Master the language of physics: base units, derived units, and dimensional analysis — the tools that underpin every calculation in A-Level Physics.
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Base SI Units State and recognise the 7 base SI units and their standard symbols.
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Derived Units Express derived units in terms of base SI units using physical definitions.
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Unit Analysis Determine the SI units of any physical quantity from its defining equation.
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Dimensional Consistency Check whether equations are dimensionally homogeneous.
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Prefixes Apply SI prefixes from pico (10⁻¹²) to tera (10¹²) correctly.
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Dimensional Formulae Write dimensional formulae using M, L, T and other base dimensions.
1. The Seven Base SI Units
The International System of Units (SI) defines seven independent base units from which all other units in physics are derived. These base units are agreed upon internationally and provide an unambiguous standard for measurement.
Base SI Unit: One of seven fundamental units that cannot be expressed in terms of other SI units. All other physical quantities are defined as combinations of these seven.
Physical Quantity
Unit Name
Symbol
Dimension
Mass
kilogram
kg
M
Length
metre
m
L
Time
second
s
T
Electric current
ampere
A
I
Temperature
kelvin
K
Θ
Amount of substance
mole
mol
N
Luminous intensity
candela
cd
J
At A-Level Physics, you will most frequently encounter M (mass), L (length), T (time), I (current), and Θ (temperature). The mole appears in thermodynamics equations such as the ideal gas law.
Notice that the base unit of mass is the kilogram, not the gram. This is an important historical quirk — the kilogram is the only base SI unit that contains a prefix in its name.
2. Derived Units
A derived unit is any unit that can be expressed as a combination of base SI units using multiplication or division. Some derived units have special names (e.g. newton, joule, pascal) while others are simply expressed as combinations of base units.
Derived Unit: A unit formed by combining base SI units according to the physical definition or equation of the quantity.
Quantity
Symbol
Special Name
In Base Units
Velocity
v
—
m s⁻¹
Acceleration
a
—
m s⁻²
Force
F
newton (N)
kg m s⁻²
Energy / Work
E, W
joule (J)
kg m² s⁻²
Power
P
watt (W)
kg m² s⁻³
Pressure
p
pascal (Pa)
kg m⁻¹ s⁻²
Charge
Q
coulomb (C)
A s
Voltage
V
volt (V)
kg m² s⁻³ A⁻¹
Resistance
R
ohm (Ω)
kg m² s⁻³ A⁻²
Frequency
f
hertz (Hz)
s⁻¹
To find the base units of a derived quantity, start from a defining equation. For example, since Force = mass × acceleration:
[F] = kg × m s⁻² = kg m s⁻²
And since Work = Force × distance:
[W] = kg m s⁻² × m = kg m² s⁻²
3. Dimensional Analysis
Dimensional analysis is a powerful technique that uses the dimensions (M, L, T, I, Θ…) of physical quantities rather than their numerical values. It serves two main purposes:
Checking equations: Every term in a valid physics equation must have the same dimensions (the equation must be dimensionally homogeneous).
Deriving relationships: Sometimes dimensional analysis can suggest the form of an equation from first principles.
Dimensional Homogeneity: An equation is dimensionally homogeneous if every term on both sides of the equation has identical dimensions.
We write dimensions using square brackets. For example:
[velocity] = L T⁻¹ [acceleration] = L T⁻²
[force] = M L T⁻² [energy] = M L² T⁻²
All three terms share dimensions L² T⁻², so the equation is dimensionally consistent.
Important limitation: dimensional analysis cannot detect dimensionless constants (such as the factor of ½ in E = ½mv²), and it cannot determine whether an equation is physically correct — only whether it is dimensionally consistent.
4. SI Prefixes
SI prefixes allow very large or very small quantities to be expressed concisely. You must be able to convert between prefixed and standard SI units confidently.
