Dimensional Analysis: Class 11 Physics Notes, Definition & Working Principle

Dimensional analysis is based on the homogeneity principle, which states that all terms in a physically valid equation must have the same dimensions. Physical quantities are expressed using seven base dimensions 

• Length: [L]
• Mass: [M]
• Time: [T]
• Electric current: [A]
• Thermodynamic temperature: [K]
• Amount of substance: [mol]
• Luminous intensity: [cd]

Derived quantities, like velocity $\left[L T^{-1}\right]$ , force $\left[M,,L,,T^{-2}\right]$ , or energy $\left[M L^2 T^{-2}\right]$ , are combinations of these. Dimensional analysis uses these dimensions for maintaining consistency or for deducing relationships which is a very important for JEE Main.

Difference Between Dimensional Formula and Dimensional Equation

The following table explains the difference between dimensional formula and dimensional equation:

Parameter	Dimensional Formula	Dimensional Equation
Definition	Expression showing how a physical quantity depends on base quantities	Equation that equates a physical quantity with its dimensional formula
Form	Uses symbols of base units (e.g., M, L, T)	Represents full relation, e.g., Force = [M¹L¹T⁻²]
Purpose	Identifies the fundamental units involved	Verifies the correctness of physical equations
Example	Dimensional formula of velocity: [M⁰L¹T⁻¹]	Dimensional equation of velocity: v = [M⁰L¹T⁻¹]
Focus	Structure of the unit (dimensions only)	Links the physical quantity directly to its dimensional representation
Used For	Unit conversion, dimensional analysis, checking unit consistency	Verifying equations, analyzing dependencies among physical quantities
Contains physical name?	No—it only gives unit structure	Yes—it includes the name of the physical quantity (e.g., "v =", "F =")

Dimensional analysis has three primary applications in JEE Main: checking dimensional consistency, deriving physical relations, and unit conversion.

2.1 Checking Dimensional Consistency

The principle of homogeneity ensures all terms in an equation have identical dimensions. For example, consider the kinematic equation:

$$v=u+at$$

Dimensions:

$v=\left[L{T}^{-1}\right]u=\left[L{T}^{-1}\right]at=\left[L{T}^{-2}\right]\times \left[T\right]=\left[L{T}^{-1}\right]$

All terms have dimensions $\left[L{T}^{-1}\right]$, confirming the equation's validity. JEE Main often includes questions asking to identify dimensionally incorrect options in multiple-choice formats.

 

2.2 Deriving Key Relations

Dimensional analysis can deduce relationships among variables. For example, derive the time period $T$ of a simple pendulum depending on length $l$, mass $m$, and gravity $g$. Assume:

$$T=k{l}^x{g}^y{m}^z$$

Dimensions:

$\left[T\right]=\left[T\right]$, $\left[l^x{g}^y{m}^z\right]$ $=\left[L^x\right]\left[{L{T}^{-2}}^y\right]\left[M^z\right]=\left[M^z{L}^{x+y}{T}^{-2y}\right]$

Equating:

$$M^0{L}^0{T}^1=M^z{L}^{x+y}{T}^{-2y}$$

Solving: $z=0,x+y=0,-2y=1$. Thus, $y=-\frac{1}{2},x=\frac{1}{2}$, so:

$$T=k\sqrt{\frac{l}{g}}$$

The constant $k=2\pi$ is determined experimentally, yielding $T=2\pi \sqrt{\frac{l}{g}}$. Such derivations are common in JEE Main numerical problems.

2.3 Unit Conversion Through Dimension Analysis

Dimensional analysis facilitates unit conversion. For example, convert 1 joule (J) to erg:

$1J=1\mathrm{kg}·m{2}^{}·s{-2}^{}$ Dimensions: $\left[ML{1}^{}T{-2}^{}\right]$

In CGS:

$1g={10}^{-3}\mathrm{kg},1\mathrm{cm}={10}^{-2}m,1s=1s$

$1J=\left({10}^{3}g\right)·\left({10}^2\mathrm{cm}\right)2^{}·s{-2}^{}={10}^{7}g·\mathrm{cm}{2}^{}·s{-2}^{}={10}^7\mathrm{erg}$

JEE Mains may require converting quantities like pressure or power across SI and CGS systems.

Problem 1: Check if $E=\frac{1}{2}m{v}^2$ is dimensionally correct for energy.

