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Physics · Chapter 2: Physical Quantities and Measurement
Dimensional Analysis
Dimensions of physical quantities, checking equations with the principle of homogeneity, deriving formulas and the limits of the method.
Key termRememberExampleWatch out
01Concept
Every physical quantity can be expressed in terms of base quantities. The dimensions of a quantity show which base quantities it is made of and to what powers. We use square brackets: [M] for mass, [L] for length, [T] for time (also [I] current, [θ] temperature, [N] amount of substance).
| Quantity | Formula | Dimensions |
|---|---|---|
| Area | length² | [L²] |
| Volume | length³ | [L³] |
| Speed / velocity | distance ÷ time | [LT⁻¹] |
| Acceleration | velocity ÷ time | [LT⁻²] |
| Force | mass × acceleration | [MLT⁻²] |
| Work / energy | force × distance | [ML²T⁻²] |
| Power | work ÷ time | [ML²T⁻³] |
| Pressure | force ÷ area | [ML⁻¹T⁻²] |
| Density | mass ÷ volume | [ML⁻³] |
Use 1: checking an equation (homogeneity)
An equation can only be correct if every term has the same dimensions. This is the principle of homogeneity. You cannot add a length to a time.
Check s = vt + ½at²:
- [s] = [L]
- [vt] = [LT⁻¹][T] = [L]
- [½at²] = [LT⁻²][T²] = [L]
- All terms are [L], so the equation is dimensionally correct
Use 2: deriving a formula
The time period T of a simple pendulum might depend on its length l, the mass of the bob m and g. Let T = k lᵃ mᵇ gᶜ.
- [T] = [L]ᵃ [M]ᵇ [LT⁻²]ᶜ
- Compare powers: M: 0 = b; T: 1 = −2c, so c = −½; L: 0 = a + c, so a = ½
- T = k √(l/g). The mass does not matter. (Experiment shows k = 2π.)
Limitations
- It cannot find dimensionless constants such as 2π or ½.
- It cannot tell whether a formula has a sum of terms (it only checks each term).
- It does not work for trigonometric, exponential or log relations.
- A dimensionally correct equation is not necessarily correct: s = 2vt is dimensionally fine but wrong.
02Key Formulas
Dimensions of common quantities
Pendulum formula from dimensions
Dimensional analysis gives the form; experiment gives k = 2π.
03Real World Examples
Engineers' quick check
Engineers check long calculations by comparing units on both sides. If the answer for a force comes out in joules, they know there is a mistake.
Model testing
Ship and aircraft designers test small models. Dimensional analysis tells them how to scale the results up to the real ship or plane.
Exam safety net
If you are not sure whether a formula is v = √(2gh) or v = 2gh, checking the dimensions shows that only the square-root form gives [LT⁻¹].
04Practice Questions
Try to answer in your own words first, then open the model answer.
Short questions
Q1.What are dimensions of a physical quantity?Show answer
Answer
The dimensions of a quantity are the powers to which the base quantities must be raised to represent it. Force has dimensions [MLT⁻²].
Q2.Find the dimensions of work and of pressure.Show answer
Answer
- Work = force × distance = [MLT⁻²][L] = [ML²T⁻²]
- Pressure = force ÷ area = [MLT⁻²] ÷ [L²] = [ML⁻¹T⁻²]
Q3.State the principle of homogeneity.Show answer
Answer
Every term on both sides of a correct physical equation must have the same dimensions.
Q4.Check the equation v = u + at dimensionally.Show answer
Answer
- [v] = [LT⁻¹]
- [u] = [LT⁻¹]
- [at] = [LT⁻²][T] = [LT⁻¹]
- All terms have the same dimensions, so it is dimensionally correct
Q5.Give two limitations of dimensional analysis.Show answer
Answer
- It cannot find dimensionless constants such as 2π.
- It cannot be used for trigonometric or exponential relations.
Long questions
Q6.Using dimensional analysis, derive the formula for the time period of a simple pendulum. State its limitations.Show answer
Answer
- Assume T = k lᵃ mᵇ gᶜ.
- Dimensions: [T] = [L]ᵃ[M]ᵇ[LT⁻²]ᶜ = [Mᵇ Lᵃ⁺ᶜ T⁻²ᶜ].
- Compare: M: b = 0; T: −2c = 1 → c = −½; L: a + c = 0 → a = ½.
- So T = k l^½ g^−½ = k √(l/g). Experiment gives k = 2π, so T = 2π√(l/g).
Limitations: it cannot give the value of k; it fails for sums of terms, and for trigonometric, exponential and log functions; a dimensionally correct equation may still be wrong.