PhysicsCBSE Class 11 · Units and Measurement

Dimensional Analysis and its Applications Class 11

Updated 17 October 2026 · Fact-checked

Dimensions express a quantity through the base quantities M, L, T (and others). The principle of homogeneity says every term of a valid equation has the same dimensions. It lets you check a formula, convert units and derive a relation up to a dimensionless constant, but it cannot give numerical constants or handle sums of trigonometric or exponential functions.

Every term in a valid equation has the same dimensions; if one term differs, as at² does here, the equation must be wrong.

Illustration drawn for QuestionBuddy.

Understand Dimensional Analysis and its Applications

The dimensions of a quantity are the powers to which the base quantities are raised in it. Mechanical quantities use $[M]$ , $[L]$ and $[T]$ . A dimensional formula is written in square brackets: speed $[L T^{- 1}]$ , force $[M L T^{- 2}]$ , work and energy $[M L^{2} T^{- 2}]$ , pressure $[M L^{- 1} T^{- 2}]$ , power $[M L^{2} T^{- 3}]$ , angular momentum $[M L^{2} T^{- 1}]$ , gravitational constant $G$ : $[M^{- 1} L^{3} T^{- 2}]$ , Planck constant $h$ : $[M L^{2} T^{- 1}]$ .

Principle of homogeneity. Only quantities of the same dimension can be added, subtracted or equated. So in $v = u + a t$ , each term is $[L T^{- 1}]$ . Arguments of $sin$ , $lo g$ and $e^{x}$ must be dimensionless.

Uses.
1. *Checking an equation.* If the two sides differ in dimension, the equation is wrong. If they agree, it may still be wrong by a numerical factor (for example $s = u t + a t^{2}$ is dimensionally fine but physically wrong; the correct factor is $\frac{1}{2}$ ).
2. *Converting units.* A quantity with dimensions $M^{a} L^{b} T^{c}$ has numerical value $n_{2} = n_{1} (\frac{M _{1}}{M _{2}})^{a} (\frac{L _{1}}{L _{2}})^{b} (\frac{T _{1}}{T _{2}})^{c}$ .
3. *Deriving a relation.* Assume the quantity depends on the others as a product of powers, equate the dimensions and solve for the exponents. The constant (such as $2 π$ ) must come from experiment.

Limits. It cannot find dimensionless constants, cannot treat relations that are sums (such as $s = u t + \frac{1}{2} a t^{2}$ ) by derivation, and fails when a quantity depends on more variables than there are base dimensions.

Dimensional Analysis and its Applications: formulas and key facts

Force: $[F] = [M L T^{- 2}]$; Unit newton.
Energy or work: $[E] = [M L^{2} T^{- 2}]$; Unit joule.
Pressure: $[P] = [M L^{- 1} T^{- 2}]$; Also stress and Young modulus.
Unit conversion: $n_{1} u_{1} = n_{2} u_{2} \Rightarrow n_{2} = n_{1} (\frac{M _{1}}{M _{2}})^{a} (\frac{L _{1}}{L _{2}})^{b} (\frac{T _{1}}{T _{2}})^{c}$; Quantity $M^{a} L^{b} T^{c}$ .
Gravitational constant: $[G] = [M^{- 1} L^{3} T^{- 2}]$; From $F = G m_{1} m_{2} / r^{2}$ .
Planck constant: $[h] = [M L^{2} T^{- 1}]$; From $E = h ν$ .

How to solve Dimensional Analysis and its Applications questions

Dimensional questions all reduce to equating powers of M, L and T on two sides.

1Write the dimensional formula of every symbol from a defining formula.
2For checking, reduce both sides to M, L, T powers and compare.
3For deriving, write $x = k a^{p} b^{q} c^{r}$ , substitute dimensions and equate the powers of M, L and T separately.
4Solve the simultaneous equations for $p$ , $q$ , $r$ .
5For unit conversion, express the old and new units in the base-unit ratios and raise to the dimensional powers.
6State that the dimensionless constant has to be found by experiment.

Dimensional Analysis and its Applications 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.

Check a recipe formula for a swing

Suppose a student guesses that the time period of a pendulum is $T = 2 π g / L$ .

Use dimensions to test it.