$\left[E\right]=\left[M{L}^2{T}^{-2}\right]$, $\left[m{v}^2\right]=\left[M\right]·\left[L{T}^{-1}\right]2^{}=\left[M{L}^2{T}^{-2}\right]$

The equation is dimensionally correct, confirming it represents energy.

 

Problem 2: The frequency $\nu$ of a vibrating string depends on tension $T$, length $l$, and mass per unit length $\mu$. Derive the relation. Assume:

$$\nu =k{T}^x{l}^y{\mu }^z$$

Dimensions:

$$\left[\nu \right]=\left[T^{-1}\right],\left[T^x{l}^y{\mu }^z\right]={\left[ML{T}^{-2}\right]}^x{\left[L\right]}^y{\left[ML{T}^{-1}\right]}^z$$

Equating:

$$T^{-1}=M^{x+z}{L}^{x+y-z}{T}^{-2x}$$

Solving: $x+z=0,x+y-z=0,-2x=-1$. Thus, $x=\frac{1}{2},z=-\frac{1}{2},y=-1$, so:

$$\nu =k\frac{\sqrt{T}}{l\sqrt{\mu }}$$

This matches the standard formula $\nu =\frac{1}{l}\sqrt{\frac{T}{\mu }}$.

1. Forgetting Base Dimensions

A lot of students accidentally leave out one of the seven SI base dimensions when they set up a dimensional equation.

• What Often Goes Wrong: You might try to find the dimension of electric charge $Q$ by saying $Q=I\times t$, but then treat current $I$ as just $\left[L\right]$, $\left[M\right]$, $\left[T\right]$ instead of remembering that current itself is a base dimension $\left[A\right]$.
• How to Fix It: Always list all relevant base dimensions first— $\left[L\right]$, $\left[M\right]$, $\left[T\right]$, $\left[A\right]$, $\left[K\right]$, $\left[\mathrm{mol}\right]$, $\left[\mathrm{cd}\right]$— before writing your equation. For charge: $$\left[Q\right]=\left[I\right]\times \left[t\right]=\left[A\right]\times \left[T\right]=\left[AT\right]$$

Make it a habit to double-check that every quantity’s base units are accounted for before you move on.

2. Expecting to Find Dimensionless Constants

Dimensional analysis can’t tell you what the number “2π” should be in the pendulum formula.

• What Students Assume: They might think they can figure out $T=k\sqrt{l/g}$ just by balancing dimensions, then guess $k=2\pi$.
• The Reality Check: Dimensionless constants come from experiments or deeper theory, not from dimensional reasoning. So write: $$T=k\sqrt{l/g},\text{where}\mathrm{ k }\text{is dimensionless (and we find }k=2\mathrm{\pi }\text{by experiment).}$$

Never try to “solve” for constant values with dimensions alone—acknowledge that dimensional analysis can’t give you numbers.

3. Overlooking Derived Quantities’ Base Units

Sometimes students write down a derived quantity without fully expanding its base units, which causes confusion.

• Typical Oversight: You might say “Pressure $P=F/A$” but then treat force $F$ as $\left[L{T}^{-2}\right]$ instead of $\left[ML{T}^{-2}\right]$. That misses the mass dimension entirely.
• How to Avoid It: Break every derived quantity into its base dimensions first. For pressure:
  1. Force $\left[F\right]=\left[ML{T}^{-2}\right]$
  2. Area $\left[A\right]=\left[L^2\right]$
  3. Pressure $\left[P\right]=\frac{\left[F\right]}{\left[A\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$. Always expand and simplify carefully so you don’t accidentally drop a dimension.

4 Misapplying Unit Conversions

When you convert units—say, joules to ergs—students sometimes convert only part of the compound unit and end up with the wrong factor.

• Where You Can Go Wrong: You might convert kilograms to grams but forget that metres must become centimetres too. For example: $1J=1\mathrm{kg}\times 1m{2}^{}\times s{-2}^{}=({10}^{3}g)\times (1m)2^{}\times s{-2}^{}={10}^{3}g·\mathrm{cm}{2}^{}·s{-2}^{}.$
  That’s incomplete—1 m 2 is actually 10 4 cm 2 .
• The Correct Way: Convert each base unit fully:
  1. $1\mathrm{kg}={10}^{3}g$
  2. $1m={10}^{-2}\mathrm{cm}$, so $1m{2}^{}={10}^4\mathrm{cm}{2}^{}$.

