Units And Dimensions

PhysicsUnits And MeasurementFor NEET aspirants

Units and dimensions form the foundation of measurement in physics, essential for both JEE and NEET aspirants. A physical quantity is anything that can be measured and expressed as a numerical value with a unit. Every physical quantity has a dimensional formula built from seven base quantities: mass $[M]$ , length $[L]$ , time $[T]$ , current $[A]$ , temperature $[K]$ , luminous intensity $[c d]$ and amount of substance $[m o l]$ . Dimensional analysis lets you convert units between systems, check the correctness of equations, and derive new formulas - all skills tested repeatedly in JEE Main, JEE Advanced and NEET.

Key Formulas - Quick Reference

Number of units in two systems: $n_{1} u_{1} = n_{2} u_{2}$
Conversion between systems: $n_{2} = n_{1} [\frac{M _{1}}{M _{2}}]^{a} [\frac{L _{1}}{L _{2}}]^{b} [\frac{T _{1}}{T _{2}}]^{c}$
Principle of homogeneity: dimensions of LHS = dimensions of RHS in any physically valid equation.
Dimensional formula of force: $[F] = [M L T^{- 2}]$
Dimensional formula of energy / work: $[E] = [M L^{2} T^{- 2}]$
Dimensional formula of power: $[P] = [M L^{2} T^{- 3}]$
Dimensional formula of pressure: $[P] = [M L^{- 1} T^{- 2}]$

1. Physical Quantity, Fundamental and Derived Units

Physical quantity: A quantity that can be measured by an instrument, is clearly defined, and has proper units is called a physical quantity.

Physical quantities are classified into two types:

Fundamental (base) quantities: quantities that do not depend on any other physical quantity, for example length, mass and time.
Derived quantities: quantities obtained by combining fundamental quantities, for example velocity (length/time) or force (mass × acceleration).

2. Systems of Units

Three systems of units are commonly used:

System	Length	Mass	Time
F.P.S. (British)	Foot	Pound	Second
C.G.S. (Gaussian)	Centimetre	Gram	Second
M.K.S. / S.I. (International)	Metre	Kilogram	Second

The SI system is the modern international standard and is used throughout JEE and NEET physics. It has seven fundamental units and two supplementary units.

2.1 SI Fundamental Units

S.No.	Physical Quantity	Unit	Symbol
1	Mass	kilogram	kg
2	Length	metre	m
3	Time	second	s
4	Electric current	ampere	A
5	Thermodynamic temperature	kelvin	K
6	Luminous intensity	candela	cd
7	Amount of substance	mole	mol

2.2 SI Supplementary Units

S.No.	Physical Quantity	Unit	Symbol
1	Plane angle	radian	rad
2	Solid angle	steradian	sr

3. Definitions of SI Base Units

3.1 Metre (m)

The metre is defined as the length of the path travelled by light in vacuum during a time interval of

1/299, 792, 458

of a second.

3.2 Kilogram (kg)

Since the 2019 SI redefinition, the kilogram is defined by fixing the numerical value of the Planck constant

h

to be exactly

6.62607015 \times 1 0^{- 34} J s

. (The older definition based on the platinum-iridium cylinder in Paris is now retired.)

3.3 Second (s)

The second is defined by fixing the frequency of the cesium-133 hyperfine transition to

9, 192, 631, 770 Hz

3.4 Ampere (A)

The ampere is defined by fixing the elementary charge

e

1.602176634 \times 1 0^{- 19} coulomb

3.5 Kelvin (K)

The kelvin is defined by fixing the Boltzmann constant

k

1.380649 \times 1 0^{- 23} J/K

. The lower limit at which molecular activity ceases corresponds to

0 K

(absolute zero).

3.6 Candela (cd)

The candela is the SI unit of luminous intensity: the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency

540 \times 1 0^{12} Hz

and has a radiant intensity of

1/683

watt per steradian.

3.7 Mole (mol)

One mole contains exactly

6.02214076 \times 1 0^{23}

elementary entities (Avogadro's number). This number is the fixed numerical value of the Avogadro constant

N_{A}

4. Dimensional Formulas of Physical Quantities

Two conventions appear in JEE/NEET problems. In the modern SI convention, current

[A]

is fundamental. In the older convention (still asked in many books), charge

[Q]

is treated as fundamental. Since

Q = A \cdot T

, you can convert freely between the two.

4.1 Basic Physical Quantities

Physical Quantity	Symbol	Dimension	SI Unit
Length	$l$	$[L]$	metre (m)
Mass	$m$	$[M]$	kilogram (kg)
Time	$t$	$[T]$	second (s)
Electric current	$I$	$[A]$	ampere (A)
Temperature	$T$	$[K]$	kelvin (K)
Luminous intensity	$I_{v}$	$[c d]$	candela (cd)
Amount of substance	$n$	$[m o l]$	mole (mol)
Plane angle	$θ$	dimensionless	radian (rad)

