---
title: "Units, Dimensions, and Measurement"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 1
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement
---

# Chapter 1 — Units, Dimensions, and Measurement

In a basement laboratory a pendulum swings, a stopwatch clicks, and a student writes $g = 9.77\,\mathrm{m}/\mathrm{s}^{2}$. Is that the “right” value? The textbook says $9.81$. Whether the two numbers agree, whether the experiment was well designed, which of its steps limited it — none of this can be decided from the digits alone. Physics measures, and a measurement without its uncertainty is a rumor. This chapter sets up the common language of every chapter to come: [units](#def-b1-units-dimensions-unit), [dimensions](#def-b1-units-dimensions-dimension), orders of magnitude, and the honest arithmetic of uncertainties.

![The NIST-4 Kibble balance, one of the instruments that now realize the kilogram from Planck’s constant: a mass is weighed against an electromagnetic force measured in electrical units. Photograph: J. L. Lee, NIST (public domain).](https://one-course.com/images/onecourse/chapters/physics-3/b1-units-dimensions/img-4f0aff410783.jpg)

*The NIST-4 Kibble balance, one of the instruments that now realize the kilogram from Planck’s constant: a mass is weighed against an electromagnetic force measured in electrical [units](#def-b1-units-dimensions-unit). Photograph: J. L. Lee, NIST (public domain).*

## 1.1 The International System of Units

**Definition 1.1 (Physical quantity and unit).**

A *physical quantity* is a property that can be measured: a length, a duration, a mass, a current. Measuring it means comparing it with a reference quantity of the same kind, the *unit*: the result is a number times a unit, $\ell = 1.257\,\mathrm{m}$. The number alone means nothing; the unit alone measures nothing.

**Definition 1.2 (Base units of the SI).**

The *International System of [Units](#def-b1-units-dimensions-unit)* (SI) rests on seven *base [units](#def-b1-units-dimensions-unit)*: the second ($\mathrm{s}$), the meter ($\mathrm{m}$), the kilogram ($\mathrm{kg}$), the ampere ($\mathrm{A}$), the kelvin ($\mathrm{K}$), the mole ($\mathrm{mol}$) and the candela ($\mathrm{cd}$). Since 2019 each is defined by fixing the numerical value of a constant of nature: the cesium hyperfine frequency $\Delta\nu_{\mathrm{Cs}} = 9\,192\,631\,770\,\mathrm{Hz}$ defines the second; the speed of light $c = 299\,792\,458\,\mathrm{m}/\mathrm{s}$ then defines the meter; the Planck constant $h = 6.626\,070\,15 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$ defines the kilogram; the elementary charge $e = 1.602\,176\,634 \times 10^{-19}\,\mathrm{C}$ the ampere; the Boltzmann constant $k_B = 1.380\,649 \times 10^{-23}\,\mathrm{J}/\mathrm{K}$ the kelvin; the Avogadro constant $N_A = 6.022\,140\,76 \times 10^{23}\,\mathrm{mol}^{-1}$ the mole; and a fixed luminous efficacy $K_{\mathrm{cd}}$ the candela. Every other [unit](#def-b1-units-dimensions-unit) is a *derived unit*, a product of powers of base [units](#def-b1-units-dimensions-unit): $1\,\mathrm{N} = 1\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}^{2}$, $1\,\mathrm{J} = 1\,\mathrm{N}\,\mathrm{m}$, $1\,\mathrm{W} = 1\,\mathrm{J}/\mathrm{s}$, $1\,\mathrm{V} = 1\,\mathrm{W}/\mathrm{A}$, $1\,\mathrm{Pa} = 1\,\mathrm{N}/\mathrm{m}^{2}$.

![The seven base units (outer ring) and the seven defining constants (inner ring). A fixed constant defines its unit only together with units already defined (dashed): c turns seconds into meters, h needs the meter and the second to define the kilogram, e needs the second to define the ampere.](https://one-course.com/images/onecourse/chapters/physics-3/b1-units-dimensions/fig-816ead3c3bb9.svg)

*The seven [base units](#def-b1-units-dimensions-si) (outer ring) and the seven defining constants (inner ring). A fixed constant defines its [unit](#def-b1-units-dimensions-unit) only together with [units](#def-b1-units-dimensions-unit) already defined (dashed): $c$ turns seconds into meters, $h$ needs the meter and the second to define the kilogram, $e$ needs the second to define the ampere.*

**Remark 1.3 (Names and symbols).**

[Units](#def-b1-units-dimensions-unit) named after people are written lowercase in full (newton, joule, pascal) and capitalized as symbols ($\mathrm{N}$, $\mathrm{J}$, $\mathrm{Pa}$); capitalized in full, the word is the person. Prefixes scale by powers of ten, from p ($10^{-12}$) through n, $\mu$, m, k, M, G to T ($10^{12}$); the kilogram is the one [base unit](#def-b1-units-dimensions-si) carrying a prefix in its name.

**Example 1.4 (Unpacking a derived unit).**

The volt: $\mathrm{V} = \mathrm{W}/\mathrm{A} = \mathrm{J}/(\mathrm{A}\,\mathrm{s}) = \mathrm{kg}\,\mathrm{m}^{2}/(\mathrm{A}\,\mathrm{s}^{3})$. The ohm: $\Omega = \mathrm{V}/\mathrm{A} = \mathrm{kg}\,\mathrm{m}^{2}/(\mathrm{A}^{2}\,\mathrm{s}^{3})$. The farad: $\mathrm{F} = \mathrm{C}/\mathrm{V} = \mathrm{A}\,\mathrm{s}/\mathrm{V} = \mathrm{A}^{2}\,\mathrm{s}^{4}/(\mathrm{kg}\,\mathrm{m}^{2})$. Such unpacking is the safest check that a formula has been remembered correctly — the next section makes it systematic.

