1.1 Introduction

Chapter 1 — Units and Measurement

On the morning of 23 September 1999, engineers at NASA’s mission control lost contact with the Mars Climate Orbiter. The spacecraft, built to study the Martian atmosphere, had travelled nearly 670 million kilometres over ten months to reach the red planet. As it slipped behind Mars to enter orbit, radio signals stopped — and never returned. A $125 million mission was gone.

The investigation that followed did not blame a faulty engine or a broken sensor. It blamed something far simpler: units. One team of engineers had calculated the thruster forces in pound-seconds, the imperial unit used in the United States. Another team, working on the navigation software, assumed the numbers were in newton-seconds, the standard SI unit. Neither team had checked. The tiny mismatch, repeated over millions of kilometres, sent the spacecraft nearly 100 kilometres too close to Mars. It broke apart in the upper atmosphere.

A single unit mix-up destroyed a decade of work. This is why physics begins here — with measurement. Every physical quantity we deal with, from the size of an atom to the mass of a star, must be measured and reported in a way that anyone, anywhere, can understand and verify.

Figure to come

Fig. 1.0 – Artist’s view of the Mars Climate Orbiter approaching Mars, with a split-screen showing ‘lb·s’ on one side and ‘N·s’ on the other, illustrating the unit mismatch.

Curiosity Corner

Q1. Why does the world use one common system of units (the SI system) instead of different national systems like the older CGS, FPS, and MKS?

Q2. How is something as basic as “one metre” or “one kilogram” defined today — and why did scientists move away from physical objects like a standard metal bar or cylinder?

Q3. When you measure the length of a rod as 2.30 cm, why is the last digit different in meaning from the first two? What do “significant figures” really tell us?

Q4. Can we check whether a physics equation is correct just by looking at the units on both sides — without solving it?

Q5. If a pendulum’s time period depends on its length, mass, and gravity, can we guess the formula without doing any experiment?

This chapter will answer each of these questions and give you the tools to measure, report, and check physical quantities with confidence.

Physics is built on measurement. When we say a rod is 2 metres long, we are comparing its length with a fixed, agreed reference called a unit.

Every measurement therefore has two parts: a number — how many times the standard fits into the quantity — and a unit — which standard was used. Without the unit, the number alone is meaningless. “The distance is 5” says nothing until we add metres, kilometres, or light years.

Real-World Application

A tablet labelled “500 mg” tells a doctor exactly how much active ingredient it contains — the number 500 alone would be dangerous. Food labels, petrol pumps, and electricity bills all rely on the same idea: a number with a unit is information; a number alone is noise.

Nature has a huge number of physical quantities — length, area, speed, force, pressure, energy, and many more. Each does not need its own independent unit, because many quantities are related to one another. Area is length × length. Speed is length ÷ time. Once units for length and time are fixed, the units for area and speed follow automatically. A small set of independent units is therefore enough for all physical quantities.

Physicists split quantities into two groups. Base (or fundamental) quantities are chosen as independent — their units cannot be derived from any others. Length, mass, and time are common examples, and their units are called base units or fundamental units.

Definition

A base (or fundamental) quantity is a physical quantity chosen as independent, with its unit fixed by an internationally agreed standard. This unit is called a base unit or fundamental unit.

Derived quantities are those that can be expressed as combinations of base quantities. Their units, formed by multiplying or dividing base units, are called derived units. If metre (m) and second (s) are the base units of length and time, then speed has the derived unit m s⁻¹, area has m², and volume has m³.

Definition

A derived quantity is a physical quantity that can be expressed in terms of base quantities. Its unit, formed by combining base units, is called a derived unit.

The full collection of base units together with all the derived units built from them is called a system of units.

Definition

A system of units is a complete set of units — both base units for fundamental quantities and derived units for all other quantities — used together for consistent measurement.

The next section describes the SI system — the system of units that scientists across the world have agreed to use today.

1.2 The International System of Units

Before scientists across the world agreed on a common system, different countries used different sets of base units. Three systems were widely used until recently:

CGS system — used the centimetre for length, gram for mass, and second for time.
FPS system (also called the British system) — used the foot, pound, and second.
MKS system — used the metre, kilogram, and second.

This diversity created constant problems. A scientist in India using MKS and one in America using FPS had to convert numbers back and forth every time they exchanged data. Every conversion introduced a chance of error — and as the chapter opener showed, even a single unit mismatch can destroy a spacecraft.

To end this confusion, an international body called the Bureau International des Poids et Mesures (BIPM, or the International Bureau of Weights and Measures) developed a common system in 1971. It is called the Système International d’Unités — French for “International System of Units” — abbreviated as SI.

The SI has been revised several times as measurement technology has improved. The most recent revision was made in November 2018 by the General Conference on Weights and Measures. Today, SI is the accepted standard for scientific, technical, industrial, and commercial work almost everywhere in the world.

Curiosity Corner

Q. Why does the world use one common system of units (the SI system) instead of different national systems like the older CGS, FPS, and MKS? A. Because when different countries used CGS, FPS, and MKS, numbers had to be converted back and forth every time data was exchanged, and each conversion brought a fresh chance of error — an error that, as the Mars Climate Orbiter showed, can be costly. To end this confusion, the BIPM developed a single common system in 1971, the Système International d’Unités (SI), which is today the accepted standard for scientific, technical, industrial, and commercial work almost everywhere in the world. It is also a decimal system, so every conversion within it involves only powers of ten, which keeps calculations simple and reduces mistakes.

One practical reason for its wide use: the SI is a decimal system. Conversions within it — kilometre to metre, gram to kilogram, and so on — involve only powers of ten, which makes calculations simple and reduces mistakes. We will follow the SI throughout this book.

The seven base units

In the SI, seven physical quantities are chosen as base quantities, and each has its own base unit. All other physical quantities are derived from these seven. The seven base quantities and their units are listed in Table 1.1.

Base quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Thermodynamic temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

Table 1.1 — The seven SI base quantities and their units

Earlier, some of these units were defined using physical objects. For example, the kilogram was defined by a specific metal cylinder — a lump of platinum-iridium — kept safely in a vault in France. The metre was once defined by a scratched line on a metal bar. But such objects can slowly wear, get damaged, or drift in mass over decades. If the reference itself changes, every measurement in the world becomes slightly wrong.

Real Incident / Discovery

Over a hundred years, the international prototype kilogram — the platinum-iridium cylinder kept near Paris — was compared with its official copies from time to time. Measurements suggested that its mass had drifted very slightly relative to the copies (by a few tens of micrograms). Since it defined the kilogram, no one could tell which had actually changed. This uncertainty was one of the main reasons scientists decided in 2018 to redefine the kilogram (and several other base units) using constants of nature instead of physical objects.

The 2018 revision fixed this problem elegantly. Instead of defining units by objects, the SI now defines them by fixed numerical values of fundamental constants of nature — such as the speed of light, the Planck constant, and the elementary charge. These constants do not drift or wear out; they are the same everywhere in the universe. As a result, any laboratory with the right equipment can realise a base unit without needing to visit Paris.

Curiosity Corner

Q. How is something as basic as “one metre” or “one kilogram” defined today — and why did scientists move away from physical objects like a standard metal bar or cylinder? A. Since the 2018 revision, the base units are defined by fixing the numerical values of fundamental constants of nature. The metre is fixed through the speed of light in vacuum, $c = 299\,792\,458$ m s⁻¹, and the kilogram through the Planck constant, $h = 6.62607015 \times 10^{-34}$ J s. Physical objects were abandoned because they can wear, be damaged, or drift over decades: the platinum-iridium prototype kilogram appeared to have drifted by a few tens of micrograms relative to its copies, and because it defined the kilogram, nobody could say which had actually changed. Constants of nature do not drift and are the same everywhere in the universe, so any suitably equipped laboratory can realise a base unit for itself.

