Dimensional Analysis — Deriving and Checking Formulas

Dimensional analysis is one of the most underrated tools in physics. In 5 seconds, it can tell you whether an equation is definitely wrong. In 2 minutes, it can derive an entire formula. It catches the most common errors in calculation — wrong unit conversions, flipped fractions, forgotten factors.

At its core, dimensional analysis says: physics equations must balance dimensionally just as they balance numerically. Mass cannot equal velocity. Force cannot be added to energy. If both sides of an equation don’t have the same dimensions, the equation is wrong.

This is tested directly in CBSE Class 11 and JEE Main. But more importantly, it’s a habit of mind that makes you a better problem-solver.

Key Terms & Definitions

Physical Quantity: Anything measurable — length, mass, time, temperature, force, energy.

Fundamental Quantities: The seven base quantities: length [L], mass [M], time [T], temperature [θ], electric current [A], amount of substance [N], luminous intensity [J].

Dimensional Formula: Expression showing the powers of fundamental dimensions for a derived quantity. Example: $[\text{Force}] = [MLT^{-2}]$ .

Dimensionless Quantity: A quantity with dimensional formula $[M^0L^0T^0]$ — a pure number. Examples: angle, strain, relative density, refractive index.

Principle of Homogeneity: Every term added, subtracted, or equated in a physics equation must have the same dimensions.

Dimensional Constant: A constant that has dimensions, like $G$ (gravitational constant, $[M^{-1}L^3T^{-2}]$ ), $h$ (Planck’s constant, $[ML^2T^{-1}]$ ).

Dimensional Formulas of Common Quantities

Quantity	Formula	Dimensional Formula
Velocity	distance/time	$[LT^{-1}]$
Acceleration	velocity/time	$[LT^{-2}]$
Force	mass × acceleration	$[MLT^{-2}]$
Work/Energy	force × distance	$[ML^2T^{-2}]$
Power	work/time	$[ML^2T^{-3}]$
Pressure	force/area	$[ML^{-1}T^{-2}]$
Momentum	mass × velocity	$[MLT^{-1}]$
Torque	force × distance	$[ML^2T^{-2}]$
Frequency	1/time	$[T^{-1}]$
Angular velocity	angle/time	$[T^{-1}]$
Gravitational constant	$F = Gm_1m_2/r^2$	$[M^{-1}L^3T^{-2}]$
Planck’s constant	$E = h\nu$	$[ML^2T^{-1}]$
Boltzmann constant	$E = kT$	$[ML^2T^{-2}\theta^{-1}]$

Three Uses of Dimensional Analysis

Use 1 — Checking Dimensional Consistency

Any correct physics equation must be dimensionally homogeneous. If the dimensions don’t match on both sides, the equation is wrong.

Procedure:

Find the dimensional formula of each term
Verify they are all identical
If any term differs, the equation is incorrect

Note: Dimensional consistency is necessary but not sufficient — an equation can be dimensionally correct but still wrong (wrong numerical constant, for instance).

Use 2 — Deriving Formulas (Dimensional Method)

Given that a quantity depends on certain variables, dimensional analysis can determine how they relate.

Procedure:

Assume $Q = k \cdot x^a y^b z^c$ (where $k$ is dimensionless)
Write dimensions of both sides
Compare powers of M, L, T separately (three equations)
Solve for $a$ , $b$ , $c$

Limitation: Cannot find the dimensionless constant $k$ , and cannot work if the formula involves sums (like $v^2 = u^2 + 2as$ — can’t derive this form dimensionally).

Use 3 — Unit Conversion Between Systems

To convert a quantity from one unit system to another:

n_2 = n_1 \times \left(\frac{M_1}{M_2}\right)^a \left(\frac{L_1}{L_2}\right)^b \left(\frac{T_1}{T_2}\right)^c

Where $a$ , $b$ , $c$ are the powers in the dimensional formula, and subscripts 1, 2 refer to the two unit systems.

Solved Examples

Example 1 — Easy: Check Consistency

Is $v = u + at^2$ dimensionally correct?

LHS: $[v] = [LT^{-1}]$

RHS: $[u] + [at^2] = [LT^{-1}] + [LT^{-2}][T^2] = [LT^{-1}] + [L]$

$[LT^{-1}] \neq [L]$ → Dimensionally INCORRECT.

(The correct equation is $v = u + at$ , where $[at] = [LT^{-2}][T] = [LT^{-1}]$ ✓)

Example 2 — Medium: Derive Time Period

Time period $T$ of a simple pendulum depends on length $L$ and $g$ . Find $T$ .

