Units and Measurement

This chapter has four connected strands, and each later strand leans on the one before it. It opens by asking what a "unit" even is, and answers with the internationally agreed SI system of seven base units — now defined through fixed constants of nature rather than physical prototypes. It then turns to the everyday discipline of reporting a measured number correctly: counting significant figures, using scientific notation to remove ambiguity, and following firm rules when such numbers are added, subtracted, multiplied, divided or rounded off. Next it introduces dimensions — the "algebra of quantity" that strips away units and numbers to reveal how a quantity is built from mass, length and time. Finally it shows what that algebra buys you: a fast, powerful check on whether an equation could possibly be correct (the principle of homogeneity), and, remarkably, a method for guessing the shape of a physical law — such as the period of a pendulum — using nothing but dimensional bookkeeping.

The idea of a unit and the SI system

A measurement is always a comparison against a reference. Saying a rope is "5 metres long" states that it is five times as long as the internationally agreed reference length, the metre. Physical quantities split into two kinds: base (or fundamental) quantities, whose units are chosen independently and arbitrarily, and derived quantities, whose units combine base units algebraically. Speed is not given its own independent unit — it is length divided by time, so metres per second is a derived unit. A complete set of base and derived units is a system of units; the CGS, FPS and MKS systems once competed for this role, and today the SI (Système International d'Unités) is the single system accepted worldwide. The deepest change in the modern SI, formalised in 2018, is philosophical as much as technical. Older definitions tied units to physical artefacts — a specific metal bar or cylinder — that could change with time or be destroyed. The 2018 SI instead fixes the exact numerical value of a fundamental constant of nature for each base unit, so the unit is reproducible anywhere, by anyone with the right apparatus.

Two further units are recognised for angles but are not counted among the seven base units, because they are dimensionless — each is a ratio of a length (or area) to a length (or area) of the same kind, so the units cancel and no new dimension is introduced:

where: $ds$ is a small arc length, $dA$ a small patch of spherical surface, and $r$ the radius to the apex in each case — you will meet this "dimensionless because it is a ratio" idea again when dimensional formulae are built below. The numerical values in the table above need not be memorised; they only show how precisely each unit is now fixed, and are periodically refined as measurement technology improves.

Significant figures: how many digits can you trust

Every real measurement carries some uncertainty, because no instrument reads with infinite precision. The digits you report should say so honestly: the significant figures of a measurement are all the digits known with certainty, plus the one additional digit that is estimated or uncertain. A pendulum period reported as $1.62\ \text{s}$ has three significant figures — the $1$ and $6$ are reliable, the final $2$ is the uncertain estimate. Writing more digits than this would falsely claim a precision the instrument never delivered; writing fewer would throw away information the instrument actually gave you. Counting significant figures follows a small set of rules, all reducing to one idea: switching units must never change how many significant figures a measurement has. All non-zero digits are significant. Zeros sandwiched between two non-zero digits are always significant, whatever the position of the decimal point — so $123\ \text{m} = 12300\ \text{cm}$ conceals a subtlety about trailing zeros, but the digits $1$, $2$, $3$ remain the significant ones either way. A zero used only to fix the decimal point in a number smaller than one, as in $0.00\underline{2}308$, is never significant — it is scaffolding, not data. Trailing zeros without a decimal point are not significant (the zeros in $12300\ \text{cm}$ above are not), but trailing zeros with a decimal point are always significant, since they can only mean the instrument resolved that digit: $4.700\ \text{m}$ has four significant figures, not two. This last rule is exactly why a plain decimal number can mislead after a unit change: $4.700\ \text{m} = 4700\ \text{mm}$, and read carelessly the trailing zeros of $4700$ look insignificant, seeming to shrink four significant figures to two — a contradiction, since a mere change of units cannot alter precision. Scientific notation removes the ambiguity for good: every measurement can be written as $a \times 10^{b}$, with $1 \le a < 10$; then $4.700\ \text{m} = 4.700 \times 10^{3}\ \text{mm}$, and all four digits of $a$ stay unambiguously significant regardless of the power of ten. The exponent $b$ is the order of magnitude of the quantity. Earth's diameter, about $1.28\times10^{7}\ \text{m}$, is of order $10^{7}\ \text{m}$; a hydrogen atom's diameter, about $1.06\times10^{-10}\ \text{m}$, is of order $10^{-10}\ \text{m}$ — Earth is $17$ orders of magnitude larger, one number conveying an almost incomprehensible span of scale.