The formal definitions of the seven base units, as given in the 2018 SI, are summarised below. You do not need to memorise the exact numerical values — they are shown only to indicate the level of accuracy achieved today.

Metre (m) — the SI unit of length. Defined by fixing the speed of light in vacuum $c = 299\,792\,458$ m s⁻¹.
Kilogram (kg) — the SI unit of mass. Defined by fixing the Planck constant $h = 6.62607015 \times 10^{-34}$ J s.
Second (s) — the SI unit of time. Defined by fixing the caesium frequency $\Delta \nu_{\mathrm{Cs}} = 9\,192\,631\,770$ Hz (the frequency of a specific transition in the caesium-133 atom).
Ampere (A) — the SI unit of electric current. Defined by fixing the elementary charge $e = 1.602176634 \times 10^{-19}$ C.
Kelvin (K) — the SI unit of thermodynamic temperature. Defined by fixing the Boltzmann constant $k = 1.380649 \times 10^{-23}$ J K⁻¹.
Mole (mol) — the SI unit of amount of substance. Defined by fixing the Avogadro constant $N_A = 6.02214076 \times 10^{23}$ mol⁻¹. One mole contains exactly this many elementary entities.
Candela (cd) — the SI unit of luminous intensity. Defined by fixing the luminous efficacy of monochromatic radiation of frequency $540 \times 10^{12}$ Hz at $K_{\mathrm{cd}} = 683$ lm W⁻¹.

Side Note

The measuring techniques for these constants improve as technology improves. Whenever they improve, the SI definitions are updated so that the base units keep step with the best possible precision — without changing their everyday value. A metre remains a metre; only the way we realise it in the laboratory becomes sharper.

Two more units: plane angle and solid angle

Besides the seven base units, the SI defines two more units for angles.

The plane angle $d\theta$ is the ratio of the length of an arc $ds$ to the radius $r$ of the circle, as shown in Fig. 1.2(a):

\[d\theta = \frac{ds}{r}\]

Its unit is the radian (symbol: rad).

The solid angle $d\Omega$ is the ratio of the intercepted area $dA$ on a spherical surface to the square of its radius $r$, as shown in Fig. 1.2(b):

\[d\Omega = \frac{dA}{r^2}\]

Its unit is the steradian (symbol: sr).

Figure to come

Fig. 1.2 – (a) A wedge of a circle with radius r and arc length ds showing the plane angle dθ = ds/r. (b) A cone from apex O intersecting a spherical surface, marking the intercepted area dA and radius r, showing the solid angle dΩ = dA/r².

Since each is a ratio of two lengths (or of an area to an area), both plane angle and solid angle are dimensionless quantities — they carry no unit of length, mass, or time. Radian and steradian are simply names given to these pure ratios for convenience.

Derived units and units retained for general use

All other physical quantities can be expressed as combinations of the seven base units. The units so formed are called derived units — for example, the newton (N) for force, the joule (J) for energy, and the watt (W) for power. Many derived units carry special names in honour of scientists (Isaac Newton, James Prescott Joule, James Watt). Detailed lists of these derived units are given in the appendices for ready reference.

Along with SI units, some older units are retained for practical or historical reasons because they are convenient in everyday work. These are listed in Table 1.2.

Name	Symbol	Value in SI unit
minute	min	60 s
hour	h	60 min = 3600 s
day	d	24 h = 86400 s
year	y	365.25 d = $3.156 \times 10^{7}$ s
degree	°	$1° = (\pi/180)$ rad
litre	L	$1 \text{ dm}^3 = 10^{-3}$ m³
tonne	t	$10^{3}$ kg
carat	c	200 mg
bar	bar	0.1 MPa = $10^{5}$ Pa
curie	Ci	$3.7 \times 10^{10}$ s⁻¹
roentgen	R	$2.58 \times 10^{-4}$ C/kg
quintal	q	100 kg
barn	b	$100$ fm² = $10^{-28}$ m²
are	a	$1 \text{ dam}^2 = 10^{2}$ m²
hectare	ha	$1 \text{ hm}^2 = 10^{4}$ m²
standard atmospheric pressure	atm	$101325$ Pa = $1.013 \times 10^{5}$ Pa

Table 1.2 — Some units retained for general use (though outside SI)

These are not part of the SI, but they are so common in daily life (hours, litres, hectares) or in specific fields (barn in nuclear physics, carat for gemstones) that they are still accepted for general use.

Prefixes and symbols

To handle very large or very small quantities without writing many zeros, the SI uses standard prefixes — such as kilo (k) for $10^{3}$, milli (m) for $10^{-3}$, mega (M) for $10^{6}$, and nano (n) for $10^{-9}$. These prefixes attach to any base or derived unit. So a distance of $5000$ m can be written simply as $5$ km. The complete list of SI prefixes is given in the appendix.

Real-World Application

The prefixes appear everywhere in daily life — your Wi-Fi router’s speed in Mbps (megabits per second), your phone’s storage in GB (gigabytes), a tablet’s dose in mg (milligrams), and a chip’s transistor size in nm (nanometres). Each prefix carries the same meaning across every SI unit, which is why an engineer, a doctor, and a chemist can read each other’s numbers instantly.

The SI also has clear rules for writing unit symbols — for example, always use “m” (not “mtr” or “M”) for metre, “kg” (not “Kg”) for kilogram, and “s” (not “sec”) for second. These rules keep scientific writing consistent worldwide and are covered in the appendix for your reference.

1.3 Significant Figures

Every measurement has a limit to how precisely it can be made. No instrument in the world can give you a “perfectly exact” reading — there is always some smallest division below which we cannot be sure. A metre scale can read up to a millimetre, a screw gauge to a hundredth of a millimetre, but no instrument goes on giving reliable digits forever.

So when we report a measurement, we should not write more digits than the instrument can honestly support. We report all the digits we know reliably, and then one more digit — the first digit that we are uncertain about. These digits together are called the significant digits or significant figures of the measurement.

Definition

The significant figures of a measurement are all the digits known reliably from the instrument, plus the first digit that is uncertain.

For example, suppose you time the swing of a simple pendulum and find its period to be $1.62$ s. The digits $1$ and $6$ are reliable — the stopwatch clearly showed them. The digit $2$ is uncertain — it is your best guess of the next place. So the measurement has three significant figures.

Similarly, if the length of an object is reported as $287.5$ cm, the digits $2$, $8$, and $7$ are certain, while the $5$ is uncertain. This value has four significant figures.

Curiosity Corner

Q. When you measure the length of a rod as 2.30 cm, why is the last digit different in meaning from the first two? What do “significant figures” really tell us? A. Because a measurement is reported as all the digits that the instrument gives reliably, plus one more — the first digit we are uncertain about. In 2.30 cm the 2 and the 3 are the reliable digits, while the final 0 is that first uncertain digit, our best estimate of what lies below the smallest division of the instrument. Significant figures therefore tell us how precisely the quantity was actually measured; how many there are depends on the least count of the instrument, and writing more digits than the instrument can support only creates a false impression of precision.

Reporting more digits than the instrument justifies is not just useless — it is misleading. It would give a false impression that our measurement is more precise than it really is. If a metre scale can only read to the nearest millimetre, writing $287.523$ cm is dishonest, because the instrument cannot support the last two digits.

The number of significant figures depends on the least count of the instrument — the smallest value it can measure directly. A better instrument (smaller least count) gives more significant figures.

An important point: changing the unit of a measurement does not change the number of significant figures. Whether we write a length in metres, centimetres, or millimetres, the underlying precision of the measurement stays the same. This idea helps clear up most of the doubts about zeros in a number.

