Chapter 1: Units and Measurements

In December 1998, NASA launched the Mars Climate Orbiter, a spacecraft built to circle Mars and study its atmosphere. After a journey of more than nine months and hundreds of millions of kilometres, the orbiter reached Mars in September 1999. It fired its engines to slip into orbit around the planet — and was never heard from again.

The investigation that followed did not find a broken engine, a cracked circuit, or a failed computer. It found a mismatch of units. One team had supplied thruster data using the pound-second, a unit belonging to the British engineering system. The team writing the navigation software expected the same data in newton-second, the SI unit. Neither number was wrong by itself. Each was a perfectly good measurement — but they were measurements in different units, and nobody converted between them.

Because of that silent mismatch, the spacecraft’s path was slightly off. It approached Mars far lower than planned, entered the thin upper atmosphere, and was destroyed. A mission costing hundreds of crores was lost not to bad physics, but to careless handling of units.

Physics is a quantitative science. Every measurement we make is a comparison with a standard, and that standard must be one that everybody agrees upon. This chapter is about those standards — how we build them, how we use them to measure things as large as a galaxy and as small as an atom, and how we honestly state what we do not know.

Diagram

Fig. 1.1 – A spacecraft approaching Mars, with two data labels feeding into its trajectory: one tagged “pound-second (FPS)” and one tagged “newton-second (SI)”, the mismatch bending the path too close to the planet.

NoteCuriosity Corner

Q1. Why do scientists across the world insist on one common system of units instead of each country using its own convenient one? Q2. If we cannot stretch a measuring tape from the Earth to a distant star, how do astronomers still state that star’s distance in metres? Q3. For 130 years the kilogram was a metal cylinder locked in a vault near Paris. Why did the world decide, in 2019, that this was no longer good enough? Q4. Can you tell that a physics equation must be wrong without substituting a single number into it? Q5. If ten students measure the same piece of cloth with the same metre scale, their answers will differ slightly. Does that mean nine of them are careless? Q6. Your calculator shows 5.13333 cm³ for a volume. Why is it wrong to write down all of those digits?

This chapter will answer each of these questions.

1.1 Introduction

  1. What is a unit?
  2. Which units have you used in the laboratory for measuring (i) length (ii) mass (iii) time (iv) temperature?
  3. Which system of units have you used?

Physics is a quantitative science. This single word carries a lot of meaning. It tells us that physics is not satisfied with saying that an object is “heavy” or that a journey is “long” — it insists on saying how heavy and how long, in numbers. Every experiment in physics therefore ends in a measurement of some physical quantity.

You already measure things constantly, even outside the laboratory. You judge the size of objects, the volume of a liquid, the amount of matter in a packet, the weight of vegetables or fruits at a shop, your body temperature when you are unwell, and the length of cloth needed for a shirt. Each of these is a measurement, whether or not you call it one.

What a Measurement Really Is

A measurement is always a comparison. When you measure something, you are silently comparing it with a standard that has been agreed upon and accepted internationally.

Consider buying fruit. To find the mass of the fruit, the shopkeeper places standard masses of 1 kg, 500 g and so on in the other pan of the balance. The fruit is not being measured against nothing — it is being measured against those standards. These accepted standards of comparison are what we call units.

NoteDefinition

(A unit is the standard measure of a physical quantity, internationally accepted, with which any measurement of that quantity is compared.)

Why a Number Alone Is Never Enough

Because measurement is a comparison, the result of a measurement always has two parts: a number, and the unit in which that number is counted.

For example, the length of a wire is written as 5 m. Here ‘m’ (metre) is the unit, and 5 is the value of the length in that unit. The same wire could equally well be written as 500 cm — the wire has not changed, only the standard we compared it with has changed, so the number changed too.

Q. If a question tells you that the length of a rod is “0.75”, is that a complete piece of information?

A. No. Without a unit, 0.75 could mean 0.75 m, 0.75 cm, or 0.75 km — three completely different rods. In physics, a numerical value written without its unit is treated as meaningless. This is also why board and entrance examinations deduct marks for a final answer written without its unit.

Different quantities are measured against different standards, because you cannot compare a length with a mass. Length is measured in metre (m), time in second (s), mass in kilogram (kg), and so on for every quantity.

Every packaged item you buy — a 1 litre bottle of milk, a 500 g packet of biscuits, a 5 m roll of wire — carries a number and a unit printed on it, and in most countries this printing is enforced by law. Legal metrology departments periodically verify shopkeepers’ weights and measures against master standards, precisely because trade collapses if two people mean different things by “one kilogram.” The idea of a standard unit is not only a scientific requirement; it is an everyday commercial one.

Diagram

Fig. 1.2 – A beam balance with a fruit in one pan and standard 1 kg and 500 g masses in the other, labelled to show that measurement is a comparison with a standard.

With this idea in place — that every measurement is a comparison with a standard unit — the natural next question is: which standards should the whole world use? That is what the next section takes up.

1.2 System of Units

Once we accept that every measurement is a comparison with a standard, a practical problem appears immediately. Different regions of the world, at different times, chose different standards. A group of standard units chosen together for length, mass and time is called a system of units.

In your earlier standards you have already come across several such systems:

  1. CGS: Centimetre Gram Second system
  2. MKS: Metre Kilogram Second system
  3. FPS: Foot Pound Second system
  4. SI: System International

Notice how each name is simply a list of the standards it uses for length, mass and time, in that order. In the CGS system, length is measured in centimetre, mass in gram and time in second. In the MKS system the same three quantities are measured in metre, kilogram and second. The FPS system, used mainly in Britain and the United States, measures length in foot and mass in pound.

From Many Systems to One

The first three systems — CGS, MKS and FPS — were used extensively till recently. Having several systems in use at the same time is workable, but clumsy: every number crossing from one system to another has to be converted, and every conversion is an opportunity for error.

To settle the matter, in 1971 the 14th International General Conference on Weights and Measures recommended the use of the ‘International system’ of units. This international system of units is called the SI units, and it is the system used throughout this book.

The great practical advantage of SI is that it uses the decimal system. Larger and smaller units are related to the base unit by simple powers of ten — a kilometre is \(10^{3}\) m, a millimetre is \(10^{-3}\) m. Conversion within the system therefore needs no awkward factors and is very simple and convenient.

Q1. Why do scientists across the world insist on one common system of units instead of each country using its own convenient one?

A. As long as CGS, MKS and FPS were all in use, the same physical quantity could be reported in three different ways, and every exchange of data needed a conversion. To remove this confusion, the 14th International General Conference on Weights and Measures in 1971 recommended a single international system, the SI. SI is more convenient because it is built on the decimal system: units within it differ only by powers of ten, so converting between them needs no strange multiplying factors such as 12 inches to a foot.

Q. The FPS system converts 1 foot into 12 inches, and 1 pound into 16 ounces. Why does this make it harder to work with than SI?

A. Because 12 and 16 are not powers of ten, every conversion requires you to remember and apply a separate factor, and mental checking becomes difficult. In SI, moving from metre to centimetre only shifts the decimal point. Fewer conversion factors means fewer chances of a mistake — which matters more in physics than saving a little arithmetic.

Having chosen SI, we must now decide which quantities deserve their own standards, and which can simply be built out of the others. That is the distinction taken up next.

1.2.1 Fundamental Quantities and Units

Look carefully at the quantities used in physics and you will find that they are not all independent of one another. Speed, for instance, cannot be defined without first knowing what length and time mean. But length itself does not lean on anything else — you can define it on its own, without borrowing from any other quantity.

Quantities of this second kind, which stand on their own feet, are called fundamental quantities.

NoteDefinition

(Fundamental quantities are those physical quantities which do not depend on any other physical quantities for their measurement.)

There are seven fundamental quantities in all: length, mass, time, temperature, electric current, luminous intensity and amount of substance. Every other quantity in the whole of physics — force, energy, pressure, resistance, charge — is built out of these seven.

The Units That Go With Them

Since these seven quantities are independent, each one needs its own independently chosen standard. The units used to measure the fundamental quantities are called fundamental units.

NoteDefinition

(Fundamental units are the units used to measure the fundamental quantities.)

The seven fundamental quantities, their SI units and the symbols for those units are listed in Table 1.1. You should learn this table thoroughly, because the symbols in the last column are the building blocks of every dimensional formula you will write later in this chapter.

Table 1.1: Fundamental Quantities with their SI Units and Symbols

Fundamental quantity SI units Symbol
1) Length metre m
2) Mass kilogram kg
3) Time second s
4) Temperature kelvin K
5) Electric current ampere A
6) Luminous Intensity candela cd
7) Amount of substance mole mol

Q. Temperature clearly feels like something that “depends” on the energy of molecules. Why is it still called a fundamental quantity?

A. “Fundamental” here has a narrow, practical meaning: it refers to the measurement, not to deeper physical theory. To measure temperature you do not first have to measure any other quantity — you can compare it directly against a standard. Whether a quantity has a deeper microscopic explanation is a separate question altogether.

The candela, the unit of luminous intensity, is the one fundamental unit most students never meet again — yet it decides what you buy. LED bulbs are sold by the light they actually deliver, not by the electrical power they consume, which is why packaging quotes lumens rather than only watts. Lighting standards for classrooms, operation theatres and roads are all fixed in units derived from the candela.

Diagram

Fig. 1.3 – Seven labelled blocks (length, mass, time, temperature, current, luminous intensity, amount of substance) forming the base of a pyramid, with derived quantities such as force, energy and pressure built on top of them.

1.2.2 Derived Quantities and Units

The seven fundamental quantities are only the beginning. In physics we constantly meet quantities such as speed, momentum, resistance and conductivity. None of these is on the list of seven — and none of them needs to be, because each can be built out of some or all of the seven.

Speed, for example, is length divided by time. Momentum is mass multiplied by velocity, which is itself length divided by time. These quantities depend on the fundamental ones and can be expressed entirely in terms of them.

NoteDefinition

(Derived quantities are physical quantities which depend on some or all of the seven fundamental quantities and can be expressed in terms of them.)

Naturally, if the quantity is built out of fundamental quantities, its unit must be built out of fundamental units in exactly the same way.

NoteDefinition

(Derived units are the units of derived quantities, which can be expressed in terms of the fundamental units.)

Building a Derived Unit

The rule for constructing a derived unit is simple: write down the defining relation of the quantity, and replace each quantity in it by its unit.

Take velocity. Velocity is displacement divided by time, so

\[\text{SI unit of velocity} = \frac{\text{Unit of displacement}}{\text{Unit of time}} = \frac{\text{m}}{\text{s}} = \text{m s}^{-1}\]

Here m is the metre (unit of length) and s is the second (unit of time). Notice how the division of the two quantities has become a division of the two units.

Now take momentum, which is mass multiplied by velocity:

\[\text{Unit of momentum} = (\text{Unit of mass}) \times (\text{Unit of velocity})\] \[= \text{kg m/s} = \text{kg m s}^{-1}\]

Both of the above units — m s\(^{-1}\) and kg m s\(^{-1}\) — are derived units, because each is written purely in terms of the fundamental units metre, kilogram and second.

Q. Is the newton (N), the SI unit of force, a fundamental unit or a derived unit?

A. It is a derived unit. Force is mass times acceleration, so its unit is kg × m s\(^{-2}\) = kg m s\(^{-2}\). The name “newton” is only a convenient short label given to this combination of fundamental units; it does not make force a fundamental quantity.

Supplementary Units

Besides the seven fundamental or basic units, there are two more units called supplementary units: (i) plane angle \(d\theta\) and (ii) solid angle \(d\Omega\).

These two are placed in a category of their own because they are angles — pure ratios of one length to another, or of an area to an area. Being ratios, they have no fundamental unit hiding inside them, and yet we still need standard names for measuring them.

(i) Plane angle (dθ): This is the ratio of the length of an arc of a circle to the radius of the circle, as shown in Fig. 1.4 (a). Thus

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

is the angle subtended by the arc at the centre of the circle. Here \(ds\) is the length of the small arc AB and \(r\) is the radius, both measured in metre. It is measured in radian (rad). An angle \(\theta\) in radian is denoted as \(\theta^{c}\).

Diagram

Fig. 1.4 (a) – A circular arc AB of length ds subtending a small plane angle dθ at the centre O, with radius r marked.

(ii) Solid angle (dΩ): This is the three-dimensional analogue of \(d\theta\). Where a plane angle opens out in a flat plane, a solid angle opens out as a cone in space — think of the cone of sky that a telescope collects light from. It is defined as the ratio of the area of a portion of the surface of a sphere to the square of the radius of the sphere. Thus

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

is the solid angle subtended by area \(ds\) at O, as shown in Fig. 1.4 (b). Here \(dA\) is the patch of area on the sphere (in m\(^2\)) and \(r\) is the radius of the sphere (in m). It is measured in steradians (sr).

Diagram

Fig. 1.4 (b) – A patch of area dA on the surface of a sphere of radius r subtending a solid angle dΩ at the centre O.

A useful check follows at once. A sphere of radius \(r\) has surface area \(4\pi r^{2}\). Therefore the solid angle subtended by the entire sphere at its centre is

\[\Omega = \frac{4\pi r^{2}}{r^{2}} = 4\pi \ \text{sr}\]

This is the largest solid angle that can exist, just as \(2\pi\) radian is the largest plane angle around a point.

Q. Both \(d\theta\) and \(d\Omega\) are ratios of quantities with units — length/length and area/area. So do they have any dimensions?

A. No. In each case the units cancel completely, so a plane angle and a solid angle are dimensionless. The names “radian” and “steradian” are therefore not units in the ordinary sense; they are labels that tell us which kind of dimensionless ratio we are talking about. This is exactly why they are kept apart as supplementary units.

What is the solid angle subtended by the moon at any point of the Earth, given that the diameter of the moon is 3474 km and its distance from the Earth is \(3.84 \times 10^{8}\) m?

Solution: The moon, seen from the Earth, appears as a flat disc. So the “area on the sphere” in the definition of solid angle is the area of the moon’s disc, and the “radius” is the moon–Earth distance. The diameter of the moon is 3474 km, so its radius is 1737 km \(= 1.737 \times 10^{3}\) km, and the distance is \(3.84 \times 10^{5}\) km.

Solid angle subtended by the moon at the Earth

\[= \frac{\text{Area of the disc of the moon}}{(\text{moon - earth distance})^{2}}\]

\[= \frac{\pi \times (1.737 \times 10^{3})^{2}}{(3.84 \times 10^{5})^{2}}\]

\[= 6.425 \times 10^{-5} \ sr\]

Note that both the radius and the distance have been written in the same unit (kilometre), so the units cancel and the answer comes out as a pure number in steradian, as it must.