Assume $T = k L^a g^b$ :

$[T] = [L]^a [LT^{-2}]^b = [L^{a+b} T^{-2b}]$

Comparing powers:

$T^1 = T^{-2b} \Rightarrow b = -1/2$
$L^0 = L^{a+b} \Rightarrow a = 1/2$

T = k\sqrt{\frac{L}{g}}

(Physics gives $k = 2\pi$ .)

Example 3 — Hard: Find Dimensions of G

Newton’s law: $F = \dfrac{Gm_1 m_2}{r^2}$

[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}]

Example 4 — Hard: Unit Conversion

Convert 1 joule to erg.

$\text{Energy} = [ML^2T^{-2}]$

SI: $M_1 = 1$ kg, $L_1 = 1$ m, $T_1 = 1$ s.

CGS: $M_2 = 1$ g, $L_2 = 1$ cm, $T_2 = 1$ s.

n_2 = n_1 \times \left(\frac{M_1}{M_2}\right)^1 \left(\frac{L_1}{L_2}\right)^2 \left(\frac{T_1}{T_2}\right)^{-2}

= 1 \times \frac{1 \text{ kg}}{1 \text{ g}} \times \left(\frac{1 \text{ m}}{1 \text{ cm}}\right)^2 \times 1 = 1000 \times 10000 \times 1 = 10^7

$1 \text{ J} = 10^7 \text{ erg}$

Example 5 — Advanced (JEE Style)

Van der Waals equation: $\left(P + \frac{a}{V^2}\right)(V - b) = nRT$

Find dimensions of $a$ and $b$ .

Solution: $a/V^2$ must have dimensions of pressure: $[a] = [P][V^2] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$

$b$ has dimensions of volume: $[b] = [L^3]$ (or $[L^3 mol^{-1}]$ if $V$ is molar volume)

Exam-Specific Tips

CBSE Class 11: The standard 2-mark question: “Check if [equation] is dimensionally correct.” Show LHS, RHS separately, then state: “Since LHS and RHS have the same dimension [X], the equation is dimensionally correct/incorrect.” This template scores full marks.

JEE Main 2024 asked: “Which of the following has the same dimension as Planck’s constant?” Answer: angular momentum ( $[ML^2T^{-1}]$ ). Always know $[h] = [ML^2T^{-1}]$ and its look-alikes: angular momentum, action.

Pairs of quantities with the same dimensional formula are favourite MCQ options:

Force and surface tension × length: both $[MLT^{-2}]$
Energy and torque: both $[ML^2T^{-2}]$
Pressure and energy density: both $[ML^{-1}T^{-2}]$
Planck’s constant and angular momentum: both $[ML^2T^{-1}]$

Common Mistakes to Avoid

Mistake 1: Thinking dimensional analysis can prove a formula correct. It can only prove it WRONG. A dimensionally correct equation may still be wrong (missing or wrong constant). Dimensional analysis is a necessary, not sufficient, test.

Mistake 2: Forgetting that angles are dimensionless. Angles in radians = arc/radius = L/L = dimensionless. Therefore sin θ, cos θ, tan θ are all dimensionless. Arguments of trigonometric and exponential functions must always be dimensionless.

Mistake 3: Treating the numerical constant as part of dimensional analysis. The number 2π, ½, etc. are dimensionless — ignore them when checking dimensions. Write $[v^2 = u^2 + 2as]$ as $[v^2] = [u^2]$ and $[2as] = [as]$ , not $[2as]$ treating 2 as a separate dimensional entity.

Mistake 4: Wrong powers in unit conversion. When converting energy (dimension $[ML^2T^{-2}]$ ), the length must be squared: $(L_1/L_2)^2$ , not $(L_1/L_2)^1$ . Using the power 1 instead of 2 for a squared dimension is a very common calculation error.

Mistake 5: Using dimensions of force for pressure or vice versa. $[F] = [MLT^{-2}]$ , $[P] = [ML^{-1}T^{-2}]$ . The difference is $L^{-2}$ — because pressure is force per unit area. Many students confuse these in complex dimensional problems.

Practice Questions

Q1. Find the dimensional formula of Boltzmann constant $k_B$ from $E = k_B T$ .

$[k_B] = [E/T] = [ML^2T^{-2}] / [\theta] = [ML^2T^{-2}\theta^{-1}]$

Q2. Check: $E = mc^2$ (where $c$ = speed of light)

$[mc^2] = [M][LT^{-1}]^2 = [ML^2T^{-2}]$ = $[E]$ ✓ Dimensionally correct.