Rules for counting significant figures

Consider the length $2.308$ cm. In different units, the same value can be written as:

$2.308$ cm
$0.02308$ m
$23.08$ mm
$23080$ μm

All four represent the same measurement. So all four must have the same number of significant figures — namely four (the digits 2, 3, 0, 8). This shows that the position of the decimal point does not decide how many significant figures a number has.

From this, we get the following rules:

Rule 1. All non-zero digits are significant.

Rule 2. All zeros between two non-zero digits are significant, no matter where the decimal point is (or whether there is one at all). In $2.308$, the zero sits between $3$ and $8$, so it is significant.

Rule 3. If the number is less than 1, the zeros written just to the right of the decimal point but to the left of the first non-zero digit are not significant. In $0.002308$, the underlined zeros (0.002308) only mark the position of the decimal — they are not part of the measurement’s precision.

Rule 4. In a number without a decimal point, trailing zeros are not significant.

For example: $123$ m $= 12300$ cm $= 123000$ mm. All three represent the same length, so all three must have the same number of significant figures — three (1, 2, 3). The trailing zeros in $12300$ and $123000$ are only there because of the change of unit; they carry no extra precision.

Rule 5. In a number with a decimal point, trailing zeros are significant.

For example, $3.500$ and $0.06900$ each have four significant figures. The trailing zeros here are written on purpose to show that the measurement is precise to that many places.

The trailing-zero trap

Now consider a subtle case. Suppose a length is reported as $4.700$ m. The trailing zeros are meant to convey the precision — if they were not, the person would simply have written $4.7$ m. So $4.700$ m has four significant figures.

Now let us change the unit: \[4.700 \text{ m} = 470.0 \text{ cm} = 4700 \text{ mm} = 0.004700 \text{ km}\]

Look at $4700$ mm. Applying Rule 4 (no decimal point, trailing zeros not significant), we would wrongly conclude that it has only two significant figures. But we already know from the original measurement that there are four. A mere change of units cannot change how precise the measurement actually is.

This is the trailing-zero ambiguity: a number like $4700$ can mean two, three, or four significant figures depending on the context, and the writing does not tell us which.

Scientific notation removes the ambiguity

The clean way to remove this confusion is to write every measurement in scientific notation — as a number between 1 and 10, multiplied by a power of 10:

\[\text{value} = a \times 10^b\]

where $a$ is a number between 1 and 10, and $b$ is any positive or negative integer.

In this form, the same measurement $4.700$ m becomes:

\[4.700 \text{ m} = 4.700 \times 10^{2} \text{ cm} = 4.700 \times 10^{3} \text{ mm} = 4.700 \times 10^{-3} \text{ km}\]

Now every version clearly shows four significant figures. The power of 10 only shifts the decimal point; it does not affect the number of significant figures. All the digits in $a$ are always significant. This is why scientists prefer scientific notation for reporting measurements.

Real-World Application

Scientific notation is used everywhere in physics and engineering — the speed of light is written as $3.00 \times 10^{8}$ m s⁻¹, the mass of an electron as $9.11 \times 10^{-31}$ kg, and Avogadro’s number as $6.022 \times 10^{23}$ mol⁻¹. Writing these without scientific notation would need many zeros, and would immediately raise the trailing-zero ambiguity.

Order of magnitude

Once a number is written in scientific notation as $a \times 10^b$, we can get a rough idea of its size by looking only at the power of 10. To do this, we round $a$ to $1$ if $a \leq 5$, or to $10$ if $5 < a \leq 10$. The number is then approximately $10^b$, and the exponent $b$ is called the order of magnitude of the physical quantity.

Definition

The order of magnitude of a physical quantity is the exponent $b$ of 10 when the quantity is expressed approximately as $10^b$ in scientific notation.

For example, the diameter of the Earth is about $1.28 \times 10^{7}$ m. Since $a = 1.28$ rounds to $1$, the order of magnitude is $7$ — the Earth is “of the order of $10^{7}$ m” in diameter.

The diameter of a hydrogen atom is about $1.06 \times 10^{-10}$ m. Rounding again, the order of magnitude is $-10$.

Comparing the two, the Earth is $7 - (-10) = 17$ orders of magnitude larger than a hydrogen atom — that is, about $10^{17}$ times bigger. Order-of-magnitude thinking is a quick way to compare very different physical scales without worrying about precise numbers.

Real-World Application

Astronomers, engineers, and physicists constantly use order-of-magnitude reasoning: is a proposed satellite orbit at $10^{6}$ m altitude or $10^{7}$ m? Is a virus $10^{-7}$ m or $10^{-9}$ m across? Getting the power of 10 right is often more important than the leading digit, because it decides which physical laws or instruments are relevant.

If scientific notation is not used

Sometimes measurements are written without scientific notation. In that case, we use these rules to count significant figures:

For a number greater than 1 with no decimal point, trailing zeros are not significant.
For a number with a decimal point, trailing zeros are significant.

For a number less than 1, such as $0.1250$, the leading zero written before the decimal is only for style — it is never significant. But the trailing zero at the end (the $0$ after $125$) is significant, since the writer added it deliberately to show precision.

Exact numbers

Some numbers that appear in formulas are not measurements at all — they are exact by definition. For example, in $r = d/2$ (radius = diameter divided by 2) or $s = 2 \pi r$ (circumference = 2 times $\pi$ times radius), the factor $2$ is an exact number. It is not a rounded measurement.

Exact numbers have an infinite number of significant figures. The $2$ in $r = d/2$ can be treated as $2$, $2.0$, $2.00$, or $2.0000$ — whatever is needed for the calculation. Similarly, in the pendulum formula $T = 2\pi \sqrt{l/g}$, the factor $2$ (and $\pi$ itself, in principle) has infinite significant figures. Exact numbers never limit the precision of a calculation.

1.3.1 Rules for Arithmetic Operations with Significant Figures

When we combine measured values by adding, subtracting, multiplying, or dividing, the answer cannot be more precise than the numbers we started with. If one input is only known to three significant figures, we cannot claim the answer is good to ten.

Consider an example. A block’s mass is measured as $4.237$ g (four significant figures), and its volume as $2.51$ cm³ (three significant figures). By simple division, its density comes out to $1.68804780876$ g/cm³ up to eleven decimal places. Writing all these digits pretends the density is known to a fantastic precision — but the original measurements do not support this at all.

So we need rules that keep the precision of the answer honest. There are two such rules, one for multiplication/division, and one for addition/subtraction.

Rule 1 (Multiplication and Division):

Principle / Law

In multiplication or division, the final result should have as many significant figures as the input value with the least significant figures.

Applying this to the density example, the volume $2.51$ has three significant figures — the smallest. So the answer must be rounded to three significant figures:

\[\text{Density} = \frac{4.237 \text{ g}}{2.51 \text{ cm}^3} = 1.69 \text{ g cm}^{-3}\]

Another example. The speed of light is given as $3.00 \times 10^{8}$ m s⁻¹ (three significant figures), and one year is $3.1557 \times 10^{7}$ s (five significant figures). The distance covered by light in one year (a “light year”) is:

\[9.47 \times 10^{15} \text{ m}\]

Because the speed has only three significant figures, the answer is rounded to three significant figures too, even though the year was given more precisely.

Rule 2 (Addition and Subtraction):

Principle / Law

In addition or subtraction, the final result should have as many decimal places as the input value with the least decimal places.

Notice the difference: for adding and subtracting, we look at decimal places, not at total significant figures.