## 1.2 Dimensional analysis

**Definition 1.5 (Dimension).**

The *dimension* of a quantity $X$, written $[X]$, records how $X$ is built from the seven base quantities, independently of the [units](#def-b1-units-dimensions-unit) chosen: length $L$, mass $M$, time $T$, electric current $I$, temperature $\Theta$, amount of substance $N$, luminous intensity $J$. A speed has dimension $[v] = L\,T^{-1}$, a force $[F] = M\,L\,T^{-2}$, an energy $[E] = M\,L^2\,T^{-2}$. A quantity of dimension $1$ (an angle, a ratio, a refractive index) is *dimensionless*.

**Theorem 1.6 (Principle of dimensional homogeneity).**

In a physical law, the two sides of an equality have the same [dimension](#def-b1-units-dimensions-dimension), and every term of a sum has the [dimension](#def-b1-units-dimensions-dimension) of the whole. The argument of an exponential, a logarithm, a sine or a cosine is [dimensionless](#def-b1-units-dimensions-dimension).

**Proof.** The laws of physics do not depend on the [units](#def-b1-units-dimensions-unit) humans choose. Changing the [unit](#def-b1-units-dimensions-unit) of length by a factor $\lambda$ multiplies every quantity of [dimension](#def-b1-units-dimensions-dimension) $L^a \cdots$ by $\lambda^{-a}$: an equality between terms of different [dimensions](#def-b1-units-dimensions-dimension) would hold in one system of [units](#def-b1-units-dimensions-unit) and fail in another. For a function such as $\exp$, the series $1 + x + x^2/2 + \dots$ adds powers of $x$ of different [dimensions](#def-b1-units-dimensions-dimension) unless $x$ is [dimensionless](#def-b1-units-dimensions-dimension). ∎

**Method 1.7 (Checking a formula).**

1. Write the [dimension](#def-b1-units-dimensions-dimension) of every symbol.
2. Reduce each side (each term) to a product $L^a M^b T^c \cdots$ .
3. Equal exponents on both sides: the formula *may* be right. Unequal: it is certainly wrong — a missing factor $g$ , a squared quantity that should be plain, an exponent with [dimensions](#def-b1-units-dimensions-dimension) .

A homogeneous formula can still be wrong by a [dimensionless](#def-b1-units-dimensions-dimension) factor ($2$, $\pi$, $\tfrac12$): [dimensions](#def-b1-units-dimensions-dimension) never see those.

**Example 1.8 (The period of a pendulum).**

A pendulum of length $\ell$ and mass $m$ swings in gravity $g$. Suppose its period is $\tau = k\,\ell^\alpha m^\beta g^\gamma$ with $k$ [dimensionless](#def-b1-units-dimensions-dimension). Then $T = L^\alpha M^\beta (L T^{-2})^\gamma =
L^{\alpha+\gamma} M^\beta T^{-2\gamma}$, so $\beta = 0$, $\gamma = -\tfrac12$, $\alpha = \tfrac12$:

$$
\tau = k\sqrt{\ell/g} .
$$

Without solving a single equation of motion: the mass drops out, and doubling the length multiplies the period by $\sqrt2$. Dynamics ([Chapter 14](https://one-course.com/books/physics/3/en/chapter/14-mechanical-oscillators-damping-and-resonance#ch-b1-oscillators-resonance)) will supply $k = 2\pi$ for small swings.

**Proposition 1.9 (Dimensionless groups).**

If a quantity $Y$ depends on $n$ quantities $X_1, \dots, X_n$ that involve $r$ independent base [dimensions](#def-b1-units-dimensions-dimension), the relation can be rewritten between $n + 1 - r$ [dimensionless](#def-b1-units-dimensions-dimension) combinations of the $X_i$ and $Y$. When $n + 1 - r = 1$, the single combination must be a constant, and dimensional analysis gives the law up to a numerical factor.

**Proof.** *Admitted at this level.* ∎

**Remark 1.10 (Using the proposition).**

The general statement (the Vaschy–Buckingham theorem) is a piece of linear algebra on the exponent vectors. What matters here is the recipe: list the relevant quantities, count [dimensions](#def-b1-units-dimensions-dimension), form the [dimensionless groups](#prop-b1-units-dimensions-pi). In [Example 1.8](#ex-b1-units-dimensions-pendulum), $n = 3$ ($\ell, m, g$), $r = 3$ ($L, M, T$), so a single group $\tau\sqrt{g/\ell}$ exists and must be constant. Drag on a sphere of radius $R$ moving at speed $v$ in a fluid of density $\rho$ and viscosity $\eta$ involves $n = 4$ quantities and $r = 3$ [dimensions](#def-b1-units-dimensions-dimension): two groups, $F/(\rho v^2 R^2)$ and $\rho v R/\eta$ (the Reynolds number), and [dimensions](#def-b1-units-dimensions-dimension) alone cannot say how the first depends on the second — experiment must.

## 1.3 Orders of magnitude

**Definition 1.11 (Order of magnitude).**

The *order of magnitude* of a quantity is the power of ten nearest to it: $6.4 \times 10^{6}\,\mathrm{m}$ (the Earth’s radius) has order of magnitude $10^7$; $3 \times 10^{3}\,\mathrm{m}$ has order $10^3$; $3.2 \times 10^{3}\,\mathrm{m}$ is closer to $10^3$ than to $10^4$ on a logarithmic scale (the boundary is $\sqrt{10} \approx 3.16$). Two quantities are “of the same order” when their ratio is below about $3$.

**Method 1.12 (Estimating).**

To estimate a quantity nobody has measured for you:

1. break it into factors you can estimate to within a factor of $2$ or $3$ each;
2. estimate each factor, keeping one significant figure;
3. multiply, keeping the power of ten and one digit.

Errors in different factors partly cancel; the result is usually right to within a factor of $3$ to $10$, which is often all one needs to rule a hypothesis in or out.

