Physics
Units and Measurements
Dimensional Analysis

Units and Measurements · JEE & NEET Physics

Dimensional Analysis: notes and previous year questions

Using dimensions to check equations, derive formulas, find unknown constants and convert units, and where the method stops working.

115 JEE Main questions (2002–2026)
14 JEE Advanced questions (1998–2026)
25 NEET questions (2000–2026)

Dimensional Analysis in short

Dimensional analysis checks equations, derives formulas and converts units, using only dimensions.
Principle of homogeneity: every term of a correct equation has the same dimensions.
Only like quantities can be added, subtracted or equated.
What sits inside sin, cos, log or eˣ must be dimensionless.

1What dimensional analysis does

Dimensional analysis means working with the dimensions of quantities, their powers of $[\mathrm{M}]$ , $[\mathrm{L}]$ and $[\mathrm{T}]$ , instead of their numbers. It needs no experiment: only the dimensional formulas you already know. It does three jobs:

Check whether an equation could be right.
Derive how one quantity depends on others.
Convert a value from one system of units to another.

2Checking equations

The principle of homogeneity says that every term of a correct equation has the same dimensions. You can only add, subtract or equate like quantities: metres to metres, never metres to seconds.

A = B + C \;\Rightarrow\; [A] = [B] = [C]

pure numbers like ½, 2 and π do not count

Whatever sits inside $\sin$ , $\cos$ , $\log$ or $e^x$ must be dimensionless. In $y = A\sin\omega t$ , the product $\omega t$ has no dimensions, so $[\omega] = [\mathrm{T}^{-1}]$ .

3Deriving a formula

When you know which quantities something depends on, dimensional analysis can find how. The steps:

List the quantities it depends on.
Assume a product of powers, $Q = k\,A^a B^b C^c$ , where $k$ is a pure number.
Write the dimensions of every quantity.
Match the powers of M, L and T on both sides. This gives one equation for each.
Solve for $a$ , $b$ , $c$ and write the formula, keeping $k$ .

ExampleSpeed of sound in a gas

Assume $v = k\,P^a\rho^b$ with $[P] = [\mathrm{ML}^{-1}\mathrm{T}^{-2}]$ and $[\rho] = [\mathrm{ML}^{-3}]$ . Then $[\mathrm{LT}^{-1}] = [\mathrm{M}^{a+b}\mathrm{L}^{-a-3b}\mathrm{T}^{-2a}]$ . T: $a = \tfrac12$ . M: $b = -a = -\tfrac12$ . L: $-\tfrac12 + \tfrac32 = 1$ , which checks.

v = k\sqrt{P/\rho}

For air, $P = 1.01 \times 10^5$ Pa and $\rho = 1.29$ kg/m³ give $\sqrt{P/\rho} \approx 280$ m/s, but sound travels at about 331 m/s. The missing number is $k = \sqrt{\gamma} = \sqrt{1.4} \approx 1.18$ , and $280 \times 1.18 \approx 331$ . Dimensions give the form; experiment gives $k$ .

4Finding the dimensions of constants

When a formula contains unknown constants, homogeneity fixes their dimensions: each term must have the dimensions of the whole.

Some groups of quantities have no dimensions at all. The Reynolds number tells smooth flow from swirling flow, and is the same pure number in any unit system:

Re = \frac{\rho v d}{\eta}

ρ density, v speed, d diameter, η viscosity

\frac{[\mathrm{ML}^{-3}][\mathrm{LT}^{-1}][\mathrm{L}]}{[\mathrm{ML}^{-1}\mathrm{T}^{-1}]} = [\mathrm{M}^0\mathrm{L}^0\mathrm{T}^0]

5Converting units

A quantity is the same whatever units you use, so $n_1 u_1 = n_2 u_2$ : a bigger unit needs a smaller number. For a quantity with dimensions $[\mathrm{M}^a\mathrm{L}^b\mathrm{T}^c]$ :

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

M₁, L₁, T₁ are the units of the system you start in

6What it cannot do

It cannot find pure numbers such as $2\pi$ , $6\pi$ or $\tfrac12$ .
It cannot derive a formula that is a sum of terms, such as $s = ut + \tfrac12 at^2$ or $v = u + at$ . It only finds a single product of powers.
It cannot make trigonometric, logarithmic or exponential functions, as in $y = A\sin\omega t$ or $N = N_0 e^{-\lambda t}$ .
In mechanics, M, L and T give only three equations, so it can find at most three unknown powers. If a quantity depends on four or more others, it cannot fix them all.
It cannot tell apart quantities with the same dimensions: work and torque are both $[\mathrm{ML}^2\mathrm{T}^{-2}]$ .
A dimensionally correct equation can still be wrong by a pure number.