For example, add $436.32$ g, $227.2$ g, and $0.301$ g. Simple arithmetic gives $663.821$ g. But the least precise input, $227.2$ g, is known only to one decimal place. So the answer must be rounded to one decimal place:

\[\text{Sum} = 663.8 \text{ g}\]

Similarly, a difference of two lengths:

\[0.307 \text{ m} - 0.304 \text{ m} = 0.003 \text{ m} = 3 \times 10^{-3} \text{ m}\]

Here, both original numbers had three decimal places (and three significant figures each). The answer has three decimal places — but only one significant figure. This shows how subtracting two close numbers can badly reduce the significant figures in the result.

Why we don’t mix these rules. In the addition example above, if we had wrongly used Rule 1 (least significant figures = three, from $227.2$), we would have written the sum as $664$ g, which does not reflect the correct precision. In the subtraction example, using Rule 1 would give $3.00 \times 10^{-3}$ m, pretending three significant figures of precision that we do not have. The two rules exist because addition/subtraction and multiplication/division propagate uncertainty differently.

1.3.2 Rounding off the Uncertain Digits

Whenever we finish a calculation, we usually need to round off the result to the correct number of significant figures. The rules for rounding are:

Rule A (dropped digit greater than 5). If the digit to be dropped is more than 5, the preceding digit is increased by 1.

Example: $2.746$ rounded to three significant figures becomes $2.75$ (the dropped $6$ is more than $5$, so the preceding $4$ becomes $5$).

Rule B (dropped digit less than 5). If the digit to be dropped is less than 5, the preceding digit is left unchanged.

Example: $1.743$ rounded to three significant figures becomes $1.74$ (the dropped $3$ is less than $5$).

Rule C (dropped digit exactly 5). If the digit to be dropped is exactly 5, the convention is:

If the preceding digit is even, the 5 is simply dropped (preceding digit stays).
If the preceding digit is odd, the preceding digit is raised by 1.

Example: $2.745$ becomes $2.74$ (preceding digit $4$ is even, so drop the $5$). Example: $2.735$ becomes $2.74$ (preceding digit $3$ is odd, so raise it to $4$).

This convention is called the “round half to even” or “banker’s rounding” rule. It stops long calculations from drifting upward, because roughly half the times a $5$ is dropped it rounds up, and half the times it rounds down.

Rule D (intermediate steps). In a multi-step calculation, keep one digit more than the final significant figures in each intermediate step. Only at the end of the calculation, round off to the correct number of significant figures. This prevents small rounding errors from piling up during a long calculation.

Similarly, a value known very precisely — like the speed of light $2.99792458 \times 10^{8}$ m s⁻¹ — is often rounded to $3 \times 10^{8}$ m s⁻¹ in ordinary use, because the other data in the problem rarely justifies more precision.

And as noted earlier, exact numbers that appear in formulas — like the $2$ and $\pi$ in $T = 2\pi \sqrt{l/g}$ — have infinite significant figures and never limit the answer’s precision.

Solved Example 1.1

Each side of a cube is measured to be $7.203$ m. Find the total surface area and the volume of the cube to the correct number of significant figures.

Answer

The measured side length $7.203$ has four significant figures. Both surface area and volume are products of measured quantities, so by Rule 1 they must also be rounded to four significant figures.

Surface area of the cube $= 6 (7.203)^2$ m²

\[= 311.299254 \text{ m}^2 = 311.3 \text{ m}^2\]

Volume of the cube $= (7.203)^3$ m³

\[= 373.714754 \text{ m}^3 = 373.7 \text{ m}^3\]

Both answers carry four significant figures, matching the precision of the original measurement.

Solved Example 1.2

A substance has mass $5.74$ g and occupies a volume of $1.2$ cm³. Express its density with the correct number of significant figures.

Answer

The mass has three significant figures, but the volume has only two. By Rule 1 (multiplication/division), the answer must be rounded to the smaller of the two — that is, two significant figures.

\[\text{Density} = \frac{5.74}{1.2} \text{ g cm}^{-3} = 4.8 \text{ g cm}^{-3}\]

1.3.3 Rules for Determining the Uncertainty in the Results of Arithmetic Calculations

Every measured value has a small uncertainty, which we usually write with a $\pm$ symbol. For example, $l = 16.2 \pm 0.1$ cm means the true length lies somewhere between $16.1$ cm and $16.3$ cm. The uncertainty can also be written as a percentage. Now we ask: when we combine two such measurements — say, multiply length and breadth to get area — how does the uncertainty of the answer depend on the uncertainties of the inputs?

Case 1: Multiplication and division — add relative errors.

Suppose a thin rectangular sheet has: - length $l = 16.2 \pm 0.1$ cm $= 16.2$ cm $\pm 0.6 \%$ - breadth $b = 10.1 \pm 0.1$ cm $= 10.1$ cm $\pm 1 \%$

When we multiply them to find the area, the relative (percentage) errors add:

\[lb = 163.62 \text{ cm}^2 \pm 1.6 \%\]

Converting back to an absolute error:

\[lb = 163.62 \pm 2.6 \text{ cm}^2\]

Since the uncertainty $\pm 2.6$ cm² is already in the units column, keeping the value to two decimal places gives a false sense of precision. So we round the value to match the uncertainty:

\[lb = 164 \pm 3 \text{ cm}^2\]

The number $3$ cm² is the uncertainty in the estimated area of the sheet.

Case 2: Combining data with the same significant figures.

Principle / Law

If a set of experimental data is specified to $n$ significant figures, any result obtained by multiplying or dividing them is also valid to $n$ significant figures.

But if we subtract two such numbers, the significant figures can drop sharply, as we already saw in Section 1.3.1. For instance, $12.9$ g $- 7.06$ g cannot be written as $5.84$ g. Both numbers are known to one decimal place at best (since $12.9$ has only one decimal place), so the correct answer is $5.8$ g. This is because subtraction and addition follow the decimal-place rule, not the significant-figure rule.

Case 3: Relative error depends on the value itself.

Two measurements can have the same absolute uncertainty but very different relative uncertainties.

For example: a mass measured as $1.02$ g $\pm 0.01$ g has:

\[\text{Relative error} = \pm \frac{0.01}{1.02} \times 100 \% = \pm 1 \%\]

Another mass measured as $9.89$ g $\pm 0.01$ g has:

\[\text{Relative error} = \pm \frac{0.01}{9.89} \times 100 \% = \pm 0.1 \%\]

The absolute error is the same, but the second measurement is ten times more accurate in relative terms, because the same $\pm 0.01$ g is a much smaller fraction of a bigger value.

Case 4: One extra digit in intermediate steps.

Principle / Law

Intermediate results in a multi-step calculation should be kept to one more significant figure than the number of digits in the least precise measurement. Round off only at the end.

Here is why. Take the reciprocal of $9.58$ rounded to three significant figures: $1/9.58 = 0.104$. Now take the reciprocal of $0.104$ rounded again to three significant figures: $1/0.104 = 9.62$. We started at $9.58$ but ended at $9.62$ — a real change, caused only by rounding twice.

But if we had kept the first reciprocal as $1/9.58 = 0.1044$ (four significant figures — one extra), then $1/0.1044 = 9.579$, which rounds back correctly to $9.58$. The extra digit absorbs the intermediate rounding and prevents the drift.

That is why in multi-step problems, we keep one guard digit through the calculation and round only the final answer. It is a small habit that keeps physics numerics honest.

Numerical 1.1

A student measures the diameter of a cylindrical wire as $d = 0.86$ mm using a screw gauge whose least count is $0.01$ mm, and its length as $L = 24.5$ cm using a metre scale whose least count is $0.1$ cm. Calculate the volume of the wire and express the result with the correct number of significant figures and its estimated uncertainty (in $\pm$ form).

Numerical 1.2

Using the values from the previous problem, if the mass of the wire is measured as $m = 3.42$ g $\pm 0.01$ g, calculate the density of the material of the wire to the correct number of significant figures and state its relative error as a percentage.