Arithmetic with significant figures and rounding off

A calculated result can never be more precise than the least precise measurement that went into it. When quantities are multiplied or divided, the answer should carry as many significant figures as the input with the fewest. Dividing a mass of $4.237\ \text{g}$ (four significant figures) by a volume of $2.51\ \text{cm}^{3}$ (three) gives $1.68804780876\ \text{g/cm}^{3}$ on a calculator, but only three digits are honest, so the density must be reported as $1.69\ \text{g cm}^{-3}$. When quantities are added or subtracted, the rule changes: the result keeps as many decimal places as the least precise input, not as many significant figures. Adding $436.32\ \text{g} + 227.2\ \text{g} + 0.301\ \text{g}$ gives $663.821\ \text{g}$ by raw arithmetic, but $227.2\ \text{g}$ is known only to one decimal place, so the honest answer is $663.8\ \text{g}$. Subtraction can be more dramatic: $12.9\ \text{g} - 7.06\ \text{g}$ must be reported as $5.8\ \text{g}$, not the naive $5.84\ \text{g}$ — subtracting nearly equal numbers can genuinely destroy significant figures.

Rounding off a dropped digit above or below $5$ follows the ordinary convention — round up above $5$, leave unchanged below $5$. The special case is a dropped digit of exactly $5$ with nothing after it: the convention is "round to even," raising the preceding digit only if that makes it even. So $2.745$ rounds to $2.74$ (the preceding $4$ is already even) while $2.735$ also rounds to $2.74$ (the preceding $3$ is odd, so it is raised) — this prevents rounding from statistically biasing results upward over many calculations. In a multi-step calculation, carry one extra digit through the intermediate steps and round only at the end; rounding too early lets small errors compound, exactly as $1/9.58 \to 0.104 \to 1/0.104 = 9.62$ fails to recover the original $9.58$, while keeping $1/9.58 = 0.1044$ before inverting does. Exact numbers in formulae — the $2$ in $r = d/2$, the $n$ in $T = t/n$, or $\pi$ itself — are not measurements and carry an unlimited number of significant figures; $\pi$ may be used as $3.14$ or $3.142$ as required precision demands. The same logic extends to combining uncertainties. If a length $l = 16.2 \pm 0.1\ \text{cm}$ (about $0.6\%$) and a breadth $b = 10.1 \pm 0.1\ \text{cm}$ (about $1\%$) are multiplied, the product's percentage error is approximately the sum of the two:

where: $\Delta l/l$ and $\Delta b/b$ are the relative (fractional) errors in each measured length. This combination-of-errors idea, extended to powers of several variables, is worked in full generality in the third worked example below. Relative error also depends on the number itself, not just the significant figures: $1.02\ \text{g} \pm 0.01\ \text{g}$ has relative error about $1\%$, while $9.89\ \text{g} \pm 0.01\ \text{g}$, an equally precise reading, has relative error only about $0.1\%$ — the same absolute uncertainty means far less against a larger value.

Dimensions and dimensional formulae

Strip a physical quantity of both its numerical value and its choice of unit, and what remains is its dimension — the statement of which base quantities it is built from, and to what powers. In mechanics, every quantity can be expressed using only three dimensions: mass $[\text{M}]$, length $[\text{L}]$ and time $[\text{T}]$ (the square-bracket notation always means "the dimension of," never the quantity's numerical value). Volume, being length times length times length, has dimension $[\text{L}^{3}]$ and is independent of both mass and time, so its full dimensional statement is $[\text{M}^{0}\text{L}^{3}\text{T}^{0}]$. Force, as mass times acceleration, is mass times length divided by time squared, giving the dimensional formula

where: the exponents of M, L and T record exactly how many powers of mass, length and time combine to build the quantity — here, one power of mass, one of length, and $-2$ of time, with the electric-current, temperature, amount-of-substance and luminous-intensity dimensions all appearing to the zero power. An equation formed by setting a quantity equal to its dimensional formula, such as $[F] = [\text{M}\,\text{L}\,\text{T}^{-2}]$, is called its dimensional equation. The dimension of a quantity captures its type, not its magnitude: a change in velocity, an initial velocity, an average velocity and a plain speed are all physically distinct, but they share exactly the same dimension, $[\text{L}\,\text{T}^{-1}]$, because dimensionally the algebra of "type" does not see those distinctions.

