Exercises — Physical quantities — fundamental and derived

Ask one question of each: "Is this one of the seven starter tools, or is it built from them?"

Four fundamental (mass, temperature, current, luminous intensity), five derived — and note the last one, refractive index, is derived yet carries no units at all because it is a ratio of two speeds.

${\text{m}}^2=\text{m}\times \text{m}$ = length × length = area. It can't be a length, because a length has units of ${\text{m}}^1$, not ${\text{m}}^2$. The exponent on the metre is the fingerprint of the quantity: power 1 = length, power 2 = area, power 3 = volume.

WHAT: substitute the units of each factor. WHY: because units multiply exactly as the quantities do. $[p]=[m][v]=(\text{kg})\left({\text{m s}}^{-1}\right)=\boxed{{\text{kg m s}}^{-1}}$ Velocity is length ÷ time $={\text{m s}}^{-1}$; multiply by mass in kg. That's it — no memorising, just tracing.

The whole method is: multiply by conversion factors, each of which equals 1 (its top and bottom are the same physical amount, just written in different units). Because every factor is really the number 1, the physical density never changes — only its label does. Now we pick each factor to cancel an unwanted unit and introduce a wanted one. $1\frac{\text{g}}{{\text{cm}}^3}\times \underset{\text{WHY: kills g, brings kg}}{\underbrace{\frac{1\text{kg}}{1000\text{g}}}}\times \underset{{\text{WHY: kills cm}}^3,{\text{brings m}}^{-3}}{\underbrace{\frac{(100\text{cm}{)}^3}{(1\text{m}{)}^3}}}$

• First factor — why $\frac{1\text{kg}}{1000\text{g}}$? We want grams gone and kilograms in. Since $1\text{kg}=1000\text{g}$, this fraction is $1$; putting $\text{g}$ on the bottom cancels the $\text{g}$ on top of the density.
• Second factor — why $\frac{(100\text{cm}{)}^3}{(1\text{m}{)}^3}$? The density has ${\text{cm}}^3$ in its denominator, so to cancel it we need ${\text{cm}}^3$ in the numerator of the factor. Because $1\text{m}=100\text{cm}$, one cubic metre is $(100\text{cm}{)}^3=10^6{\text{cm}}^3$ — the cube is essential since we are converting a volume, not a length.

Carrying out the arithmetic: $=1\times \frac{1}{1000}\times 10^6\frac{\text{kg}}{{\text{m}}^3}=1000{\text{kg m}}^{-3}$ Water is $1000{\text{kg m}}^{-3}$ — the same substance, just a different label.

WHAT: solve for $G$ symbolically, then read units. WHY: $G$ is defined by this equation, so its units are forced by dimensional balance — both sides must match. $G=\frac{F{r}^2}{m_1{m}_2}$ $[G]=\frac{[F][r{]}^2}{[m_1][m_2]}=\frac{({\text{kg m s}}^{-2})({\text{m}}^2)}{(\text{kg})(\text{kg})}$ Numerator: $\text{kg}\cdot \text{m}\cdot {\text{m}}^2={\text{kg m}}^3{\text{ s}}^{-2}$. Divide by ${\text{kg}}^2$ (subtract exponents on kg: $1-2=-1$): $[G]=\boxed{{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}}$

A pure number like $\frac{1}{2}$ has no units, so we ignore it. $[\frac{1}{2}m{v}^2]=(\text{kg})({\text{m s}}^{-1}{)}^2={\text{kg m}}^2{\text{ s}}^{-2}$ $[mgh]=(\text{kg})({\text{m s}}^{-2})(\text{m})={\text{kg m}}^2{\text{ s}}^{-2}$ Both are ${\text{kg m}}^2{\text{ s}}^{-2}$ = the joule ($\text{J}$). They match, so the two formulas can describe the same quantity (kinetic vs potential energy).

But now torque: $[\tau ]=[r][F]=(\text{m})({\text{kg m s}}^{-2})={\text{kg m}}^2{\text{ s}}^{-2}$ — identical base units, yet torque is not energy. In the figure below, the picture makes the difference visible: energy involves a force acting along the direction it pushes ($F$ and displacement parallel), while torque involves a force acting perpendicular to a lever arm ($F$ and $r$ at right angles). Same units, genuinely different physics. That is why we write torque as $\text{N m}$ and energy as $\text{J}$ even though $\text{N m}=\text{J}$ dimensionally — the name records the physical meaning that the units alone throw away.