The relation between radian and degree is \[\pi \ \text{radians} = \pi^{c} = 180^{\circ}\] \[\therefore \ 1 \ \text{radian} = \frac{180}{\pi} = \frac{180}{3.1415} = 57.297^{\circ}\] Similarly, \[1^{\circ} = \frac{\pi}{180} = \frac{3.1415}{180} = 1.745 \times 10^{-2} \ \text{rad}\] \[1^{\circ} = 60', \quad 1' = 2.91 \times 10^{-4} \ \text{rad}\] and \[1' = 60'', \quad 1'' = 4.847 \times 10^{-6} \ \text{rad}\] These conversion factors are worth marking: every problem on parallax and angular diameter later in this chapter begins by converting an angle given in degrees, minutes or seconds of arc into radian.

The Sun has a diameter of about \(1.39 \times 10^{9}\) m and is at a distance of about \(1.50 \times 10^{11}\) m from the Earth. Calculate the solid angle subtended by the Sun at a point on the Earth, in steradian.

1.2.3 Conventions for the use of SI Units

A unit is a piece of information, and like any piece of information it can be corrupted by careless writing. Over the years the international bodies that maintain the SI have therefore fixed a set of writing conventions. These are not decorative rules of neatness — each one exists to remove a specific ambiguity that has confused people in the past.

The conventions are as follows.

(1) The unit of every physical quantity should be represented by its symbol. So we write 5 m, not 5 metres, in a calculation.

(2) The full name of a unit always starts with a small letter, even if the unit is named after a person — for example, 1 newton, 1 joule. But the symbol for a unit named after a person is written with a capital letter — for example, N after the scientist Newton, and J after the scientist Joule.

There is a quiet logic here. The lower-case full name reminds us that the unit is a public standard, not a personal possession; the capital symbol preserves the honour paid to the scientist.

Q. Should the SI unit of temperature be written as 10 kelvins, 10 Kelvin, or 10 K?

A. The correct forms are 10 K (symbol, capital because Kelvin was a person) or 10 kelvin (full name, small letter, no plural ‘s’). “10 Kelvins” breaks two conventions at once — a capital in the full name, and a plural. Examiners do notice this.

(3) Symbols for units do not take the plural form. We write a force of 20 N, and not 20 newtons or 20 Ns. This one matters a great deal, because “Ns” already has a meaning of its own — newton second, the unit of impulse. Adding a plural ‘s’ would silently turn a force into an impulse.

(4) Symbols for units do not contain any full stop at the end of the recommended letter — for example, 25 kg and not 25 kg. (A full stop after a symbol can easily be mistaken for a decimal point in a printed table.)

(5) The units of physical quantities appearing in the numerator and the denominator should be written as one ratio. For example, the SI unit of acceleration is written m/s\(^{2}\) or m s\(^{-2}\), but not m/s/s. A double division sign is ambiguous — it is never clear which division is meant to be carried out first.

(6) Use of a combination of the full name and the symbol is avoided when a physical quantity is expressed as a combination of two units. For example, the unit J/kg K is correct, while joule/kg K is not correct. Either write the whole thing in symbols or write the whole thing in words; do not mix the two languages in one expression.

Prefixes and the Powers of Ten

(7) A prefix symbol is used before the symbol of the unit. The prefix symbol and the unit symbol together constitute a new symbol for the unit, which can be raised to a positive or negative power of 10.

For example, \[1\ \text{ms} = 1 \ \text{millisecond} = 10^{-3}\ \text{s}\] \[1\ \mu\text{s} = 1 \ \text{microsecond} = 10^{-6}\ \text{s}\] \[1\ \text{ns} = 1 \ \text{nanosecond} = 10^{-9}\ \text{s}\]

Use of double prefixes is avoided when a single prefix is available. Thus \[10^{-6}\ \text{s} = 1\ \mu\text{s} \quad \text{and not 1 mms}\] \[10^{-9}\ \text{s} = 1\ \text{ns} \quad \text{and not 1 m}\mu\text{s}\]

The reason is simply that a chain of prefixes forces the reader to do arithmetic before they can even understand the number. If a single prefix says the same thing, it should be used.

(8) A space or a hyphen must be introduced while indicating the multiplication of two units. For example, m/s should be written as m s\(^{-1}\) or m-s\(^{-1}\), and not as ms\(^{-1}\) — because “ms” will be read as millisecond.

This is the most dangerous of all the conventions, and it is worth pausing on. The symbol ms\(^{-1}\) could be read in two completely different ways: as metre per second (a speed), or as “per millisecond” (a frequency). One missing space changes the meaning of the quantity entirely.

Q. A student writes the speed of sound as 340 ms⁻¹. What has gone wrong, and how should it be written?

A. Written that way, the symbol reads as “per millisecond”, not “metre per second”. The multiplication of the two units metre and second⁻¹ must be shown by a space or a hyphen: 340 m s⁻¹ or 340 m-s⁻¹. The number is right; only the notation has destroyed the meaning.

Hospital drug charts are one place where these conventions are treated with deadly seriousness. A dose written as “mg” (milligram) and one written as “µg” (microgram) differ by a factor of one thousand, and health services worldwide have issued instructions on how such symbols must be written precisely because misreading a prefix has caused real overdoses. The habit of writing units exactly, which you build now in a physics notebook, is the same habit that matters later in a laboratory, a workshop or a ward.

Diagram

Fig. 1.5 – A two-column chart of “Correct” and “Incorrect” ways of writing units, illustrating conventions (2), (3), (5), (6) and (8) side by side.

1.3 Measurement of Length

One fundamental quantity which we have already discussed is length. To measure length or distance, the SI unit used is the metre (m).

But what exactly is one metre? A unit is only as good as the standard behind it, and the standard for the metre has been sharpened twice within living memory.

From a Coloured Light to a Beam of Light

In 1960, a standard for the metre based on the wavelength of orange-red light emitted by atoms of krypton was adopted. The advantage of such a standard is that it is reproducible anywhere: a krypton atom in Paris and a krypton atom in Pune emit exactly the same light, so any properly equipped laboratory in the world could recreate the metre for itself without borrowing anyone’s metal bar.

By 1983 a still more precise measurement was developed. It says that a metre is the length of the path travelled by light in vacuum during a time interval of 1/299792458 second.

NoteDefinition

(One metre is the length of the path travelled by light in vacuum during a time interval of 1/299792458 second.)

This definition became possible only because, by that time, the speed of light in vacuum could be measured extremely precisely as \(c = 299792458\) m/s. Once the measurement became that good, scientists made a bold decision: instead of continuing to measure the speed of light more and more accurately, they fixed its value by definition and used it to define the metre instead. From that moment onwards, the speed of light in vacuum is not an experimental result at all — it is an exact number, with no uncertainty, by international agreement.

Q. If the speed of light is now fixed by definition, does that mean no one can ever measure it again and get a slightly different answer?

A. Correct — and that is the point. Any experiment that once measured \(c\) now becomes an experiment that measures length instead. The uncertainty has not disappeared; it has been moved out of the constant and into the measurement of distance, where it belongs. This is exactly the strategy the SI later used for the kilogram, as you will see in Section 1.4.

Notice how the standard has become steadily less physical. A metal bar can be scratched, a wavelength must be produced by a real lamp, but a definition written in terms of light in vacuum and a fixed constant can never be lost, stolen or damaged.

The Range of Lengths in Nature

Lengths in physics do not come in a comfortable, human-sized range. They stretch from the almost unimaginably vast to the almost unimaginably tiny, and the same unit — the metre — must cover all of it. Some typical distances and lengths are given in Table 1.2.

Table 1.2: Some Useful Distances

Measurement Length in metre
Distance to Andromeda Galaxy (from Earth) \(2 \times 10^{22}\) m
Distance to nearest star (after Sun) Proxima Centuari (from Earth) \(4 \times 10^{16}\) m
Distance to Pluto (from Earth) \(6 \times 10^{12}\) m
Average Radius of Earth \(6 \times 10^{6}\) m
Height of Mount Everest \(9 \times 10^{3}\) m
Thickness of this paper \(1 \times 10^{-4}\) m
Length of a typical virus \(1 \times 10^{-8}\) m
Radius of hydrogen atom \(5 \times 10^{-11}\) m
Radius of proton \(1 \times 10^{-15}\) m

Read the table from top to bottom and notice the span: from \(10^{22}\) m down to \(10^{-15}\) m — a range of some thirty-seven powers of ten. This is why every entry is written in scientific notation, in powers of ten, rather than as a string of zeros.

Diagram

Fig. 1.6 – A single logarithmic scale line running from \(10^{-15}\) m to \(10^{22}\) m, with the entries of Table 1.2 marked as labelled points along it.

This enormous range also explains what follows. No single instrument can measure across all of it. A metre scale is useless for the distance to Pluto and equally useless for the radius of a proton. Different regions of the scale demand entirely different methods — and the next few subsections take them one at a time.

1.3.1 Measurements of Large Distance

Parallax method

Large distances, such as the distance of a planet or a star from the Earth, cannot be measured directly with a metre scale. Nobody can lay a scale from here to Jupiter. For such distances, an indirect method called the parallax method is used.

The remarkable thing about this method is that it converts a distance we cannot reach into an angle we can measure comfortably from where we stand.

An Experiment You Can Do Right Now

Let us do a simple experiment to understand what parallax is.

Hold your hand in front of you and look at it with your left eye closed. Now open the left eye and close the right eye instead. You will find that your hand appears to shift sideways against the background — a distant wall, a window, a tree.

Your hand has not moved. The background has not moved. Only the position of the observer — the eye doing the looking — has changed. This apparent shift is called parallax.

NoteDefinition

(Parallax is the apparent change in the position of an object due to a change in the position of the observer.)

Q. Repeat the experiment, but this time look at a distant tree instead of your hand. What do you notice?

A. The tree shifts much less than your hand did. The farther the object, the smaller the parallax shift. This is not an inconvenience — it is precisely the fact the method is built on. Since the shift shrinks in a predictable way with distance, measuring the shift tells us the distance.

Now look at Fig. 1.7. The two eyes are at \(E_1\) and \(E_2\), and the object is at P. The angle \(\theta\) at P, between the two lines of sight, is called the parallax angle. By measuring this parallax angle \(\theta\) and knowing the distance \(E_1E_2\) between the two eyes, we can determine the distance of the object from us, that is, OP as \(E_1E_2/\theta\).

Diagram

Fig. 1.7 – Parallax method for determining distance: an object at P viewed from two eye positions E₁ and E₂, with the parallax angle θ marked at P and O the midpoint of E₁E₂.

Scaling the Method Up to the Sky

There is an obvious limitation. The distances of planets from the Earth are enormous, and the gap between your two eyes is a few centimetres. The parallax angle produced would be far too small to detect. So the two-eyes method, as it stands, cannot be used.

The remedy is to make the “eyes” much farther apart. In order to make simultaneous observations of an astronomical object, we select two distant points on the Earth and use two observers instead of two eyes.

Consider two positions A and B on the surface of the Earth, separated by a straight line of length \(b\), as shown in Fig. 1.8. Two observers at these two points observe a distant planet S at the same moment. We measure the angle \(\angle ASB\) between the two directions along which the planet is viewed at these two points. This angle, represented by the symbol \(\theta\), is the parallax angle.

Diagram

Fig. 1.8 – Measurement of distances of planets: points A and B on the Earth separated by basis b, both sighting a planet S at distance D, with parallax angle θ at S.

The observations must be simultaneous, because both the Earth and the planet are in motion. If one observer looks in the morning and the other in the evening, the geometry of the triangle has already changed and the angle measured is meaningless.

Turning the Triangle into an Arc

Now comes the step that makes the mathematics simple. The planet is far away — that is, the distance of the planet from the Earth is very large in comparison to \(b\). In symbols,

\[\frac{b}{D} \ll 1\]

and therefore \(\theta\) is very small.

When the angle at the apex of a triangle is very small, the base of that triangle is almost indistinguishable from a small arc of a circle drawn with the apex as centre. We can thus consider AB as the arc of length \(b\) of a circle, and \(D\) as its radius.

Once we do that, we are allowed to use the definition of plane angle from Section 1.2.2 — angle = arc/radius:

\[AB = b \quad \text{and} \quad AS = BS = D \quad \text{and} \quad \theta \cong \frac{AB}{D},\]

where \(\theta\) is in radian.

\[\therefore \ D = \frac{b}{\theta} \qquad \text{--- (1.1)}\]

Here \(b\) is the distance between the two observation points on Earth, called the basis (in metre), \(\theta\) is the measured parallax angle (in radian), and \(D\) is the distance of the planet from the Earth (in metre).

Equation (1.1) says something worth stating in words: for a fixed basis \(b\), the distance is inversely proportional to the parallax angle. A small angle means a large distance. This is the same fact you noticed with the distant tree, now written as an equation.

Q. In a numerical problem, the parallax angle is given as 1° 30′. Can you substitute this directly into \(D = b/\theta\)?

A. No. Equation (1.1) was derived using the arc–radius definition of an angle, which is valid only when \(\theta\) is expressed in radian. The angle must first be converted — using the conversions given in Section 1.2.2 — before it is substituted. Forgetting this conversion is the single most common mistake in parallax numericals.

Two observers stationed at points on the Earth’s surface separated by a basis of \(6.4 \times 10^{6}\) m observe a planet simultaneously. The parallax angle is found to be \(2.13 \times 10^{-4}\) rad. Calculate the distance of the planet from the Earth.

1.3.2 Measurement of Distance to Stars

The Sun is the star which is closest to the Earth. The next closest star is at a distance of 4.29 light years — a distance so large that the parallax method, as used for planets, runs into trouble.

Recall equation (1.1): \(D = b/\theta\), so \(\theta = b/D\). For a fixed basis \(b\), the parallax angle shrinks as the distance \(D\) grows. Stars are hundreds of thousands of times farther than the planets, so the parallax measured from even the two most distant points on the Earth will be too small to be measured.

Using the Earth’s Orbit as the Basis

If the angle is too small, there is only one thing left to do: make the basis \(b\) bigger. And the largest basis available to us is not on the Earth at all — it is the Earth’s own orbit.

For this purpose we measure the parallax between two farthest points along the orbit of the Earth around the Sun. These two points, observed six months apart, are 2 AU apart (see the box below for what an AU is). The basis has jumped from a few thousand kilometres to about \(3 \times 10^{11}\) m, and the parallax angle becomes large enough to detect.

Diagram

Fig. 1.9 – The Earth at two diametrically opposite points A and B of its orbit around the Sun S, each 1 AU from the Sun, both sighting a distant star; the parallax angle marked at the star.

Q2. If we cannot stretch a measuring tape from the Earth to a distant star, how do astronomers still state that star’s distance in metres?

A. By using parallax — the apparent shift of an object when the observer moves. A star is viewed twice from two widely separated positions, and the small angle between the two lines of sight is measured. Since \(D = b/\theta\), dividing the known separation of the two viewing points by this angle gives the distance. For stars the separation used is the Earth’s own orbit: the same star is observed six months apart, when the Earth is at two opposite ends of its orbit, 2 AU apart. Nobody has to travel anywhere; the Earth carries the observer for us.