Prefix
Symbol
Multiplier
Power of 10
tera
T
1 000 000 000 000
10¹²
giga
G
1 000 000 000
10⁹
mega
M
1 000 000
10⁶
kilo
k
1 000
10³
milli
m
0.001
10⁻³
micro
μ
0.000 001
10⁻⁶
nano
n
0.000 000 001
10⁻⁹
pico
p
0.000 000 000 001
10⁻¹²
Always convert to base SI units before substituting into equations. For example, 5 mA = 5 × 10⁻³ A, and 200 nm = 200 × 10⁻⁹ m = 2 × 10⁻⁷ m.
When a prefixed unit is raised to a power, the prefix is also raised to that power:
(1 cm)² = (1 × 10⁻² m)² = 1 × 10⁻⁴ m²
This is a very common source of errors — always expand the prefix with the unit before squaring or cubing.
5. Finding Units of Constants from Equations
A key exam skill is finding the SI units of a constant that appears in a physical equation. The method is always the same: rearrange the equation to make the constant the subject, then substitute in the units of all other quantities.
For example, consider the equation for gravitational force:
F = G m₁ m₂ / r²
Rearranging: G = F r² / (m₁ m₂)
[G] = (kg m s⁻²)(m²) / (kg × kg)
[G] = kg m³ s⁻² kg⁻² = kg⁻¹ m³ s⁻²
Similarly, for the equation for electrical resistivity ρ = R A / L:
[ρ] = Ω × m² / m = Ω m = kg m³ s⁻³ A⁻²
Step-by-step method: (1) Write the defining equation. (2) Rearrange for the unknown constant. (3) Replace every quantity with its SI unit. (4) Simplify by cancelling and combining powers.
Example 1: Express the unit of pressure (pascal) in terms of base SI units only, using a defining equation.
1Write the defining equation for pressure: Pressure = Force ÷ Area, so P = F / A
2Write the units of force. Since F = ma: [F] = kg × m s⁻² = kg m s⁻²
3Write the units of area: [A] = m²
4Combine: [P] = kg m s⁻² / m² = kg m s⁻² × m⁻² = kg m⁻¹ s⁻²
Pressure in base SI units: kg m⁻¹ s⁻² (This is the pascal: 1 Pa = 1 kg m⁻¹ s⁻²)
Example 2: Check whether the equation for kinetic energy, E = ½ mv², is dimensionally homogeneous.
1Identify the dimensions of the left-hand side: [E] = M L² T⁻² (energy has units kg m² s⁻²)
2Find the dimensions of the right-hand side. The factor ½ is dimensionless, so we ignore it: [mv²] = M × (L T⁻¹)²
3Expand the bracket: [mv²] = M × L² T⁻² = M L² T⁻²
4Compare both sides: LHS: M L² T⁻² RHS: M L² T⁻² ✓
Both sides have dimensions M L² T⁻², so the equation is dimensionally homogeneous. ✓
Example 3: Find the SI units of the gravitational constant G, given that F = Gm₁m₂ / r².
1Rearrange to make G the subject: G = F r² / (m₁ m₂)
2Write the units of each quantity on the right-hand side: [F] = kg m s⁻², [r²] = m², [m₁] = kg, [m₂] = kg
3Substitute units: [G] = (kg m s⁻²)(m²) / (kg × kg)
4Simplify — multiply numerator: kg m³ s⁻² Divide by kg²: kg m³ s⁻² ÷ kg² = kg⁻¹ m³ s⁻²
Units of G: kg⁻¹ m³ s⁻² (Numerically G = 6.67 × 10⁻¹¹ kg⁻¹ m³ s⁻²)
Example 4: A student proposes the equation: v = √(F/m) where v is velocity, F is force and m is mass. Check if this equation is dimensionally consistent.
1Write dimensions of the left-hand side: [v] = L T⁻¹
2Write dimensions inside the square root on the right-hand side: [F/m] = (M L T⁻²) / M = L T⁻²
3Apply the square root — halve all powers: [√(F/m)] = (L T⁻²)^(1/2) = L^(1/2) T⁻¹
4Compare: LHS = L T⁻¹ RHS = L^(1/2) T⁻¹ These are NOT equal — L ≠ L^(1/2)
The equation is NOT dimensionally homogeneous ✗ It cannot be a valid physics equation (at least not in this form).