Route 1 — pressure: $[P]=\frac{[F]}{[\text{area}]}=\frac{{\text{kg m s}}^{-2}}{{\text{m}}^2}={\text{kg m}}^{-1}{\text{ s}}^{-2}$ Route 2 — energy density: $[u]=\frac{[\text{energy}]}{[\text{volume}]}=\frac{{\text{kg m}}^2{\text{ s}}^{-2}}{{\text{m}}^3}={\text{kg m}}^{-1}{\text{ s}}^{-2}$ Subtract exponents on the metre: $2-3=-1$. Both give ${\text{kg m}}^{-1}{\text{ s}}^{-2}$ = one pascal ($\text{Pa}$). Two different physical stories, one unit — and here, unlike torque-vs-energy, they really do describe the same underlying thing.

(a) $P=\frac{180000\text{J}}{12\text{s}}=15000\text{W}$. Base units: $\frac{[\text{J}]}{[\text{s}]}=\frac{{\text{kg m}}^2{\text{ s}}^{-2}}{\text{s}}=\boxed{{\text{kg m}}^2{\text{ s}}^{-3}}$ (add $-2-1=-3$ on the second). This pile of base units is exactly what the name watt ($\text{W}$) abbreviates. (b) Multiply by the conversion factor (which equals 1, because $746\text{W}$ and $1\text{hp}$ are the same power): $15000\text{W}\times \frac{1\text{hp}}{746\text{W}}=20.1\text{hp}(\text{3 s.f.})$ (See Errors and Significant Figures for why we stop at 3 significant figures.)

Strategy: peel the units into recognisable chunks. Recall from L4 that power is ${\text{kg m}}^2{\text{ s}}^{-3}$ (the watt, $\text{W}$). So ${\text{kg m}}^2{\text{ s}}^{-3}{\text{ A}}^{-2}=\frac{[\text{power}]}{{\text{A}}^2}=\frac{[P]}{[I{]}^2}.$ The formula $P=I^{2}R$ (heat dissipated in a resistor of resistance $R$) rearranges to $R=P/I^2$. So $Q$ is electrical resistance, unit the ohm ($\Omega$). Every base-unit fingerprint points back to a real quantity if you decompose it patiently.

(a) Rearrange: $\rho =\frac{RA}{\ell }$. Using $[R]=\Omega ={\text{kg m}}^2{\text{ s}}^{-3}{\text{ A}}^{-2}$ from L5·Q1: $[\rho ]=\frac{[R][A]}{[\ell ]}=\frac{({\text{kg m}}^2{\text{ s}}^{-3}{\text{ A}}^{-2})({\text{m}}^2)}{\text{m}}={\text{kg m}}^3{\text{s}}^{-3}{\text{A}}^{-2}$ Metre exponent: $2+2-1=3$. In shorthand this is $\Omega \text{m}$ — matching the given units. ✓

(b) First the area. Convert radius: $r=0.50\text{mm}=0.50\times 10^{-3}\text{m}=5.0\times 10^{-4}\text{m}$. $A=\pi r^2=\pi (5.0\times 10^{-4}{)}^2=\pi \times 2.5\times 10^{-7}=7.854\times 10^{-7}{\text{m}}^2$ Now the resistance: $R=\frac{\rho \ell }{A}=\frac{(1.7\times 10^{-8})(2.0)}{7.854\times 10^{-7}}=\boxed{0.0433\Omega }(\text{3 s.f.})$ (c) As $r\to 0$, the area $A=\pi r^2\to 0$, and $R=\rho \ell /A\to \infty$. The figure below plots $R$ against $r$: the curve rockets upward as the radius shrinks toward zero. Physically sensible: an infinitely thin wire is an infinitely narrow pipe for charge — it chokes the current, so resistance blows up. The amber dot marks our worked point $r=0.50\text{mm}$, sitting on the gentle part of the curve; slide left toward $r=0$ and you climb the wall. The formula's limiting behaviour matches intuition, a good sign it's the right shape.

1. Base units of momentum, and the formula they came from.
2. Why must the length-conversion factor be cubed when converting a density?
3. State one thing dimensional analysis can do and one thing it cannot do.
4. Base units of $G$ (gravitation constant).
5. Explain physically why $R\to \infty$ as a wire's radius $\to 0$.
6. Give two quantities that share the base units ${\text{kg m}}^2{\text{ s}}^{-2}$ yet are physically different.
7. Name three dimensionless derived quantities and say why they carry no units.