For measuring large distances, astronomers use the following units.

\[1 \ \text{astronomical unit (AU)} = 1.496 \times 10^{11}\ \text{m}\] \[1 \ \text{light year} = 9.467 \times 10^{15}\ \text{m}\] \[1 \ \text{parsec (pc)} = 3.08 \times 10^{16}\ \text{m} \cong 3.26 \ \text{light years}\]

A light year is the distance travelled by light in one year. Note carefully that in spite of the word “year”, it is a unit of distance, not of time.

The astronomical unit (AU) is the mean distance between the centre of the Earth and the centre of the Sun.

A parsec (pc) is the distance from where 1 AU subtends an angle of 1 second of arc. The name itself is a compression of “parallax of one second”. Its value follows directly from the arc–radius relation, using \(1'' = 4.847 \times 10^{-6}\) rad:

\[r = \frac{1\ \text{AU}}{(1'')^{c}} = \frac{1.496 \times 10^{11}}{4.847 \times 10^{-6}} = 3.086 \times 10^{16}\ \text{m}\]

Diagram

Fig. 1.10 – A right triangle with the Sun at one corner, a baseline of 1 AU, and the opposite vertex at a distance of 1 pc subtending an angle of 1″.

Q. Which is larger — a light year or a parsec — and why is the parsec the astronomer’s working unit?

A. A parsec is larger; 1 pc ≅ 3.26 light years. Astronomers prefer it because it comes straight out of the measurement itself. A star whose parallax is 1″ is at 1 pc, a star whose parallax is 0.5″ is at 2 pc, and so on — the distance in parsec is simply the reciprocal of the parallax in arcsec. No conversion is needed between what is measured and what is reported.

A star is 5.5 light years away from the Earth. How much parallax, in arcsec, will it subtend when viewed from two opposite points along the orbit of the Earth?

Solution: Two opposite points A and B along the orbit of the Earth are 2 AU apart. This separation is the basis \(b\). Rearranging equation (1.1) as \(\theta = b/D\), the angle subtended by AB at the position of the star is

\[= \frac{2\,\text{AU}}{5.5\ ly} = \frac{2 \times 1.496 \times 10^{11}\ \text{m}}{5.5 \times 9.46 \times 10^{15}\ \text{m}} = 5.75 \times 10^{-6}\ \text{rad}\]

Both the numerator and the denominator have been converted into metre before dividing, so that the units cancel and the angle emerges in radian. The question, however, asks for the answer in arcsec, so we now convert radian into arcsec using \(1\) radian \(= 57.297^{\circ}\), \(1^{\circ} = 60'\) and \(1' = 60''\):

\[= 5.75 \times 10^{-6} \times 57.297 \times 60 \times 60\ \text{arcsec}\]

\[= 1.186\ \text{arcsec}\]

Even with the whole diameter of the Earth’s orbit as a basis, the shift of a nearby star is only about one second of arc — roughly the angle subtended by a one-rupee coin seen from four kilometres away. This is why stellar parallax defeated astronomers for centuries until telescopes became good enough.

A star shows a parallax of 0.742 arcsec when observed from two opposite points of the Earth’s orbit, 2 AU apart. Calculate its distance from the Earth in metre, and express it in light years. (Take 1 AU \(= 1.496 \times 10^{11}\) m and 1 light year \(= 9.467 \times 10^{15}\) m.)

1.3.3 Measurement of the Size of a Planet or a Star

Knowing how far away a planet is only half the story. We would also like to know how big it is — and once again, no scale can be laid across it.

The way out is the same idea used for distance, but turned around. Instead of viewing one point of the object from two places, we now view two opposite points of the object from one place.

Angular Diameter

If \(d\) is the diameter of a planet, the angle subtended by it at any single point on the Earth is called the angular diameter of the planet.

NoteDefinition

(The angular diameter of a planet is the angle subtended by its diameter at any single point on the Earth.)

Let \(\alpha\) be the angle between the two directions along which two diametrically opposite points of the planet are viewed through a telescope, as shown in Fig. 1.11. The telescope is turned from one edge of the planet’s disc to the other, and the angle through which it has been turned is \(\alpha\).

Diagram

Fig. 1.11 – Measurement of the size of a planet: a planet of diameter d at distance D from the Earth, subtending angular diameter α at the observer.

Since the planet is very far away, the same small-angle reasoning of Section 1.3.1 applies: the diameter \(d\) can be treated as a small arc, and the distance \(D\) as the radius of the circle. Using angle = arc/radius,

\[\alpha = \frac{d}{D}\]

The distance \(D\) of the planet is large, but we assume it has already been measured — by the parallax method of the previous subsections. With \(D\) known and \(\alpha\) measured, the diameter of the planet follows at once:

\[\therefore \ d = \alpha D \qquad \text{--- (1.2)}\]

Here \(d\) is the diameter of the planet (in metre), \(D\) is its distance from the Earth (in metre), and \(\alpha\) is the angular diameter in radian.

Q. Why must the distance \(D\) be measured first, before the size can be found at all?

A. Because an angle alone cannot distinguish a small nearby object from a large distant one. A coin held at arm’s length and the Moon both subtend roughly half a degree. The angle \(\alpha\) fixes only the ratio \(d/D\); it is the independently measured \(D\) that converts that ratio into an actual size. This is exactly why the parallax method had to come first.

A total solar eclipse is possible only because of a coincidence in angular diameters. The Sun is about 400 times larger than the Moon, but it is also about 400 times farther away, so both subtend almost the same angle — roughly half a degree — at a point on the Earth. Their angular diameters being nearly equal, the Moon’s disc can just cover the Sun’s disc. If the Moon’s angular diameter were slightly smaller, we would never see totality at all.

The moon is at a distance of \(3.84 \times 10^{8}\) m from the Earth. If viewed from two diametrically opposite points on the Earth, the angle subtended at the moon is 1° 54′. What is the diameter of the Earth?

Solution: Notice that this problem is equation (1.2) used in reverse. Here it is the Earth whose diameter is unknown, and the angle it subtends is measured at the Moon. The Earth’s diameter plays the role of \(d\), and the Earth–Moon distance plays the role of \(D\).

The angle must first be converted into radian, since equation (1.2) is valid only for \(\alpha\) in radian. Using \(1° = 60'\) and \(1' = 2.91 \times 10^{-4}\) rad:

Angle subtended

\[\theta = 1^{\circ}\ 54' = 114' = 114 \times 2.91 \times 10^{-4}\ \text{rad}\]

\[= 3.317 \times 10^{-2}\ \text{rad}\]

Now apply \(d = \alpha D\):

Diameter of the Earth \(= \theta \times\) distance to the moon from the Earth

\[= 3.317 \times 10^{-2} \times 3.84 \times 10^{8}\ \text{m}\]

\[= 1.274 \times 10^{7}\ \text{m}\]

This is about \(1.27 \times 10^{4}\) km, which agrees well with the accepted diameter of the Earth — a reassuring check that the small-angle method is sound.

The angular diameter of the Sun measured from the Earth is \(9.31 \times 10^{-3}\) rad. If the Sun is at a distance of \(1.496 \times 10^{11}\) m from the Earth, calculate the diameter of the Sun in metre.

1.3.4 Measurement of Very Small Distances

At the other end of Table 1.2 lies the opposite problem. When we intend to measure the size of atoms and molecules, the conventional apparatus like a Vernier calliper or a screw gauge will not be useful.

The reason is a matter of scale. A Vernier calliper can resolve about \(10^{-4}\) m and a screw gauge about \(10^{-5}\) m, but an atom has a radius of the order of \(10^{-10}\) m. Asking such an instrument to measure an atom is like asking a metre scale to measure the thickness of a soap film — the object is smaller than the smallest division the instrument can register. Worse, these instruments work by touching the object, and there is no jaw or screw fine enough to grip a single atom.

Therefore, to measure the size of atoms we use an electron microscope or a tunnelling electron microscope.

Diagram

Fig. 1.12 – Comparison of measuring ranges: a metre scale, a Vernier calliper, a screw gauge, an optical microscope and an electron microscope placed along a logarithmic length scale from \(10^{0}\) m down to \(10^{-10}\) m.

Small distances are measured in units of (i) fermi (\(1\ \text{F} = 10^{-15}\) m) in the SI system — thus 1 F is one femtometre (fm) — or (ii) Angstrom (\(1\ \overset{\circ}{\text{A}} = 10^{-10}\) m).

For measuring sizes using a microscope, we need to select the wavelength of light to be used in the microscope such that it is smaller than the size of the object to be measured. Thus visible light, whose wavelength ranges from \(4000\ \overset{\circ}{\text{A}}\) to \(7000\ \overset{\circ}{\text{A}}\), can measure sizes down to about \(4000\ \overset{\circ}{\text{A}}\).

If we want to measure even smaller sizes, we need to use an even smaller wavelength, and so the use of an electron microscope becomes necessary. As you will study in the XII\(^{\text{th}}\) standard, each material particle corresponds to a wave. The approximate wavelength of the electrons in an electron microscope is about \(0.6\ \overset{\circ}{\text{A}}\), so that one can measure atomic sizes \(\approx 1\ \overset{\circ}{\text{A}}\) using this microscope.

Q. Why can a wave never reveal detail finer than its own wavelength?

A. A wave “feels” an obstacle only if the obstacle is comparable to or larger than its wavelength; anything much smaller is simply passed over, much as an ocean swell rolls past a floating pin without being disturbed by it. Light of wavelength \(4000\ \overset{\circ}{\text{A}}\) therefore cannot resolve an atom of size \(1\ \overset{\circ}{\text{A}}\). The only remedy is to use a wave with a shorter wavelength — and the electron, whose wave is about \(0.6\ \overset{\circ}{\text{A}}\) long, provides it.

The scanning tunnelling microscope was developed in the early 1980s by Gerd Binnig and Heinrich Rohrer, working at an IBM laboratory in Zurich. Instead of shining anything on the sample, their instrument brought an extremely fine metal tip so close to a surface that electrons “tunnelled” across the gap, and the tiny current that flowed revealed the contour of the surface atom by atom. The work earned them a Nobel Prize within a few years. Later, researchers at IBM used the same kind of tip not merely to see individual atoms but to push them into chosen positions — the first time human beings had arranged matter atom by atom.

Q. A student writes the radius of a proton as \(1\ \overset{\circ}{\text{A}}\). What is wrong?

A. From Table 1.2, the radius of a proton is about \(1 \times 10^{-15}\) m, which is 1 fermi, not 1 Angstrom. The two units differ by a factor of \(10^{5}\) — the Angstrom is the natural unit for atoms, the fermi for nuclei and nuclear particles. Choosing the wrong one misplaces the answer by five orders of magnitude.

1.4 Measurement of Mass

The SI unit of mass is the kilogram, and for well over a century its standard was an object you could actually hold.

Since 1889, a kilogram was the mass of a shiny piece of platinum–iridium alloy kept in a special glass case at the International Bureau of Weights and Measures. Every kilogram used anywhere on the Earth was, in the end, a copy of a copy of that one cylinder.

Why a Metal Cylinder Was Not Good Enough

This definition of mass has been modified on 20th May 2019. The reason is unsettling in its simplicity: the carefully kept platinum–iridium piece is seen to pick up micro particles of dirt, and is also affected by the atmosphere, causing its mass to change.

Think about what that means. The cylinder is the kilogram, by definition. So if the cylinder gains a few micrograms of dirt, the cylinder cannot be said to have become heavier — instead, the kilogram itself has quietly changed, and with it every mass measurement in the world. A standard that drifts is no standard at all.

Q3. For 130 years the kilogram was a metal cylinder locked in a vault near Paris. Why did the world decide, in 2019, that this was no longer good enough?

A. Because the cylinder was not perfectly stable. Over the decades it picked up microscopic particles of dirt and was affected by the surrounding atmosphere, so its mass slowly changed. Since the cylinder defined the kilogram, any change in it changed the unit itself, and there was no way to detect this from inside the definition. On 20th May 2019 the kilogram was therefore redefined in terms of a fixed physical constant instead of an object — something that can never be scratched, contaminated or lost.

The New Kilogram

The new measure of the kilogram is defined in terms of the magnitude of electric current. At first this sounds strange — how can a current define a mass? The link is the electromagnet.

We know that electric current can be used to make an electromagnet. An electromagnet attracts magnetic materials, and is thus used in research and in industrial applications such as cranes to lift heavy pieces of iron and steel. The larger the current, the stronger the magnetic pull.

Now imagine an extremely sensitive balance. One side holds one standard kilogram mass. On the other side, instead of placing metal weights, we let an electromagnet pull the pan down. We then ask: exactly how much current must be passed through the electromagnet so that its pull balances the standard kilogram?

That current is a precisely measurable electrical quantity. Thus the kilogram mass can be described in terms of the amount of current which has to be passed through an electromagnet so that it can pull down one side of an extremely sensitive balance to balance the other side which holds one standard kg mass.

Diagram

Fig. 1.13 – A sensitive balance with a standard 1 kg mass in one pan and a current-carrying electromagnet pulling down the other pan, with an ammeter in the circuit.

Q. What has really been gained by this change? A current has to be measured either way — where is the improvement?

A. The gain is that the definition no longer depends on any single object anywhere in the world. Electrical quantities can be tied to fixed constants of nature, and those constants are the same in every laboratory, in every century. A well-equipped laboratory in India can now realise the kilogram for itself, without ever consulting a cylinder in Paris — and no accident to that cylinder can ever change the unit again.

Mass on the Atomic Scale

While dealing with the mass of atoms and molecules, the kilogram is an inconvenient unit. An atom has a mass of the order of \(10^{-26}\) kg, and writing every atomic mass with such an exponent is clumsy and invites error.

It is much easier to compare the mass of any atom in terms of the mass of some standard atom. The standard atom has been decided internationally to be the C\(^{12}\) atom. The \((1/12)^{\text{th}}\) mass of an unexcited atom of C\(^{12}\) is called the atomic mass unit (amu).

NoteDefinition

(One atomic mass unit (amu) is one-twelfth of the mass of an unexcited atom of carbon-12.)

\[1 \ \text{amu} = 1.6605402 \times 10^{-27}\ \text{kg}\]

with an uncertainty of 10 in the last two decimal places.

The one-twelfth is not arbitrary. A carbon-12 atom contains 6 protons and 6 neutrons, so taking a twelfth of it makes 1 amu come out very close to the mass of a single proton or neutron. On this scale, the mass number of an atom is very nearly its mass in amu — which is exactly what makes the unit convenient.

Why is only carbon used, and not any other element, for defining the atomic mass unit? Carbon-12 (C\(^{12}\)) is the most abundant isotope of carbon and the most stable one. Around 98% of the available carbon is the C\(^{12}\) isotope. Earlier, oxygen and hydrogen were used as the standard atoms. But various isotopes of oxygen and hydrogen are present in higher proportion, and therefore it is more accurate to use C\(^{12}\).