1.4 Dimensions of Physical Quantities

Every physical quantity has a certain “nature” — it is fundamentally a length, or a mass, or some combination of the base quantities. This nature is captured by what we call the dimensions of the quantity.

Recall from the earlier sections that all physical quantities can be built from the seven base quantities of the SI system. In the same way, the nature of any quantity can be described by stating how it is built from these seven. We treat these seven base quantities as the seven dimensions of the physical world, and we write each one inside square brackets:

length → $[\text{L}]$
mass → $[\text{M}]$
time → $[\text{T}]$
electric current → $[\text{A}]$
thermodynamic temperature → $[\text{K}]$
luminous intensity → $[\text{cd}]$
amount of substance → $[\text{mol}]$

When we write a quantity inside square brackets, it means we are talking about the dimensions of that quantity — not its numerical value, and not its unit. It is a way of asking, “What kind of quantity is this, and how is it made from the base quantities?”

Definition

The dimensions of a physical quantity are the powers (or exponents) to which the base quantities must be raised to represent that quantity. Writing a quantity inside square brackets $[\ ]$ denotes “the dimensions of” that quantity.

In mechanics — the study of motion and forces — every quantity can be expressed using just three of the seven dimensions: length $[\text{L}]$, mass $[\text{M}]$, and time $[\text{T}]$. Electric current, temperature, and the others simply do not appear in purely mechanical quantities.

An example: the dimensions of volume

Consider the volume of an object. Volume is found by multiplying three lengths together — for a box, length × breadth × height. Since each of these is a length, the dimensions of volume are:

\[[\text{L}] \times [\text{L}] \times [\text{L}] = [\text{L}]^3 = [\text{L}^3]\]

Volume does not depend on mass or time at all. So we say it has zero dimension in mass $[\text{M}^0]$, zero dimension in time $[\text{T}^0]$, and three dimensions in length. Writing all three together, the complete dimensional description of volume is $[\text{M}^0 \text{L}^3 \text{T}^0]$.

Here $[\text{M}^0]$ and $[\text{T}^0]$ are both equal to 1 (any quantity raised to the power zero is 1), so they do not change the value. We still write them to make it clear that we have checked mass and time and found their powers to be zero — it shows the description is complete.

Figure to come

Fig. 1.3 – A rectangular box labelled with length, breadth, and height (each an [L]), with an arrow showing the product [L]×[L]×[L] = [L³], and a note “M⁰, T⁰” indicating no dependence on mass or time.

Another example: the dimensions of force

Force is a slightly richer example because it involves all three mechanical dimensions. From Newton’s second law, force is the product of mass and acceleration:

\[\text{Force} = \text{mass} \times \text{acceleration}\]

Now, acceleration is a change in velocity per unit time, and velocity is length per unit time. So acceleration is length divided by time squared:

\[\text{acceleration} = \frac{\text{length}}{(\text{time})^2}\]

Putting this into the expression for force:

\[\text{Force} = \text{mass} \times \frac{\text{length}}{(\text{time})^2}\]

Replacing each quantity by its dimension:

\[[\text{Force}] = [\text{M}] \, \frac{[\text{L}]}{[\text{T}]^2} = [\text{M L T}^{-2}]\]

So force has one dimension in mass, one dimension in length, and minus two dimensions in time. The dimensions in all the other base quantities (current, temperature, and so on) are zero.

Dimensions describe the “type”, not the amount

An important idea to grasp here: dimensions do not care about magnitude. They describe only the kind of physical quantity, not how big or small it is.

This means that a change in velocity, an initial velocity, an average velocity, a final velocity, and speed are all treated as the same in this context. They differ in value and in physical meaning, but each of them is fundamentally a length divided by a time. So all of them share the same dimensions:

\[\frac{[\text{L}]}{[\text{T}]} = [\text{L T}^{-1}]\]

Real-World Application

Engineers and physicists use this “type-matching” idea constantly. Pressure, stress, and energy density all reduce to the same dimensions, which is why the same mathematical tools reappear across mechanics, fluids, and material science. Recognising that two seemingly different quantities share dimensions often reveals a hidden physical connection — and, as the next sections show, lets us check equations and even guess relationships without doing a single experiment.

Numerical 1.3

Express the dimensions of the following mechanical quantities in the form $[\text{M}^a \text{L}^b \text{T}^c]$: (i) area, (ii) momentum (mass × velocity), (iii) work (force × displacement), and (iv) power (work ÷ time).

1.5 Dimensional Formulae and Dimensional Equations

In the previous section we learned that every physical quantity has dimensions — a set of powers of the base quantities that describes its nature. We now put this idea into a compact written form. There are two closely related ways to record the dimensions of a quantity: the dimensional formula and the dimensional equation.

Dimensional formula

The dimensional formula of a physical quantity is the expression that shows which base quantities go into it and to what power each one is raised. It is simply the quantity’s dimensions written out in square brackets.

Definition

The dimensional formula of a physical quantity is the expression that shows how and which of the base quantities represent the dimensions of that quantity.

Here are the dimensional formulae of a few common quantities:

Volume: $[\text{M}^0 \text{L}^3 \text{T}^0]$
Speed or velocity: $[\text{M}^0 \text{L} \text{T}^{-1}]$
Acceleration: $[\text{M}^0 \text{L} \text{T}^{-2}]$
Mass density: $[\text{M} \text{L}^{-3} \text{T}^0]$

To read a dimensional formula, look at the power on each base symbol. In the formula for mass density, $[\text{M} \text{L}^{-3} \text{T}^0]$, mass has power $+1$, length has power $-3$ (meaning length appears three times in the denominator), and time has power $0$ (density does not depend on time).

Figure to come

Fig. 1.4 – The dimensional formula [M L⁻³ T⁰] with labelled callouts pointing to each base symbol and its power: “mass, power +1”, “length, power −3 (in denominator)”, “time, power 0 (no dependence)”.

Notice that we keep the zero-power terms like $[\text{M}^0]$ and $[\text{T}^0]$ in the written formula even though they equal 1. This makes the formula complete and shows at a glance that every base quantity has been accounted for.

Dimensional equation

When we write the symbol of a physical quantity inside square brackets and set it equal to its dimensional formula, we get a dimensional equation.

Definition

A dimensional equation is an equation obtained by equating a physical quantity (written in square brackets) with its dimensional formula, thereby expressing its dimensions in terms of the base quantities.

For example, using $[V]$ for volume, $[v]$ for speed, $[F]$ for force, and $[\rho]$ for mass density, the dimensional equations are:

\[[V] = [\text{M}^0 \text{L}^3 \text{T}^0]\] \[[v] = [\text{M}^0 \text{L} \text{T}^{-1}]\] \[[F] = [\text{M} \text{L} \text{T}^{-2}]\] \[[\rho] = [\text{M} \text{L}^{-3} \text{T}^0]\]

The difference between the two ideas is small but worth being clear about. The dimensional formula is only the right-hand side — the bracketed expression of powers. The dimensional equation is the full statement that puts the quantity’s symbol on the left and its formula on the right, joined by an equals sign.

Obtaining a dimensional equation

A dimensional equation is not something we memorise separately — it can always be worked out from the equation that defines the quantity in terms of others.

Take mass density as an example. By definition, density is mass divided by volume:

\[\rho = \frac{\text{mass}}{\text{volume}}\]

Replacing each quantity by its dimensions — mass by $[\text{M}]$ and volume by $[\text{L}^3]$ — gives:

\[[\rho] = \frac{[\text{M}]}{[\text{L}^3]} = [\text{M} \text{L}^{-3} \text{T}^0]\]

This matches the dimensional formula listed above. The same method works for any derived quantity: start from its defining relation, substitute the dimensions of each quantity on the right, and simplify.