Job	What it can do	What it cannot do
Check equations	Show that an equation is wrong	Prove that an equation is right
Derive formulas	Find a product of powers	Find pure numbers, sums, sin, log, eˣ
Find constants	Give the dimensions of a constant	Tell apart quantities with the same dimensions
Convert units	Move a value between unit systems	—

7A quick plan for problems

Checking: find the dimensions of each term separately, and check that they all match. Look inside sin, cos, log and exp: that part must be dimensionless.
Deriving: list the quantities, assume a product of powers with a pure number $k$ , write one equation each for M, L and T, solve, and keep $k$ in the answer.
Converting: write the dimensional formula, note the ratio of each base unit, and apply the conversion rule with the right powers.

Summary

Key ideas

Dimensional analysis checks equations, derives formulas and converts units, using only dimensions.
Principle of homogeneity: every term of a correct equation has the same dimensions.
Only like quantities can be added, subtracted or equated.
What sits inside sin, cos, log or eˣ must be dimensionless.
A failed check proves an equation wrong; a pass only means it might be right.
To derive a formula, assume Q = k Aᵃ Bᵇ Cᶜ and match the powers of M, L and T.
A pendulum's period is T = k√(l/g): it does not depend on the mass of the bob.
The unknown constant k is a pure number, found by experiment (2π for the pendulum, 6π in Stokes' law).
Homogeneity gives the dimensions of unknown constants in a formula.
To convert units, multiply by each base-unit ratio raised to its power in the dimensional formula.
The method cannot find pure numbers, sums, trigonometric, log or exponential forms, or more than three unknown powers.
Same dimensions do not mean the same quantity: work and torque are both [ML²T⁻²].

Every equation

Homogeneity: $A = B + C \Rightarrow [A] = [B] = [C]$
Check s = ut + ½at²: $[s] = [ut] = [\tfrac12 at^2] = [\mathrm{L}]$
Check v² = u² + 2as: $[v^2] = [u^2] = [2as] = [\mathrm{L}^2\mathrm{T}^{-2}]$
Assumed form: $Q = k\,A^a B^b C^c$
Pendulum: $T = k\sqrt{l/g},\ k = 2\pi$
Stokes' law: $F = k\,\eta r v,\ k = 6\pi$
Centripetal force: $F = mv^2/r$
Speed of sound: $v = k\sqrt{P/\rho},\ k = \sqrt{\gamma}$
Liquid drop: $T = k\sqrt{\rho r^3/S}$
Spring: $T = 2\pi\sqrt{m/k}$
Constant A in F = Av + Bv³: $[A] = [\mathrm{MT}^{-1}]$
Constant B in F = Av + Bv³: $[B] = [\mathrm{ML}^{-2}\mathrm{T}]$
Constant a in P = a/V − b/V²: $[a] = [\mathrm{ML}^2\mathrm{T}^{-2}]$
Constant b in P = a/V − b/V²: $[b] = [\mathrm{ML}^5\mathrm{T}^{-2}]$
Planck's constant: $[h] = [E]/[\nu] = [\mathrm{ML}^2\mathrm{T}^{-1}]$
Reynolds number: $[\rho v d/\eta] = [\mathrm{M}^0\mathrm{L}^0\mathrm{T}^0]$
Mass with F, L, T as base: $[m] = [\mathrm{FL}^{-1}\mathrm{T}^2]$
Conversion rule: $n_2 = n_1 (M_1/M_2)^a (L_1/L_2)^b (T_1/T_2)^c$
Dyne: $1\ \text{dyne} = 10^{-5}\ \text{N}$
Erg: $1\ \text{J} = 10^7\ \text{erg}$
Watt: $1\ \text{W} = 10^7\ \text{erg/s}$
G in CGS: $G = 6.67 \times 10^{-8}\ \text{(CGS)}$

Previous year questions with solutions

Real JEE and NEET questions on dimensional analysis. Try each one before you open the solution.

Q1JEE Main 2026One correct option

The potential energy of a particle changes with distance $x$ from a fixed origin as $V=\frac{A\sqrt{x}}{x+B}$ , where $A$ and $B$ are constant with appropriate dimensions. The dimensions of $AB$ are $\_\_\_\_$

A $[M^{1}L^{5/2}T^{-2}]$
B $[M^{3/2}L^{5/2}T^{-2}]$
C $[M^{1}L^{2}T^{-2}]$
D $[M^{1}L^{7/2}T^{-2}]$

Show answer and solution

Answer: Option D

Take the addition first: $B$ is added to $x$ , so $[B] = [\mathrm{L}]$ and the whole bracket is $[\mathrm{L}]$ .