A certain molecule has a mass of 180 amu. Express its mass in kilogram, using \(1\) amu \(= 1.6605402 \times 10^{-27}\) kg.

1.5 Measurement of Time

The SI unit of time is the second (s). As with length and mass, the interesting question is not what the unit is called, but what natural event we choose to count.

The Second from the Turning Earth

For many years, the duration of one mean solar day was considered as the reference. A mean solar day is the average time interval from one noon to the next noon.

NoteDefinition

(A mean solar day is the average time interval from one noon to the next noon.)

The word “average” is doing quiet work here. The interval from noon to noon is not exactly the same on every day of the year, so the standard is built from the average taken over the year rather than from any single day.

The average duration of a day is taken as 24 hours. One hour is of 60 minutes and each minute is of 60 seconds. Thus a mean solar day

\[= 24 \ \text{hours} = 24 \times 60 \times 60 = 86400\ \text{s}\]

Accordingly, a second was defined as 1/86400 of a mean solar day.

This is a satisfying definition, because it ties our unit of time to the rotation of the Earth — the most familiar clock human beings have ever had.

Why the Earth Was Dismissed as a Clock

It was later observed that the length of a solar day varies gradually, due to the gradual slowing down of the Earth’s rotation. The Earth is very slightly losing spin, so the day is very slightly lengthening.

The consequence is the same as with the platinum–iridium cylinder: if the second is defined as a fraction of a day, and the day is stretching, then the second itself is stretching. The unit is drifting, and we cannot detect the drift from inside the definition.

Hence, to get a more standard and non-varying (constant) unit for the measurement of time, a cesium atomic clock is used. It is based on the periodic vibrations produced in the cesium atom. In a cesium atomic clock, a second is taken as the time needed for 9,192,631,770 vibrations of the radiation (wave) emitted during a transition between two hyperfine states of the Cs\(^{133}\) atom.

NoteDefinition

(One second is the time needed for 9,192,631,770 vibrations of the radiation emitted during a transition between two hyperfine states of the caesium-133 atom.)

Diagram

Fig. 1.14 – Two clocks side by side: the Earth rotating from one noon to the next (mean solar day, 86400 s) and a caesium atom emitting radiation whose 9,192,631,770 vibrations make one second.

Q. Why is an atom a better clock than a planet?

A. Because an atom cannot wear out, slow down, or be worn away. The frequency of the radiation emitted in a transition of Cs\(^{133}\) is fixed by the internal structure of the atom itself, so every caesium atom in the universe ticks at exactly the same rate, today and a thousand years from now. The Earth, being a massive body subject to tides and friction, is gradually slowing — and a clock that slowly slows is precisely what a standard must never be.

Every satellite of a navigation system such as GPS carries atomic clocks on board. Your position is worked out from the tiny differences in the arrival times of the signals from several satellites, and since radio waves travel at \(3 \times 10^{8}\) m s\(^{-1}\), an error of even one microsecond in timekeeping would shift your calculated position by a few hundred metres. Navigation on a phone is, at bottom, an exercise in extremely accurate time measurement.

Because atomic clocks now keep time better than the Earth does, the two slowly fall out of step. To keep our civil clocks in agreement with the rotating Earth, timekeeping authorities occasionally insert an extra “leap second” into the year. It is a curious reversal: we no longer correct our clocks against the Earth — we correct the Earth’s account against our clocks.

1.6 Dimensions and Dimensional Analysis

As mentioned earlier, a derived physical quantity can be expressed in terms of some combination of the seven basic or fundamental quantities. Velocity is length divided by time; force is mass times length divided by time squared. Strip away the numbers and the names, and what is left is this skeleton of fundamental quantities. That skeleton is what we call the dimensions of the quantity.

For convenience, the basic quantities are represented by symbols:

  • ‘L’ for length
  • ‘M’ for mass
  • ‘T’ for time
  • ‘K’ for temperature
  • ‘I’ for current
  • ‘C’ for luminous intensity
  • ‘mol’ for amount of substance
NotePrinciple / Law

(The dimensions of a physical quantity are the powers to which the concerned fundamental units must be raised in order to obtain the unit of the given physical quantity.)

Read that definition slowly, because every word in it is doing work. Dimensions are powers — not units, not numbers. To say that force has dimensions \([L^{1}M^{1}T^{-2}]\) is to say that its unit is obtained by raising metre to the power 1, kilogram to the power 1 and second to the power \(-2\), and multiplying them together.

Writing a Dimensional Formula

When we represent any derived quantity with appropriate powers of the symbols of the fundamental quantities, such an expression is called a dimensional formula.

NoteDefinition

(A dimensional formula is the expression that shows a derived quantity written as the appropriate powers of the symbols of the fundamental quantities.)

Two points of notation must be respected. The dimensional formula is expressed inside a square bracket, and no comma is written in between any of the symbols. So we write \([L^{1}M^{0}T^{-1}]\) and never \([L^{1}, M^{0}, T^{-1}]\).

Diagram

Fig. 1.15 – The dimensional formula \([L^{1}M^{1}T^{-2}]\) of force, with callouts labelling the square bracket, the base symbols L, M, T and the powers 1, 1, −2.

Illustration

The method is always the same: write the defining relation of the quantity, replace each quantity by its dimensions, and simplify the powers.

(i) Dimensional formula of velocity

\[\text{Velocity} = \frac{\text{displacement}}{\text{time}}\]

\[\text{Dimensions of velocity} = \frac{[L]}{[T]} = [L^{1}M^{0}T^{-1}]\]

Notice that mass does not enter into velocity at all. We do not simply omit M — we write it with the power zero. Recording the missing quantities as zero powers keeps every dimensional formula in the same standard shape, so that two of them can be compared term by term.

(ii) Dimensional formula of velocity gradient

\[\text{velocity gradient} = \frac{\text{velocity}}{\text{distance}}\]

\[\text{Dimensions of velocity gradient} = \frac{[LT^{-1}]}{[L]} = [L^{0}M^{0}T^{-1}]\]

Here the two lengths have cancelled, leaving a quantity that depends only on time. Its SI unit is therefore s\(^{-1}\).

(iii) Dimensional formula for charge

\[\text{charge} = \text{current} \times \text{time}\]

\[\text{Dimensions of charge} = [I][T] = [L^{0}M^{0}T^{1}I^{1}]\]

This is the first quantity we have met whose dimensional formula needs a symbol beyond L, M and T. Electrical quantities bring in the current I, exactly as thermal quantities bring in the temperature K.

Q. What are the dimensions of a pure ratio, such as relative density or the plane angle in radian?

A. They are \([L^{0}M^{0}T^{0}]\) — every power is zero. Such quantities are called dimensionless. This is a useful thing to be able to spot: a plane angle, a solid angle, a refractive index and a strain all have no dimensions, and therefore no dimensional formula worth writing. Keep this in mind — it explains an important limitation of dimensional analysis that appears in Section 1.6.2.

Table 1.3 gives the dimensions of various physical quantities commonly used in mechanics. Study it carefully and confirm each entry for yourself from its formula, rather than memorising the last column.

Table 1.3: Some Common Physical Quantities, their SI Units and Dimensions

S. No Physical quantity Formula SI unit Dimensional formula
1 Density \(\rho = M/V\) kilogram per cubic metre (kg/m\(^{3}\)) \([L^{-3}M^{1}T^{\circ}]\)
2 Acceleration \(a = v/t\) metre per second square (m/s\(^{2}\)) \([L^{1}M^{\circ}T^{-2}]\)
3 Momentum \(P = mv\) kilogram metre per second (kg m/s) \([L^{1}M^{1}T^{-1}]\)
4 Force \(F = ma\) kilogram metre per second square (kg m/s\(^{2}\)) or newton (N) \([L^{1}M^{1}T^{-2}]\)
5 Impulse \(J = F.t\) newton second (Ns) \([L^{1}M^{1}T^{-1}]\)
6 Work \(W = F.s\) joule (J) \([L^{2}M^{1}T^{-2}]\)
7 Kinetic Energy \(KE = 1/2\ mv^{2}\) joule (J) \([L^{2}M^{1}T^{-2}]\)
8 Pressure \(P = F/A\) kilogram per metre second square (kg/ms\(^{2}\)) \([L^{-1}M^{1}T^{-2}]\)

Two entries in this table deserve a second look. Momentum and impulse have the same dimensional formula, \([L^{1}M^{1}T^{-1}]\); so do work and kinetic energy, \([L^{2}M^{1}T^{-2}]\). This is not a coincidence — impulse is the change in momentum, and work is the change in kinetic energy. Quantities that can be equated to one another must share the same dimensions.

The converse, however, does not hold: two quantities having the same dimensions need not be the same physical quantity. Work and torque, for instance, share a dimensional formula but mean quite different things. Dimensions tell us about the structure of a quantity, not about its identity.

1.6.1 Uses of Dimensional Analysis

Dimensions are not merely a bookkeeping device. Once a quantity’s skeleton of L, M and T is known, that skeleton can be put to work in three distinct ways.

(i) To Check the Correctness of Physical Equations

In any equation relating different physical quantities, if the dimensions of all the terms on both the sides are the same, then that equation is said to be dimensionally correct. This is called the principle of homogeneity of dimensions.

NotePrinciple / Law

(Principle of homogeneity of dimensions: in a correct physical equation, every term on both sides must have the same dimensions.)

The reasoning behind it is common sense dressed in symbols. You cannot add a length to a time, any more than you can add three metres to five seconds. Terms may only be added, subtracted or equated if they are the same kind of quantity — and “the same kind of quantity” is precisely what having the same dimensions means.

Consider the first equation of motion.

\[v = u + at\]

Dimension of L.H.S \(= [v] = [LT^{-1}]\)

Now examine the right-hand side, term by term:

\[[u] = [LT^{-1}]\] \[[at] = [LT^{-2}][T] = [LT^{-1}]\]

Dimension of R.H.S \(= [LT^{-1}] + [LT^{-1}]\)

\[[L.H.S] = [R.H.S]\]

As the dimensions of L.H.S and R.H.S are the same, the given equation is dimensionally correct.

Note carefully that we did not add the two right-hand terms to get \(2[LT^{-1}]\). Dimensions are not added like numbers; we merely observe that both terms carry the same dimensions, and that is enough.

Q4. Can you tell that a physics equation must be wrong without substituting a single number into it?

A. Yes — by checking its dimensions. Every term in a correct physical equation must have the same dimensions, because quantities of different kinds cannot be added or equated. So if you write out the dimensional formula of each term and find that one of them does not match the others, the equation is certainly wrong, and no numerical value can rescue it. This is the principle of homogeneity of dimensions, and it is the fastest error-check available in physics.

Q. An equation passes the dimensional check. Does that prove it is correct?

A. No. Dimensional correctness is a necessary condition, not a sufficient one. The equation \(v = u + 3at\) is dimensionally perfect and physically wrong, because dimensions are blind to pure numbers. So the test can prove an equation wrong with certainty, but can never prove it right.

(iii) To Find the Conversion Factor Between the Units of the Same Physical Quantity in Two Different Systems of Units

The third use is the most practical of all. A quantity keeps its dimensions whichever system you measure it in; only the size of the base units changes. That single fact lets us convert between systems without knowing anything about the quantity itself.

Let us use dimensional analysis to determine the conversion factor between the joule (SI unit of work) and the erg (CGS unit of work).

Let \(1\ \text{J} = x\) erg

The dimensional formula for work is \([L^{2}M^{1}T^{-2}]\). Substituting in the above equation, we can write

\[[L_{1}^{2}M_{1}^{1}T_{1}^{-2}] = x\ [L_{2}^{2}M_{2}^{1}T_{2}^{-2}]\]

\[x = \frac{[L_{1}^{2}M_{1}^{1}T_{1}^{-2}]}{[L_{2}^{2}M_{2}^{1}T_{2}^{-2}]}\]

\[\text{or,} \quad x = \left(\frac{L_{1}}{L_{2}}\right)^{2}\left(\frac{M_{1}}{M_{2}}\right)^{1}\left(\frac{T_{1}}{T_{2}}\right)^{-2}\]

where suffix 1 indicates SI units and 2 indicates CGS units.

In SI units, L, M, T are expressed in m, kg and s, and in the CGS system L, M, T are represented in cm, g and s respectively.

\[\therefore \ x = \left(\frac{\text{kg}}{\text{g}}\right)^{1}\left(\frac{\text{m}}{\text{cm}}\right)^{2}\left(\frac{\text{s}}{\text{s}}\right)^{-2}\]

\[\text{or}\quad x = \left(10^{3}\ \frac{\text{g}}{\text{g}}\right)^{1}\left((100)\ \frac{\text{cm}}{\text{cm}}\right)^{2}(1)^{-2}\]

\[\therefore \ x = (10^{3})(10^{4}) = 10^{7}\]

\[\therefore \ 1\ \text{joule} = 10^{7}\ \text{erg}\]

Observe the logic of the ratio. Since 1 kg contains 1000 g, the number measuring a given mass becomes 1000 times larger when expressed in grams. Each base unit is converted in this way, raised to the power that the dimensional formula demands, and the product of all three gives the conversion factor.

Q. Which of the three base units contributed most to the factor of \(10^{7}\)?

A. Length. The metre-to-centimetre ratio is only 100, but the dimensional formula of work carries \(L^{2}\), so it enters squared and contributes \(10^{4}\) — more than the \(10^{3}\) contributed by mass. This illustrates a general rule you will meet again in Section 1.8.2: the quantity raised to the highest power dominates the result.

A calorie is a unit of heat and it equals 4.2 J, where 1 J = kg m\(^{2}\) s\(^{-2}\). A distant civilisation employs a system of units in which the units of mass, length and time are \(\alpha\) kg, \(\beta\) m and \(\gamma\) s. Also J′ is their unit of energy. What will be the magnitude of a calorie in their units?

Solution: Let us write the new units as A, B and C for mass, length and time respectively. We are given

\[1\ \text{A} = \alpha\ \text{kg}\]

\[1\ \text{B} = \beta\ \text{m}\]

\[1\ \text{C} = \gamma\ \text{s}\]

The strategy is the same as in the joule–erg conversion: express the calorie in terms of the SI base units, then replace each SI base unit by the new one. Since \(1\ \text{A} = \alpha\) kg, it follows that \(1\) kg \(= (1/\alpha)\) A, and similarly for the others — which is why each factor below appears divided by its Greek letter.

\[1\ \text{cal} = 4.2\ \text{J} = 4.2\ \text{kg m}^{2}\ \text{s}^{-2}\]

\[= 4.2\left(\frac{\text{A}}{\alpha}\right)\left(\frac{\text{B}}{\beta}\right)^{2}\left(\frac{\text{C}}{\gamma}\right)^{-2}\]

\[= \frac{4.2\,\gamma^{2}}{\alpha\beta^{2}}\ \text{AB}^{2}\ \text{C}^{-2}\]

\[= \frac{4.2\,\gamma^{2}}{\alpha\beta^{2}}\ \text{J}'\]

Thus in the new units, 1 calorie is \(\dfrac{4.2\gamma^{2}}{\alpha\beta^{2}}\) J′.