  Then

  $1J=({10}^{3}g)\times ({10}^4\mathrm{cm}{2}^{})\times s{-2}^{}={10}^{7}g·\mathrm{cm}{2}^{}·s{-2}^{}={10}^7\mathrm{erg}.$

Always check that you’ve converted every single base unit, not just one.

5. Slipping Up on Exponent Balancing

It’s easy to mix up exponents when you match dimensions on both sides of an equation.

• Where Students Trip Up: When deriving the pendulum period $T=k{l}^x{g}^y$, some set $x+y=1$ and $-2y=1$ without checking mass, or they equate exponents out of order. This leads to incorrect powers.
• How to Do It Right: Write down the full dimension equality and match each base exponent one by one. For the pendulum:
  1. Start with $T^1=l^x{g}^y$ = L x L T − 2 y = M 0 L x + y T − 2 y .
  2. Equate exponents:
     • Mass ( $M$): $0=0$ (no new condition here)
     • Length ( $L$): $0=x+y$
     • Time ( $T$): $1=-2y$, so $y=-\frac{1}{2}$.
     • Then $x=-y=\frac{1}{2}$.
  3. Conclude $$T=k\sqrt{l/g}$$. Going through each dimension systematically prevents mistakes.

Following points highlight the limitations of dimensional formula:

1. Cannot Determine Numerical Constants
Dimensional analysis tells you how physical quantities relate in terms of their base units, but it cannot yield the exact numerical multiplier. For instance, you can show that the pendulum period $T\propto \sqrt{\frac{l}{g}}$, but dimensional reasoning alone gives no clue that the constant is $2\pi$. Relying solely on dimensions might mislead you into thinking you can “derive” that $2\pi$ factor—when in reality, it comes from solving the differential equation or measuring the system experimentally.

JEE Example: If a problem asks you to prove dimensionally that $T=k\sqrt{\frac{l}{g}}$, it’s correct to conclude $T\propto \sqrt{\frac{l}{g}}$. However, assuming $k=2$ (or $1$, or any other number) is wrong—only experiment or deeper physics shows $k=2\pi$.

2. Inapplicable to Sums and Differences of Different Dimensions
Whenever you have an equation that adds or subtracts terms with different dimensions, dimensional analysis cannot verify it unless you inspect each term individually. For example, if you see $v=u+at$, dimensional analysis can confirm each term has units of $\left[L{T}^{-1}\right]$, but if you tried something like $E=mgh+\frac{1}{2}m{v}^2$, you must check dimensions separately for $mgh$ and $\frac{1}{2}m{v}^2$. You can’t treat the sum as a single dimensional “group”; each term needs its own check, or you risk overlooking an inconsistency.

JEE Example: Consider a hypothetical formula $F=A+B$, where $\left[ML{T}^{-2}\right]$ (dimensions of A) and $\left[M{L}^2{T}^{-2}\right]$ (dimensions of B). Dimensional analysis alone cannot validate F unless you separately confirm that A and B should combine—often they shouldn’t. You must handle each term’s dimension independently.

3. Limited to Quantities with Dimensions
Dimensional analysis only applies when you’re dealing with physical quantities that have units. It cannot address purely dimensionless parameters—such as angles in radians or coefficients in an empirical fit. If a problem asks for, say, the Reynolds number in fluid flow or the geometric factor in a lens formula, you need nondimensional arguments or additional physical reasoning. Dimensional checks won’t help you find those dimensionless quantities.

JEE Example: When constructing the Reynolds number $\mathrm{Re}=\frac{\rho vL}{\mu }$, dimensional analysis alone cannot tell you that combining density $\rho$, velocity $v$, length $L$, and viscosity $\mu$ yields a dimensionless ratio critical for flow characterization. You must rely on fluid‐dynamics principles, not just unit balancing.

Please note that the below-given tips are useful for NEET exam aspirants as well:

1. Master Base Dimensions: Memorize dimensions of common quantities (force, work, pressure).

2. Simplify Equations: Break complex expressions into base units for analysis.

3. Practice Derivations: Solve past JEE Main questions on deriving relations (e.g., pendulum, frequency).

4. Unit Conversions: Be proficient in SI-CGS conversions for quick calculations.