Q3. Is $\frac{1}{2}mv^2 + mgh$ dimensionally valid?

$[\frac{1}{2}mv^2] = [M][LT^{-1}]^2 = [ML^2T^{-2}]$

$[mgh] = [M][LT^{-2}][L] = [ML^2T^{-2}]$

Both terms have $[ML^2T^{-2}]$ ✓ The sum is valid.

Q4. By dimensional analysis, find how velocity of a wave on a string depends on tension $T$ (dimensions $[MLT^{-2}]$ ) and linear density $\mu$ ( $[ML^{-1}]$ ).

$[v] = [T^a][\mu^b] = [MLT^{-2}]^a [ML^{-1}]^b = [M^{a+b} L^{a-b} T^{-2a}]$

$L: 1 = a - b$ , $T: -1 = -2a \Rightarrow a = 1/2$ , then $b = -1/2$

$v = k\sqrt{T/\mu}$ (the physics gives $k = 1$ , so $v = \sqrt{T/\mu}$ )

FAQs

Can dimensional analysis work for all physics problems?

No. It cannot determine dimensionless constants. It cannot distinguish between quantities with the same dimensions (force and surface tension × length are both $[MLT^{-2}]$ ). It fails for equations with sums or differences of variables. But as a quick consistency check, it works universally.

What does it mean for a quantity to be dimensionless?

A dimensionless quantity has the same value regardless of the unit system. Examples: refractive index (ratio of two speeds), strain (ratio of two lengths), specific gravity (ratio of two densities). They are pure ratios, independent of measurement units.

Why can we not add force and energy?

Force has dimension $[MLT^{-2}]$ and energy has $[ML^2T^{-2}]$ . Adding them is like adding meters to kilograms — physically meaningless. The principle of homogeneity mathematically encodes this physical truth.

Is dimensional analysis the same as unit analysis?

Very similar. Dimensional analysis works with abstract dimensions (M, L, T). Unit analysis works with specific units (kg, m, s). Unit analysis is more concrete; dimensional analysis is more general. For checking formulas, dimensional analysis is preferred; for numerical calculations, unit analysis is cleaner.

Additional Solved Examples

Example 6 — JEE Main Style: Finding which quantity has given dimensions

Which of the following has dimensions $[ML^2T^{-3}]$ ? (a) Force (b) Energy (c) Power (d) Momentum

$[Force] = [MLT^{-2}]$ , $[Energy] = [ML^2T^{-2}]$ , $[Power] = [ML^2T^{-3}]$ , $[Momentum] = [MLT^{-1}]$ .

Answer: (c) Power.

Example 7 — Deriving viscosity dimensions

Viscous force: $F = 6\pi\eta rv$ (Stokes’ law). Find $[\eta]$ .

[\eta] = \frac{[F]}{[r][v]} = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = \frac{[MLT^{-2}]}{[L^2T^{-1}]} = [ML^{-1}T^{-1}]

Unit: Pa s (pascal second) or poise (CGS).

Pairs with same dimensions (JEE favourite)

Pair	Common dimension
Work and torque	$[ML^2T^{-2}]$
Pressure and stress and energy density	$[ML^{-1}T^{-2}]$
Angular momentum and Planck’s constant	$[ML^2T^{-1}]$
Impulse and momentum	$[MLT^{-1}]$
Force and surface tension $\times$ length	$[MLT^{-2}]$
Frequency and decay constant	$[T^{-1}]$

When JEE asks “which pair has same dimensions?”, check these standard pairs. They recur almost every year.

Additional Practice Questions

Q5. Find the dimensions of the coefficient of thermal conductivity from $Q = \frac{kA\Delta T \cdot t}{d}$ .

$[k] = \frac{[Q][d]}{[A][\Delta T][t]} = \frac{[ML^2T^{-2}][L]}{[L^2][\theta][T]} = [MLT^{-3}\theta^{-1}]$ .

Q6. Show that $\frac{e^2}{4\pi\varepsilon_0 m_e c^2}$ has dimensions of length.

$[e^2/\varepsilon_0] = [Force \times r^2] = [MLT^{-2}][L^2] = [ML^3T^{-2}]$ .

$[m_e c^2] = [M][LT^{-1}]^2 = [ML^2T^{-2}]$ .

$\frac{[ML^3T^{-2}]}{[ML^2T^{-2}]} = [L]$ . Confirmed — this is the classical electron radius.