Real-World Application

Reference tables of dimensional formulae are used across physics and engineering to catch mistakes quickly. If a newly derived expression for, say, energy does not reduce to the known dimensional formula of energy, the derivation must contain an error — the mismatch is spotted long before any numbers are plugged in. The dimensional formulae of a large variety of physical quantities are collected in the appendix at the end of the book for ready reference.

Numerical 1.4

Starting from their defining relations, obtain the dimensional formula of each of the following: (i) pressure (force ÷ area), (ii) kinetic energy $\tfrac{1}{2}mv^2$, and (iii) the universal gravitational constant $G$, given that gravitational force is $F = \dfrac{G m_1 m_2}{r^2}$.

1.6 Dimensional Analysis and its Applications

We now come to the real power of dimensions. Knowing the dimensions of a quantity is not just a way of classifying it — it is a practical tool. Dimensional analysis lets us check whether an equation could be correct, and sometimes even lets us guess the form of a relationship without doing any experiment.

The starting idea is simple but strict: only quantities with the same dimensions can be added to or subtracted from one another. It makes no sense to add a length to a mass, or to subtract a time from a temperature — the result would be meaningless. This restriction is what makes dimensions so useful for testing equations.

There is one more key idea. When we multiply or divide physical quantities, we can treat their units — and their dimensions — exactly like ordinary algebraic symbols. Identical units in the numerator and denominator cancel, just as $x$ cancels with $x$ in algebra.

For example, if we divide a distance in metres by a time in seconds, the result carries the unit m s⁻¹, formed by the same cancelling and combining we do with algebraic symbols. The same is true for dimensions: $[\text{L}]$ divided by $[\text{T}]$ gives $[\text{L T}^{-1}]$.

Because of this, whenever two quantities are joined by an equals sign in a physics equation, the symbols on both sides must reduce to the same dimensions. This single requirement is the foundation of everything in this section.

1.6.1 Checking the Dimensional Consistency of Equations

As stated above, two physical quantities can be added or subtracted only if they have the same dimensions. We can add two lengths, or subtract one time from another, because they are “similar” quantities. But we cannot add a velocity to a force, nor subtract an electric current from a temperature — these are quantities of different kinds.

This rule, applied to a full equation, is called the principle of homogeneity of dimensions. It says that every term in a valid physical equation must have the same dimensions.

Principle / Law

Principle of homogeneity of dimensions — In any correct physical equation, all terms that are added, subtracted, or set equal must have the same dimensions. If the dimensions of all the terms are not the same, the equation cannot be correct.

This principle is a quick and powerful way to check the correctness of an equation. If even one term has different dimensions from the others, we know immediately — without solving anything — that the equation is wrong.

It also gives us a check on derivations. Suppose we derive an expression for the length (or distance) of an object. No matter what symbols appear in the working, when all the dimensions are simplified, the final result must have the dimension of length, $[\text{L}]$. If instead we derive an expression for a speed, the two sides must simplify to $[\text{L T}^{-1}]$. If they do not, there is a mistake somewhere in the derivation.

Dimensionless quantities and special functions. Not everything in physics carries dimensions. A pure number, or a ratio of two similar quantities, has no dimensions at all. For example, a plane angle is the ratio of arc length to radius — length divided by length — so it is dimensionless. Refractive index is the ratio of the speed of light in vacuum to the speed of light in a medium — speed divided by speed — so it too is dimensionless.

This matters because the arguments of special mathematical functions — trigonometric functions like $\sin$ and $\cos$, logarithms, and exponentials — must always be dimensionless. You can take the sine of an angle (dimensionless), but it is meaningless to take the sine of “5 metres.” Keep this in mind when checking equations that contain such functions.

A worked check. Let us test the well-known equation of motion for a body starting from position $x_0$ with initial velocity $v_0$ and moving with uniform acceleration $a$ for a time $t$:

\[x = x_0 + v_0 t + \frac{1}{2} a t^2\]

Here $x$ is the position of the body after time $t$. To check consistency, we find the dimensions of each term separately.

\[[x] = [\text{L}]\] \[[x_0] = [\text{L}]\] \[[v_0 t] = [\text{L T}^{-1}] \, [\text{T}] = [\text{L}]\] \[\left[\frac{1}{2} a t^2\right] = [\text{L T}^{-2}] \, [\text{T}^2] = [\text{L}]\]

Every term on the right-hand side works out to $[\text{L}]$, which matches the dimension of the left-hand side. So the equation is dimensionally consistent — it passes the test. (Note that the factor $\tfrac{1}{2}$ is a dimensionless number and does not affect the dimensions.)

Curiosity Corner

Q. Can we check whether a physics equation is correct just by looking at the units on both sides — without solving it? A. Partly, yes. By the principle of homogeneity of dimensions, every term that is added, subtracted, or set equal in a correct physical equation must have the same dimensions, so an equation whose terms do not match can be declared wrong at once, without solving anything. Working with dimensions rather than units has the added advantage that no particular system of units, and no conversion between multiples and sub-multiples, is needed. But the test is only preliminary: a dimensionally inconsistent equation must be wrong, while a dimensionally consistent one is not thereby proved right, because dimensional analysis is blind to missing dimensionless factors and cannot tell apart two quantities that share the same dimensions.

Figure to come

Fig. 1.5 – The equation x = x₀ + v₀t + ½at² with each term boxed and a tag beneath showing its dimension all equal to [L], illustrating that every term matches.

Dimensions versus units. A test of dimensional consistency is really the same as a test of consistency of units — but it has one big advantage. We do not have to commit to any particular system of units, and we do not have to worry about converting between multiples and sub-multiples (kilometres to metres, grams to kilograms, and so on). The dimensions stay the same whatever units we choose.

An important warning. Dimensional consistency is only a preliminary test. It can tell us an equation is wrong, but it can never fully prove an equation is right. This asymmetry is worth stating carefully.

Principle / Law

If an equation fails the dimensional consistency test, it is proved wrong. But if it passes, it is not proved right. A dimensionally correct equation need not be an exact (correct) equation, but a dimensionally wrong or inconsistent equation must be wrong.

Why the one-sidedness? Because dimensional analysis is blind to anything without dimensions. It cannot detect a missing dimensionless factor (like a $\tfrac{1}{2}$ or a $2\pi$), and it cannot tell apart two different quantities that happen to have the same dimensions. So an equation can be perfectly balanced in dimensions and still be physically wrong.

Solved Example 1.3

Consider the equation $\dfrac{1}{2} m v^2 = m g h$, where $m$ is the mass of a body, $v$ its velocity, $g$ the acceleration due to gravity, and $h$ the height. Check whether this equation is dimensionally correct.

Answer

We find the dimensions of each side separately and compare.

The dimensions of the left-hand side (LHS):

\[[\text{M}] \, [\text{L T}^{-1}]^2 = [\text{M}] \, [\text{L}^2 \text{T}^{-2}] = [\text{M L}^2 \text{T}^{-2}]\]

The dimensions of the right-hand side (RHS):

\[[\text{M}] \, [\text{L T}^{-2}] \, [\text{L}] = [\text{M}] \, [\text{L}^2 \text{T}^{-2}] = [\text{M L}^2 \text{T}^{-2}]\]

The dimensions of LHS and RHS are the same, so the equation is dimensionally correct.

Solved Example 1.4

The SI unit of energy is $\text{J} = \text{kg m}^2 \text{s}^{-2}$; that of speed $v$ is $\text{m s}^{-1}$; and that of acceleration $a$ is $\text{m s}^{-2}$. On the basis of dimensional arguments alone, which of the following formulae for kinetic energy $K$ can be ruled out? ($m$ stands for the mass of the body.)