Note where the powers have landed: \(\gamma\) appears squared in the numerator because time carries the power \(-2\) in the dimensional formula of energy, and \(\beta\) appears squared in the denominator because length carries the power \(+2\). Every exponent in the answer is dictated by the dimensional formula \([L^{2}M^{1}T^{-2}]\).

1.6.2 Limitations of Dimensional Analysis

Powerful as it is, dimensional analysis is not a substitute for physics. It knows only about the powers of L, M and T, and is entirely blind to everything else. That blindness produces four definite limitations, and an examiner is as likely to ask about these as about the uses.

1) The value of a dimensionless constant can be obtained with the help of experiments only.

We have already met this. The method gave us \(T = k\sqrt{l/g}\) for the simple pendulum, but the value \(k = 2\pi\) had to come from the laboratory. Pure numbers have no dimensions, so no amount of dimensional reasoning can pin them down.

2) Dimensional analysis cannot be used to derive relations involving trigonometric, exponential and logarithmic functions, as these quantities are dimensionless.

The argument of \(\sin\theta\), of \(e^{x}\) or of \(\log x\) must itself be a dimensionless number — you cannot take the sine of three metres. And the value these functions return is likewise a pure number. Since dimensional analysis works only by balancing powers of L, M and T, an equation such as \(y = a\sin(\omega t)\) lies completely outside its reach.

Q. Why must the argument of a function such as \(\sin(\omega t)\) be dimensionless?

A. Because such a function is defined by an infinite series whose terms are \(\omega t\), \((\omega t)^{3}\), \((\omega t)^{5}\) and so on, all added together. Terms of different powers can only be added if the quantity itself has no dimensions — otherwise we would be adding a length to a volume. This is why \(\omega\) must carry the dimension \([T^{-1}]\), so that \(\omega t\) comes out as a pure number.

3) This method is not useful if the constant of proportionality is not a dimensionless quantity.

Illustration: The gravitational force between two point masses is directly proportional to the product of the two masses and inversely proportional to the square of the distance between the two.

\[\therefore F \propto \frac{m_{1}m_{2}}{r^{2}}\]

Let

\[F = G\ \frac{m_{1}m_{2}}{r^{2}}\]

The constant of proportionality ‘G’ is NOT dimensionless. Thus the earlier method will not work.

Look at what goes wrong. In the pendulum derivation we could equate dimensions on both sides only because \(k\) contributed nothing. Here, \(G\) carries dimensions of its own, and those unknown dimensions sit on the right-hand side of the equation exactly where we needed clarity. The balancing of powers can no longer be completed, and the relation cannot be derived by dimensions alone — it had to be discovered from observation.

4) If the correct equation contains some more terms of the same dimension, it is not possible to know about their presence using a dimensional equation.

For example, with standard symbols, the equation

\[S = \frac{1}{2}at^{2}\]

is dimensionally correct. However, the complete equation is

\[S = ut + \frac{1}{2}at^{2}\]

Both \(ut\) and \(\frac{1}{2}at^{2}\) have the dimension of length, so the homogeneity test is passed whether or not the first term is present. Dimensional analysis therefore cannot tell us that a term has gone missing, and an incomplete equation can sail through the check unnoticed.

Q. Given these four limitations, what exactly is dimensional analysis good for?

A. It is a filter, not a proof. It can reject a wrong equation instantly, suggest the form of an unknown relation up to a constant, and convert units between systems — all without any experiment. What it cannot do is supply pure numbers, handle trigonometric or exponential relations, cope with dimensional constants, or detect a missing term. Used with that understanding, it is one of the sharpest tools in physics; used as a substitute for physical reasoning, it will mislead.

Diagram

Fig. 1.16 – A two-column summary chart: “Dimensional analysis CAN” (check equations, find relations up to a constant, convert units) and “Dimensional analysis CANNOT” (find pure numbers, handle sin/exp/log, handle dimensional constants like G, detect missing terms).

1.7 Accuracy, Precision and Uncertainty in Measurement

Physics is a science based on observations and experiments. Observations of various physical quantities are made during an experiment. For example, during an atmospheric study we measure atmospheric pressure, wind velocity, humidity, and so on.

Every one of these observations is a number produced by a real instrument, handled by a real person, under real conditions. And no such number is ever perfect. Two separate questions must therefore be asked about any measurement: is it close to the truth? and is it repeatable? These are two different questions, and physics gives them two different names.

Accuracy

All the measurements may be accurate, meaning that the measured values are the same as the true values. Accuracy is how close a measurement is to the actual value of that quantity.

NoteDefinition

(Accuracy is how close a measurement is to the actual (true) value of the quantity being measured.)

Precision

These measurements may be precise, meaning that multiple measurements give nearly identical values — that is, reproducible results.

NoteDefinition

(Precision is the closeness of repeated measurements of the same quantity to one another, that is, the reproducibility of the result.)

Notice what precision does not mention: the true value. A precise instrument is one that agrees with itself, not necessarily one that agrees with nature.

Imagine a weighing balance that has been wrongly zeroed, so that it reads 200 g too high. A student weighs the same object five times and gets 700.1 g, 700.0 g, 700.2 g, 700.1 g, 700.1 g. The readings agree beautifully with one another — this is high precision. But every one of them is 200 g away from the truth — this is poor accuracy. No amount of repeating will reveal the fault, because the error is the same every time.

In actual measurements, an observation may be both accurate and precise, or neither accurate nor precise. All four combinations are possible, as shown in Fig. 1.17.

Diagram

Fig. 1.17 – Four target boards showing shots that are (a) accurate and precise, (b) precise but not accurate, (c) accurate but not precise, (d) neither accurate nor precise.

The goal of the observer should be to get accurate as well as precise measurements. Precision alone gives a confident wrong answer; accuracy alone gives an answer that cannot be trusted to repeat.

Q. Which of the two — poor accuracy or poor precision — is harder to detect from the readings alone?

A. Poor accuracy. Poor precision announces itself: the readings scatter, and you can see the spread immediately. Poor accuracy can hide completely, because a consistently wrong instrument produces beautifully consistent numbers. It is exposed only by checking the instrument against a known standard — which is exactly why calibration exists.

Where the Uncertainty Comes From

Possible uncertainties in an observation may arise due to the following reasons:

  1. Quality of the instrument used. A metre scale marked only in centimetres cannot deliver a reading to the nearest millimetre, however careful the observer.

  2. Skill of the person doing the experiment. Two students using the same Vernier calliper may not tighten it identically or read the coincidence of the divisions with equal care.

  3. The method used for measurement. Some methods are inherently blunter than others; measuring the volume of an irregular solid by displacement is more reliable than estimating it from its shape.

  4. External or internal factors affecting the result of the experiment. Temperature, pressure, humidity, vibration or a fluctuating power supply can all shift a reading without the observer noticing.

Study this list carefully, because it is really a preview of the next section. Reasons 1, 2 and 3 tend to push the reading the same way every time and can, in principle, be identified and removed. Reason 4 pushes it now this way, now that. This distinction is the seed of the classification of errors that follows.

If ten students are asked to measure the length of a piece of cloth up to a mm, using a metre scale, do you think their answers will be identical? Give reasons.

1.8 Errors in Measurements

Faulty measurements of a physical quantity can lead to errors. In physics, the word “error” does not mean a blunder or a mistake in arithmetic. It means the unavoidable difference between the value we obtain and the value that is actually there.

Errors are broadly divided into the following two categories.

Systematic Errors

a) Systematic errors: Systematic errors are errors that are not determined by chance but are introduced by an inaccuracy — involving either the observation or the measurement process — inherent to the system.

NoteDefinition

(Systematic errors are errors that arise not by chance but from an inaccuracy inherent in the observation or measurement process itself.)

Sources of systematic error may be due to imperfect calibration of the instrument, and sometimes an imperfect method of observation.

The defining feature of these errors is stated in a single sentence, and it is worth underlining: each of these errors tends to be in one direction, either positive or negative. They do not scatter. They push every single reading the same way by the same amount, which is why repeating the experiment a hundred times will not reveal them.

The sources of systematic errors are as follows.

(i) Instrumental error: This type of error arises due to defective calibration of an instrument. For example, an incorrect zeroing of an instrument will lead to such a kind of error — the ‘zero’ of a thermometer not graduated at the proper place; the pointer of a weighing balance in the laboratory already indicating some value instead of showing zero when no load is kept on it; an ammeter showing a current of 0.5 amp even when it is not connected in a circuit.

In each of these examples, the instrument is lying by a fixed amount before the experiment has even begun, and it will go on telling the same lie in every reading.

(ii) Error due to imperfection in experimental technique: This is an error due to defective setting of an instrument. For example, the measured volume of a liquid in a graduated tube will be inaccurate if the tube is not held vertical.

Here the instrument itself is perfectly sound. It is the way it has been set up that introduces the fault — and since the tube stays tilted throughout the experiment, the error again repeats identically in every observation.

(iii) Personal error: Such errors are introduced due to the fault of the observer. Bias of the observer, carelessness in taking observations, and so on, could result in such errors.

For example, while measuring the length of an object with a ruler, it is necessary to look at the ruler directly from above. If the observer looks at it from an angle, the measured length will be wrong due to parallax — the same apparent shift of position with the observer’s position that was used to advantage in Section 1.3.1, but which here works against us. And because an observer usually sits in the same place throughout an experiment, the viewing angle stays the same, and so does the error.

Diagram

Fig. 1.18 – An observer’s eye viewing a scale from directly above (correct reading) and from an angle (parallax error), with the two apparent readings marked on the scale.

Q. A student takes fifty readings of the same length with an ammeter whose needle rests at 0.5 A when disconnected, and carefully averages them. Has the averaging helped?

A. Not at all. Averaging removes errors that fall on both sides of the truth, because the positive and negative deviations cancel. A zero error pushes every reading the same way, so the average is wrong by exactly the same 0.5 A as each individual reading. Systematic errors survive averaging untouched — this is what makes them so dangerous.

Systematic errors can be minimized by using the correct instrument, following the proper experimental procedure, and removing personal error. Notice the phrasing: they are removed by thinking about the setup, never by taking more readings.

Random Errors

b) Random errors: These are the errors which are introduced even after following all the procedures to minimize systematic errors.

NoteDefinition

(Random errors are errors that persist even after all systematic errors have been minimized, and which may be positive or negative in an unpredictable way.)

These types of errors may be positive or negative. They cannot be eliminated completely, but we can minimize them by taking repeated observations and then taking their mean (average).

This is exactly the opposite of the situation with systematic errors, and the reason lies in the word “random”. Since a random error is as likely to push a reading up as down, the deviations tend to cancel one another when many readings are averaged. The more readings we take, the more thoroughly they cancel.

Random errors occur due to variation in the conditions in which the experiment is performed. For example, the temperature may change during the course of an experiment, the pressure of any gas used in the experiment may change, or the voltage of the power supply may change randomly.

Q. How can you tell, from the readings alone, whether you are looking at a random error or a systematic one?

A. By their scatter. Random errors make the readings spread out around some central value — you can see them. Systematic errors leave the scatter untouched and simply shift the whole cluster away from the truth, so they are invisible in the readings themselves. That is why the treatment differs: random errors are handled by mathematics (averaging), systematic errors only by better experimental practice.

The next subsection takes up the mathematical treatment of random errors: given a set of scattered readings, what single value do we report, and how far from the truth might it be?

1.8.1 Estimation of error

Suppose we have done the honest thing — checked the instrument, set it up properly, removed personal bias — and yet the readings still refuse to agree with one another. Random error remains. What number do we then report, and how much can we trust it?

Suppose the readings recorded repeatedly for a physical quantity during a measurement are

\[a_{1},\ a_{2},\ a_{3},\ \ldots\ldots\ldots\ a_{n}\]

The Most Probable Value

The arithmetic mean \(a_{mean}\) is given by

\[a_{mean} = \frac{a_{1} + a_{2} + a_{3} + \ldots\ldots\ldots + a_{n}}{n}\]

\[a_{mean} = \frac{1}{n}\sum_{i=1}^{n} a_{i} \qquad \text{--- (1.3)}\]

This is the most probable value of the quantity. The justification comes straight from the previous section: random errors are as likely to be positive as negative, so when the readings are added, the positive deviations tend to cancel the negative ones. What survives the averaging is our best estimate of the truth.

NoteDefinition

(The arithmetic mean of a set of repeated readings is taken as the most probable value of the measured quantity.)

Absolute Error

We now ask how far each individual reading has strayed from this most probable value. The magnitude of the difference between the mean value and each individual value is called the absolute error in the observation.

NoteDefinition

(The absolute error in an observation is the magnitude of the difference between the mean value and that individual observed value.)

Thus, for ‘\(a_{1}\)’, the absolute error \(\Delta a_{1}\) is given by

\[\Delta a_{1} = |\ a_{mean} - a_{1}\ |,\]

for \(a_{2}\),

\[\Delta a_{2} = |\ a_{mean} - a_{2}\ |\]

and so for \(a_{n}\) it will be

\[\Delta a_{n} = |\ a_{mean} - a_{n}\ |\]

The modulus signs are essential. We are interested in how far a reading has strayed, not in which side it strayed to — and if the signs were kept, the deviations would cancel and give zero, which would tell us nothing.

Mean Absolute Error

The arithmetic mean of all the absolute errors is called the mean absolute error in the measurement of the physical quantity.

\[\Delta a_{mean} = \frac{\Delta a_{1} + \Delta a_{2} + \ldots\ldots + \Delta a_{n}}{n}\]

\[= \frac{1}{n}\sum_{i=1}^{n}\Delta a_{i} \qquad \text{--- (1.4)}\]

NoteDefinition

(The mean absolute error is the arithmetic mean of the absolute errors of all the individual observations.)

The measured value of the physical quantity \(a\) can then be represented by

\[a = a_{mean} \pm \Delta a_{mean}\]

which tells us that the actual value of ‘\(a\)’ could be between \(a_{mean} - \Delta a_{mean}\) and \(a_{mean} + \Delta a_{mean}\).

This is the honest way to report a measurement. The single number \(a_{mean}\) is our best guess; the \(\pm\ \Delta a_{mean}\) is the width of the band inside which the truth is expected to lie.

Q5. If ten students measure the same piece of cloth with the same metre scale, their answers will differ slightly. Does that mean nine of them are careless?