**Example 1.13 (The mass of the atmosphere).**

Atmospheric pressure $P_0 \approx 1 \times 10^{5}\,\mathrm{Pa}$ is the weight of the air column above one square meter, $P_0 = m_{\text{col}}\,g$, so $m_{\text{col}} \approx 10^5/10 = 1 \times 10^{4}\,\mathrm{kg}$ per square meter. The Earth’s surface is $4\pi R^2 \approx 4 \times 3 \times (6\times10^6)^2
\approx 5 \times 10^{14}\,\mathrm{m}^{2}$; the atmosphere weighs about $5 \times 10^{18}\,\mathrm{kg}$. The accepted value is $5.1 \times 10^{18}\,\mathrm{kg}$ — the estimate needed only $P_0$ and $R$.

**Definition 1.14 (Significant figures).**

The *significant figures* of a written number are its digits from the first non-zero one: $0.0820$ has three, $8.20 \times 10^{-2}$ the same three, $820$ is ambiguous (two or three) unless written $8.2 \times 10^{2}$ or $8.20 \times 10^{2}$. A result is written with the significant figures its uncertainty allows ([Method 1.20](#met-b1-units-dimensions-writing) below): not “$g = 9.7723\,\mathrm{m}/\mathrm{s}^{2}$” when the fourth digit is unknown.

## 1.4 Measurement uncertainty

**Definition 1.15 (Measurand, error, uncertainty).**

The *measurand* is the quantity one intends to measure. Repeating a measurement gives scattered values: the measurement is *variable*. The *error* of one result is its unknown difference to the true value; the *standard uncertainty* $u(x)$ of a result $x$ is the standard deviation of the values one could reasonably attribute to the measurand — a quantity one *can* evaluate. Its sources: the operator (reaction time, parallax), the environment (temperature, vibrations), the instrument (resolution, calibration), the method (a model that is only approximately true).

**Proposition 1.16 (Type A evaluation).**

From $n$ independent repeated values $x_1, \dots, x_n$ of mean $\bar x = \frac1n \sum x_i$ and experimental standard deviation

$$
s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar x)^2},
$$

the best estimate of the [measurand](#def-b1-units-dimensions-uncertainty) is $\bar x$ and its [standard uncertainty](#def-b1-units-dimensions-uncertainty) is

$$
u(\bar x) = \frac{s}{\sqrt n} .
$$

**Partial proof.** That $s$ estimates the spread of a single measurement is the definition of the experimental standard deviation ($n - 1$ rather than $n$ because the mean itself was estimated from the data). That the mean of $n$ independent values scatters $\sqrt n$ times less than one value is the addition of variances: the variance of a sum of $n$ independent variables is $n$ times the variance of one, so the variance of the mean is $s^2 n / n^2 = s^2/n$. The probability course of this year proves both statements. ∎

![Fifty timings of the same pendulum period, binned by 0.01\, s. Their spread is s 0.016\, s (one timing); the mean of the fifty is known to s/√50 0.002\, s. The bell curve is the Gaussian of the same mean and spread.](https://one-course.com/images/onecourse/chapters/physics-3/b1-units-dimensions/fig-505825c9c7fe.svg)

*Fifty timings of the same pendulum period, binned by $0.01\,\mathrm{s}$. Their spread is $s \approx 0.016\,\mathrm{s}$ (one timing); the mean of the fifty is known to $s/\sqrt{50} \approx 0.002\,\mathrm{s}$. The bell curve is the Gaussian of the same mean and spread.*

**Proposition 1.17 (Type B evaluation).**

When a value is read once, its uncertainty is evaluated from what is known of the instrument. If the true value can lie anywhere between $x - a$ and $x + a$ with no preference (a *uniform* distribution of half-width $a$),

$$
u(x) = \frac{a}{\sqrt 3} .
$$

For a scale graduated in steps $\delta$ (half-width $a = \delta/2$), $u = \delta/(2\sqrt3) = \delta/\sqrt{12}$; the same applies to the last digit of a digital display. A manufacturer’s tolerance quoted as “$\pm a$” is treated the same way unless the data sheet says otherwise.

**Proof.** The variance of a uniform distribution on $[x - a, x + a]$ is $\frac{1}{2a}\int_{-a}^{a} t^2\,\dd t = \frac{1}{2a}\cdot\frac{2a^3}{3}
= \frac{a^2}{3}$; its standard deviation is $a/\sqrt3$. ∎

![A reading known only to lie within ± a: uniform density 1/(2a), standard deviation a/√3 0.58\,a — smaller than a, because the extremes are no likelier than the center.](https://one-course.com/images/onecourse/chapters/physics-3/b1-units-dimensions/fig-0a5bb55f1710.svg)

*A reading known only to lie within $\pm a$: uniform density $1/(2a)$, standard deviation $a/\sqrt3 \approx 0.58\,a$ — smaller than $a$, because the extremes are no likelier than the center.*

**Theorem 1.18 (Combined uncertainty).**

Let $y = f(x_1, \dots, x_n)$ be computed from independent measured values with standard uncertainties $u(x_i)$. Then

$$
u(y)^2 = \sum_{i=1}^n \left(\frac{\partial f}{\partial x_i}\right)^2 u(x_i)^2 .
$$

In particular:

- sum or difference $y = x_1 \pm x_2$ : $u(y)^2 = u(x_1)^2 + u(x_2)^2$ ;
- product or quotient $y = x_1 x_2$ or $x_1/x_2$ : $\left(\dfrac{u(y)}{y}\right)^2 = \left(\dfrac{u(x_1)}{x_1}\right)^2  + \left(\dfrac{u(x_2)}{x_2}\right)^2$ ;
- power $y = k\,x^\alpha$ : $\dfrac{u(y)}{\abs y} = \abs\alpha\,\dfrac{u(x)}{\abs x}$ .