$V$ here is a potential energy, $[\mathrm{ML^{2}T^{-2}}]$ , so $[A] = \dfrac{[V][\mathrm{L}]}{[\sqrt{x}]} = \dfrac{[\mathrm{ML^{2}T^{-2}}][\mathrm{L}]}{[\mathrm{L^{1/2}}]} = [\mathrm{ML^{5/2}T^{-2}}]$ .

Multiplying by $[B] = [\mathrm{L}]$ raises the power of length by one more: $[AB] = [\mathrm{ML^{7/2}T^{-2}}]$ , which is option D.

Q2JEE Advanced 2026Numerical answer

In a new system of units, the units of mass, length, time and current are $5 \mathrm{kg},5 m,5 s$ and 5 A , respectively. If ${\mu}_{0}$ and ${\epsilon}_{0}$ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $\sqrt{{\mu}_{0}/{\epsilon}_{0}}$ , is :

Show answer and solution

Answer: 25

First find the dimensional formula. $\dfrac{[\mu_{0}]}{[\epsilon_{0}]} = \dfrac{[\mathrm{MLT^{-2}A^{-2}}]}{[\mathrm{M^{-1}L^{-3}T^{4}A^{2}}]} = [\mathrm{M^{2}L^{4}T^{-6}A^{-4}}]$ , and the square root of that is $[\mathrm{ML^{2}T^{-3}A^{-2}}]$ — a resistance.

Now change systems. A quantity's number goes up when the unit gets smaller, so $n_{2} = n_{1}\left(\dfrac{M_{1}}{M_{2}}\right)^{1}\left(\dfrac{L_{1}}{L_{2}}\right)^{2}\left(\dfrac{T_{1}}{T_{2}}\right)^{-3}\left(\dfrac{A_{1}}{A_{2}}\right)^{-2}$ , with every ratio equal to $\tfrac{1}{5}$ .

The powers add up to $1 + 2 - 3 - 2 = -2$ , so $n_{2} = \left(\tfrac{1}{5}\right)^{-2} = 25$ .

Q3NEET 2026One correct option

The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is:

A $3\times {10}^{8}$
B500
C $3\times {10}^{10}$
D400

Show answer and solution

Answer: Option D

Taking the speed of light as $1$ makes the new unit of length the distance light covers in one second. The travel time is $6 \times 60 + 40 = 400\ \mathrm{s}$ , so the Sun is $400$ of those units away: distance $= 1 \times 400 = 400$ , option D.

Practice questions, easy to hard

Three questions from the dimensional analysis practice ladder: one easy, one medium, one hard.

Q4One or more correct options

Select every number below that is smaller than $1$ .

A $5 \times 10^{-1}$
B $5 \times 10^{1}$
C $2 \times 10^{-6}$
D $1.6 \times 10^{-19}$

Show answer and solution

Answer: Options A, C, D

The minus sign on the power is the whole test here. $5 \times 10^{-1}$ is $0.5$ and $5 \times 10^{1}$ is $50$ : same $a$ , opposite sizes. That last one, $1.6 \times 10^{-19}$ , is worth remembering the look of, because it turns up again before this rung is over.

Q5One correct option

Mass, length and time cover most of mechanics, but not all of physics. Temperature is a kind of its own, written $[\mathrm{K}]$ after the kelvin. Electric current has one too, $[\mathrm{A}]$ after the ampere, and it turns up in a later rung; here we stay with $\mathrm{M}$ , $\mathrm{L}$ , $\mathrm{T}$ and $\mathrm{K}$ .

What is the dimensional formula of the temperature of a cup of tea?

A $[\mathrm{T}]$
B $[\mathrm{M}]$
C $[\mathrm{K}]$
D $[\mathrm{TK}]$

Show answer and solution

Answer: Option C

Temperature is $[\mathrm{K}]$ . The trap is $[\mathrm{T}]$ , which time had already taken long before temperature came along, so temperature borrows the first letter of the kelvin instead.

Q6One or more correct options

That $\frac{1}{2}$ is worth stopping on, because it marks the honest limit of this tool. A dimension check looks only at the powers of $[\mathrm{M}]$ , $[\mathrm{L}]$ and $[\mathrm{T}]$ , and plain numbers like $\frac{1}{2}$ , $2\pi$ and $7$ have none of those, so the check cannot see them.

Select every statement that is true.

AA dimension check can show that a formula is wrong.
BA dimension check can tell you the constant in front is $\frac{1}{2}$ .
CA formula that passes a dimension check is certainly correct.
D $s = ut + 7at^{2}$ would pass a dimension check even though it is wrong.

Show answer and solution

Answer: Options A, D

A term with the wrong powers is caught at once, so the check can condemn a formula outright. But it is blind to plain numbers, so a $7$ slips through where $\frac{1}{2}$ belongs — passing the check means a formula is possible, never that it is right.