the substance. To make such
estimates, we should learn how errors combine
in various mathematical operations. For this,
we use the following procedure.
(a) Error of a sum or a difference
Suppose two physical quantities A and B have
measured values A ± ∆A, B ± ∆B respectively
where ∆A and ∆B are their absolute errors. We
wish to find the error ∆Z in the sum
Z = A + B.
We have by addition, Z ± ∆Z
= (A ± ∆A) + (B ± ∆B).
The maximum possible error in Z
∆Z = ∆A + ∆B
For the difference Z = A – B, we have
Z ± ∆ Z = (A ± ∆A) – (B ± ∆B)
= (A – B) ± ∆A ± ∆B
or, ± ∆Z =
± ∆A ± ∆B
The maximum value of the error ∆Z is again
∆A + ∆B.
Hence the rule : When two quantities are
added or subtracted, the absolute error in the
final result is the sum of the absolute errors
in the individual quantities.
Example 2.8 The temperatures of two
bodies measured by a thermometer are
t
1
= 20
0
C ± 0.5
0
C and t
2
= 50
0
C ± 0.5
0
C.
Calculate the temperature difference and
the error theirin.
Answer t′ = t
2
–t
1
= (50
0
C±0.5
0
C)– (20
0
C±0.5
0
C)
t′ = 30
0
C ± 1
0
C
t
(b) Error of a product or a quotient
Suppose Z = AB and the measured values of A
and B are A ± ∆A and B ± ∆B. Then
Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= AB ± B ∆A ± A ∆B ± ∆A ∆B.
Dividing LHS by Z and RHS by AB we have,
1±(∆Z/Z) = 1 ± (∆A/A) ± (∆B/B) ± (∆A/A)(∆B/B).
Since ∆A and ∆B are small, we shall ignore their
product.
Hence the maximum relative error
∆Z/ Z = (∆A/A) + (∆B/B).
You can easily verify that this is true for division
also.
Hence the rule : When two quantities are
multiplied or divided, the relative error in the
result is the sum of the relative errors in the
multipliers.
How will you measure the length of a line?
What a naïve question, at this stage, you might
say! But what if it is not a straight line? Draw
a zigzag line in your copy, or on the blackboard.
Well, not too difficult again. You might take a
thread, place it along the line, open up the
thread, and measure its length.
Now imagine that you want to measure the
length of a national highway, a river, the railway
track between two stations, or the boundary
between two states or two nations. If you take
a string of length 1 metre or 100 metre, keep it
along the line, shift its position every time, the
arithmetic of man-hours of labour and expenses
on the project is not commensurate with the
outcome. Moreover, errors are bound to occur
in this enormous task. There is an interesting
fact about this. France and Belgium share a
common international boundary, whose length
mentioned in the official documents of the two
countries differs substantially!
Go one step beyond and imagine the
coastline where land meets sea. Roads and rivers
have fairly mild bends as compared to a
coastline. Even so, all documents, including our
school books, contain information on the length
of the coastline of Gujarat or Andhra Pradesh,
or the common boundary between two states,
etc. Railway tickets come with the distance
between stations printed on them. We have
‘milestones’ all along the roads indicating the
distances to various towns. So, how is it done?
One has to decide how much error one can
tolerate and optimise cost-effectiveness. If you
want smaller errors, it will involve high
technology and high costs. Suffice it to say that
it requires fairly advanced level of physics,
mathematics, engineering and technology. It
belongs to the areas of fractals, which has lately
become popular in theoretical physics. Even
then one doesn’t know how much to rely on
the figure that props up, as is clear from the
story of France and Belgium. Incidentally, this
story of the France-Belgium discrepancy
appears on the first page of an advanced Physics
book on the subject of fractals and chaos!
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UNITS AND MEASUREMENT 27
t
t
t
t
Example 2.9 The resistance R = V/I where
V = (100 ± 5)V and I = (10 ± 0.2)A. Find the
percentage error in R.
Answer The percentage error in V is 5% and in
I it is 2%. The total error in R would therefore
be 5% + 2% = 7%. t
Example 2.10 Two resistors of resistances
R
1
= 100±3 ohm and R
2
= 200 ± 4 ohm are
connected (a) in series, (b) in parallel. Find
the equivalent resistance of the (a) series
combination, (b) parallel combination. Use
for (a) the relation R =R
1
+ R
2,
and for (b)
and
Answer (a) The equivalent resistance of series
combination
R =R
1
+ R
2
= (100 ± 3) ohm + (200 ± 4) ohm
= 300 ± 7 ohm.
(b) The equivalent resistance of parallel
combination
1 2
1 2
200
3
R R
R
R R
′
= =
+
= 66.7 ohm
Then, from
1 2
1 1 1
R R R
= +
′
we get,
1 2
2 2 2
1 2
R R
R
R R R
′
∆ ∆
∆
= +
′
( ) ( )
2 2
1 2
2 2
1 2
R R
R R R
R R
∆ ∆
′ ′ ′
∆ = +
2 2
66.7 66.7
3 4
100 200
= +
= 1.8
Then,
66.7 1.8 ohm
R
′
= ±
(Here, ∆R is expresed as 1.8 instead of 2 to
keep in confirmity with the rules of significant
figures.)
t
(c) Error in case of a measured quantity
raised to a power
Suppose Z = A
2
,
Then,
∆Z/Z = (∆A/A) + (∆A/A) = 2 (∆A/A).