**Proof.** For small deviations $\delta x_i$, the first-order Taylor expansion gives $\delta y = \sum_i (\partial f/\partial x_i)\,\delta x_i$. The deviations are independent and of zero mean, so the variance of the sum is the sum of the variances, each scaled by the square of its coefficient — the displayed formula. The special cases follow: $\partial(x_1 x_2)/\partial x_1 = x_2$ gives $u(y)^2 = x_2^2 u(x_1)^2 + x_1^2 u(x_2)^2$, divide by $y^2 = x_1^2 x_2^2$; for $y = kx^\alpha$, $\partial y/\partial x = \alpha y/x$. ∎

**Example 1.19 (The density of a cylinder).**

A cylinder: $m = 47.32\,\mathrm{g}$ ($u = 0.01\,\mathrm{g}$), $d = 12.00\,\mathrm{mm}$ ($u = 0.02\,\mathrm{mm}$), $h = 50.2\,\mathrm{mm}$ ($u = 0.1\,\mathrm{mm}$). $\rho = 4m/(\pi d^2 h)$, so the relative uncertainty squared adds $(0.01/47.32)^2 = 4.5 \times 10^{-8}$, $(2 \times 0.02/12.00)^2 = 1.1 \times 10^{-5}$ and $(0.1/50.2)^2 = 4.0 \times 10^{-6}$: the diameter dominates, because it is squared and measured to $0.17\%$. $u(\rho)/\rho = \sqrt{1.5 \times 10^{-5}}
= 0.39\%$; $\rho = 8334\,\mathrm{kg}/\mathrm{m}^{3}$, so $u(\rho) = 33\,\mathrm{kg}/\mathrm{m}^{3}$: $\rho = (8.33 \pm 0.03)\times10^3\ \mathrm{kg}/\mathrm{m}^{3}$ — brass.

**Method 1.20 (Writing a result).**

1. Round the uncertainty to one significant figure (two if the first is $1$ or $2$ ).
2. Round the value to the same decimal place.
3. Write value and uncertainty with the [unit](#def-b1-units-dimensions-unit) : $g = (9.77 \pm 0.03)\,\mathrm{m}/\mathrm{s}^{2}$ , or $g = 9.77\,\mathrm{m}/\mathrm{s}^{2}$ , $u(g) = 0.03\,\mathrm{m}/\mathrm{s}^{2}$ .

Keep extra digits *during* the calculation; round only at the end.

**Definition 1.21 (Normalized deviation).**

Two results $x_1 \pm u_1$ and $x_2 \pm u_2$ of the same [measurand](#def-b1-units-dimensions-uncertainty) are compared through their *normalized deviation*

$$
z = \frac{\abs{x_1 - x_2}}{\sqrt{u_1^2 + u_2^2}} .
$$

They are *compatible* when $z \leq 2$; a larger $z$ points to an unaccounted effect — or an underestimated uncertainty. When $x_2$ is a reference value with no quoted uncertainty, take $u_2 = 0$.

**Remark 1.22 (Why 2).**

If the two results scatter as Gaussians, their difference has standard deviation $\sqrt{u_1^2 + u_2^2}$, and a Gaussian exceeds twice its standard deviation in about $5\%$ of cases. “$z \leq 2$” is a convention, not a theorem: it accepts roughly one good measurement in twenty as a false disagreement.

## 1.5 Fitting a model to data

**Proposition 1.23 (Least-squares line).**

Given $n$ points $(x_i, y_i)$ expected to obey $y = a x + b$, the straight line minimizing the sum of squared vertical distances $\sum_i (y_i - a x_i - b)^2$ has

$$
a = \frac{\sum_i (x_i - \bar x)(y_i - \bar y)}{\sum_i (x_i - \bar x)^2},
\qquad
b = \bar y - a\,\bar x .
$$

**Proof.** The sum is a quadratic function of $(a, b)$; setting both partial derivatives to zero gives two linear equations: $\sum_i (y_i - ax_i - b) = 0$, which is $\bar y = a \bar x + b$, and $\sum_i x_i (y_i - ax_i - b) = 0$. Substituting $b$ into the second and centering the variables yields $a$. The minimization itself is carried out in the mathematics course on functions of two variables. ∎

**Method 1.24 (Validating a model graphically).**

1. Choose variables that make the expected law a straight line: $\tau^2$ against $\ell$ for the pendulum, $\ln y$ against $\ln x$ for a power law $y = Cx^\alpha$ (slope $\alpha$ ).
2. Plot the points *with their uncertainty bars* .
3. Fit the line; the model is acceptable if the bars straddle the line without a systematic trend of the residuals, and if the intercept is what the model predicts (zero, here).
4. Read the physics off the slope, with its uncertainty.

![2 against for five pendulum lengths, with uncertainty bars; the least-squares line passes through the origin within uncertainty and its slope 4π2/g gives g.](https://one-course.com/images/onecourse/chapters/physics-3/b1-units-dimensions/fig-56a516d0cd8b.svg)

*$\tau^2$ against $\ell$ for five pendulum lengths, with uncertainty bars; the least-squares line passes through the origin within uncertainty and its slope $4\pi^2/g$ gives $g$.*

**Example 1.25 (Reading ggg from a slope).**

The five points of the figure have $\bar\ell = 0.800\,\mathrm{m}$, $\overline{\tau^2} = 3.232\,\mathrm{s}^{2}$, and the sums $\sum (\ell_i - \bar\ell)(\tau_i^2 - \overline{\tau^2}) = 1.614\,\mathrm{m}\,\mathrm{s}^{2}$, $\sum (\ell_i - \bar\ell)^2 = 0.400\,\mathrm{m}^{2}$: slope $a = 4.035\,\mathrm{s}^{2}/\mathrm{m}$, intercept $b = 0.004\,\mathrm{s}^{2}$, negligible. With $\tau^2 = (4\pi^2/g)\,\ell$, $g = 4\pi^2/a = 9.78\,\mathrm{m}/\mathrm{s}^{2}$.