Try it first, then show the solutionSolution

[g/L] = [LT⁻²]/[L] = [T⁻²], so √(g/L) has dimension [T⁻¹].
T should have dimension [T], so the guess is dimensionally wrong.
The corrected form T = 2π√(L/g) has [√(L/g)] = [T], which is consistent.

Answer: The guess fails; T = 2π√(L/g) passes.

Memory hook: If the units do not match, the formula cannot be right.

Pressure units in a tyre gauge

Suppose a tyre gauge reads 2.0 × 10⁵ Pa and the student must quote it in dyne/cm² (CGS).

Convert 2.0 × 10⁵ Pa to dyne/cm².

Try it first, then show the solutionSolution

[P] = ML⁻¹T⁻². 1 Pa = 1 kg m⁻¹ s⁻² = (1000 g)(100 cm)⁻¹ s⁻² = 10 g cm⁻¹ s⁻² = 10 dyne cm⁻².
2.0 × 10⁵ Pa = 2.0 × 10⁶ dyne/cm².

Answer: 2.0 × 10⁶ dyne/cm².

Memory hook: 1 Pa = 10 dyne/cm².

Tricks to remember Dimensional Analysis and its Applications

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

F, E, P ladder
Force MLT⁻² → energy adds an L → power divides by T.
Use it when: Recalling dimensions quickly.
Same dimension family
Work, energy, torque (M L² T⁻²); pressure, stress, energy density (M L⁻¹T⁻²).
Use it when: Matching quantities.
Dimensionless arguments
sin, log and eˣ take only dimensionless arguments, so ωt, kx and t/τ are dimensionless.
Use it when: Checking wave and decay equations.
Check, convert, derive
Three uses; none gives the constant.
Use it when: Listing applications.

Dimensional Analysis and its Applications: solved examples

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

Example 1: Check an equation · 2 marks · Very short answerBoard level

Check the dimensional correctness of $v^{2} = u^{2} + 2 a s$ .

Try it first, then show the solutionSolution

[v²] = [L²T⁻²], [u²] = [L²T⁻²], [as] = [LT⁻²][L] = [L²T⁻²]. (1)
All terms have the same dimension, so the equation is dimensionally consistent. (1)

Answer: Dimensionally correct.

Example 2: Dimension of a constant · 2 marks · Very short answerBoard level

In $F = \frac{G m _{1} m _{2}}{r ^{2}}$ , find the dimensional formula of G.

Try it first, then show the solutionSolution

G = Fr²/(m₁m₂). (1)
[G] = [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²]. (1)

Answer: [M⁻¹L³T⁻²].

Example 3: Unit conversion · 2 marks · Very short answerBoard level

Convert 5 J into erg using dimensions (1 J = 10⁷ erg).

Try it first, then show the solutionSolution

[E] = ML²T⁻². n₂ = 5 × (1000 g/1 g)(100 cm/1 cm)²(1 s/1 s)⁻² = 5 × 10³ × 10⁴. (1)
n₂ = 5 × 10⁷ erg. (1)

Answer: 5 × 10⁷ erg.

Example 4: Derive the time period of a pendulum · 3 marks · Short answerHarder board level

Assuming T depends on the length L, the mass m and g, derive the form of T.

Try it first, then show the solutionSolution

T = k mᵃ Lᵇ gᶜ. [T] = [M]ᵃ[L]ᵇ[LT⁻²]ᶜ = MᵃLᵇ⁺ᶜT⁻²ᶜ. (1)
M: a = 0; T: −2c = 1 so c = −1/2; L: b + c = 0 so b = 1/2. (1)
T = k√(L/g); k = 2π comes from experiment, not from dimensions. (1)

Answer: T = k√(L/g), independent of m.

Example 5: Dimensions of a constant in a given relation · 4 marks · Long answerHOTS / competency

The viscous force on a sphere is $F = 6 π η r v$ . (a) Find [η]. (b) By dimensions, show that the speed v of ripples of wavelength λ on a liquid of surface tension S and density ρ varies as $v \propto S / (ρ λ)$ .