Hence, the relative error in A
2
is two times the
error in A.
In general, if Z = A
p
B
q
/C
r
Then,
∆Z/Z = p (∆A/A) + q (∆B/B) + r (∆C/C).
Hence the rule : The relative error in a
physical quantity raised to the power k is the
k times the relative error in the individual
quantity.
Example 2.11 Find the relative error in
Z, if Z = A
4
B
1/3
/CD
3/2
.
Answer The relative error in Z is ∆Z/Z =
4(∆A/A) +(1/3) (∆B/B) + (∆C/C) + (3/2) (∆D/D).
t
Example 2.12 The period of oscillation of
a simple pendulum is
T L/g.= 2π
Measured value of L is 20.0 cm known to 1
mm accuracy and time for 100 oscillations
of the pendulum is found to be 90 s using
a wrist watch of 1 s resolution. What is the
accuracy in the determination of g ?
Answer g = 4π
2
L/T
2
Here, T =
t
n
and
t
T
n
∆
∆ =
. Therefore,
T t
T t
∆ ∆
=
.
The errors in both L and t are the least count
errors. Therefore,
(∆g/g) = (∆L/L) + 2(∆T/T )
=
0 1
20 0
2
1
90
0 027
.
.
.+
=
Thus, the percentage error in g is
100 (∆g/g) = 100(∆L/L) + 2 × 100 (∆T/T )
= 3% t
2.7 SIGNIFICANT FIGURES
As discussed above, every measurement
involves errors. Thus, the result of
measurement should be reported in a way that
indicates the precision of measurement.
Normally, the reported result of measurement
is a number that includes all digits in the
number that are known reliably plus the first
digit that is uncertain. The reliable digits plus
2020-21
PHYSICS28
the first uncertain digit are known as
significant digits or significant figures. If we
say the period of oscillation of a simple
pendulum is 1.62 s, the digits 1 and 6 are
reliable and certain, while the digit 2 is
uncertain. Thus, the measured value has three
significant figures. The length of an object
reported after measurement to be 287.5 cm has
four significant figures, the digits 2, 8, 7 are
certain while the digit 5 is uncertain. Clearly,
reporting the result of measurement that
includes more digits than the significant digits
is superfluous and also misleading since it would
give a wrong idea about the precision of
measurement.
The rules for determining the number of
significant figures can be understood from the
following examples. Significant figures indicate,
as already mentioned, the precision of
measurement which depends on the least count
of the measuring instrument. A choice of
change of different units does not change the
number of significant digits or figures in a
measurement. This important remark makes
most of the following observations clear:
(1) For example, the length 2.308 cm has four
significant figures. But in different units, the
same value can be written as 0.02308 m or 23.08
mm or 23080 µm.
All these numbers have the same number of
significant figures (digits 2, 3, 0, 8), namely four.
This shows that the location of decimal point is
of no consequence in determining the number
of significant figures.
The example gives the following rules :
• All the non-zero digits are significant.
• All the zeros between two non-zero digits
are significant, no matter where the
decimal point is, if at all.
• If the number is less than 1, the zero(s)
on the right of decimal point but to the
left of the first non-zero digit are not
significant. [In
0.00 2308, the underlined
zeroes are not significant].
• The terminal or trailing zero(s) in a
number without a decimal point are not
significant.
[Thus 123 m = 12300 cm = 123000 mm has
three significant figures, the trailing zero(s)
being not significant.] However, you can also
see the next observation.
• The trailing zero(s) in a number with a
decimal point are significant.
[The numbers 3.500 or 0.06900 have four
significant figures each.]
(2) There can be some confusion regarding the
trailing zero(s). Suppose a length is reported to
be 4.700 m. It is evident that the zeroes here
are meant to convey the precision of
measurement and are, therefore, significant. [If
these were not, it would be superfluous to write
them explicitly, the reported measurement
would have been simply 4.7 m]. Now suppose
we change units, then
4.700 m = 470.0 cm = 4700 mm = 0.004700 km
Since the last number has trailing zero(s) in a
number with no decimal, we would conclude
erroneously from observation (1) above that the
number has two significant figures, while in
fact, it has four significant figures and a mere
change of units cannot change the number of
significant figures.
(3) To remove such ambiguities in
determining the number of significant
figures, the best way is to report every
measurement in scientific notation (in the
power of 10). In this notation, every number is
expressed as a × 10
b
, where a is a number
between 1 and 10, and b is any positive or
negative exponent (or power) of 10. In order to
get an approximate idea of the number, we may
round off the number a to 1 (for a
≤
5) and to 10
(for 5<a
≤
10). Then the number can be
expressed approximately as 10
b
in which the
exponent (or power) b of 10 is called order of
magnitude of the physical quantity. When only
an estimate is required, the quantity is of the
order of 10
b
. For example, the diameter of the
earth (1.28×10
7
m) is of the order of 10
7
m with
the order of magnitude 7. The diameter of
hydrogen atom (1.06 ×10
–10
m) is of the order of
10
–10
m, with the order of magnitude
–10. Thus, the diameter of the earth is 17 orders
of magnitude larger than the hydrogen atom.