## 1.6 Exercises

**Exercise 1.1 ★.**

Express the joule, the watt, the pascal and the coulomb in [base units](#def-b1-units-dimensions-si). Deduce that $\mathrm{Pa}\,\mathrm{m}^{3}$ is a [unit](#def-b1-units-dimensions-unit) of energy.

**Solution of Exercise 1.1.**

$\mathrm{J} = \mathrm{kg}\,\mathrm{m}^{2}\,\mathrm{s}^{-2}$; $\mathrm{W} = \mathrm{kg}\,\mathrm{m}^{2}\,\mathrm{s}^{-3}$; $\mathrm{Pa} = \mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2}$; $\mathrm{C} = \mathrm{A}\,\mathrm{s}$. $\mathrm{Pa}\,\mathrm{m}^{3} = \mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2}\,\mathrm{m}^{3} = \mathrm{kg}\,\mathrm{m}^{2}\,\mathrm{s}^{-2} = \mathrm{J}$ (the work of pressure forces, $P\,\dd V$).

**Exercise 1.2 ★.**

Give the [dimensions](#def-b1-units-dimensions-dimension) of a pressure, a power, an electric charge, a voltage, and of the constants $G$ (in $F = Gm_1m_2/r^2$), $h$ (in $E = h\nu$) and $k_B$ (in $E = k_B T$).

**Solution of Exercise 1.2.**

Pressure $M L^{-1} T^{-2}$; power $M L^2 T^{-3}$; charge $I\,T$; voltage $=$ power/current $= M L^2 T^{-3} I^{-1}$. $[G] = [F r^2/m^2] = M L T^{-2} L^2 M^{-2} = L^3 M^{-1} T^{-2}$; $[h] = [E/\nu] = M L^2 T^{-2} \cdot T = M L^2 T^{-1}$; $[k_B] = [E/T] = M L^2 T^{-2} \Theta^{-1}$.

**Exercise 1.3 ★.**

Are these formulas homogeneous? $v = \sqrt{2gh}$; $E = \tfrac12 m v$; $P = \rho g h$ for a pressure; $x = v_0 t +
\tfrac12 g t^2$; $T = 2\pi\sqrt{g/\ell}$; $I = I_0 \eu^{-t/RC}$.

**Solution of Exercise 1.3.**

$\sqrt{2gh}$: $\sqrt{L T^{-2} \cdot L} = L T^{-1}$, homogeneous. $\tfrac12 mv$: $M L T^{-1} \neq M L^2 T^{-2}$, not an energy. $\rho g h$: $M L^{-3} \cdot L T^{-2} \cdot L = M L^{-1} T^{-2}$, a pressure. $v_0 t + \tfrac12 g t^2$: both terms $L$, homogeneous. $2\pi\sqrt{g/\ell}$: $\sqrt{T^{-2}} = T^{-1}$ — a frequency, not a period. $I_0\eu^{-t/RC}$: $RC$ in $\Omega\,\mathrm{F} = \mathrm{s}$, argument [dimensionless](#def-b1-units-dimensions-dimension), homogeneous.

**Exercise 1.4 ★.**

A ruler is graduated in millimeters; a digital balance displays $0.01\,\mathrm{g}$ steps; a voltmeter’s data sheet promises “$\pm 0.5\%$ of the reading”. Give the type B [standard uncertainty](#def-b1-units-dimensions-uncertainty) of a $12.7\,\mathrm{cm}$ length, a $5.43\,\mathrm{g}$ mass and a $4.80\,\mathrm{V}$ reading.

**Solution of Exercise 1.4.**

Ruler: $\delta = 1\,\mathrm{mm}$, $u = 1/\sqrt{12} = 0.29\,\mathrm{mm}$. Balance: $u = 0.01/\sqrt{12} = 0.003\,\mathrm{g}$. Voltmeter: $a = 0.005 \times 4.80 = 0.024\,\mathrm{V}$, $u = 0.024/\sqrt3 = 0.014\,\mathrm{V}$.

**Exercise 1.5 ★★.**

The speed $v$ of waves on a stretched string depends on its tension $F$ (a force) and its mass per [unit](#def-b1-units-dimensions-unit) length $\mu$. Find $v$ up to a [dimensionless](#def-b1-units-dimensions-dimension) factor. Same question for the speed of sound in a gas of pressure $P$ and density $\rho$.

**Solution of Exercise 1.5.**

$v = k F^\alpha \mu^\beta$: $L T^{-1} = (M L T^{-2})^\alpha (M L^{-1})^\beta$, so $\alpha + \beta = 0$, $\alpha - \beta = 1$, $-2\alpha = -1$: $\alpha = \tfrac12$, $\beta = -\tfrac12$, $v = k\sqrt{F/\mu}$ (in fact $k = 1$). Sound: $P/\rho$ has [dimension](#def-b1-units-dimensions-dimension) $M L^{-1} T^{-2} / (M L^{-3}) = L^2 T^{-2}$, so $v = k\sqrt{P/\rho}$ ($k = \sqrt\gamma \approx 1.2$ for air).

**Exercise 1.6 ★★.**

Eight timings of the same fall, in milliseconds: $452$, $447$, $461$, $455$, $449$, $458$, $444$, $454$. Compute the mean, the experimental standard deviation and the [standard uncertainty](#def-b1-units-dimensions-uncertainty) of the mean; write the result.

**Solution of Exercise 1.6.**

Sum $= 3620\,\mathrm{ms}$, $\bar t = 452.5\,\mathrm{ms}$. Deviations $-0.5, -5.5, 8.5, 2.5, -3.5, 5.5, -8.5, 1.5$; squares sum to $226$; $s = \sqrt{226/7} = 5.7\,\mathrm{ms}$; $u(\bar t) = 5.7/\sqrt8 = 2.0\,\mathrm{ms}$. Result: $t = (452.5 \pm 2.0)\,\mathrm{ms}$.