A. No — the spread is not carelessness but random error, which persists even after every systematic error has been minimized and which may be positive or negative in an unpredictable way, because the conditions of the measurement and the judgement of the observer vary slightly from one reading to the next. None of the ten readings is by itself the true one. The best estimate is their arithmetic mean, called the most probable value, because random errors push readings above and below the truth roughly equally and so cancel in the average. How far the truth may still lie from it is measured by the mean absolute error — the average of the deviations of the individual readings from that mean — so the result is reported not as a single number but as \(a_{mean} \pm \Delta a_{mean}\), a band within which the true value is expected to lie.

Relative Error and Percentage Error

An error of 1 cm means very different things when you are measuring a pencil and when you are measuring a cricket pitch. To judge how serious an error is, it must be compared with the size of the quantity being measured.

The ratio of the mean absolute error to its arithmetic mean value is called the relative error.

\[\text{Relative error} = \frac{\Delta a_{mean}}{a_{mean}} \qquad \text{--- (1.5)}\]

NoteDefinition

(Relative error is the ratio of the mean absolute error to the mean (most probable) value of the measured quantity.)

Since it is a ratio of two quantities of the same kind, the relative error has no unit and no dimension — it is a pure number, and this is precisely what allows the quality of two completely different measurements to be compared.

When the relative error is represented as a percentage, it is called the percentage error.

\[\text{Percentage error} = \frac{\Delta a_{mean}}{a_{mean}} \times 100 \qquad \text{--- (1.6)}\]

NoteDefinition

(Percentage error is the relative error expressed as a percentage, that is, multiplied by 100.)

Q. Measurement A gives \((2.00 \pm 0.02)\) m and measurement B gives \((500.0 \pm 0.5)\) m. Which measurement is of better quality?

A. B. Its absolute error, 0.5 m, is twenty-five times larger than A’s, yet its relative error is \(0.5/500.0 = 0.001\), or 0.1%, against A’s \(0.02/2.00 = 0.01\), or 1%. Absolute error alone can never settle such a comparison; only the relative or percentage error can.

The radius of a sphere measured repeatedly yields the values 5.63 m, 5.54 m, 5.44 m, 5.40 m and 5.35 m. Determine the most probable value of the radius and the mean absolute, relative and percentage errors.

Solution: The most probable value of the radius is its arithmetic mean, by equation (1.3):

\[= \frac{5.63 + 5.54 + 5.44 + 5.40 + 5.35}{5}\ \text{m}\]

\[= 5.472\ \text{m}\]

Next, the absolute error of each reading is its deviation from this mean, taken without sign, and the mean absolute error is the average of those five deviations, by equation (1.4):

Mean absolute error

\[= \frac{1}{5}\left\{ \begin{array}{l} |5.63 - 5.472| + |5.54 - 5.472| \\ + |5.44 - 5.472| + |5.40 - 5.472| \\ + |5.35 - 5.472| \end{array} \right\}\ \text{m}\]

\[= \frac{0.452}{5} = 0.0904\ \text{m}\]

The relative error now follows from equation (1.5), as the ratio of the mean absolute error to the mean value:

\[\text{Relative error} = \frac{0.0904}{5.472} = 0.017\]

\[\%\ \text{error} = 1.7\%\]

The radius may therefore be reported as \((5.472 \pm 0.0904)\) m — the mean value, followed by the band of uncertainty around it.

Perform an experiment using a Vernier callipers of least count 0.01 cm to measure the external diameter of a hollow cylinder. Take 3 readings at different positions on the cylinder and find (i) the mean diameter (ii) the absolute mean error and (iii) the percentage error in the measurement of the diameter.

The time period of a simple pendulum is measured five times and the readings are 2.06 s, 2.02 s, 2.10 s, 2.04 s and 2.08 s. Calculate the most probable value of the time period, the mean absolute error, the relative error and the percentage error.

1.8.2 Combination of errors

When we do an experiment and measure various physical quantities associated with the experiment, we must know how the errors from the individual measurements combine to give the error in the final result.

For example, in the measurement of the resistance of a conductor using Ohm’s law, there will be an error in the measurement of the potential difference and also in that of the current. Neither quantity is measured perfectly, and the resistance is calculated from both. It is important to study how these errors combine to give the error in the measurement of resistance.

The rules that follow are always derived on the same principle: we ask for the worst case. Errors could partially cancel if we are lucky, but a careful physicist never counts on luck. We therefore quote the largest error the result could possibly carry.

a) Errors in sum and in difference

Suppose two physical quantities A and B have measured values \(A \pm \Delta A\) and \(B \pm \Delta B\) respectively, where \(\Delta A\) and \(\Delta B\) are their mean absolute errors. We wish to find the absolute error \(\Delta Z\) in their sum.

\[Z = A + B\] \[Z \pm \Delta Z = (A \pm \Delta A) + (B \pm \Delta B)\] \[= (A + B) \pm \Delta A \pm \Delta B\] \[\pm \Delta Z = \pm \Delta A \pm \Delta B,\]

For the difference, i.e., if \(Z = A - B\),

\[Z \pm \Delta Z = (A \pm \Delta A) - (B \pm \Delta B)\] \[= (A - B) \pm \Delta A \mp \Delta B\] \[\pm \Delta Z = \pm \Delta A \pm \Delta B,\]

There are four possible values for \(\Delta Z\), namely \((+\Delta A - \Delta B)\), \((+\Delta A + \Delta B)\), \((-\Delta A - \Delta B)\) and \((-\Delta A + \Delta B)\). Hence the maximum value of the absolute error is \(\Delta Z = \Delta A + \Delta B\) in both the cases.

The signs must be allowed to take every combination, because we do not know in which direction each individual measurement has gone wrong. Among the four possibilities, the largest magnitude of \(\Delta Z\) is obtained when both errors happen to act the same way — and that worst case is the one we report.

NotePrinciple / Law

(When two quantities are added or subtracted, the maximum absolute error in the final result is the sum of the absolute errors in the individual quantities.)

Read the statement again and note the surprise buried in it: even for a difference, the errors are added. Subtracting two measured quantities does not subtract their uncertainties; it accumulates them.

Q. Two lengths are measured as \((20.0 \pm 0.1)\) cm and \((19.5 \pm 0.1)\) cm. Why is their difference such a poor measurement?

A. The difference is \((0.5 \pm 0.2)\) cm. The absolute error has doubled to 0.2 cm while the result itself has collapsed to 0.5 cm, so the relative error is a huge 40% — even though each original measurement was accurate to about 0.5%. Taking the difference of two nearly equal quantities is one of the most treacherous operations in experimental physics, and good experiments are designed to avoid it.

b) Errors in product and in division

Suppose \(Z = AB\) and the measured values of A and B are \((A \pm \Delta A)\) and \((B \pm \Delta B)\). Then

\[Z \pm \Delta Z = (A \pm \Delta A)\ (B \pm \Delta B)\] \[= AB \pm A\Delta B \pm B\Delta A \pm \Delta A \Delta B\]

Dividing the L.H.S by \(Z\) and the R.H.S by \(AB\) — which is legitimate because \(Z = AB\) — we get

\[\left(1 \pm \frac{\Delta Z}{Z}\right) = \left(1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm \left(\frac{\Delta A}{A}\right)\left(\frac{\Delta B}{B}\right)\right)\]

Since \(\Delta A/A\) and \(\Delta B/B\) are very small, we shall neglect their product. If each relative error is of the order of one hundredth, their product is of the order of one ten-thousandth — utterly negligible beside the terms we are keeping. Hence the maximum relative error in Z is

\[\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \qquad \text{--- (1.7)}\]

This formula also applies to the division of two quantities.

NotePrinciple / Law

(When two quantities are multiplied or divided, the maximum relative error in the result is the sum of the relative errors in each quantity.)

Q. Why do sums and differences combine absolute** errors, while products and quotients combine relative errors?**

A. Because of what is being combined. In a sum, the quantities themselves are added, so their uncertainties add in the same units. In a product, the factors are multiplied, and multiplying a quantity by a factor that is 1% too large makes the product 1% too large — the error travels as a fraction, not as a fixed amount. Mixing up the two rules is the single most common error in this chapter’s numericals: never add a relative error to an absolute one.

c) Errors due to the power (index) of a measured quantity

Suppose

\[Z = A^{3} = A.A.A\]

This is just a product of three identical quantities, so the multiplication rule above applies three times over:

\[\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta A}{A} + \frac{\Delta A}{A}\]

Hence the maximum relative error in \(Z = A^{3}\) is three times the relative error in A. So if \(Z = A^{n}\),

\[\frac{\Delta Z}{Z} = n\ \frac{\Delta A}{A} \qquad \text{--- (1.8)}\]

In general, if \(Z = \dfrac{A^{p}B^{q}}{C^{r}}\),

\[\frac{\Delta Z}{Z} = p\ \frac{\Delta A}{A} + q\ \frac{\Delta B}{B} + r\ \frac{\Delta C}{C} \qquad \text{--- (1.9)}\]

Look carefully at equation (1.9). The power \(r\) of the quantity in the denominator enters with a plus sign, exactly like \(p\) and \(q\). Errors never subtract, whether the quantity divides or multiplies; every one of them makes the result worse.

The quantity in the formula which has a large power is responsible for the maximum error. This is a practical instruction for the laboratory: if a quantity appears cubed in your formula, its measurement deserves three times the care, because its error is tripled before it reaches the answer.

Diagram

Fig. 1.19 – A flow chart: “Are the quantities added/subtracted?” → add absolute errors; “Are they multiplied/divided/raised to powers?” → add relative errors weighted by their powers.

In an experiment to determine the volume of an object, the mass and the density are recorded as \(m = (5 \pm 0.15)\) kg and \(\rho = (5 \pm 0.2)\) kg m\(^{-3}\) respectively. Calculate the percentage error in the measurement of volume.

Solution : We know,

\[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]

\[\therefore \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{M}{\rho}\]

Volume is a quotient of two measured quantities, so by equation (1.7) the relative errors in mass and density are added — not subtracted, even though the density divides.

Maximum percentage error in volume

\[= \left(\frac{\Delta m}{m} + \frac{\Delta \rho}{\rho}\right) \times 100\]

\[= \left(\frac{0.15}{5} + \frac{0.2}{5}\right) \times 100\]

\[= (0.03 + 0.04) \times 100\]

\[= (0.07) \times 100 = 7\%\]

The acceleration due to gravity is determined by using a simple pendulum of length \(l = (100 \pm 0.1)\) cm. If its time period is \(T = (2 \pm 0.01)\) s, find the maximum percentage error in the measurement of \(g\).

Solution: The time period of oscillation of a simple pendulum is given by

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

We must first rearrange this to make \(g\) the subject, because the error formula can only be applied to the quantity we are actually calculating.

Squaring both sides,

\[T^{2} = 4\pi^{2}l/g\]

\[\therefore\ g = 4\pi^{2}\ \frac{l}{T^{2}}\]

Now \(g\) has the form of equation (1.9): \(l\) appears to the power 1 and \(T\) to the power 2, while \(4\pi^{2}\) is a pure number and carries no error at all. Hence the relative error in \(T\) enters doubled.

Now \(\Delta l = 0.1,\ l = 100\) cm, \(\Delta T = 0.01\) s, \(T = 2\) s

\[\text{Maximum Percentage error} = \frac{\Delta g \times 100}{g}\]

\[= \left(\frac{\Delta l}{l} + \frac{2\Delta T}{T}\right) \times 100\]

\[= \left(\frac{0.1}{100} + \frac{2 \times 0.01}{2}\right) \times 100\]

\[= (0.001 + 0.01) \times 100 = 1.1\]

Maximum percentage error in the measurement of \(g\) is 1.1%.

Notice which measurement dominated. The length was measured to 0.1%, the time only to 0.5% — and because the time is squared, its contribution became 1%, nine times the contribution of the length. If this experiment is to be improved, it is the timing that must be improved, not the metre scale.

The radius of a sphere is measured as \(r = (3.0 \pm 0.1)\) cm. Calculate the percentage error in the volume of the sphere, given \(V = \frac{4}{3}\pi r^{3}\).

1.9 Significant Figures

In the previous sections, we have studied various types of errors, their origins and the ways to minimize them. One source of limitation, however, can never be argued away: our accuracy is limited to the least count of the instrument used during the measurement.

Least count is the smallest measurement that can be made using the given instrument. For example, with the usual metre scale, one can measure 0.1 cm as the least value. Hence its least count is 0.1 cm.

NoteDefinition

(The least count of an instrument is the smallest measurement that can be made with that instrument.)

How Many Digits Are We Entitled To?

Suppose we measure the length of a metal rod using a metre scale of least count 0.1 cm. The measurement is done three times and the readings are 15.4, 15.4, and 15.5 cm. The most probable length, which is the arithmetic mean as per our earlier discussion, is 15.43.

Now look hard at that number. Out of it, we are certain about the digits 1 and 5, but we are not certain about the last 2 digits because of the least count limitation. The scale itself could not resolve beyond 0.1 cm, so the 4 was already an estimate, and the 3 that appeared from the averaging is beyond any claim we can honestly make.

This leads to the definition that governs how a measured value must be written.

NoteDefinition

(The number of digits in a measurement about which we are certain, plus one additional digit — the first one about which we are not certain — is known as significant figures or significant digits.)

Thus in the above example, we have 3 significant digits: 1, 5 and 4.

Note what is being permitted and what is being forbidden. We are allowed one doubtful digit — the first estimated one — because it still carries genuine information. We are not allowed a second, because it carries none. Writing 15.43 claims a precision the metre scale never possessed.

The larger the number of significant figures obtained in a measurement, the greater is the accuracy of the measurement. If one uses an instrument of smaller least count, the number of significant digits increases. Significant figures are therefore not a rule of arithmetic; they are an honest statement about the quality of the instrument that produced the number.

Q. A calculator displays the mean of three readings as 15.43333333 cm. How many of those digits may you write down?

A. Only three: 15.4 cm. The calculator does not know what instrument you used, so it happily manufactures digits out of arithmetic. Precision cannot be created by division; it can only come from the measurement itself. Carrying extra digits into a final answer is treated as an error in board examinations, and rightly so — it is a false claim.

Rules for determining significant figures

1) All the nonzero digits are significant. For example, if the volume of an object is 178.43 cm\(^{3}\), there are five significant digits, which are 1, 7, 8, 4 and 3.

2) All the zeros between two nonzero digits are significant. For example, m = 165.02 g has 5 significant digits. Such a zero is a genuine measured value — it says the instrument read zero in that place, and that is real information.

3) If the number is less than 1, the zero or zeroes on the right of the decimal point and to the left of the first nonzero digit are not significant. For example, in 0.001405, the underlined zeros are not significant. Thus the above number has four significant digits.

These leading zeros are only placeholders that fix the position of the decimal point. Writing the same length as \(1.405 \times 10^{-3}\) makes this obvious: the zeros vanish, but nothing has been lost.