It is often customary to write the decimal after
the first digit. Now the confusion mentioned in
(a) above disappears :
4.700 m = 4.700 × 10
2
cm
= 4.700 × 10
3
mm = 4.700 × 10
–3
km
The power of 10 is irrelevant to the
determination of significant figures. However, all
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UNITS AND MEASUREMENT 29
zeroes appearing in the base number in the
scientific notation are significant. Each number
in this case has four significant figures.
Thus, in the scientific notation, no confusion
arises about the trailing zero(s) in the base
number a. They are always significant.
(4) The scientific notation is ideal for reporting
measurement. But if this is not adopted, we use
the rules adopted in the preceding example :
• For a number greater than 1, without any
decimal, the trailing zero(s) are not
significant.
• For a number with a decimal, the trailing
zero(s) are significant.
(5) The digit 0 conventionally put on the left of a
decimal for a number less than 1 (like 0.1250)
is never significant. However, the zeroes at the
end of such number are significant in a
measurement.
(6) The multiplying or dividing factors which are
neither rounded numbers nor numbers
representing measured values are exact and
have infinite number of significant digits. For
example in
2
d
r
=
or s = 2πr, the factor 2 is an
exact number and it can be written as 2.0, 2.00
or 2.0000 as required. Similarly, in
t
T
n
=
, n is
an exact number.
2.7.1 Rules for Arithmetic Operations with
Significant Figures
The result of a calculation involving approximate
measured values of quantities (i.e. values with
limited number of significant figures) must reflect
the uncertainties in the original measured values.
It cannot be more accurate than the original
measured values themselves on which the result
is based. In general, the final result should not
have more significant figures than the original
data from which it was obtained. Thus, if mass of
an object is measured to be, say, 4.237 g (four
significant figures) and its volume is measured to
be 2.51 cm
3
, then its density, by mere arithmetic
division, is 1.68804780876 g/cm
3
upto 11 decimal
places. It would be clearly absurd and irrelevant
to record the calculated value of density to such a
precision when the measurements on which the
value is based, have much less precision. The
following rules for arithmetic operations with
significant figures ensure that the final result of
a calculation is shown with the precision that is
consistent with the precision of the input
measured values :
(1) In multiplication or division, the final
result should retain as many significant
figures as are there in the original number
with the least significant figures.
Thus, in the example above, density should
be reported to three significant figures.
Density
4.237g
2.51 cm
1.69 g cm
3
-3
= =
Similarly, if the speed of light is given as
3 × 10
8
m s
-1
(one significant figure) and one
year (1y = 365.25 d) has 3.1557 × 10
7
s (five
significant figures), the light year is 9.47 × 10
15
m
(three significant figures).
(2) In addition or subtraction, the final result
should retain as many decimal places as are
there in the number with the least decimal
places.
For example, the sum of the numbers
436.32 g, 227.2 g and 0.301 g by mere arithmetic
addition, is 663.821 g. But the least precise
measurement (227.2 g) is correct to only one
decimal place. The final result should, therefore,
be rounded off to 663.8 g.
Similarly, the difference in length can be
expressed as :
0.307 m – 0.304 m = 0.003 m = 3 × 10
–3
m.
Note that we should not use the rule (1)
applicable for multiplication and division and
write 664 g as the result in the example of
addition and 3.00 × 10
–3
m in the example of
subtraction. They do not convey the precision
of measurement properly. For addition and
subtraction, the rule is in terms of decimal
places.
2.7.2 Rounding off the Uncertain Digits
The result of computation with approximate
numbers, which contain more than one
uncertain digit, should be rounded off. The rules
for rounding off numbers to the appropriate
significant figures are obvious in most cases. A
number 2.74
6 rounded off to three significant
figures is 2.75, while the number 2.743 would
be 2.74. The rule by convention is that the
preceding digit is raised by 1 if the
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PHYSICS30
t
t
insignificant digit to be dropped (the
underlined digit in this case) is more than
5, and is left unchanged if the latter is less
than 5. But what if the number is 2.74
5 in
which the insignificant digit is 5. Here, the
convention is that if the preceding digit is
even, the insignificant digit is simply
dropped and, if it is odd, the preceding digit
is raised by 1. Then, the number 2.74
5 rounded
off to three significant figures becomes 2.74. On
the other hand, the number 2.73
5 rounded off
to three significant figures becomes 2.74 since
the preceding digit is odd.
