Dimensional Analysis — Checking Answers by Units

A student derives a formula for the time period of a pendulum as $T = 2\pi\sqrt{g/L}$ instead of $T = 2\pi\sqrt{L/g}$ . Without a calculator, without solving equations, dimensional analysis immediately flags the error — because $\sqrt{g/L}$ has units $\sqrt{(m/s^2)/m} = \sqrt{1/s^2} = 1/s$ , which would give $T$ in seconds per (1/s) = s², not seconds.

Dimensional analysis is the art of using units as an algebra system. It’s fast, powerful, and catches errors that numerical checking might miss. In physics, it’s called dimensional analysis. In chemistry, it’s unit factor method. In everyday problem-solving, it’s just good sense.

For JEE, dimensional analysis is a standalone chapter with direct exam questions. For CBSE, it underpins how you verify your answers.

Key Terms & Definitions

Dimension: The physical nature of a quantity — expressed in terms of fundamental dimensions: Mass [M], Length [L], Time [T], Temperature [θ], Current [A], Amount [N], Luminosity [J].

Dimensional Formula: The expression showing which fundamental dimensions appear in a quantity and with what power. Example: velocity = distance/time = [L][T]⁻¹ = [LT⁻¹].

Dimensionally Consistent: A equation where both sides have identical dimensions. A necessary (but not sufficient) condition for a correct equation.

Dimensionless Quantity: A quantity with dimension [M⁰L⁰T⁰] — pure numbers. Examples: π, angles (in radians), relative density, refractive index, strain.

Principle of Homogeneity: In any valid physical equation, every term must have the same dimensions. You cannot add speed to acceleration — they have different dimensions.

Fundamental and Derived Dimensions

Fundamental (Base) Dimensions:

Quantity	Symbol	SI Unit
Mass	[M]	kg
Length	[L]	m
Time	[T]	s
Temperature	[θ]	K
Electric current	[A]	A

Common Derived Dimensions:

Quantity	Dimensional Formula
Velocity	[LT⁻¹]
Acceleration	[LT⁻²]
Force	[MLT⁻²]
Energy/Work	[ML²T⁻²]
Power	[ML²T⁻³]
Pressure	[ML⁻¹T⁻²]
Momentum	[MLT⁻¹]
Torque	[ML²T⁻²]

Three Applications of Dimensional Analysis

Application 1 — Checking Dimensional Consistency

To check if a formula is dimensionally correct, verify that both sides have the same dimension.

Example: Check if $v^2 = u^2 + 2as$ is dimensionally correct.

LHS: $[v^2] = [LT^{-1}]^2 = [L^2T^{-2}]$

RHS: $[u^2] = [L^2T^{-2}]$ ✓

$[2as] = [LT^{-2}][L] = [L^2T^{-2}]$ ✓

All terms are $[L^2T^{-2}]$ → dimensionally consistent.

Application 2 — Deriving Formulas

If we suspect a quantity depends on certain variables, dimensional analysis can find the formula (up to a dimensionless constant).

Example: Time period $T$ of a simple pendulum — find how it depends on $L$ (length) and $g$ (acceleration due to gravity).

Let $T = kL^a g^b$ where $k$ is a dimensionless constant.

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

Comparing powers:

Time: $1 = -2b \Rightarrow b = -1/2$
Length: $0 = a + b \Rightarrow a = 1/2$

Therefore: $T = k\sqrt{L/g}$

(The constant $k = 2\pi$ comes from solving the differential equation — dimensional analysis can’t determine it.)

Application 3 — Unit Conversion

Dimensional analysis is the most reliable method for converting between unit systems.

Example: Convert $60$ km/h to m/s.

60 \text{ km/h} = 60 \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{60000}{3600} \text{ m/s} = \frac{50}{3} \approx 16.67 \text{ m/s}

The unit fractions $\frac{1000 \text{ m}}{1 \text{ km}}$ and $\frac{1 \text{ h}}{3600 \text{ s}}$ are both equal to 1 — multiplying by them doesn’t change the value, only the units.