**Exercise 1.7 ★★.**

A resistance is measured as $R = U/I$ with $U = 4.80\,\mathrm{V}$, $u(U) = 0.03\,\mathrm{V}$ and $I = 12.4\,\mathrm{mA}$, $u(I) = 0.2\,\mathrm{mA}$. Compute $R$ and $u(R)$; which measurement should be improved first?

**Solution of Exercise 1.7.**

$R = 4.80/0.0124 = 387\,\Omega$. Relative uncertainties $0.03/4.80 =
0.63\%$ and $0.2/12.4 = 1.6\%$; combined $\sqrt{0.63^2 + 1.6^2} = 1.7\%$, $u(R) = 7\,\Omega$: $R = (387 \pm 7)\,\Omega$. The current dominates — improve the ammeter (or its range) first.

**Exercise 1.8 ★★.**

Two groups measure the speed of sound: $(343 \pm 4)\,\mathrm{m}/\mathrm{s}$ and $(336 \pm 3)\,\mathrm{m}/\mathrm{s}$. Are they compatible? A third group finds $(352 \pm 2)\,\mathrm{m}/\mathrm{s}$: compatible with the first? with the second?

**Solution of Exercise 1.8.**

$z = 7/\sqrt{16 + 9} = 1.4 \leq 2$: compatible. Third vs first: $z = 9/\sqrt{4 + 16} = 2.0$: at the limit, barely compatible. Third vs second: $z = 16/\sqrt{4 + 9} = 4.4$: incompatible — at least one of the two has a systematic error or an underestimated uncertainty.

**Exercise 1.9 ★★.**

The volume of a sphere is computed from its diameter $d = 25.40\,\mathrm{mm}$ measured with a caliper, $u(d) = 0.02\,\mathrm{mm}$. Compute $V$, its relative and absolute uncertainty, and write the result.

**Solution of Exercise 1.9.**

$V = \pi d^3/6 = \pi \times 25.40^3/6 = 8580\,\mathrm{mm}^{3}$. $u(V)/V = 3\,u(d)/d = 3 \times 0.02/25.40 = 0.24\%$, $u(V) = 20\,\mathrm{mm}^{3}$: $V = (8.58 \pm 0.02)\,\mathrm{cm}^{3}$.

**Exercise 1.10 ★★★.**

A small sphere of radius $R$ falls slowly in a viscous liquid; the drag depends on $R$, on the speed $v$ and on the viscosity $\eta$, of [dimension](#def-b1-units-dimensions-dimension) $M L^{-1} T^{-1}$. Show that $F = k\,\eta R v$. At high speed the drag depends instead on $R$, $v$ and the fluid density $\rho$: find the new law. Estimate, for a raindrop ($R = 1\,\mathrm{mm}$, $v = 5\,\mathrm{m}/\mathrm{s}$, air: $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$, $\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$), which regime applies by comparing the two estimates.

**Solution of Exercise 1.10.**

$F = k\eta^a R^b v^c$: $M$: $a = 1$; $T$: $-a - c = -2$, $c = 1$; $L$: $-a + b + c = 1$, $b = 1$: $F = k\eta R v$ (Stokes, $k = 6\pi$). With $\rho$ instead: $M$: $a = 1$; $T$: $-c = -2$, $c = 2$; $L$: $-3 + b + 2 = 1$, $b = 2$: $F = k\rho R^2 v^2$. Raindrop: $\eta R v = 1.8\times10^{-5} \times 10^{-3} \times 5 \approx
1 \times 10^{-7}\,\mathrm{N}$; $\rho R^2 v^2 = 1.2 \times 10^{-6} \times 25 =
3 \times 10^{-5}\,\mathrm{N}$, three hundred times larger: the inertial (high-speed) regime applies — the ratio $\rho v R/\eta \approx 330$ is the Reynolds number.

**Exercise 1.11 ★★★.**

Four measurements of the voltage across a resistor for increasing currents: $(I, U) = (2.0\,\mathrm{mA}, 0.41\,\mathrm{V})$, $(4.0\,\mathrm{mA},
0.78\,\mathrm{V})$, $(6.0\,\mathrm{mA}, 1.22\,\mathrm{V})$, $(8.0\,\mathrm{mA},
1.59\,\mathrm{V})$. Compute the least-squares slope and intercept; deduce $R$; is the intercept compatible with zero if each $U$ carries $u = 0.02\,\mathrm{V}$?

**Solution of Exercise 1.11.**

$\bar I = 5.0\,\mathrm{mA}$, $\bar U = 1.000\,\mathrm{V}$; deviations in $I$: $-3, -1, 1, 3$; in $U$: $-0.59, -0.22, 0.22, 0.59$. $\sum (I_i - \bar I)(U_i - \bar U) = 3.98$, $\sum (I_i - \bar I)^2 = 20$: $a = 0.199\,\mathrm{V}/\mathrm{mA} = 199\,\Omega$, $b = 1.000 - 0.199 \times 5 =
0.005\,\mathrm{V}$. With $u(U) = 0.02\,\mathrm{V}$ per point, an intercept of $0.005\,\mathrm{V}$ is well within uncertainty: compatible with zero, the resistor is ohmic, $R \approx 199\,\Omega$.

**Exercise 1.12 ★★★.**

From $G = 6.67 \times 10^{-11}\,\mathrm{m}^{3}/(\mathrm{kg}\,\mathrm{s}^{2})$, $h = 6.63 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$ and $c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, build by dimensional analysis a length, a time and a mass (the Planck [units](#def-b1-units-dimensions-unit)). Compute them and compare with the size of a proton ($1 \times 10^{-15}\,\mathrm{m}$) and with a human mass.