In any involved or complex multi-step
calculation, you should retain, in intermediate
steps, one digit more than the significant digits
and round off to proper significant figures at the
end of the calculation. Similarly, a number
known to be within many significant figures,
such as in 2.99792458 × 10
8
m/s for the speed
of light in vacuum, is rounded off to an
approximate value 3 × 10
8
m/s , which is often
employed in computations. Finally, remember
that exact numbers that appear in formulae like
2 π
in
T
L
g
= 2π ,
have a large (infinite) number
of significant figures. The value of π =
3.1415926.... is known to a large number of
significant figures. You may take the value as
3.142 or 3.14 for π, with limited number of
significant figures as required in specific
cases.
Example 2.13 Each side of a cube is
measured to be 7.203 m. What are the
total surface area and the volume of the
cube to appropriate significant figures?
Answer The number of significant figures in
the measured length is 4. The calculated area
and the volume should therefore be rounded off
to 4 significant figures.
Surface area of the cube = 6(7.203)
2
m
2
= 311.299254 m
2
= 311.3 m
2
Volume of the cube = (7.203)
3
m
3
= 373.714754 m
3
= 373.7 m
3
t
Example 2.14 5.74 g of a substance
occupies 1.2 cm
3
. Express its density by
keeping the significant figures in view.
Answer There are 3 significant figures in the
measured mass whereas there are only 2
significant figures in the measured volume.
Hence the density should be expressed to only
2 significant figures.
Density =
−
5.74
1.2
g cm
3
= 4.8 g cm
--3
. t
2.7.3 Rules for Determining the Uncertainty
in the Results of Arithmatic
Calculations
The rules for determining the uncertainty or
error in the number/measured quantity in
arithmetic operations can be understood from
the following examples.
(1) If the length and breadth of a thin
rectangular sheet are measured, using a metre
scale as 16.2 cm and, 10.1 cm respectively, there
are three significant figures in each
measurement. It means that the length l may
be written as
l = 16.2 ± 0.1 cm
= 16.2 cm ± 0.6 %.
Similarly, the breadth b may be written as
b = 10.1 ± 0.1 cm
= 10.1 cm ± 1 %
Then, the error of the product of two (or more)
experimental values, using the combination of
errors rule, will be
l b = 163.62 cm
2
+ 1.6%
= 163.62 + 2.6 cm
2
This leads us to quote the final result as
l b = 164 + 3 cm
2
Here 3 cm
2
is the uncertainty or error in the
estimation of area of rectangular sheet.
(2) If a set of experimental data is specified
to n significant figures, a result obtained by
combining the data will also be valid to n
significant figures.
However, if data are subtracted, the number of
significant figures can be reduced.
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UNITS AND MEASUREMENT 31
For example, 12.9 g – 7.06 g, both specified to three
significant figures, cannot properly be evaluated
as 5.84 g but only as 5.8 g, as uncertainties in
subtraction or addition combine in a different
fashion (smallest number of decimal places rather
than the number of significant figures in any of
the number added or subtracted).
(3) The relative error of a value of number
specified to significant figures depends not
only on n but also on the number itself.
For example, the accuracy in measurement of
mass 1.02 g is ± 0.01 g whereas another
measurement 9.89 g is also accurate to ± 0.01 g.
The relative error in 1.02 g is
= (± 0.01/1.02) × 100 %
= ± 1%
Similarly, the relative error in 9.89 g is
= (± 0.01/9.89) × 100 %
= ± 0.1 %
Finally, remember that intermediate results in
a multi-step computation should be
calculated to one more significant figure in
every measurement than the number of
digits in the least precise measurement.
These should be justified by the data and then
the arithmetic operations may be carried out;
otherwise rounding errors can build up. For
example, the reciprocal of 9.58, calculated (after
rounding off) to the same number of significant
figures (three) is 0.104, but the reciprocal of
0.104 calculated to three significant figures is
9.62. However, if we had written 1/9.58 = 0.1044
and then taken the reciprocal to three significant
figures, we would have retrieved the original
value of 9.58.
This example justifies the idea to retain one
more extra digit (than the number of digits in
the least precise measurement) in intermediate
steps of the complex multi-step calculations in
order to avoid additional errors in the process
of rounding off the numbers.
2.8 DIMENSIONS OF PHYSICAL QUANTITIES
The nature of a physical quantity is described
by its dimensions. All the physical quantities
represented by derived units can be expressed
in terms of some combination of seven
fundamental or base quantities. We shall call
these base quantities as the seven dimensions
of the physical world, which are denoted with
square brackets [ ]. Thus, length has the
dimension [L], mass [M], time [T], electric current
[A], thermodynamic temperature [K], luminous
intensity [cd], and amount of substance [mol].
The dimensions of a physical quantity are the
powers (or exponents) to which the base
quantities are raised to represent that
quantity. Note that using the square brackets
[ ] round a quantity means that we are dealing
with ‘the dimensions of’ the quantity.
In mechanics, all the physical quantities can
be written in terms of the dimensions [L], [M]
and [T]. For example, the volume occupied by
an object is expressed as the product of length,
breadth and height, or three lengths. Hence the
dimensions of volume are [L] × [L] × [L] = [L]
3
= [L
3
].
As the volume is independent of mass and time,
it is said to possess zero dimension in mass [M°],
zero dimension in time [T°] and three
dimensions in length.