$K = m^2 v^3$ (b) $K = \tfrac{1}{2} m v^2$ (c) $K = ma$ (d) $K = \tfrac{3}{16} m v^2$ (e) $K = \tfrac{1}{2} m v^2 + ma$

Answer

Every correct formula must have the same dimensions on both sides, and only quantities with the same dimensions can be added together. So we find the dimensions of each right-hand side and compare them with the known dimensions of kinetic energy.

The dimensions of the right side of each option are:

1. $m^2 v^3$: $[\text{M}^2] \, [\text{L T}^{-1}]^3 = [\text{M}^2 \text{L}^3 \text{T}^{-3}]$
1. $\tfrac{1}{2} m v^2$: $[\text{M}] \, [\text{L T}^{-1}]^2 = [\text{M L}^2 \text{T}^{-2}]$
1. $ma$: $[\text{M}] \, [\text{L T}^{-2}] = [\text{M L T}^{-2}]$
1. $\tfrac{3}{16} m v^2$: $[\text{M L}^2 \text{T}^{-2}]$ (same as (b), since the fraction is dimensionless)
1. $\tfrac{1}{2} m v^2 + ma$: here two terms of different dimensions are added — $[\text{M L}^2 \text{T}^{-2}]$ and $[\text{M L T}^{-2}]$ — so this expression has no proper dimensions at all.

Kinetic energy $K$ has the dimensions of energy, $[\text{M L}^2 \text{T}^{-2}]$. Comparing, formulae (a), (c), and (e) do not match and are ruled out.

Notice, however, that dimensional arguments cannot tell which of (b) and (d) is the correct formula — both have exactly the right dimensions $[\text{M L}^2 \text{T}^{-2}]$. They differ only by the dimensionless factor ($\tfrac{1}{2}$ versus $\tfrac{3}{16}$), and dimensions are blind to such factors. To settle it, we must use the actual definition of kinetic energy (studied in a later chapter), which gives the correct formula as (b), $K = \tfrac{1}{2} m v^2$. This example shows exactly why passing the dimensional test does not prove an equation right.

Numerical 1.5

Check whether the equation $v^2 = u^2 + 2as$ is dimensionally consistent, where $u$ and $v$ are velocities, $a$ is acceleration, and $s$ is displacement. State the dimension each term reduces to.

1.6.2 Deducing Relation among the Physical Quantities

Dimensional analysis can do more than check equations — it can sometimes help us build a relation from scratch. If we know which quantities a certain physical quantity depends on, we can often find how they combine, without any experiment.

The method works under two conditions. First, we must know the quantities on which the result depends — and there should be at most three of them (more precisely, three linearly independent variables). Second, we assume the dependence is of the product type: the quantity equals a constant times each variable raised to some power. We then use dimensions to find those powers.

Solved Example 1.5

Consider a simple pendulum — a bob attached to a string that swings under gravity. Suppose the time period of oscillation $T$ depends on the length of the string $l$, the mass of the bob $m$, and the acceleration due to gravity $g$. Derive an expression for the time period using the method of dimensions.

Answer

We assume a product-type dependence, with unknown powers $x$, $y$, and $z$:

\[T = k \, l^x \, g^y \, m^z\]

Here $k$ is a dimensionless constant, and $x$, $y$, $z$ are the exponents we need to find.

Now we write the dimensions of both sides. Time period $T$ has dimension $[\text{T}]$. Length $l$ has $[\text{L}]$, acceleration $g$ has $[\text{L T}^{-2}]$, and mass $m$ has $[\text{M}]$:

\[[\text{L}^0 \text{M}^0 \text{T}^1] = [\text{L}^1]^x \, [\text{L}^1 \text{T}^{-2}]^y \, [\text{M}^1]^z\]

Collecting the powers of each base quantity on the right-hand side:

\[[\text{L}^0 \text{M}^0 \text{T}^1] = [\text{L}^{x+y} \, \text{T}^{-2y} \, \text{M}^{z}]\]

For the two sides to be equal, the power of each base quantity must match. Comparing the powers of L, T, and M separately gives three equations:

\[x + y = 0; \qquad -2y = 1; \qquad z = 0\]

Solving these:

\[x = \frac{1}{2}, \qquad y = -\frac{1}{2}, \qquad z = 0\]

Substituting back:

\[T = k \, l^{1/2} \, g^{-1/2}\]

which can be written more neatly as:

\[T = k \sqrt{\frac{l}{g}}\]

Note two things about this result. First, the mass $m$ has dropped out completely, because $z = 0$ — the period of a simple pendulum does not depend on the mass of the bob, and dimensional analysis reveals this automatically. Second, the value of the constant $k$ cannot be found by this method. Dimensional analysis does not care what dimensionless number multiplies the right side, because a pure number has no dimensions to match.

The actual value, found by proper theory or experiment, is $k = 2\pi$, so the complete formula is:

\[T = 2\pi \sqrt{\frac{l}{g}}\]

Curiosity Corner

Q. If a pendulum’s time period depends on its length, mass, and gravity, can we guess the formula without doing any experiment? A. To a large extent, yes. Assuming a product-type dependence $T = k \, l^x \, g^y \, m^z$ and matching the powers of L, T and M on both sides gives $x = 1/2$, $y = -1/2$ and $z = 0$, so $T = k \sqrt{l/g}$ — and the mass of the bob drops out of its own accord. What the method cannot supply is the dimensionless constant $k$; its value, $2\pi$, has to come from proper theory or experiment, giving the complete formula $T = 2\pi \sqrt{l/g}$.

Real Incident / Discovery

In the 1940s, the British physicist G. I. Taylor used dimensional analysis to estimate the energy released by the first atomic bomb explosion. Working only from published photographs that showed how the radius of the fireball grew with time, and knowing the density of the surrounding air, he reasoned about how the blast energy must combine with radius, time, and air density on dimensional grounds. His estimate of the enormous energy released was remarkably close to the then-secret official figure — a striking demonstration of how much dimensional reasoning can reveal from very little data.

Dimensional analysis, then, is very useful for deducing relations among interdependent physical quantities. But it has clear limits, which we should state honestly:

It cannot find dimensionless constants (like the $2\pi$ above); these must come from theory or experiment.
It can only test the dimensional validity of a relation, not the exact relationship — it never proves a formula is complete.
It cannot distinguish between two physical quantities that share the same dimensions (like work and torque, or energy and moment of force).

The exercises at the end of this chapter will give you plenty of practice in both uses of dimensional analysis — checking equations and deducing relations.

Numerical 1.6

The speed $v$ of a wave on a stretched string is thought to depend on the tension $F$ in the string (dimensions $[\text{M L T}^{-2}]$) and the mass per unit length $\mu$ (dimensions $[\text{M L}^{-1}]$). Using the method of dimensions, find how $v$ depends on $F$ and $\mu$ (that is, find the powers), leaving the dimensionless constant undetermined.