Solved Examples

Example 1 — Easy (Check Dimensional Consistency)

Check: $s = ut + \frac{1}{2}at^2$

$[s] = [L]$

$[ut] = [LT^{-1}][T] = [L]$ ✓

$[\frac{1}{2}at^2] = [LT^{-2}][T^2] = [L]$ ✓

All terms have dimension [L] → dimensionally consistent.

Example 2 — Medium (Derive Formula)

The time of oscillation $T$ of a spring-mass system depends on mass $m$ and spring constant $k$ (force per unit length, dimension $[MT^{-2}]$ ). Find the formula.

Let $T = m^a k^b$ .

$[T] = [M]^a [MT^{-2}]^b = [M^{a+b} T^{-2b}]$

T: $1 = -2b \Rightarrow b = -1/2$
M: $0 = a + b \Rightarrow a = 1/2$

$T = C\sqrt{m/k}$ (where $C = 2\pi$ from physics)

Example 3 — Hard (JEE Level)

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

Find the dimensions of $a$ and $b$ .

Solution: For dimensional consistency, $\frac{a}{V^2}$ must have the same dimensions as $P$ :

$[a] = [P][V^2] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$

For $V - b$ : $b$ must have the same dimensions as $V$ :

$[b] = [V] = [L^3]$

(where $V$ here is volume — if it’s molar volume, $[V] = [L^3 \text{ mol}^{-1}]$ and $[b] = [L^3\text{mol}^{-1}]$ )

Example 4 — Unit Conversion Chain

Convert $9.8 \text{ m/s}^2$ to $\text{cm/s}^2$ :

9.8 \text{ m/s}^2 \times \frac{100 \text{ cm}}{1 \text{ m}} = 980 \text{ cm/s}^2

Exam-Specific Tips

JEE Main: Dimensional analysis questions ask you to: (1) find dimensions of a physical constant (like the Van der Waals constants), (2) derive a formula, (3) identify which combination of given quantities has the dimension of a specified quantity. The dimensional formulas of energy, power, momentum, and pressure are the most used.

CBSE Class 11: The typical question is “check dimensional consistency of [formula]” — worth 2 marks. Show LHS and RHS dimensions separately and compare. Always state the conclusion: “Since both sides have dimension [X], the equation is dimensionally consistent.”

Memorise these dimensional equivalences: $[Force] = [MLT^{-2}]$ , $[Energy] = [ML^2T^{-2}]$ , $[Pressure] = [ML^{-1}T^{-2}]$ , $[Planck's\ constant] = [ML^2T^{-1}]$ . These appear repeatedly across problems.

Common Mistakes to Avoid

Mistake 1: Treating dimensionless constants as having dimensions. Numbers like 2, π, ½ are dimensionless. Do not include them in dimensional analysis — they don’t affect the dimension of a term.

Mistake 2: Forgetting that a dimensionally consistent equation may still be wrong. Dimensional analysis can prove an equation WRONG but cannot prove it right. The equation $s = 2ut$ is dimensionally consistent ( $[L] = [L]$ ) but physically incorrect. The constant factor (2 instead of the correct value) is invisible to dimensional analysis.

Mistake 3: Mixing dimensions with units. Dimension [M] is “mass” — it’s the same whether in kg, g, or pounds. Unit conversion is about changing the scale; dimensional analysis is about the fundamental nature.

Mistake 4: Forgetting that you cannot add quantities with different dimensions. In any equation, every term that is added or subtracted must have the same dimension. Speed + acceleration is not meaningful — they have different dimensions.

Practice Questions

Q1. Find the dimension of pressure.

Pressure = Force / Area = $[MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$

Q2. Is the equation $v = u + at^2$ dimensionally correct?

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

$[u] = [LT^{-1}]$ ✓ but $[at^2] = [LT^{-2}][T^2] = [L]$ ✗

$[LT^{-1}] \neq [L]$ → The equation is dimensionally INCORRECT. (The correct equation is $v = u + at$ .)