4) The zeros on the right hand side of the last nonzero number are significant — but for this, the number must be written with a decimal point. For example, 1.500 or 0.01500 have 4 significant figures each.

A trailing zero after the decimal point is a deliberate statement. By writing 1.500 rather than 1.5, the experimenter is declaring that the instrument was good enough to establish those last two zeros.

On the contrary, if a measurement yields a length L given as

\[L = 125\ \text{m} = 12500\ \text{cm} = 125000\ \text{mm},\]

it has only three significant digits.

This is the ambiguity that the whole system stumbles on. The zeros in 125000 mm arrived only because the unit was changed; they were never measured. Yet the way the number is written, no one can tell whether they were measured or not — and the quality of a measurement obviously cannot depend on whether we chose to report it in metre or in millimetre.

Scientific Notation and Order of Magnitude

To avoid the ambiguities in determining the number of significant figures, it is necessary to report every measurement in scientific notation (i.e., in powers of 10), that is, by using the concept of order of magnitude.

The magnitude of any physical quantity can be expressed as \(A \times 10^{n}\), where ‘A’ is a number such that \(0.5 \le A < 5\), and ‘n’ is an integer called the order of magnitude.

NoteDefinition

(In the representation \(A \times 10^{n}\) with \(0.5 \le A < 5\), the integer n is called the order of magnitude of the physical quantity.)

The advantage is immediate: the digits of \(A\) carry the significant figures, and the power of ten carries the size. The two jobs are separated, and no zero has to do both at once.

(i) radius of Earth \(= 6400\) km \[= 0.64 \times 10^{7}\ \text{m}\]

The order of magnitude is 7 and the number of significant figures is 2.

Note why the mantissa is written as 0.64 and not 6.4 — the definition here requires \(A\) to lie between 0.5 and 5, and 6.4 lies outside that band, so one more power of ten is moved into the exponent.

(ii) Magnitude of the charge on an electron \[e = 1.6 \times 10^{-19}\ \text{C}\]

Here the order of magnitude is \(-19\) and the number of significant digits is 2.

Q6. Your calculator shows 5.13333 cm³ for a volume. Why is it wrong to write down all of those digits?

A. Because a calculated result can never be more precise than the measurement it came from. If the diameter was measured with an instrument of limited least count, only a certain number of its digits are trustworthy — the certain ones plus a single doubtful one. Division and multiplication inside a calculator will cheerfully produce ten more digits, but none of those digits was ever measured. Reporting them claims an accuracy the instrument never had, so the answer must be rounded back to the correct number of significant figures.

Q. The diameter of a sphere is measured as 2.14 cm. To how many significant figures should its calculated volume be reported?

A. Three — because the measured diameter has only three significant figures (2, 1 and 4), and no arithmetic performed on it can increase that number. The volume must be rounded to three significant figures, no matter how many digits the calculator displays.

Internet my friend 1. videolectures.net/mit801f99_lewin_lec01/ 2. hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Definitions of SI Units

Till May 20, 2019 the kilogram did not have a definition; it was the mass of the prototype cylinder kept under controlled conditions of temperature and pressure at the SI museum at Paris. Rigorous and meticulous experimentation has shown that the mass of the standard prototype for the kilogram has changed in the course of time. This shows the acute necessity for standardisation of units. The new definitions aim to improve the SI without changing the size of any units, thus ensuring continuity with existing measurements. In November 2018, the 26th General Conference on Weights and Measures (CGPM) unanimously approved these changes, which the International Committee for Weights and Measures (CIPM) had proposed earlier that year. These definitions came in force from 20 May 2019.

  1. As per the new SI units, each of the seven fundamental units (metre, kilogram, etc.) uses one of the following 7 constants, which are proposed to be having exact values as given below:

The Planck constant, \[h = 6.62607015 \times 10^{-34}\ \text{joule-second (J s or kg m}^{2}\ \text{s}^{-1})\]

The elementary charge, \[e = 1.602176634 \times 10^{-19}\ \text{coulomb (C or A s)}\]

The Boltzmann constant, \[k = 1.380649 \times 10^{-23}\ \text{joule per kelvin (J K}^{-1}\ \text{or kg m}^{2}\ \text{s}^{-2}\ \text{K}^{-1})\]

The Avogadro constant (number), \[N_{A} = 6.02214076 \times 10^{23}\ \text{reciprocal mole (mol}^{-1})\]

The speed of light in vacuum, \[c = 299792458\ \text{metre per second (m s}^{-1})\]

The ground state hyperfine structure transition frequency of Caesium-133 atom, \[\Delta \nu_{Cs} = 9192631770\ \text{hertz (Hz or s}^{-1})\]

The luminous efficacy of monochromatic radiation of frequency \(540 \times 10^{12}\) Hz, \[K_{cd} = 683\ \text{lumen per watt (lm}\cdot\text{W}^{-1}) = 683\ \text{cd sr s}^{3}\ \text{kg}^{-1}\ \text{m}^{-2},\] where sr is steradian, the SI unit of solid angle.

  1. Definitions of the units second and mole are based only upon their respective constants. For example, (a) the second uses the ground state hyperfine structure transition frequency of the Caesium-133 atom to be exactly 9192631770 hertz. Thus, the second is defined as 9192631770 periods of that transition. (b) The mole uses Avogadro’s number to be \(N_{A} = 6.02214076 \times 10^{23}\). Thus, one mole is that amount of substance which contains exactly \(6.02214076 \times 10^{23}\) molecules.

  2. Definitions of all the other fundamental units use one constant each and at least one other fundamental unit. For example, the metre makes use of the speed of light in vacuum as a constant and the second as a fundamental unit. Thus, the metre is defined as the distance travelled by light in vacuum in exactly 1/299792458 second.

  3. The figures show the dependency of various units upon their respective constants and other units (wherever used). The arrows arriving at a unit refer to the constant and the fundamental unit (or units, wherever used) used for defining that unit. The arrows going away from a unit indicate other units which use this unit for their definition.

For example, as described above, in Fig. 1.20, (i) the arrows directed to the metre are from the second and \(c\). The arrows going away from the metre indicate that the metre is used in defining the kilogram, the candela and the kelvin. (ii) The newly defined unit kilogram uses the Planck constant, the metre and the second, while the kilogram itself is used in defining the kelvin and the candela. This definition relates the kilogram to the equivalent mass of the energy of a photon given its frequency, via the Planck constant.

Fig. 1.20 refers to the new definitions, while Fig. 1.21 refers to the corresponding definitions before 20 May 2019. Interested students may compare them to know which definitions are modified and how.

Diagram

Fig. 1.20 – New SI: a network of the seven base units (s, m, kg, A, K, mol, cd) with arrows from the defining constants Δν_Cs, c, h, e, k_B, N_A and K_cd.

Diagram

Fig. 1.21 – Old SI: the corresponding network of base units before 20 May 2019, with arrows from Δν_Cs, c, M_IPK, μ₀, T_TPW, m(¹²C) and K_cd.

Write down the number of significant figures in each of the following, and express each in scientific notation, stating the order of magnitude: (i) 0.00302 m (ii) 4.700 kg (iii) 1200 s (iv) 6.626 × 10⁻³⁴ J s.

1.10 Summary

Units and Systems of Units
  1. Physics is a quantitative science. Every measurement is a comparison of a quantity with an internationally accepted standard called a unit, and a measured value is always written as a number followed by its unit.
  2. The systems of units used earlier are CGS (centimetre, gram, second), MKS (metre, kilogram, second) and FPS (foot, pound, second). In 1971 the 14th International General Conference on Weights and Measures recommended the SI (System International), which uses the decimal system and so makes conversion simple.
Fundamental, Derived and Supplementary Units
  1. Fundamental quantities do not depend on any other physical quantity for their measurement. There are seven: length (metre, m), mass (kilogram, kg), time (second, s), temperature (kelvin, K), electric current (ampere, A), luminous intensity (candela, cd) and amount of substance (mole, mol).
  2. Derived quantities depend on some or all of the seven fundamental quantities, and their units are derived units, e.g. velocity in m s\(^{-1}\) and momentum in kg m s\(^{-1}\).
  3. Two supplementary units exist:
    1. Plane angle, \(d\theta = \dfrac{ds}{r}\), measured in radian (rad).
    2. Solid angle, \(d\Omega = \dfrac{dA}{r^{2}}\), measured in steradian (sr). The whole sphere subtends \(\Omega = \dfrac{4\pi r^{2}}{r^{2}} = 4\pi\) sr at its centre.
  4. Key angle conversions: \(\pi^{c} = 180^{\circ}\); \(1\) radian \(= 57.297^{\circ}\); \(1^{\circ} = 1.745 \times 10^{-2}\) rad; \(1' = 2.91 \times 10^{-4}\) rad; \(1'' = 4.847 \times 10^{-6}\) rad.
Conventions for SI Units
  1. Units are written by their symbols; full names begin with a small letter (newton), but symbols of units named after persons are capitalised (N, J).
  2. Symbols take no plural and no full stop: 20 N, 25 kg.
  3. Numerator and denominator are written as one ratio: m s\(^{-2}\), not m/s/s.
  4. Do not mix names and symbols: J/kg K is correct, joule/kg K is not.
  5. Prefixes stand for powers of ten: 1 ms \(= 10^{-3}\) s, 1 µs \(= 10^{-6}\) s, 1 ns \(= 10^{-9}\) s. Double prefixes are avoided.
  6. A space or hyphen shows multiplication of units: m s\(^{-1}\), not ms\(^{-1}\) (ms means millisecond).
Measurement of Length, Mass and Time
  1. Metre: the length of the path travelled by light in vacuum during a time interval of 1/299792458 second, the speed of light being fixed as \(c = 299792458\) m/s.
  2. Parallax is the apparent change in the position of an object due to a change in the position of the observer. For a basis \(b\) and parallax angle \(\theta\) (in radian), \[D = \frac{b}{\theta} \qquad \text{--- (1.1)}\]
  3. For stars, the parallax is measured between two farthest points on the Earth’s orbit, 2 AU apart.
  4. The angular diameter \(\alpha\) of a planet of diameter \(d\) at distance \(D\) gives \[d = \alpha D \qquad \text{--- (1.2)}\]
  5. Astronomical units of distance: \[1\ \text{AU} = 1.496 \times 10^{11}\ \text{m}, \quad 1\ \text{light year} = 9.467 \times 10^{15}\ \text{m},\] \[1\ \text{parsec (pc)} = 3.08 \times 10^{16}\ \text{m} \cong 3.26\ \text{light years}\]
  6. Very small distances use the fermi (\(1\ \text{F} = 10^{-15}\) m) and the Angstrom (\(1\ \overset{\circ}{\text{A}} = 10^{-10}\) m). Atomic sizes (\(\approx 1\ \overset{\circ}{\text{A}}\)) are measured with an electron microscope, whose electron wavelength is about \(0.6\ \overset{\circ}{\text{A}}\).
  7. Kilogram: from 1889 it was the mass of a platinum–iridium cylinder; since 20 May 2019 it is defined through the magnitude of electric current needed in an electromagnet to balance one standard kilogram.
  8. For atoms, \(1\) amu \(= (1/12)^{\text{th}}\) of the mass of an unexcited C\(^{12}\) atom \(= 1.6605402 \times 10^{-27}\) kg.
  9. Second: earlier 1/86400 of a mean solar day (24 × 60 × 60 = 86400 s); now the time needed for 9,192,631,770 vibrations of the radiation emitted in a transition between two hyperfine states of the Cs\(^{133}\) atom.
Dimensions and Dimensional Analysis
  1. The dimensions of a physical quantity are the powers to which the concerned fundamental units must be raised to obtain the unit of that quantity. The dimensional formula is written in square brackets with no commas, e.g. velocity \(= [L^{1}M^{0}T^{-1}]\), velocity gradient \(= [L^{0}M^{0}T^{-1}]\), charge \(= [L^{0}M^{0}T^{1}I^{1}]\).
  2. Common dimensional formulae: force \([L^{1}M^{1}T^{-2}]\), momentum and impulse \([L^{1}M^{1}T^{-1}]\), work and kinetic energy \([L^{2}M^{1}T^{-2}]\), pressure \([L^{-1}M^{1}T^{-2}]\), density \([L^{-3}M^{1}T^{\circ}]\), acceleration \([L^{1}M^{\circ}T^{-2}]\).
  3. Uses of dimensional analysis:
    1. To check the correctness of an equation, using the principle of homogeneity of dimensions — all terms on both sides must have the same dimensions (e.g. \(v = u + at\)).
    2. To establish a relation between related quantities, e.g. for a simple pendulum \(T = k\sqrt{l/g}\), with \(k = 2\pi\) found experimentally, so that \(T = 2\pi\sqrt{l/g}\).
    3. To find the conversion factor between two systems, e.g. \(1\ \text{joule} = 10^{7}\ \text{erg}\).
  4. Limitations: dimensionless constants can be found only by experiment; relations with trigonometric, exponential or logarithmic functions cannot be derived; the method fails when the constant of proportionality has dimensions (e.g. G in \(F = G\,m_{1}m_{2}/r^{2}\)); and missing terms of the same dimension go undetected (e.g. \(S = \frac{1}{2}at^{2}\) passes the test although the complete equation is \(S = ut + \frac{1}{2}at^{2}\)).
Accuracy, Precision and Errors
  1. Accuracy is closeness of a measurement to the true value; precision is closeness of repeated measurements to one another. A measurement should be both.
  2. Uncertainty arises from the quality of the instrument, the skill of the observer, the method used, and external or internal factors.
  3. Systematic errors act in one direction and arise from instrumental error (faulty zero/calibration), imperfection in experimental technique, and personal error. They are removed by better instruments and procedures, not by averaging.
  4. Random errors may be positive or negative and arise from varying conditions. They are minimized by taking repeated observations and their mean.
Estimation and Combination of Errors
  1. Most probable value: \[a_{mean} = \frac{1}{n}\sum_{i=1}^{n} a_{i} \qquad \text{--- (1.3)}\]
  2. Absolute error of a reading: \(\Delta a_{i} = |\ a_{mean} - a_{i}\ |\)
  3. Mean absolute error: \[\Delta a_{mean} = \frac{1}{n}\sum_{i=1}^{n}\Delta a_{i} \qquad \text{--- (1.4)}\]
  4. The result is reported as \(a = a_{mean} \pm \Delta a_{mean}\).
  5. Relative error and percentage error: \[\text{Relative error} = \frac{\Delta a_{mean}}{a_{mean}} \qquad \text{--- (1.5)}\] \[\text{Percentage error} = \frac{\Delta a_{mean}}{a_{mean}} \times 100 \qquad \text{--- (1.6)}\]
  6. Sum or difference: the maximum absolute error is the sum of the absolute errors, \(\Delta Z = \Delta A + \Delta B\) — errors add even for a difference.
  7. Product or division: the maximum relative error is the sum of the relative errors, \[\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \qquad \text{--- (1.7)}\]
  8. Powers: if \(Z = A^{n}\), \[\frac{\Delta Z}{Z} = n\ \frac{\Delta A}{A} \qquad \text{--- (1.8)}\]
  9. In general, if \(Z = \dfrac{A^{p}B^{q}}{C^{r}}\), \[\frac{\Delta Z}{Z} = p\ \frac{\Delta A}{A} + q\ \frac{\Delta B}{B} + r\ \frac{\Delta C}{C} \qquad \text{--- (1.9)}\]
  10. The quantity carrying the largest power contributes the largest error.
Significant Figures and Order of Magnitude
  1. Least count is the smallest measurement possible with an instrument (0.1 cm for a metre scale).
  2. Significant figures are the digits about which we are certain, plus one additional doubtful digit.
  3. Rules: all nonzero digits are significant; zeros between nonzero digits are significant; for numbers less than 1, zeros to the right of the decimal point and to the left of the first nonzero digit are not significant (0.001405 has four); zeros to the right of the last nonzero digit are significant only if a decimal point is written (1.500 and 0.01500 have four each). \(L = 125\) m \(= 12500\) cm \(= 125000\) mm still has only three.
  4. Every measurement should be reported in scientific notation as \(A \times 10^{n}\), where \(0.5 \le A < 5\) and the integer \(n\) is the order of magnitude. For the radius of the Earth, \(6400\) km \(= 0.64 \times 10^{7}\) m — order of magnitude 7, two significant figures. For the electronic charge, \(e = 1.6 \times 10^{-19}\) C — order of magnitude \(-19\), two significant figures.