Similarly, force, as the product of mass and
acceleration, can be expressed as
Force = mass × acceleration
= mass × (length)/(time)
2
The dimensions of force are [M] [L]/[T]
2
=
[M L T
–2
]. Thus, the force has one dimension in
mass, one dimension in length, and –2
dimensions in time. The dimensions in all other
base quantities are zero.
Note that in this type of representation, the
magnitudes are not considered. It is the quality
of the type of the physical quantity that enters.
Thus, a change in velocity, initial velocity,
average velocity, final velocity, and speed are
all equivalent in this context. Since all these
quantities can be expressed as length/time,
their dimensions are [L]/[T] or [L T
–1
].
2.9 DIMENSIONAL FORMULAE AND
DIMENSIONAL EQUATIONS
The expression which shows how and which of
the base quantities represent the dimensions
of a physical quantity is called the dimensional
formula of the given physical quantity. For
example, the dimensional formula of the volume
is [M° L
3
T°], and that of speed or velocity is
[M° L T
-1
]. Similarly, [M° L T
–2
] is the dimensional
formula of acceleration and [M L
–3
T°] that of
mass density.
An equation obtained by equating a physical
quantity with its dimensional formula is called
the dimensional equation of the physical
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PHYSICS32
quantity. Thus, the dimensional equations are
the equations, which represent the dimensions
of a physical quantity in terms of the base
quantities. For example, the dimensional
equations of volume [V], speed [v], force [F] and
mass density [
ρ
] may be expressed as
[V] = [M
0
L
3
T
0
]
[v] = [M
0
L T
–1
]
[F] = [M L T
–2
]
[
ρ
] = [M L
–3
T
0
]
The dimensional equation can be obtained
from the equation representing the relations
between the physical quantities. The
dimensional formulae of a large number and
wide variety of physical quantities, derived from
the equations representing the relationships
among other physical quantities and expressed
in terms of base quantities are given in
Appendix 9 for your guidance and ready
reference.
2.10 DIMENSIONAL ANALYSIS AND ITS
APPLICATIONS
The recognition of concepts of dimensions, which
guide the description of physical behaviour is
of basic importance as only those physical
quantities can be added or subtracted which
have the same dimensions. A thorough
understanding of dimensional analysis helps us
in deducing certain relations among different
physical quantities and checking the derivation,
accuracy and dimensional consistency or
homogeneity of various mathematical
expressions. When magnitudes of two or more
physical quantities are multiplied, their units
should be treated in the same manner as
ordinary algebraic symbols. We can cancel
identical units in the numerator and
denominator. The same is true for dimensions
of a physical quantity. Similarly, physical
quantities represented by symbols on both sides
of a mathematical equation must have the same
dimensions.
2.10.1 Checking the Dimensional
Consistency of Equations
The magnitudes of physical quantities may be
added together or subtracted from one another
only if they have the same dimensions. In other
words, we can add or subtract similar physical
quantities. Thus, velocity cannot be added to
force, or an electric current cannot be subtracted
from the thermodynamic temperature. This
simple principle called the principle of
homogeneity of dimensions in an equation is
extremely useful in checking the correctness of
an equation. If the dimensions of all the terms
are not same, the equation is wrong. Hence, if
we derive an expression for the length (or
distance) of an object, regardless of the symbols
appearing in the original mathematical relation,
when all the individual dimensions are
simplified, the remaining dimension must be
that of length. Similarly, if we derive an equation
of speed, the dimensions on both the sides of
equation, when simplified, must be of length/
time, or [L T
–1
].
Dimensions are customarily used as a
preliminary test of the consistency of an
equation, when there is some doubt about the
correctness of the equation. However, the
dimensional consistency does not guarantee
correct equations. It is uncertain to the extent
of dimensionless quantities or functions. The
arguments of special functions, such as the
trigonometric, logarithmic and exponential
functions must be dimensionless. A pure
number, ratio of similar physical quantities,
such as angle as the ratio (length/length),
refractive index as the ratio (speed of light in
vacuum/speed of light in medium) etc., has no
dimensions.
Now we can test the dimensional consistency
or homogeneity of the equation
(
)
2
1/2
0 0
x x v t a t
= + +
for the distance x travelled by a particle or body
in time t which starts from the position x
0
with
an initial velocity v
0
at time t = 0 and has uniform
acceleration a along the direction of motion.
The dimensions of each term may be written as
[x] = [L]
[x
0
] = [L]
[v
0
t] = [L T
–1
] [T]
= [L]
[(1/2) a t
2
] = [L T
–2
] [T
2
]
= [L]
As each term on the right hand side of this
equation has the same dimension, namely that
of length, which is same as the dimension of
left hand side of the equation, hence this
equation is a dimensionally correct equation.
It may be noted that a test of consistency of
dimensions tells us no more and no less than a
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UNITS AND MEASUREMENT 33
t
t
t
test of consistency of units, but has the
advantage that we need not commit ourselves
to a particular choice of units, and we need not
worry about conversions among multiples and
sub-multiples of the units. It may be borne in
mind that if an equation fails this consistency
test, it is proved wrong, but if it passes, it is
not proved right. Thus, a dimensionally correct
equation need not be actually an exact
(correct) equation, but a dimensionally wrong
(incorrect) or inconsistent equation must be
wrong.