**Solution of Exercise 1.12.**

$\ell = G^a h^b c^c$: $M$: $-a + b = 0$; $T$: $-2a - b - c = 0$; $L$: $3a + 2b + c = 1$. So $b = a$, $c = -3a$, $2a = 1$: $\ell_P = \sqrt{Gh/c^3} = \sqrt{4.42\times10^{-44}/2.7\times10^{25}}
= 4.1 \times 10^{-35}\,\mathrm{m}$; $t_P = \ell_P/c = 1.4 \times 10^{-43}\,\mathrm{s}$; $m_P = \sqrt{hc/G} = \sqrt{1.99\times10^{-25}/6.67\times10^{-11}}
= 5.5 \times 10^{-8}\,\mathrm{kg}$. The Planck length is twenty orders of magnitude below a proton; the Planck mass, $55\,\text{µ}\mathrm{g}$, is a speck of dust — huge for a particle, tiny for a person ($10^9$ times lighter).

![The raw material of the chapter: a caliper, a stopwatch, a ball, a rule, and a column of repeated readings that never quite agree.](https://one-course.com/images/onecourse/chapters/physics-3/b1-units-dimensions/img-013d91118919.jpg)

*The raw material of the chapter: a caliper, a stopwatch, a ball, a rule, and a column of repeated readings that never quite agree.*

## 1.7 Problem: Measuring $g$ with a pendulum

**Problem 1.1.**

Weekend problem — a string, a bob, a stopwatch and a tape: how well can a kitchen table measure the gravitational field of the Earth, and which of its four numbers is the weak one

A pendulum is a steel ball hung from a fixed point by a light string; small swings have period $\tau = 2\pi\sqrt{\ell/g}$ where $\ell$ is the distance from the suspension point to the ball’s center. The aim is a value of $g$ with a defensible uncertainty.

**Part I — The model.**

1. Check the homogeneity of $\tau = 2\pi\sqrt{\ell/g}$ .
2. The ball has mass $m$ . Argue by [dimensions](#def-b1-units-dimensions-dimension) alone that the period cannot depend on $m$ if it depends only on $\ell$ , $g$ and $m$ .
3. The period also depends, slightly, on the amplitude $\theta_0$ of the swing. Why does dimensional analysis allow this, and what must one do experimentally to make the formula applicable?
4. Express $g$ as a function of $\ell$ and $\tau$ .

**Part II — Measuring the length.** The tape measure is graduated in millimeters; the string is $99.0\,\mathrm{cm}$ from suspension to the top of the ball, the ball’s diameter is $2.00\,\mathrm{cm}$, both read once.

5. Compute $\ell$ .
6. Evaluate the type B uncertainty of a single reading on the tape.
7. Locating the suspension point and the ball’s center adds an estimated $\pm5\,\mathrm{mm}$ of possible error (uniform). Combine with the previous item into $u(\ell)$ , and give the relative uncertainty $u(\ell)/\ell$ .

**Part III — Measuring the period.** To reduce the effect of reaction time, one times ten periods. Eight such timings, in seconds: $20.12$, $20.05$, $20.18$, $20.09$, $20.15$, $20.02$, $20.11$, $20.08$.

8. Why time ten periods rather than one? Why repeat the timing eight times?
9. Compute the mean $\overline{t_{10}}$ and the experimental standard deviation $s$ of the eight timings.
10. Deduce the [standard uncertainty](#def-b1-units-dimensions-uncertainty) of the mean, then the period $\tau$ and its uncertainty $u(\tau)$ .
11. Compare $u(\tau)/\tau$ with $u(\ell)/\ell$ .

**Part IV — The result and its weak point.**

12. Compute $g$ .
13. Write $u(g)/g$ in terms of $u(\ell)/\ell$ and $u(\tau)/\tau$ , and compute it.
14. Write the result in the standard form.
15. Compare with the reference value $9.81\,\mathrm{m}/\mathrm{s}^{2}$ through the [normalized deviation](#def-b1-units-dimensions-zscore) . Verdict?
16. Which of the two measured quantities limits the precision of $g$ ? Propose two concrete improvements and say by how much each would reduce the relative uncertainty.
17. Suppose that, by a careless alignment, every length was overestimated by $5\,\mathrm{mm}$ . Is this accounted for in the uncertainty budget? What name does this kind of error carry, and how would one detect it?

**Part V — Five lengths and a straight line.** The experiment is repeated for $\ell = 0.400$, $0.600$, $0.800$, $1.000$ and $1.200\,\mathrm{m}$, giving $\tau^2 = 1.62$, $2.41$, $3.25$, $4.04$ and $4.84\,\mathrm{s}^{2}$, each with $u(\tau^2) \approx 1\%$.

18. Why plot $\tau^2$ against $\ell$ rather than $\tau$ against $\ell$ ?
19. Compute $\bar\ell$ and $\overline{\tau^2}$ .
20. Compute the least-squares slope $a$ and intercept $b$ .
21. Deduce $g$ from the slope.
22. Interpret the intercept: what would a clearly non-zero $b$ reveal?
23. The computer’s fit reports $u(a) = 0.05\,\mathrm{s}^{2}/\mathrm{m}$ . Deduce $u(g)$ and write the result.
24. Compare the single-length result of Part IV with this one by the [normalized deviation](#def-b1-units-dimensions-zscore) .
25. A student proposes to measure instead one swing of a $10\,\mathrm{m}$ pendulum in a stairwell with the same tape and stopwatch. Estimate the relative uncertainty on $g$ that this would give, and conclude on the best strategy.

**Solution of Problem 1.1.**

**1.** $\sqrt{L/(L T^{-2})} = \sqrt{T^2} = T$.

**2.** $\tau = k\ell^\alpha m^\beta g^\gamma$ gives $T = L^{\alpha+\gamma} M^\beta T^{-2\gamma}$, so $\beta = 0$: no combination of $\ell$ and $g$ can cancel a mass.

**3.** $\theta_0$ is [dimensionless](#def-b1-units-dimensions-dimension), so any factor $f(\theta_0)$ is invisible to [dimensions](#def-b1-units-dimensions-dimension). Keep the amplitude small (below about $10^\circ$), where $f \approx 1$ to better than $0.2\%$.