Question 1: Which of the following is a base SI unit?
Question 2: What are the base SI units of electric resistance (Ω)?
Question 3: Convert 450 nm into metres. Give your answer in standard form.
Question 4: The equation s = ut + ½at² is checked dimensionally. Which option correctly shows the dimensions of the term ½at²?
Question 5: Using the equation P = IV, show that the unit of power (watt) in base SI units is kg m² s⁻³. State the base units of I and V separately, then combine them.
Challenge 1: The period T of a simple pendulum is thought to depend on its length L and the gravitational field strength g. Using dimensional analysis, show that T ∝ √(L/g) and hence state why the mass of the bob cannot appear in the equation.
Solution:
Assume T = k Lᵃ gᵇ where k is a dimensionless constant.
Dimensions: [T] = T¹ [L] = L [g] = L T⁻²
So: T¹ = Lᵃ (L T⁻²)ᵇ = L^(a+b) T^(−2b)
Comparing powers of T: 1 = −2b → b = −½
Comparing powers of L: 0 = a + b → a = ½
Therefore T ∝ L^(1/2) g^(−1/2) = √(L/g) ✓
Mass has dimensions M. Since neither T nor L/g contains M, any factor of mᶜ would require M^c = M⁰, meaning c = 0. Mass cannot appear in the equation.
Challenge 2: The Planck constant h appears in the equation E = hf, where E is energy and f is frequency. (a) Determine the SI base units of h. (b) A student claims h could also be expressed as units of "momentum × length". Verify whether this is dimensionally consistent.
Part (a):
h = E / f
[h] = (kg m² s⁻²) / (s⁻¹) = kg m² s⁻² × s = kg m² s⁻¹ SI base units of h: kg m² s⁻¹
Part (b):
[momentum] = M L T⁻¹ = kg m s⁻¹
[momentum × length] = kg m s⁻¹ × m = kg m² s⁻¹
This equals kg m² s⁻¹ — the same as h ✓
So yes, the claim is dimensionally consistent. (h is indeed angular momentum / action.)
Challenge 3: A physicist proposes an equation for the speed of sound in a medium: v = √(E/ρ), where E is the Young modulus and ρ is the density. Show by dimensional analysis that this equation is dimensionally homogeneous. State the base SI units of Young modulus E.
Base units of E (Young modulus = stress = Force/Area):
[E] = kg m s⁻² / m² = kg m⁻¹ s⁻²
Base units of ρ (density = mass/volume):
[ρ] = kg m⁻³
Dimensional check of E/ρ:
[E/ρ] = (kg m⁻¹ s⁻²) / (kg m⁻³) = kg m⁻¹ s⁻² × m³ kg⁻¹ = m² s⁻²
Applying the square root:
[√(E/ρ)] = (m² s⁻²)^(1/2) = m s⁻¹
[v] = m s⁻¹ = [√(E/ρ)] ✓
The equation is dimensionally homogeneous.
Challenge 4 (Extended): The Stokes drag force on a sphere is given by F = 6πηrv, where η is the viscosity of the fluid, r is the radius of the sphere, and v is its velocity. (a) Find the SI base units of dynamic viscosity η. (b) A student rewrites the equation as η = F/(6πrv). State whether the factor 6π affects the dimensional analysis, and explain why.
Part (a):
η = F / (r × v) [ignoring dimensionless constants]
[η] = (kg m s⁻²) / (m × m s⁻¹)
[η] = kg m s⁻² / m² s⁻¹
[η] = kg m⁻¹ s⁻¹ SI base units of dynamic viscosity η: kg m⁻¹ s⁻¹
(This is sometimes written as Pa s, or 1 poiseuille.)
Part (b):
The factor 6π is a pure number — it has no dimensions (it is dimensionless).
Dimensional analysis only tracks powers of M, L, T etc., not numerical values.
Therefore 6π does NOT affect the dimensional analysis: it contributes M⁰L⁰T⁰ = 1 dimensionally.
Dimensional analysis can verify whether an equation is consistent, but it cannot determine or verify dimensionless numerical prefactors.