SI units, prefixes and homogeneity
Key idea: Derived units come from SI base units. Every term added or equated must have matching units; this can rule out a wrong equation but cannot prove an equation correct.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Connect each physical quantity to its SI base unit.
- Track each power of ten, including every length factor in an area or volume.
- Convert the mass and area independently, then check both values.
- Follow the defining equations to build derived units from base units.
- Derive the ohm from voltage divided by current, then compare your exponents.
- Check every term and explain why matching units cannot determine a numerical factor.
- Determine the units of the pendulum constant and the limits of your conclusion.
- Infer both unknown units from the homogeneous sum; show why one has a fractional exponent.
- Convert the square micrometres to square metres without the prefix table.
- Derive the units of the energy constant and decide whether it could be a pure number.
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ContinueA number on its own says nothing in physics: 5 could be five metres, five seconds or five kilograms. Every measured quantity is a number multiplied by a unit, and every unit used in physics can be built from a handful of SI base units. This lesson sets out the base units and prefixes, shows how to write any derived unit in base units, and uses base units to check whether an equation can be right.
SI base quantities and units
The SI system is built on a small set of base quantities, each with its own unit. You need these six:
| Base quantity | Base unit | Symbol |
|---|---|---|
| mass | kilogram | kg |
| length | metre | m |
| time | second | s |
| electric current | ampere | A |
| thermodynamic temperature | kelvin | K |
| amount of substance | mole | mol |
SI also has luminous intensity as a base quantity, with unit candela; it is not needed here. A derived quantity is defined by an equation linking it to base quantities. Its unit follows from that equation.
Prefixes
Prefixes stand for powers of ten, so that very large and very small values can be written compactly.
| Prefix | Symbol | Multiplier | Prefix | Symbol | Multiplier |
|---|---|---|---|---|---|
| pico | p | 10⁻¹² | deci | d | 10⁻¹ |
| nano | n | 10⁻⁹ | kilo | k | 10³ |
| micro | μ | 10⁻⁶ | mega | M | 10⁶ |
| milli | m | 10⁻³ | giga | G | 10⁹ |
| centi | c | 10⁻² | tera | T | 10¹² |
To convert, replace the prefix by its power of ten: 7.2 GW = 7.2 × 10⁹ W. Capital and small letters mean different things: m is milli, M is mega.
When a prefixed unit is squared or cubed, the prefix is squared or cubed too, because it belongs to every length in the product:
The base unit of mass already carries a prefix: it is the kilogram, not the gram. So 1 g = 10⁻³ kg and 1 mg = 10⁻⁶ kg.
Guided practice 1
Milligrams and square millimetres
Problem
Convert 35 mg to the SI base unit of mass. A rectangle has sides 3.0 mm and 2.0 mm: find its area in m².
Convert both quantities
Show the steps Optional working
Hints
Hint 1
mass
The base unit is the kilogram, so convert milligrams to grams and then grams to kilograms.
Hint 2
area
An area has two length factors, so the factor 10⁻³ appears twice.
Show solution step by step
Convert the mass
Reason
Each prefix is a power of ten, and the base unit is the kilogram.Working
Answer
35 mg = 35 × 10⁻³ g = 35 × 10⁻⁶ kg = 3.5 × 10⁻⁵ kg
Convert the area
Working
Answer
A = (3.0 × 10⁻³ m)(2.0 × 10⁻³ m) = 6.0 × 10⁻⁶ m²
Derived units in base units
To write a derived unit in base units, take the equation that defines the quantity and replace each quantity by its unit.
| Quantity | Defining equation | Named unit | In SI base units |
|---|---|---|---|
| force | F = ma | newton, N | kg m s⁻² |
| work, energy | W = Fs | joule, J | kg m² s⁻² |
| power | P = W/t | watt, W | kg m² s⁻³ |
| pressure | p = F/A | pascal, Pa | kg m⁻¹ s⁻² |
| charge | Q = It | coulomb, C | A s |
| potential difference | V = W/Q | volt, V | kg m² s⁻³ A⁻¹ |
| frequency | f = 1/T | hertz, Hz | s⁻¹ |
Work down the table: each row uses units already found above it. For the volt, J C⁻¹ = (kg m² s⁻²)/(A s) = kg m² s⁻³ A⁻¹.