**4.** $g = 4\pi^2\ell/\tau^2$.

**5.** $\ell = 99.0 + 1.00 = 100.0\,\mathrm{cm} = 1.000\,\mathrm{m}$.

**6.** $u = 1\,\mathrm{mm}/\sqrt{12} = 0.29\,\mathrm{mm}$.

**7.** Alignment: $5/\sqrt3 = 2.9\,\mathrm{mm}$; combined $\sqrt{0.29^2 + 2.9^2} = 2.9\,\mathrm{mm}$ — the tape reading is negligible. $u(\ell)/\ell = 0.29\%$.

**8.** The reaction-time error (about $0.1\,\mathrm{s}$) is the same whether one or ten periods are timed, so its effect on one period is ten times smaller. Repeating reveals the scatter (hence $s$) and divides the uncertainty of the mean by $\sqrt8$.

**9.** Sum $= 160.80\,\mathrm{s}$, $\overline{t_{10}} = 20.100\,\mathrm{s}$. Deviations $0.02, -0.05, 0.08, -0.01, 0.05, -0.08, 0.01, -0.02$; squares sum to $188 \times 10^{-4}$; $s = \sqrt{188\times10^{-4}/7} = 0.052\,\mathrm{s}$.

**10.** $u(\overline{t_{10}}) = 0.052/\sqrt8 = 0.018\,\mathrm{s}$; $\tau = 2.0100\,\mathrm{s}$, $u(\tau) = 0.0018\,\mathrm{s}$.

**11.** $u(\tau)/\tau = 0.09\%$, three times smaller than $u(\ell)/\ell = 0.29\%$.

**12.** $g = 4\pi^2 \times 1.000/2.0100^2 = 39.48/4.040 =
9.772\,\mathrm{m}/\mathrm{s}^{2}$.

**13.** $g \propto \ell\,\tau^{-2}$: $\dfrac{u(g)}{g} = \sqrt{\left(\dfrac{u(\ell)}{\ell}\right)^2
+ \left(2\dfrac{u(\tau)}{\tau}\right)^2} = \sqrt{0.29^2 + 0.18^2}\,\%
= 0.34\%$.

**14.** $u(g) = 0.0034 \times 9.772 = 0.033\,\mathrm{m}/\mathrm{s}^{2}$: $g = (9.77 \pm 0.03)\,\mathrm{m}/\mathrm{s}^{2}$.

**15.** $z = \abs{9.772 - 9.81}/0.033 = 1.2 \leq 2$: compatible.

**16.** The length ($0.29\%$ against $0.18\%$ for the period term). Locating both ends to $\pm1\,\mathrm{mm}$ brings $u(\ell)/\ell$ to $0.06\%$ and $u(g)/g$ to $\sqrt{0.06^2 + 0.18^2} = 0.19\%$; doubling the length halves $u(\ell)/\ell$ and lengthens $\tau$, helping both terms. Timing twenty periods instead of ten halves the period term only: $u(g)/g = \sqrt{0.29^2 + 0.09^2} = 0.30\%$ — a small gain while the length is the bottleneck.

**17.** No: a shift common to every reading is a *systematic error* (a bias); the budget covers random scatter and declared tolerances, not a consistent offset. It is detected by changing method or instrument, or — Part V — by a non-zero intercept of the $\tau^2$–$\ell$ line.

**18.** $\tau^2 = (4\pi^2/g)\,\ell$ is a straight line through the origin: linearity and the zero intercept are both testable by eye and by [least squares](#prop-b1-units-dimensions-regression); $\tau \propto \sqrt\ell$ is not.

**19.** $\bar\ell = 0.800\,\mathrm{m}$; $\overline{\tau^2} = 16.16/5 =
3.232\,\mathrm{s}^{2}$.

**20.** Deviations in $\ell$: $-0.4, -0.2, 0, 0.2, 0.4$; in $\tau^2$: $-1.612, -0.822, 0.018, 0.808, 1.608$. Cross sum $= 1.614$, $\sum(\ell_i - \bar\ell)^2 = 0.400$: $a = 4.035\,\mathrm{s}^{2}/\mathrm{m}$, $b = 3.232 - 4.035 \times 0.800 = 0.004\,\mathrm{s}^{2}$.

**21.** $g = 4\pi^2/a = 39.48/4.035 = 9.78\,\mathrm{m}/\mathrm{s}^{2}$.

**22.** $b \approx 0$, as the model predicts. A clearly non-zero $b$ would betray a systematic offset $\delta$ in every length ($\tau^2 = a(\ell_{\text{meas}} - \delta)$ gives $b = -a\delta$), e.g. a mislocated ball center — or an amplitude effect.

**23.** $u(g)/g = u(a)/a = 0.05/4.035 = 1.2\%$, $u(g) =
0.12\,\mathrm{m}/\mathrm{s}^{2}$: $g = (9.78 \pm 0.12)\,\mathrm{m}/\mathrm{s}^{2}$.

**24.** $z = \abs{9.772 - 9.784}/\sqrt{0.033^2 + 0.12^2} = 0.1$: fully compatible. The single careful measurement is more precise; the line validates the model and guards against a length offset.

**25.** $\tau = 2\pi\sqrt{10/9.8} = 6.3\,\mathrm{s}$; one swing timed with a reaction time of about $0.2\,\mathrm{s}$ gives $u(\tau)/\tau \approx
3\%$, hence $u(g)/g \approx 6\%$, while $u(\ell)/\ell$ drops to $0.03\%$: twenty times worse than Part IV. The length gain is wasted by timing a single period. Best strategy: long pendulum *and* many periods, repeated — ten swings of the $10\,\mathrm{m}$ pendulum, eight times, gives $u(\tau)/\tau \approx 0.03\%$ and $u(g)/g \approx 0.07\%$.