Dimensional analysis: checking equations and deriving relations

Dimensions earn their keep through the principle of homogeneity: quantities can be added to, subtracted from, or set equal to one another only if they share the same dimension — you can no more add a velocity to a force than three apples to two hours. This gives a fast, purely algebraic test of whether a proposed equation is even possible: reduce every term to its dimensional formula, and if they do not all match, the equation is certainly wrong. If they do match, the equation has passed a necessary but not sufficient test — dimensional consistency says nothing about dimensionless numerical factors, so a dimensionally correct equation need not be the physically correct one, though a dimensionally inconsistent equation is guaranteed wrong. Arguments of $\sin$, $\log$ and $e^{x}$ must always be dimensionless, since they are ratios or infinite sums that only make algebraic sense that way. Beyond checking equations, dimensions can deduce the form of a relationship, provided you know which quantities a result depends on (up to about three) and assume the dependence is a simple product of powers. Writing the unknown relation as a product, matching dimensional formulae on both sides, and equating the exponents of M, L and T one base quantity at a time turns a physics question into simultaneous algebra — the method used below for the simple pendulum. What it can never deliver is a pure number: dimensionless constants such as $2\pi$ or $\tfrac{1}{2}$ carry no dimension, so no amount of bookkeeping can pin down their value — that comes from the actual physics, not from dimensions alone.

Board level - Significant figures in a multi-part calculation

Each side of a cube is measured to be $7.203\ \text{m}$. Find the total surface area and the volume of the cube to appropriate significant figures.

Step 1. Count the significant figures in the given data: $7.203\ \text{m}$ has four significant figures (7, 2, 0, 3), so both the surface area and the volume must be reported to four significant figures — no more, no fewer.

Step 2. Compute the surface area of a cube, six times the square of a side: $6 \times (7.203)^{2} = 6 \times 51.883209 = 311.299254\ \text{m}^{2}$. Multiplication and division retain the least number of significant figures among the inputs, so this rounds to four significant figures.

Step 3. Compute the volume, the cube of a side: $(7.203)^{3} = 373.714754\ \text{m}^{3}$, which likewise must be rounded to four significant figures. Answer: surface area $= 311.3\ \text{m}^{2}$ and volume $= 373.7\ \text{m}^{3}$.

Board level - Dimensional consistency and ruling out formulas

Check whether $\tfrac{1}{2}mv^{2} = mgh$ is dimensionally consistent, where $m$ is mass, $v$ is speed, $g$ is the acceleration due to gravity and $h$ is height. Then use dimensions to identify which of the following proposed formulas for kinetic energy $K$ cannot be correct: (a) $K = m^{2}v^{3}$, (b) $K = \tfrac{1}{2}mv^{2}$, (c) $K = ma$, (d) $K = \tfrac{3}{16}mv^{2}$, (e) $K = \tfrac{1}{2}mv^{2} + ma$.

Step 1. Find the dimension of the left side, $\tfrac{1}{2}mv^{2}$: $[\text{M}][\text{L}\,\text{T}^{-1}]^{2} = [\text{M}\,\text{L}^{2}\,\text{T}^{-2}]$, since the dimensionless $\tfrac{1}{2}$ contributes nothing.

Step 2. Find the dimension of the right side, $mgh$: $[\text{M}][\text{L}\,\text{T}^{-2}][\text{L}] = [\text{M}\,\text{L}^{2}\,\text{T}^{-2}]$. Both sides match exactly, so the equation is dimensionally consistent — a genuine physical result you will re-derive from energy conservation later in the course.

Step 3. Test each candidate against dimension $[\text{M}\,\text{L}^{2}\,\text{T}^{-2}]$: (a) $m^{2}v^{3}$ gives $[\text{M}^{2}\text{L}^{3}\text{T}^{-3}]$ — ruled out. (c) $ma$ gives $[\text{M}\,\text{L}\,\text{T}^{-2}]$, the dimension of force, not energy — ruled out. (e) adds $\tfrac{1}{2}mv^{2}$ to $ma$, two different dimensions, which can never be added — ruled out on that basis alone.

Step 4. Both (b) $\tfrac{1}{2}mv^{2}$ and (d) $\tfrac{3}{16}mv^{2}$ have the correct dimension, and dimensional analysis, blind to numerical factors, cannot distinguish them. Answer: (a), (c) and (e) are ruled out by dimensions; only the actual definition of kinetic energy (met later) identifies (b), $K = \tfrac{1}{2}mv^{2}$, as correct.