1.11 Check Your Concepts

  1. Why is it said that a measurement is essentially a comparison? Explain with the example of finding the mass of a fruit on a beam balance.

  2. A student reports the length of a rod simply as “12.5”. Explain why this statement carries no physical information at all.

  3. Why did the international community finally adopt the SI in preference to the CGS, MKS and FPS systems? State the single feature of SI that makes conversion within it easy.

  4. Temperature can be explained in terms of the motion of molecules. Why, then, is it still classified as a fundamental quantity and not as a derived one?

  5. The newton is the SI unit of force, and force is not one of the seven fundamental quantities. Is the newton a fundamental unit or a derived unit? Justify your answer by writing the unit in terms of base units.

  6. Why are the plane angle and the solid angle placed in a separate category called supplementary units rather than among the seven fundamental units?

  7. A student writes the speed of sound as 340 ms\(^{-1}\). What has gone wrong with this notation, and how should it be corrected? Why does this convention matter?

  8. Why can a Vernier calliper or a screw gauge not be used to measure the size of an atom? What property of the measuring “probe” decides the smallest size that can be resolved?

  9. What happens to the parallax angle of a star as its distance from the Earth increases? Why is the Earth’s orbit, and not the Earth’s diameter, used as the basis while measuring stellar distances?

  10. Both a coin held at arm’s length and the Moon subtend nearly the same angle at the eye. What does this tell you about the information contained in an angular diameter, and why must the distance be known independently before a size can be found?

  11. The kilogram was defined for over a century by a platinum–iridium cylinder. Explain why the slow contamination of that cylinder was a far more serious problem than it might appear at first sight.

  12. Why was the mean solar day abandoned as the standard for the second, and why is a caesium atom considered a better clock than the rotating Earth?

  13. “A dimensionally correct equation need not be physically correct, but a dimensionally incorrect equation is certainly wrong.” Justify this statement with one example of each kind.

  14. Dimensional analysis gives the period of a simple pendulum as \(T = k\sqrt{l/g}\), but cannot supply the value of \(k\). Why not? What does this failure tell you about the nature of dimensionless constants?

  15. Explain why the relation \(F = G\,m_{1}m_{2}/r^{2}\) cannot be derived using dimensional analysis, even though it is a perfectly correct physical law.

  16. State whether the following can be derived by dimensional analysis, giving a reason in each case: (a) the displacement of a particle in simple harmonic motion, \(y = a\sin(\omega t)\); (b) the relation between the velocity of a wave on a string, its tension and its mass per unit length.

  17. An ammeter reads 0.5 A even when it is disconnected from the circuit. A student takes fifty readings with it and carefully averages them. Has the averaging improved the result? Explain, and classify the error involved.

  18. Distinguish between accuracy and precision. Can a set of measurements be highly precise and yet highly inaccurate? Illustrate your answer with a laboratory example.

  19. Why are random errors reduced by taking the mean of many readings, while systematic errors are not? What must be done instead to remove a systematic error?

  20. Why are the absolute errors added — and never subtracted — when two measured quantities are subtracted? Explain why finding the difference of two nearly equal measured quantities is considered a poor experimental design.

  21. In the determination of \(g\) using a simple pendulum, \(g = 4\pi^{2}l/T^{2}\). Which measurement, that of \(l\) or that of \(T\), must be made with greater care, and why?

  22. Explain why the relative error, and not the absolute error, is used to compare the quality of two measurements of different quantities.

  23. A calculator displays the mean of three readings taken with a metre scale as 15.43333 cm. How many of these digits may be reported, and why is writing all of them scientifically dishonest?

  24. A length is recorded as \(L = 125\) m \(= 12500\) cm \(= 125000\) mm. Explain the ambiguity this creates in counting significant figures, and how scientific notation removes it.

  25. Why is a trailing zero significant in 1.500 but not in 1500? What is the writer of “1.500” claiming about the instrument used?

  26. What are the dimensions of the quantity \(l\sqrt{l/g}\), \(l\) being the length and \(g\) the acceleration due to gravity?

  27. Define absolute error, mean absolute error, relative error and percentage error.

  28. Describe what is meant by significant figures and order of magnitude.

  29. If the measured values of two quantities are \(A \pm \Delta A\) and \(B \pm \Delta B\), \(\Delta A\) and \(\Delta B\) being the mean absolute errors, what is the maximum possible error in \(A \pm B\)? Show that if \(Z = \dfrac{A}{B}\), then \(\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}\).

  30. Derive the formula for kinetic energy of a particle having mass \(m\) and velocity \(v\) using dimensional analysis.

1.12 Practice with Numericals

  1. Express the following in SI base units: (a) 25 km h\(^{-1}\) (b) 5 g cm\(^{-3}\) (c) 72 µs.

  2. A circular arc of length 0.44 m is drawn on a circle of radius 0.20 m. Calculate the plane angle subtended by the arc at the centre, in radian and in degrees.

  3. A patch of area \(2.4 \times 10^{-2}\) m\(^{2}\) lies on the surface of a sphere of radius 0.40 m. Calculate the solid angle it subtends at the centre of the sphere, in steradian.

  4. Two observers separated by a distance of \(8.0 \times 10^{6}\) m on the Earth’s surface observe a planet simultaneously and measure a parallax angle of 1 minute of arc. Calculate the distance of the planet from the Earth.

  5. A star shows a parallax of 0.5 arcsec when observed from two opposite points of the Earth’s orbit, 2 AU apart. Calculate its distance from the Earth in metre and in parsec. (Take 1 AU = \(1.496 \times 10^{11}\) m.)

  6. The angular diameter of the Moon as seen from the Earth is \(9.05 \times 10^{-3}\) rad. If the Moon is at a distance of \(3.84 \times 10^{8}\) m, calculate its diameter.

  7. Convert 4.5 light years into (a) metre (b) astronomical units. (Take 1 light year = \(9.467 \times 10^{15}\) m and 1 AU = \(1.496 \times 10^{11}\) m.)

  8. The mass of one molecule of a compound is 44 amu. Calculate its mass in kilogram, and the number of such molecules present in 1 kg of the compound. (Take 1 amu = \(1.6605402 \times 10^{-27}\) kg.)

  9. Obtain the dimensional formulae of (a) frequency (b) angular velocity (c) power (d) surface tension (force per unit length) (e) coefficient of viscosity, given \(F = \eta A \dfrac{dv}{dx}\).

  10. Check the dimensional correctness of the equation \(S = ut + \dfrac{1}{2}at^{2}\), where the symbols have their usual meanings.

  11. The frequency \(n\) of a stretched wire is assumed to depend on its length \(l\), the tension \(F\) and the mass per unit length \(\mu\). Using dimensional analysis, obtain the relation between them up to a dimensionless constant.

  12. Using dimensional analysis, find the conversion factor between the SI unit of force (newton) and the CGS unit of force (dyne), and hence show how many dyne make one newton.

  13. A distant civilisation uses units in which the unit of mass is 5 kg, the unit of length is 2 m and the unit of time is 4 s. Express 1 joule of energy in their units.

  14. The period of oscillation of a liquid drop is assumed to depend on its density \(\rho\), its radius \(r\) and the surface tension \(S\) (dimensions \([L^{0}M^{1}T^{-2}]\)). Obtain, by dimensional analysis, the form of the relation for the period, up to a dimensionless constant.

  15. Five measurements of the diameter of a wire give 0.51 mm, 0.53 mm, 0.52 mm, 0.55 mm and 0.54 mm. Calculate the most probable value, the mean absolute error, the relative error and the percentage error.

  16. Two lengths are measured as \((4.50 \pm 0.02)\) cm and \((3.20 \pm 0.03)\) cm. Calculate (a) their sum and (b) their difference, each with the maximum absolute error, and find the percentage error in the difference.

  17. The current through a resistor is measured as \((2.5 \pm 0.1)\) A and the potential difference across it as \((10.0 \pm 0.2)\) V. Calculate the resistance and the maximum percentage error in it.

  18. The side of a cube is measured as \((2.00 \pm 0.02)\) cm. Calculate the percentage error in its volume and in its total surface area.

  19. In an experiment, a physical quantity is given by \(Y = \dfrac{a^{2}b^{3}}{c\sqrt{d}}\). If the percentage errors in \(a\), \(b\), \(c\) and \(d\) are 1%, 2%, 3% and 4% respectively, calculate the maximum percentage error in Y.

  20. The mass of a body is \((20.0 \pm 0.2)\) g and its velocity is \((5.0 \pm 0.1)\) m s\(^{-1}\). Calculate the percentage error in the kinetic energy of the body.

  21. State the number of significant figures in each of the following: 2.008 kg, 0.00450 m, 6.400 × 10\(^{3}\) s, 0.0250 A, 1200 m.

  22. The length, breadth and height of a rectangular block are 5.24 cm, 3.1 cm and 2.06 cm. Calculate its volume to the correct number of significant figures.

  23. Express the following in scientific notation and state the order of magnitude of each: (a) the mass of the Earth, \(5.97 \times 10^{24}\) kg (b) the radius of a hydrogen atom, \(5 \times 10^{-11}\) m (c) the speed of light, 299792458 m s\(^{-1}\).

  24. A thin sheet of area 1.0 m\(^{2}\) is made by rolling out a wire of circular cross section of diameter 0.20 mm. Neglecting gaps, estimate the total length of wire used, to the nearest order of magnitude.

  25. The acceleration due to gravity is determined with a simple pendulum of length \(l = (50.0 \pm 0.1)\) cm and time period \(T = (1.42 \pm 0.01)\) s. Calculate the maximum percentage error in the value of \(g\), and state which measurement limits the accuracy of the experiment.

  26. The masses of two bodies are measured to be \(15.7 \pm 0.2\) kg and \(27.3 \pm 0.3\) kg. What is the total mass of the two and the error in it?

  27. The distance travelled by an object in time \((100 \pm 1)\) s is \((5.2 \pm 0.1)\) m. What is the speed and its relative error?

  28. An electron with charge \(e\) enters a uniform magnetic field \(\vec{B}\) with a velocity \(\vec{v}\). The velocity is perpendicular to the magnetic field. The force on the charge \(e\) is given by \(|\vec{F}| = Bev\). Obtain the dimensions of \(\vec{B}\).

  29. A large ball 2 m in radius is made up of a rope of square cross section with edge length 4 mm. Neglecting the air gaps in the ball, what is the total length of the rope to the nearest order of magnitude?

  30. Nuclear radius \(R\) has a dependence on the mass number \((A)\) as \(R = 1.3 \times 10^{-16} A^{1/3}\) m. For a nucleus of mass number \(A = 125\), obtain the order of magnitude of \(R\) expressed in metre.

  31. In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measured value of the length.

  32. Find the percentage error in the kinetic energy of a body having mass \(60.0 \pm 0.3\) g moving with a velocity \(25.0 \pm 0.1\) cm/s.

  33. In Ohm’s experiments, the values of the unknown resistances were found to be 6.12 Ω, 6.09 Ω, 6.22 Ω, 6.15 Ω. Calculate the mean absolute error, relative error and percentage error in these measurements.

  34. An object is falling freely under the gravitational force. Its velocity after travelling a distance \(h\) is \(v\). If \(v\) depends upon gravitational acceleration \(g\) and distance, prove with dimensional analysis that \(v = k\sqrt{gh}\), where \(k\) is a constant.

  35. \(v = at + \dfrac{b}{t + c} + v_{0}\) is a dimensionally valid equation. Obtain the dimensional formula for \(a\), \(b\) and \(c\), where \(v\) is velocity, \(t\) is time and \(v_{0}\) is initial velocity.

  36. 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.

  37. If the length of a cylinder is \(l = (4.00 \pm 0.001)\) cm, radius \(r = (0.0250 \pm 0.001)\) cm and mass \(m = (6.25 \pm 0.01)\) gm, calculate the percentage error in the determination of density.

  38. When the planet Jupiter is at a distance of 824.7 million kilometer from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.

  39. If the formula for a physical quantity is \(X = \dfrac{a^{4}b^{3}}{c^{1/3}d^{1/2}}\), and if the percentage errors in the measurements of \(a\), \(b\), \(c\) and \(d\) are 2%, 3%, 3% and 4% respectively, calculate the percentage error in X.

  40. Write down the number of significant figures in the following: 0.003 m\(^{2}\), 0.1250 gm cm\(^{-2}\), 6.4 × 10\(^{6}\) m, 1.6 × 10\(^{-19}\) C, 9.1 × 10\(^{-31}\) kg.

  41. The diameter of a sphere is 2.14 cm. Calculate the volume of the sphere to the correct number of significant figures.

1.13 Objective Questions

  1. \([L^{1}M^{1}T^{-2}]\) is the dimensional formula for
    1. Velocity
    2. Acceleration
    3. Force
    4. Work
  2. The error in the measurement of the sides of a rectangle is 1%. The error in the measurement of its area is
    1. 1%
    2. (½)%
    3. 2%
    4. None of the above.
  3. Light year is a unit of
    1. Time
    2. Mass
    3. Distance
    4. Luminosity
  4. Dimensions of kinetic energy are the same as that of
    1. Force
    2. Acceleration
    3. Work
    4. Pressure
  5. Which of the following is not a fundamental unit?
    1. cm
    2. kg
    3. centigrade
    4. volt