Q3. If $E = mc^2$ , what is the dimension of energy?

$[E] = [m][c^2] = [M][LT^{-1}]^2 = [ML^2T^{-2}]$

Q4. Convert 1 joule to ergs. (1 m = 100 cm, 1 kg = 1000 g)

$1 \text{ J} = 1 \text{ kg·m}^2/\text{s}^2 = 1000 \text{ g} \times (100 \text{ cm})^2 / \text{s}^2 = 1000 \times 10000 \text{ g·cm}^2/\text{s}^2 = 10^7 \text{ ergs}$

FAQs

What is the difference between dimensions and units?

Dimensions describe the nature of a quantity (mass, length, time). Units are the specific scales for measuring those dimensions (kg for mass, m for length, s for time). The same dimension can have many units — mass can be in kg, g, lb, or tonne.

Can dimensional analysis find the value of dimensionless constants?

No. Dimensional analysis can find the form of a formula (which variables appear and with what powers) but cannot determine numerical constants like $2\pi$ , $\frac{1}{2}$ , etc. These require solving the actual physics.

What is the dimension of angle?

Angle (in radians) = arc length / radius = length / length = dimensionless, [M⁰L⁰T⁰]. This is why trigonometric functions (sin, cos, tan) are dimensionless — their argument (angle) is dimensionless, and dimensionless functions are dimensionless.

Additional Solved Examples

Example 5 — Deriving centripetal force formula

Centripetal force $F$ on a body in circular motion depends on mass $m$ , velocity $v$ , and radius $r$ . Find the formula.

Let $F = km^a v^b r^c$ .

$[MLT^{-2}] = [M]^a [LT^{-1}]^b [L]^c = [M^a L^{b+c} T^{-b}]$

Comparing: $M: a = 1$ , $T: -b = -2 \implies b = 2$ , $L: b + c = 1 \implies c = -1$ .

F = k\frac{mv^2}{r}

Physics gives $k = 1$ .

Example 6 — Dimensions of universal constants

Constant	Formula used	Dimensional formula
Gravitational constant $G$	$F = Gm_1m_2/r^2$	$[M^{-1}L^3T^{-2}]$
Planck’s constant $h$	$E = h\nu$	$[ML^2T^{-1}]$
Boltzmann constant $k_B$	$E = k_BT$	$[ML^2T^{-2}\theta^{-1}]$
Stefan’s constant $\sigma$	$P/A = \sigma T^4$	$[MT^{-3}\theta^{-4}]$
Permittivity $\varepsilon_0$	$F = q_1q_2/(4\pi\varepsilon_0 r^2)$	$[M^{-1}L^{-3}T^4A^2]$

JEE Main frequently asks: “Which of the following has the dimensions of Planck’s constant?” Common look-alikes: angular momentum $[ML^2T^{-1}]$ — same as $h$ . Energy $\times$ time also gives $[ML^2T^{-1}]$ .

Limitations of dimensional analysis

Cannot determine dimensionless constants (like $2\pi$ in $T = 2\pi\sqrt{L/g}$ )
Cannot derive formulas involving sums or differences of terms
Cannot distinguish between quantities with the same dimensions (torque vs energy, both $[ML^2T^{-2}]$ )
Fails when the quantity depends on dimensionless variables (like angle)

Additional Practice Questions

Q5. Find the dimensions of magnetic field $B$ from $F = qvB$ .

$[B] = [F/(qv)] = [MLT^{-2}]/([AT][LT^{-1}]) = [MLT^{-2}]/[ALT^{0}] = [MT^{-2}A^{-1}]$ . Unit: tesla (T) = kg/(A s $^2$ ).

Q6. By dimensional analysis, show that the formula $T = 2\pi\sqrt{m/k}$ is correct for a spring-mass system.

$[T] = [T]$ . $[\sqrt{m/k}] = \sqrt{[M]/[MT^{-2}]} = \sqrt{[T^2]} = [T]$ . Both sides have dimension $[T]$ . Dimensionally consistent.