1.7 Summary

Physics is a quantitative science — it rests on the measurement of physical quantities. A few quantities are chosen as fundamental or base quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Everything else is built from these.
Each base quantity is measured against a fixed, carefully standardised reference called its unit. The seven base units are the metre, kilogram, second, ampere, kelvin, mole, and candela. The units of the base quantities are called fundamental or base units.
All other physical quantities are derived from the base quantities, and their units — formed by combining base units — are called derived units. A complete set of units, both fundamental and derived, together forms a system of units.
The International System of Units (SI), based on the seven base units, is the system now accepted internationally and used almost everywhere in the world.
SI units are used in every physical measurement, for both base quantities and the derived quantities obtained from them. Some derived units are given special names, such as the joule (energy), newton (force), and watt (power).
SI units have clearly defined and internationally agreed symbols — for example, m for metre, kg for kilogram, s for second, A for ampere, and N for newton. Using these symbols correctly keeps scientific writing consistent worldwide.
Very large and very small quantities are usually written in scientific notation, as a number between 1 and 10 multiplied by a power of 10. Scientific notation, together with SI prefixes, keeps measurements compact, simplifies calculation, and clearly shows the precision of a number.
A set of general rules and guidelines must be followed when writing symbols for physical quantities, SI units, other accepted units, and SI prefixes, so that quantities and measurements are expressed properly and unambiguously.
When computing a physical quantity, the units of the quantities in a relation are treated like algebraic symbols — multiplied, divided, and cancelled — until the desired unit is obtained.
Only the proper number of significant figures should be kept in measured and computed quantities. The rules for counting significant figures, for arithmetic operations with them, and for rounding off the uncertain digits must be followed.
The dimensions of the base quantities, combined in the right powers, describe the nature of any physical quantity. Dimensional analysis can be used to check the dimensional consistency of an equation and to deduce relations among physical quantities. Remember, though: a dimensionally consistent equation is not necessarily correct, but a dimensionally wrong or inconsistent equation must certainly be wrong.

1.8 NCERT Questions

Note: In stating numerical answers, take care of significant figures.

Fill in the blanks:
1. The volume of a cube of side 1 cm is equal to ….. m³
2. The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to … (mm)²
3. A vehicle moving with a speed of 18 km h⁻¹ covers …. m in 1 s
4. The relative density of lead is 11.3. Its density is …. g cm⁻³ or …. kg m⁻³.
Fill in the blanks by suitable conversion of units:
1. $1 \ \text{kg m}^2 \text{s}^{-2} = \ldots \ \text{g cm}^2 \text{s}^{-2}$
2. $1 \ \text{m} = \ldots \ \text{ly}$
3. $3.0 \ \text{m s}^{-2} = \ldots \ \text{km h}^{-2}$
4. $G = 6.67 \times 10^{-11} \ \text{N m}^2 \, (\text{kg})^{-2} = \ldots \ (\text{cm})^3 \text{s}^{-2} \text{g}^{-1}$.
A calorie is a unit of heat (energy in transit) and it equals about 4.2 J, where $1 \ \text{J} = 1 \ \text{kg m}^2 \text{s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha$ kg, the unit of length equals $\beta$ m, and the unit of time is $\gamma$ s. Show that a calorie has a magnitude $4.2 \, \alpha^{-1} \beta^{-2} \gamma^{2}$ in terms of the new units.
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:
1. atoms are very small objects
2. a jet plane moves with great speed
3. the mass of Jupiter is very large
4. the air inside this room contains a large number of molecules
5. a proton is much more massive than an electron
6. the speed of sound is much smaller than the speed of light.
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?
Which of the following is the most precise device for measuring length:
1. a vernier callipers with 20 divisions on the sliding scale
2. a screw gauge of pitch 1 mm and 100 divisions on the circular scale
3. an optical instrument that can measure length to within a wavelength of light?
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?
Answer the following:
1. You are given a thread and a metre scale. How will you estimate the diameter of the thread?
2. 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?
3. 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?
The photograph of a house occupies an area of 1.75 cm² 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 m². What is the linear magnification of the projector-screen arrangement?
State the number of significant figures in the following:
1. 0.007 m²
2. $2.64 \times 10^{24}$ kg
3. 0.2370 g cm⁻³
4. 6.320 J
5. 6.032 N m⁻²
6. 0.0006032 m²
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.
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?
A famous relation in physics relates the ‘moving mass’ $m$ to the ‘rest mass’ $m_0$ 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:

\[m = \frac{m_0}{\left(1 - v^2\right)^{1/2}}\]

Guess where to put the missing $c$.

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: $1 \ \text{Å} = 10^{-10}$ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?
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 a hydrogen molecule to be about 1 Å.) Why is this ratio so large?
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.)
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 of gases? Check if your guess is correct from the following data: mass of the Sun $= 2.0 \times 10^{30}$ kg, radius of the Sun $= 7.0 \times 10^{8}$ m.

1.9 Check Your Concepts

Before 2018, the kilogram was defined by a physical platinum–iridium cylinder kept in France, but the SI now defines it using the fixed value of the Planck constant. Explain why defining base units through constants of nature is considered better than using physical objects.
The number 4700 could have two, three, or four significant figures. Explain why this ambiguity arises, and show how writing the number in scientific notation removes it.
Why must the argument of a trigonometric, logarithmic, or exponential function always be a dimensionless quantity? Support your reasoning using the principle of homogeneity of dimensions.
“A dimensionally correct equation need not be physically correct, but a dimensionally incorrect equation must always be wrong.” Explain this asymmetry with reasoning.
Two physical quantities are found to have exactly the same dimensional formula. Does this necessarily mean they are the same physical quantity? Give one example to justify your answer.
Assertion–Reason. Assertion: The order of magnitude of $8.2 \times 10^{6}$ is 7. Reason: While finding the order of magnitude, the number $a$ in $a \times 10^{b}$ is rounded to 10 when $5 < a \leq 10$. State whether the assertion is true, whether the reason is true, and whether the reason correctly explains the assertion.
Why does subtracting two nearly equal measured quantities often leave a result with far fewer significant figures than the original measurements? Illustrate your reasoning with a short example.
Using the method of dimensions, the time period of a simple pendulum turns out to be independent of the mass of the bob. Explain how the dimensional derivation reveals this, even though the derivation was set up assuming a possible mass dependence.
Why can the dimensionless constant appearing in a physical relation (such as the factor $2\pi$ in the pendulum formula) never be determined by dimensional analysis alone? What is needed to find it?
A measurement of length is converted from metres to millimetres. Does the number of significant figures change? Justify your answer with reference to the rules for significant figures.

1.10 Practice with Numericals

Starting from their defining relations, determine the dimensional formula of each of the following: (i) impulse (force × time), (ii) surface tension (force per unit length), and (iii) angular velocity (angle ÷ time).
The time period of a mass attached to a spring is given by $T = 2\pi \sqrt{\dfrac{m}{k}}$, where $m$ is the mass and $k$ is the force constant of the spring (SI unit N m⁻¹). Check whether this equation is dimensionally consistent.
A rectangular metal plate has a measured length of 2.34 m and a measured breadth of 1.2 m. Calculate its area, giving the result to the correct number of significant figures.
Three lengths are measured as 12.5 cm, 0.36 cm, and 145.23 cm. Find their sum and express it to the correct number of decimal places.
The radius of a sphere is measured as $r = 3.2 \ \text{cm} \pm 0.1 \ \text{cm}$. Calculate its volume using $V = \dfrac{4}{3}\pi r^{3}$, and find the percentage error in the calculated volume.
Round off each of the following numbers to three significant figures: (a) 4.735, (b) 4.725, (c) 2.3456, (d) 0.08725.
The density of a material is measured as 2.5 g cm⁻³. Express this density in SI units (kg m⁻³), showing the unit conversion clearly.
The escape velocity $v$ of a body from the surface of a planet is expected to depend on the planet’s radius $R$ and the acceleration due to gravity $g$ at its surface. Using the method of dimensions, find how $v$ depends on $R$ and $g$ (leave the dimensionless constant undetermined).
The energy $E$ of a photon is given by $E = h\nu$, where $h$ is Planck’s constant with dimensions $[\text{M L}^2 \text{T}^{-1}]$ and $\nu$ is the frequency with dimensions $[\text{T}^{-1}]$. Verify, using dimensional analysis, that this expression has the dimensions of energy.
The distance to a nearby star is 4.3 light years. Taking the speed of light as $3.0 \times 10^{8}$ m s⁻¹ and one year as $3.156 \times 10^{7}$ s, express this distance in metres, giving the answer in scientific notation to the correct number of significant figures.