PhysicsISC Class 11 · Units and Measurements

Dimensional Analysis: Checking, Deriving and Converting with Dimensions

Updated 17 October 2026 · Fact-checked

Dimensional analysis writes every physical quantity in terms of the base dimensions M, L, T. Using the principle of homogeneity, an equation is correct only if every term has the same dimensions. The method checks equations, derives relations between quantities and converts units, but it cannot find dimensionless constants.

Every term in v = u + at has dimensions [LT⁻¹], so the equation is homogeneous; this check cannot reveal constants like 2π or ½.

Illustration drawn for QuestionBuddy.

Understand Dimensional Analysis and Its Uses

Dimensions and dimensional formulae. The dimensions of a quantity show how it depends on the base quantities. Velocity is [LT⁻¹], force is [MLT⁻²], energy and work are [ML²T⁻²], pressure is [ML⁻¹T⁻²], and Planck's constant h (from E = hν) is [ML²T⁻¹].

Principle of homogeneity. Only quantities with the same dimensions can be added, subtracted or equated. Every term of a correct physical equation therefore has the same dimensions. Arguments of sin, cos, log and exponentials must be dimensionless.

Uses.
1. *Checking an equation.* If the two sides have different dimensions, the equation is certainly wrong.
2. *Deriving a relation.* Assume the quantity is a product of powers of the variables it depends on, write the dimensions on both sides, and equate the indices of M, L and T.
3. *Converting units.* If a quantity has dimensions Mᵃ Lᵇ Tᶜ, then n₂ = n₁ (M₁/M₂)ᵃ (L₁/L₂)ᵇ (T₁/T₂)ᶜ.

Limitations. The method cannot give dimensionless constants such as 2π or ½. It fails when a quantity depends on more than three others in a way that gives too many unknown indices, when it depends on a sum or difference of terms, and for trigonometric, exponential or logarithmic dependence. A dimensionally correct equation may still be physically wrong.

Dimensional Analysis and Its Uses: formulas and key facts

Force, energy, pressure: $F = [M L T^{- 2}] ∣ E = [M L^{2} T^{- 2}] ∣ p = [M L^{- 1} T^{- 2}]$; From F = ma, W = Fs, p = F/A.
Gravitational constant: $G = [M^{- 1} L^{3} T^{- 2}]$; From F = Gm₁m₂/r².
Planck constant: $h = [M L^{2} T^{- 1}]$; From E = hν.
Unit conversion: $n_{2} = n_{1} (M_{1} / M_{2})^{a} (L_{1} / L_{2})^{b} (T_{1} / T_{2})^{c}$; For a quantity of dimensions Mᵃ Lᵇ Tᶜ.

Dimensional Analysis and Its Uses in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Converting an acceleration

The acceleration due to gravity is 9.8 m s⁻², but a physics simulation for a long-distance projectile needs it in km h⁻².

Convert 9.8 m s⁻² to km h⁻².

Try it first, then show the solutionSolution

Dimensions [LT⁻²]: 1 m = 10⁻³ km and 1 s = 1/3600 h.
So 1 m s⁻² = 10⁻³ km / (1/3600 h)² = 10⁻³ × 3600² km h⁻² = 12960 km h⁻².
9.8 × 12960 = 127008 km h⁻².

Answer: 127008 km h⁻², about 1.27 × 10⁵ km h⁻².

Memory hook: A smaller time unit squared in the denominator means a much larger number.

Tricks to remember Dimensional Analysis and Its Uses

Short memory aids. Say each one aloud once, then use it on a question the same day.

Check each term
Reduce every term to M, L, T and compare: the same on both sides or the equation is wrong.
Use it when: Verifying a given formula in one or two lines.
Equate indices
Write a product of powers, then one equation each for M, L and T.
Use it when: Deriving formulae such as the time period of a pendulum.

Dimensional Analysis and Its Uses: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Dimensions of pressure · 1 mark · MCQBoard level

The dimensional formula of pressure is:
(a) [M L T⁻²]
(b) [M L⁻¹ T⁻²]
(c) [M L² T⁻²]
(d) [M L⁻² T⁻²]

Try it first, then show the solutionSolution

Pressure = force/area = [MLT⁻²]/[L²] = [ML⁻¹T⁻²].

Answer: (b) [M L⁻¹ T⁻²]

Example 2: Checking equations · 3 marks · Short answerBoard level

Check by dimensions whether (i) v² = u² + 2as and (ii) v = u + at² can be correct.

Try it first, then show the solutionSolution

(i) [v²] = L²T⁻²; [u²] = L²T⁻²; [as] = LT⁻² × L = L²T⁻². All terms agree, so (i) is dimensionally correct. (1.5 marks)
(ii) [v] = LT⁻¹; [at²] = LT⁻² × T² = L. They differ, so (ii) is wrong. (1.5 marks)

Answer: (i) dimensionally correct; (ii) wrong because [at²] = [L] is not [LT⁻¹].

Example 3: Period of a loaded spring · 4 marks · Short answerHarder board level

The time period T of a mass m hanging on a spring of force constant k (force per unit extension) may depend on m and k only. Use dimensions to find the relation.

Try it first, then show the solutionSolution

[k] = force/length = MLT⁻²/L = MT⁻². Let T = c mᵃ kᵇ, so [T] = MᵃMᵇT⁻²ᵇ. (1 mark)
T: −2b = 1 so b = −½. (1 mark)
M: a + b = 0 so a = ½. (1 mark)
T = c √(m/k). The constant c (equal to 2π) cannot be found by this method. (1 mark)

Answer: T = c√(m/k).

Example 4: Constants in the van der Waals equation · 3 marks · Short answerHarder board level

The van der Waals equation is (p + a/V²)(V − b) = RT. Find the dimensions of a and b, where V is the volume.