Example 2.15 Let us consider an equation
1
2
m v m g h
2
=
where m is the mass of the body, v its
velocity, g is the acceleration due to
gravity and h is the height. Check
whether this equation is dimensionally
correct.
Answer The dimensions of LHS are
[M] [L T
–1
]
2
= [M] [ L
2
T
–2
]
= [M L
2
T
–2
]
The dimensions of RHS are
[M][L T
–2
] [L] = [M][L
2
T
–2
]
= [M L
2
T
–2
]
The dimensions of LHS and RHS are the same and
hence the equation is dimensionally correct. t
Example 2.16 The SI unit of energy is
J = kg m
2
s
–2
; that of speed v is m s
–1
and
of acceleration a is m s
–2
. Which of the
formulae for kinetic energy (K) given below
can you rule out on the basis of
dimensional arguments (m stands for the
mass of the body) :
(a) K = m
2
v
3
(b) K = (1/2)mv
2
(c) K = ma
(d) K = (3/16)mv
2
(e) K = (1/2)mv
2
+ ma
Answer Every correct formula or equation must
have the same dimensions on both sides of the
equation. Also, only quantities with the same
physical dimensions can be added or
subtracted. The dimensions of the quantity on
the right side are [M
2
L
3
T
–3
] for (a); [M L
2
T
–2
] for
(b) and (d); [M L T
–2
] for (c). The quantity on the
right side of (e) has no proper dimensions since
two quantities of different dimensions have been
added. Since the kinetic energy K has the
dimensions of [M L
2
T
–2
], formulas (a), (c) and (e)
are ruled out. Note that dimensional arguments
cannot tell which of the two, (b) or (d), is the
correct formula. For this, one must turn to the
actual definition of kinetic energy (see Chapter
6). The correct formula for kinetic energy is given
by (b). t
2.10.2 Deducing Relation among the
Physical Quantities
The method of dimensions can sometimes be
used to deduce relation among the physical
quantities. For this we should know the
dependence of the physical quantity on other
quantities (upto three physical quantities or
linearly independent variables) and consider it
as a product type of the dependence. Let us take
an example.
Example 2.17 Consider a simple
pendulum, having a bob attached to a
string, that oscillates under the action of
the force of gravity. Suppose that the period
of oscillation of the simple pendulum
depends on its length (l), mass of the bob
(m) and acceleration due to gravity (g).
Derive the expression for its time period
using method of dimensions.
Answer The dependence of time period T on
the quantities l, g and m as a product may be
written as :
T = k l
x
g
y
m
z
where k is dimensionless constant and x, y
and z are the exponents.
By considering dimensions on both sides, we
have
o o 1 1 1 –2 1
[L M T ]=[L ] [L T ] [M ]
x y z
= L
x+y
T
–2y
M
z
On equating the dimensions on both sides,
we have
x + y = 0; –2y = 1; and z = 0
So that
1 1
, – , 0
2 2
x y z
= = =
Then, T = k l
½
g
–½
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PHYSICS34
SUMMARY
1. Physics is a quantitative science, based on measurement of physical quantities. Certain
physical quantities have been chosen as fundamental or base quantities (such as length,
mass, time, electric current, thermodynamic temperature, amount of substance, and
luminous intensity).
2. Each base quantity is defined in terms of a certain basic, arbitrarily chosen but properly
standardised reference standard called unit (such as metre, kilogram, second, ampere,
kelvin, mole and candela). The units for the fundamental or base quantities are called
fundamental or base units.
3. Other physical quantities, derived from the base quantities, can be expressed as a
combination of the base units and are called derived units. A complete set of units,
both fundamental and derived, is called a system of units.
4. The International System of Units (SI) based on seven base units is at present
internationally accepted unit system and is widely used throughout the world.
5. The SI units are used in all physical measurements, for both the base quantities and
the derived quantities obtained from them. Certain derived units are expressed by
means of SI units with special names (such as joule, newton, watt, etc).
6. The SI units have well defined and internationally accepted unit symbols (such as m for
metre, kg for kilogram, s for second, A for ampere, N for newton etc.).
7. Physical measurements are usually expressed for small and large quantities in scientific
notation, with powers of 10. Scientific notation and the prefixes are used to simplify
measurement notation and numerical computation, giving indication to the precision
of the numbers.
8. Certain general rules and guidelines must be followed for using notations for physical
quantities and standard symbols for SI units, some other units and SI prefixes for
expressing properly the physical quantities and measurements.
9. In computing any physical quantity, the units for derived quantities involved in the
relationship(s) are treated as though they were algebraic quantities till the desired
units are obtained.
10. Direct and indirect methods can be used for the measurement of physical quantities.
In measured quantities, while expressing the result, the accuracy and precision of
measuring instruments along with errors in measurements should be taken into account.
11. In measured and computed quantities proper significant figures only should be retained.
Rules for determining the number of significant figures, carrying out arithmetic
operations with them, and ‘rounding off ‘ the uncertain digits must be followed.