Competitive level - Percentage error in a quantity built from several measurements (JEE Main pattern)

A physical quantity $P$ is related to four measured quantities $a$, $b$, $c$ and $d$ by $P = \dfrac{a^{3}b^{2}}{\sqrt{c}\,d}$. The percentage errors in the measurement of $a$, $b$, $c$ and $d$ are $1\%$, $2\%$, $3\%$ and $4\%$ respectively. Find the maximum percentage error in $P$.

Step 1. Extend the combination-of-errors rule (illustrated in the text for a simple product of two lengths) to a general power-law relation: relative errors of a product or quotient of powers add, each weighted by the magnitude of its own exponent — taking logarithms turns the power law into a sum, and errors in a sum simply add.

Step 2. Read off the exponents in $P = a^{3}b^{2}c^{-1/2}d^{-1}$: $3$ for $a$, $2$ for $b$, $-\tfrac{1}{2}$ for $c$, $-1$ for $d$. Only the magnitude of each exponent matters, since the worst-case error always adds.

Step 3. Multiply each percentage error by the magnitude of its exponent and sum: $\dfrac{\Delta P}{P}\times100\% = 3(1\%) + 2(2\%) + \tfrac{1}{2}(3\%) + 1(4\%) = 3\% + 4\% + 1.5\% + 4\% = 12.5\%$. Answer: the maximum percentage error in $P$ is $12.5\%$. Timing note: under ninety seconds once the exponents are written down. The only physics is recognising that $\sqrt{c}$ contributes exponent $\tfrac{1}{2}$ and that a denominator quantity still contributes its exponent's magnitude, never a negative percentage — forgetting the square root's half, or dropping $d$ entirely, costs the full mark in a JEE-style question.

Miscounting significant figures across a unit change. Recounting digits after converting units often gives a wrong answer with trailing zeros in whole numbers — $4.700\ \text{m}$ as $4700\ \text{mm}$ looks like it lost significant figures. It has not: significant figures belong to the measurement, not the unit, and scientific notation is the safe check.

Using the multiplication rule for addition, or vice versa. Multiplication and division preserve significant figures; addition and subtraction preserve decimal places. Mixing these up is the single most common exam error here — always ask which operation you are doing before applying a rule.

Treating "dimensionally correct" as "physically correct." A dimensionally consistent equation has only passed a necessary test; it says nothing about numerical constants like $2\pi$ or $\tfrac{1}{2}$, and cannot distinguish between two formulas that happen to share a dimension.

Trying to find dimensionless constants by dimensional analysis. The method that deduces $T = k\sqrt{l/g}$ can never determine $k$; that needs the actual equation of motion. Do not present $T = 2\pi\sqrt{l/g}$ as "proved" by dimensions alone — only $\sqrt{l/g}$ is.

Forgetting that angles and ratios are dimensionless. Radian, steradian, refractive index and strain all carry zero net dimension, being ratios of like quantities. Wrongly assigning them a dimension wrecks a homogeneity check involving trigonometric or exponential functions, whose arguments must always be dimensionless.

Rounding intermediate steps too early. Rounding a mid-calculation result before continuing can compound errors — keep one extra digit through intermediate steps and round only the final answer.

• State the seven SI base quantities and their units, and explain in one sentence how each is now fixed by a constant of nature.
• Explain why radian and steradian are dimensionless even though they have named units.
• Count the significant figures in any given decimal or scientific-notation number, including tricky trailing-zero cases.
• Apply the correct rule (least significant figures for multiplication and division, least decimal places for addition and subtraction) when combining measured quantities.
• Round a number correctly, including the round-to-even convention for a dropped digit of exactly five.
• Convert a number to scientific notation and read off its order of magnitude.
• Write the dimensional formula of a derived quantity such as force, energy, pressure or density starting from its defining relation.
• Use the principle of homogeneity to check whether an equation is dimensionally consistent, and state clearly what such a check does and does not prove.
• Use dimensional analysis to deduce the form of a simple physical relation among up to three variables, and explain why it cannot fix a dimensionless constant.
• Convert a physical quantity's numerical value between two systems of units using its dimensional formula.

Ask the tutor: if you are unsure whether a formula you have memorised is dimensionally possible, or want to practise the significant-figure rules on your own measurements, ask the tutor to walk through it with you step by step.