Check your understanding 1
Use R = V/I and the table to express the ohm in SI base units.
Show answer
Ω = V A⁻¹ = (kg m² s⁻³ A⁻¹)/A = kg m² s⁻³ A⁻²
Checking that an equation is homogeneous
You can only add, subtract or equate quantities of the same kind: a length plus a time means nothing. So in a correct physical equation, every term added or subtracted, and the two sides of the equation, must have the same base units. Such an equation is homogeneous.
A homogeneity check can prove an equation wrong: if the base units do not match, the equation cannot be right. It cannot prove an equation right, because pure numbers such as 1/2 or 2π have no units, and an equation can have the right units but the wrong physics.
Worked example 2
Test two equations for homogeneity
Problem
Show that p = ρ gh is homogeneous, where p is pressure, ρ density, g the acceleration of free fall and h depth. Then test s = ut + at² and s = ut + (1/2)at², where s is displacement, u initial velocity, t time and a acceleration.
Show full solution
Write the base units of each side
Method
Find the base units of p, then of ρ gh.Reason
A homogeneous equation has the same base units on both sides.Working
Answer
Left: Pa = kg m⁻¹ s⁻². Right: (kg m⁻³)(m s⁻²)(m) = kg m⁻¹ s⁻². The units match, so p = ρ gh is homogeneous.
Check every term in a sum
Method
Find the base units of s, ut and at².Reason
Terms that are added must have the same units as each other and as the left-hand side.Working
Answer
s: m. ut: (m s⁻¹)(s) = m. at²: (m s⁻²)(s²) = m.
Draw the conclusion
Working
Answer
Both displacement equations are homogeneous, but only one is correct. Homogeneity cannot tell them apart, because the factor 1/2 has no units.
Try it yourself 3
The constant in a pendulum equation
Problem
The period of a pendulum is proposed to be T = C square root of (l/g), where l is the length of the pendulum and g the acceleration of free fall. Find the base units of C. What can a homogeneity check not tell you about C?
Find the units of C
Hints
Hint 1
inside the root
Find the base units of l/g first, then take the square root and compare with the unit of T.
Show solution step by step
Find the units of the square root
Reason
C must supply whatever is needed to turn square root of (l/g) into seconds.Working
Answer
l/g: m/(m s⁻²) = s², so square root of (l/g) has unit s.
Conclude
Working
Answer
square root of (l/g) already has the unit of T, so C has no units. The check cannot find its value (2π), and it cannot show whether a real pendulum obeys the equation.
Try it yourself 4
Units in a homogeneous sum
Problem
The equation z² = p + qt is homogeneous. t is a time in seconds and q has unit m s⁻¹. Find the SI units of p and of z.
Find the units
Hints
Hint 1
start inside the sum
Find the unit of qt. Every other term, and z², must have the same unit.
Show solution step by step
Find the unit of the known term
Reason
q is a speed and t a time.Working
Answer
qt: (m s⁻¹)(s) = m
Match the other terms
Working
Answer
p is added to qt, so p has unit m. z² also has unit m, so z has unit m^(1/2).
Quick check
An unusual unit such as m^(1/2) is allowed for a quantity defined this way.
Common mistakes
- Treating the gram as the base unit of mass. It is the kilogram.
- Converting cm² or cm³ with 10⁻² once, instead of squaring or cubing it.
- Mixing up m (milli) and M (mega).
- Saying an equation is correct because it is homogeneous. Homogeneity can only show that an equation is wrong.
Before you move on
Check your understanding 2
Convert 2.0 (μ m)² to m².
Show answer
The area is 2.0 × 10⁻¹² m² because the micro prefix is squared.
Check your understanding 3
A proposed energy equation is E = kmv³, where m is mass and v speed. Find the SI base units required for k. Could k be a pure number such as 1/2?
Show answer
Dividing the energy unit by the unit of mv³ gives [k] = (kg m² s⁻²)/(kg m³ s⁻³) = m⁻¹ s. A pure number has no units, so it cannot make this equation homogeneous.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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