12. The dimensions of base quantities and combination of these dimensions describe the
nature of physical quantities. Dimensional analysis can be used to check the dimensional
consistency of equations, deducing relations among the physical quantities, etc. A
dimensionally consistent equation need not be actually an exact (correct) equation,
but a dimensionally wrong or inconsistent equation must be wrong.
or, T =
l
k
g
Note that value of constant k can not be obtained
by the method of dimensions. Here it does not
matter if some number multiplies the right side
of this formula, because that does not affect its
dimensions.
Actually, k = 2π so that T =
2
l
g
π
t
Dimensional analysis is very useful in deducing
relations among the interdependent physical
quantities. However, dimensionless constants
cannot be obtained by this method. The method
of dimensions can only test the dimensional
validity, but not the exact relationship between
physical quantities in any equation. It does not
distinguish between the physical quantities
having same dimensions.
A number of exercises at the end of this
chapter will help you develop skill in
dimensional analysis.
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UNITS AND MEASUREMENT 35
EXERCISES
Note : In stating numerical answers, take care of significant figures.
2.1 Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to .....m
3
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to
...(mm)
2
(c) A vehicle moving with a speed of 18 km h
–1
covers....m in 1 s
(d) The relative density of lead is 11.3. Its density is ....g cm
–3
or ....kg m
–3
.
2.2 Fill in the blanks by suitable conversion of units
(a) 1 kg m
2
s
–2
= ....g cm
2
s
–2
(b) 1 m = ..... ly
(c) 3.0 m s
–2
= .... km h
–2
(d) G = 6.67 × 10
–11
N m
2
(kg)
–2
= .... (cm)
3
s
–2
g
–1
.
2.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J =
1 kg m
2
s
–2
. Suppose we employ a system of units in which the unit of mass equals
α
kg, the unit of length equals
β
m, the unit of time is
γ
s. Show that a calorie has a
magnitude 4.2
α
–1
β
–2
γ
2
in terms of the new units.
2.4 Explain this statement clearly :
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a
standard for comparison”. In view of this, reframe the following statements wherever
necessary :
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.
2.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What
is the distance between the Sun and the Earth in terms of the new unit if light takes
8 min and 20 s to cover this distance ?
2.6 Which of the following is the most precise device for measuring length :
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light ?
2.7 A student measures the thickness of a human hair by looking at it through a
microscope of magnification 100. He makes 20 observations and finds that the average
width of the hair in the field of view of the microscope is 3.5 mm. What is the
estimate on the thickness of hair ?
2.8 Answer the following :
(a)You are given a thread and a metre scale. How will you estimate the diameter of
the thread ?
(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do
you think it is possible to increase the accuracy of the screw gauge arbitrarily by
increasing the number of divisions on the circular scale ?
(c)The mean diameter of a thin brass rod is to be measured by vernier callipers. Why
is a set of 100 measurements of the diameter expected to yield a more reliable
estimate than a set of 5 measurements only ?
2.9 The photograph of a house occupies an area of 1.75 cm
2
on a 35 mm slide. The slide
is projected on to a screen, and the area of the house on the screen is 1.55 m
2
. What
is the linear magnification of the projector-screen arrangement.
2.10 State the number of significant figures in the following :
(a) 0.007 m
2
(b) 2.64 × 10
24
kg
(c) 0.2370 g cm
–3
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PHYSICS36
(d) 6.320 J
(e) 6.032 N m
–2
(f) 0.0006032 m
2
2.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m,
and 2.01 cm respectively. Give the area and volume of the sheet to correct significant
figures.
2.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of
masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the
box, (b) the difference in the masses of the pieces to correct significant figures ?
2.13 A physical quantity P is related to four observables a, b, c and d as follows :
(
)
=
3 2
P a b / c
d
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%,
respectively. What is the percentage error in the quantity P ? If the value of P calculated
using the above relation turns out to be 3.763, to what value should you round off
the result ?
2.14 A book with many printing errors contains four different formulas for the displacement
y of a particle undergoing a certain periodic motion :
(a) y = a sin 2π t/T
(b) y = a sin vt
(c) y = (a/T) sin t/a
(d)
y a t T t T= ( )2 (sin 2 / + cos 2 / )π π
(a = maximum displacement of the particle, v = speed of the particle. T = time-period
of motion). Rule out the wrong formulas on dimensional grounds.
2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m
o
of a
particle in terms of its speed v and the speed of light, c. (This relation first arose as
a consequence of special relativity due to Albert Einstein). A boy recalls the relation
almost correctly but forgets where to put the constant c. He writes :
( )
m
m
1 v
0
=
−
2
1/2
.
Guess where to put the missing c.
2.16 The unit of length convenient on the atomic scale is known as an angstrom and is
denoted by Å: 1 Å = 10
–10
m. The size of a hydrogen atom is about 0.5 Å. What is the
total atomic volume in m
3
of a mole of hydrogen atoms ?
2.17 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L
(molar volume). What is the ratio of molar volume to the atomic volume of a mole of
hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio
so large ?