Try it first, then show the solutionSolution

By homogeneity [a/V²] = [p], so [a] = [p][V²] = ML⁻¹T⁻² × L⁶ = [ML⁵T⁻²]. (1.5 marks)
[b] = [V] = [L³]. (1.5 marks)

Answer: [a] = ML⁵T⁻²; [b] = L³.

Example 5: Vibrating liquid drop · 5 marks · Long answerHOTS / competency

The frequency ν of small oscillations of a liquid drop is thought to depend on its density ρ, its radius r and its surface tension S (force per unit length). Derive the form of ν, and find by what factor ν changes when r is doubled.

Try it first, then show the solutionSolution

[S] = MLT⁻²/L = MT⁻². Let ν = k ρᵃ rᵇ Sᶜ: T⁻¹ = (ML⁻³)ᵃ Lᵇ (MT⁻²)ᶜ. (1 mark)
T: −2c = −1 so c = ½. (1 mark)
M: a + c = 0 so a = −½. (1 mark)
L: −3a + b = 0 so b = 3a = −3/2. Hence ν = k √(S/(ρ r³)). (1 mark)
With r doubled, ν becomes 2^(−3/2) = 0.354 times its value, roughly a third. (1 mark)

Answer: ν = k√(S/(ρr³)); doubling r multiplies ν by 2^(−3/2) ≈ 0.35.

Where students lose marks in Dimensional Analysis and Its Uses

The slips examiners see most often, and what to write instead.

Mistake: Treating a dimensionless number as having a unit
Students include 2π or ½ in the dimensions.
Do this: Pure numbers have no dimensions; leave them out.
Usually costs: 1 mark.
Mistake: Stopping at "both sides match"
The student says the formula is correct.
Do this: Say it is dimensionally correct; the method cannot confirm the numerical constant.
Usually costs: 1 mark.
Mistake: Wrong powers while solving indices
Writing −2c = −1 as c = −2.
Do this: Solve each linear equation separately and substitute back to check.
Usually costs: 1 to 2 marks.

Exam tips

Write dimensions of every term in a column and compare them.
In derivations, always start with a product of powers and a constant k.
Say clearly that k cannot be found by dimensional analysis.
For unit conversion, write the dimensions first, then convert each base unit.

Practice questions on Dimensional Analysis and Its Uses

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which pair has the same dimensions?
(a) force and energy
(b) work and torque
(c) pressure and force
(d) power and work
Hint
Both work and torque are [ML²T⁻²].
Answer
(b) work and torque
Q2 · 2 marks · Short answerBoard level
Find the dimensional formula of the coefficient of viscosity η from F = 6πηrv, where r is a radius and v a speed.
Hint
Make η the subject; 6π has no dimensions.
Answer
η = F/(6π r v) → MLT⁻²/(L × LT⁻¹) = [M L⁻¹ T⁻¹].
Q3 · 3 marks · Short answerHarder board level
Show that h/(mv) has the dimensions of length, where h is Planck's constant.
Hint
Get [h] from E = hν.
Answer
[h] = [E]/[ν] = ML²T⁻²/T⁻¹ = ML²T⁻¹. [h/(mv)] = ML²T⁻¹/(M × LT⁻¹) = L.
Q4 · 3 marks · Short answerHarder board level
Convert 9.8 m s⁻² into km h⁻² and show the dimension used.
Hint
Dimension [LT⁻²].
Answer
1 m s⁻² = 10⁻³ × 3600² km h⁻² = 12960 km h⁻². 9.8 × 12960 = 127008 km h⁻².
Q5 · 3 marks · Short answerHarder board level
Show that every term in Bernoulli's equation p + ½ρv² + ρgh has the dimensions of pressure.
Hint
Find each of p, ρv² and ρgh in terms of M, L, T.
Answer
p: ML⁻¹T⁻². ρv²: ML⁻³ × L²T⁻² = ML⁻¹T⁻². ρgh: ML⁻³ × LT⁻² × L = ML⁻¹T⁻². All three are the same, so the equation is homogeneous.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student writes the position of a body as x = a + bt + ct², with x in metres and t in seconds.
(i) Give the units of a, b and c. (1 mark)
(ii) Is x = ut + ½at² + 3 dimensionally correct? Give a reason. (1 mark)
(iii) Can dimensions tell whether the factor ½ in ½at² is correct? (1 mark)
(iv) Give one reason why a dimensionally correct equation may still be wrong. (1 mark)
Hint
Each term must have the dimension of length.
Answer
(i) a: m; b: m s⁻¹; c: m s⁻². (ii) No: ut and ½at² have dimension L, but the constant 3 is dimensionless, so it cannot be added. (iii) No; pure numbers are invisible to dimensions. (iv) For example x = ut + at² is homogeneous but wrong, since the factor ½ is missing.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the principle of homogeneity.Show answer
Every term in a correct physical equation has the same dimensions.
2.Dimensions of energy?Show answer
[M L² T⁻²].
3.Dimensions of Planck's constant?Show answer
[M L² T⁻¹].
4.Name one limitation of dimensional analysis.Show answer
It cannot determine dimensionless constants.

Dimensional Analysis and Its Uses: frequently asked questions

Can dimensional analysis prove that an equation is correct?

No. It can only show that an equation is dimensionally consistent. A wrong numerical factor, such as a missing ½, is not detected.

Why can't we use dimensions for sin, cos or exponential relations?

The argument of such functions must be dimensionless, and these functions cannot be written as a product of powers, so the index method does not apply.

Is "order of magnitude" part of this topic?

It is a nearby idea in the chapter: it estimates a quantity to the nearest power of ten. The chapter page covers it.