Try it first, then show the solutionSolution

(a) [η] = [F]/[rv] = MLT⁻²/(L·LT⁻¹) = [ML⁻¹T⁻¹]. (1)
(b) Let v = k Sᵃ ρᵇ λᶜ. [S] = F/L = MT⁻², [ρ] = ML⁻³. (1)
M: a + b = 0; T: −2a = −1 so a = 1/2, b = −1/2. (1)
L: −3b + c = 1 → 3/2 + c = 1, c = −1/2. So v ∝ √(S/(ρλ)). (1)

Answer: (a) [ML⁻¹T⁻¹]; (b) v = k√(S/(ρλ)).

Where students lose marks in Dimensional Analysis and its Applications

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

Mistake: Saying a dimensionally correct equation must be correct.
Confusing necessary with sufficient.
Do this: Dimensions cannot detect dimensionless factors or constants.
Usually costs: 1 mark.
Mistake: Adding quantities of different dimensions inside a bracket.
Missing a hidden unit mismatch.
Do this: Check each term of the sum separately.
Usually costs: The check mark.
Mistake: Writing [G] with L² instead of L³.
Careless algebra with r².
Do this: Redo: G = Fr²/(m₁m₂).
Usually costs: 1 mark.
Mistake: Putting a dimensional quantity inside sin or log.
Ignoring the dimensionless rule.
Do this: Argument = number × unit⁻¹, so it must be dimensionless.
Usually costs: Reasoning mark.

Exam tips

For a "derive using dimensions" question list the assumed form, the three equations (M, L, T) and the solution, one line each.
Always add the sentence that the numerical constant cannot be found by dimensions.
Remember the standard dimensional formulae of G, h, η, S and pressure.
Treat angles, strain and refractive index as dimensionless.
Check each homogeneity question term by term.

Practice questions on Dimensional Analysis and its Applications

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
The dimensional formula of pressure is
(a) [MLT⁻²] (b) [ML⁻¹T⁻²] (c) [ML²T⁻²] (d) [ML⁻²T⁻²]
Hint
Force over area.
Answer
(b).
Q2 · 2 marks · Very short answerHarder board level
In the van der Waals equation (P + a/V²)(V − b) = RT, find the dimensional formulae of a and b.
Hint
a/V² must have the dimension of P; b must have that of V.
Answer
[a] = [P][V²] = ML⁻¹T⁻² × L⁶ = [ML⁵T⁻²]; [b] = [V] = [L³].
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The equation s = ut + at² is dimensionally correct but physically wrong.
Reason (R): Dimensional analysis cannot determine numerical factors such as ½.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Check at, then at².
Answer
(a). [ut] = [L]; [at²] = [L], so the check passes, but the factor ½ is missing.
Q4 · 2 marks · Very short answerHarder board level
The magnitude of the force F between two charges is $F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}$ . If a force is 1 N, convert it to dyne.
Hint
Use [F] = MLT⁻².
Answer
1 N = 10³ g × 10² cm/s² = 10⁵ dyne.
Q5 · 4 marks · Long answerHOTS / competency
In the relation $y = A sin (ω t - k x)$ (a) state the dimensions of ω and k; (b) hence find the dimensions of ω/k and say what it represents.
Hint
The argument must be dimensionless.
Answer
(a) ωt dimensionless: [ω] = [T⁻¹]; kx dimensionless: [k] = [L⁻¹]. (b) ω/k has dimension [LT⁻¹], that of speed: the wave speed.

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.[E]?Show answer
ML²T⁻²
2.[G]?Show answer
M⁻¹L³T⁻²
3.[h]?Show answer
ML²T⁻¹
4.[Surface tension S]?Show answer
MT⁻²
5.[Viscosity η]?Show answer
ML⁻¹T⁻¹
6.Principle of homogeneity?Show answer
All terms in a valid equation have the same dimensions.

Dimensional Analysis and its Applications: frequently asked questions

Can dimensions prove a formula?

No. They only test consistency and give the form up to a dimensionless constant.

Why does it fail for s = ut + ½at²?

It is a sum of two terms, so no single product of powers represents it; the principle can still check each term.

Are angles dimensionless?

Yes, an angle is a ratio of two lengths (arc over radius).

What if a quantity depends on four variables?

With only three base dimensions (M, L, T) you cannot solve uniquely for four exponents.

Are dimensions the same as units?

No. Dimensions show the type of quantity; units are specific scales such as metre or foot.