2.18 Explain this common observation clearly : If you look out of the window of a fast
moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite
to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.)
seem to be stationary. (In fact, since you are aware that you are moving, these
distant objects seem to move with you).
2.19 The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances
of very distant stars. The baseline AB is the line joining the Earth’s two locations six
months apart in its orbit around the Sun. That is, the baseline is about the diameter
of the Earth’s orbit ≈ 3 × 10
11
m. However, even the nearest stars are so distant that
with such a long baseline, they show parallax only of the order of 1” (second) of arc
or so. A parsec is a convenient unit of length on the astronomical scale. It is the
distance of an object that will show a parallax of 1” (second of arc) from opposite
ends of a baseline equal to the distance from the Earth to the Sun. How much is a
parsec in terms of metres ?
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UNITS AND MEASUREMENT 37
2.20 The nearest star to our solar system is 4.29 light years away. How much is this
distance in terms of parsecs? How much parallax would this star (named Alpha
Centauri) show when viewed from two locations of the Earth six months apart in its
orbit around the Sun ?
2.21 Precise measurements of physical quantities are a need of science. For example, to
ascertain the speed of an aircraft, one must have an accurate method to find its
positions at closely separated instants of time. This was the actual motivation behind
the discovery of radar in World War II. Think of different examples in modern science
where precise measurements of length, time, mass etc. are needed. Also, wherever
you can, give a quantitative idea of the precision needed.
2.22 Just as precise measurements are necessary in science, it is equally important to be
able to make rough estimates of quantities using rudimentary ideas and common
observations. Think of ways by which you can estimate the following (where an
estimate is difficult to obtain, try to get an upper bound on the quantity) :
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.
2.23 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding
10
7
K, and its outer surface at a temperature of about 6000 K. At these high
temperatures, no substance remains in a solid or liquid phase. In what range do you
expect the mass density of the Sun to be, in the range of densities of solids and
liquids or gases ? Check if your guess is correct from the following data : mass of the
Sun = 2.0 × 10
30
kg, radius of the Sun = 7.0 × 10
8
m.
2.24 When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth,
its angular diameter is measured to be 35.72" of arc. Calculate the diameter of
Jupiter.
Additional Exercises
2.25 A man walking briskly in rain with speed v must slant his umbrella forward making
an angle θ
with the vertical. A student derives the following relation between
θ
and
v : tan
θ
= v and checks that the relation has a correct limit: as v → 0,
θ
→ 0, as
expected. (We are assuming there is no strong wind and that the rain falls vertically
for a stationary man). Do you think this relation can be correct ? If not, guess the
correct relation.
2.26 It is claimed that two cesium clocks, if allowed to run for 100 years, free from any
disturbance, may differ by only about 0.02 s. What does this imply for the accuracy
of the standard cesium clock in measuring a time-interval of 1 s ?
2.27 Estimate the average mass density of a sodium atom assuming its size to be about
2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium).
Compare it with the mass density of sodium in its crystalline phase : 970 kg m
–3
. Are
the two densities of the same order of magnitude ? If so, why ?
2.28 The unit of length convenient on the nuclear scale is a fermi : 1 f = 10
–15
m. Nuclear
sizes obey roughly the following empirical relation :
r = r
0
A
1/3
where r is the radius of the nucleus, A its mass number, and r
o
is a constant equal to
about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant
for different nuclei. Estimate the mass density of sodium nucleus. Compare it with
the average mass density of a sodium atom obtained in Exercise. 2.27.
2.29 A LASER is a source of very intense, monochromatic, and unidirectional beam of
light. These properties of a laser light can be exploited to measure long distances.
The distance of the Moon from the Earth has been already determined very precisely
using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to
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PHYSICS38
return after reflection at the Moon’s surface. How much is the radius of the lunar
orbit around the Earth ?
2.30 A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate
objects under water. In a submarine equipped with a SONAR, the time delay between
generation of a probe wave and the reception of its echo after reflection from an
enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine?
(Speed of sound in water = 1450 m s
–1
).
2.31 The farthest objects in our Universe discovered by modern astronomers are so distant
that light emitted by them takes billions of years to reach the Earth. These objects
(known as quasars) have many puzzling features, which have not yet been satisfactorily
explained. What is the distance in km of a quasar from which light takes 3.0 billion
years to reach us?
2.32 It is a well known fact that during a total solar eclipse the disk of the moon almost
completely covers the disk of the Sun. From this fact and from the information you
can gather from examples 2.3 and 2.4, determine the approximate diameter of the
moon.
2.33 A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of
Fundamental constants of nature. This led him to an interesting observation. Dirac
found that from the basic constants of atomic physics (c, e, mass of electron, mass of
proton) and the gravitational constant G, he could arrive at a number with the
dimension of time. Further, it was a very large number, its magnitude being close to
the present estimate on the age of the universe (~15 billion years). From the table of
fundamental constants in this book, try to see if you too can construct this number
(or any other interesting number you can think of ). If its coincidence with the age of
the universe were significant, what would this imply for the constancy of fundamental
constants?
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