4.2 Mechanical Derived Quantities

Quantity	Symbol	Dimensional Formula	SI Unit
Area	$A$	$[L^{2}]$	$m^{2}$
Volume	$V$	$[L^{3}]$	$m^{3}$
Velocity	$v$	$[L T^{- 1}]$	$m/s$
Angular velocity	$ω$	$[T^{- 1}]$	$rad/s$
Acceleration	$a$	$[L T^{- 2}]$	$m/s^{2}$
Angular acceleration	$α$	$[T^{- 2}]$	$rad/s^{2}$
Force	$F$	$[M L T^{- 2}]$	newton (N)
Momentum / Impulse	$p$	$[M L T^{- 1}]$	$kg m/s$ or N s
Work / Energy / Heat	$W, E, Q$	$[M L^{2} T^{- 2}]$	joule (J)
Torque	$τ$	$[M L^{2} T^{- 2}]$	$N m$
Power	$P$	$[M L^{2} T^{- 3}]$	watt (W)
Density	$ρ$	$[M L^{- 3}]$	$kg/m^{3}$
Pressure / Stress	$P, σ$	$[M L^{- 1} T^{- 2}]$	pascal (Pa)
Moment of inertia	$I$	$[M L^{2}]$	$kg m^{2}$
Entropy	$S$	$[M L^{2} T^{- 2} K^{- 1}]$	$J/K$
Volume flow rate	$Q$	$[L^{3} T^{- 1}]$	$m^{3} / s$
Kinematic viscosity	$ν$	$[L^{2} T^{- 1}]$	$m^{2} / s$
Dynamic viscosity	$η$	$[M L^{- 1} T^{- 1}]$	$Pa s$
Surface tension	$T$	$[M T^{- 2}]$	$N/m$
Frequency	$f$	$[T^{- 1}]$	hertz (Hz)
Wavelength	$λ$	$[L]$	metre (m)

4.3 Electrical and Magnetic Derived Quantities

Quantity	Symbol	Dimensional Formula	SI Unit
Electric charge	$q$	$[A T]$	coulomb (C)
Potential difference / EMF	$V, ε$	$[M L^{2} T^{- 3} A^{- 1}]$	volt (V)
Resistance	$R$	$[M L^{2} T^{- 3} A^{- 2}]$	ohm ( $Ω$ )
Capacitance	$C$	$[M^{- 1} L^{- 2} T^{4} A^{2}]$	farad (F)
Inductance	$L$	$[M L^{2} T^{- 2} A^{- 2}]$	henry (H)
Electric field	$E$	$[M L T^{- 3} A^{- 1}]$	$V/m$ or $N/C$
Magnetic field	$B$	$[M T^{- 2} A^{- 1}]$	tesla (T)
Magnetic flux	$Φ$	$[M L^{2} T^{- 2} A^{- 1}]$	weber (Wb)
Permittivity	$ε_{0}$	$[M^{- 1} L^{- 3} T^{4} A^{2}]$	$F/m$
Permeability	$μ_{0}$	$[M L T^{- 2} A^{- 2}]$	$H/m$
Dielectric constant	$K, κ$	dimensionless	none

5. Applications of Dimensional Analysis

Dimensional analysis has four main applications in JEE and NEET problems.

5.1 To find the unit of a physical quantity

If you know the dimensional formula of a quantity, you can write its SI unit directly.

Solved Example 1

The universal gravitational constant

G

has dimensional formula

[M^{- 1} L^{3} T^{- 2}]

. Find its SI unit.

Solution:

Replacing each dimension with its SI unit: $M \to kg$ , $L \to m$ , $T \to s$ .

SI unit of $G = kg^{- 1} m^{3} s^{- 2}$ , which is the same as $N m^{2} kg^{- 2}$ .

5.2 To convert a physical quantity from one system to another

If a quantity has value

n_{1}

in system 1 (with unit

u_{1}

) and

n_{2}

in system 2 (with unit

u_{2}

), then

n_{1} u_{1} = n_{2} u_{2}

. If the dimensional formula is

[M^{a} L^{b} T^{c}]

n_{2} = n_{1} [\frac{M _{1}}{M _{2}}]^{a} [\frac{L _{1}}{L _{2}}]^{b} [\frac{T _{1}}{T _{2}}]^{c}

Solved Example 2

Convert

g = 9.8 m/s^{2}

from SI to CGS system.

Solution:

Dimensional formula of acceleration is $[L T^{- 2}]$ , so $a = 0$ , $b = 1$ , $c = - 2$ .

$n_{2} = 9.8 [\frac{L _{1}}{L _{2}}]^{1} [\frac{T _{1}}{T _{2}}]^{- 2} = 9.8 [\frac{1 m}{1 cm}] [\frac{1 s}{1 s}]^{- 2}$

$= 9.8 \times \frac{100 cm}{1 cm} \times 1 = 980$

Therefore, $g = 980 cm/s^{2}$ in the CGS system.

5.3 To check the correctness of a physical relation

The principle of homogeneity of dimensions states that the dimensions of every term on both sides of a valid physical equation must be identical.

Solved Example 3

Check the dimensional correctness of the centripetal force formula

F = \frac{m v ^{2}}{r}

Solution:

LHS: $[F] = [M L T^{- 2}]$

RHS: $\frac{[ m ] [ v ] ^{2}}{[ r ]} = \frac{[ M ] [ L T ^{- 1} ] ^{2}}{[ L ]} = \frac{[ M L ^{2} T ^{- 2} ]}{[ L ]} = [M L T^{- 2}]$

LHS = RHS, so the equation is dimensionally correct.

5.4 To derive a relation between physical quantities

If a quantity depends on at most three other quantities, you can guess its form as a product of powers and find the powers by matching dimensions.

Solved Example 4

Derive the Planck length

L_{P}

in terms of the gravitational constant

G

, the speed of light

c

and Planck's constant

h

Solution:

Assume $L_{P} = k G^{x} c^{y} h^{z}$ , where $k$ is a dimensionless constant.

Dimensions: $[G] = [M^{- 1} L^{3} T^{- 2}]$ , $[c] = [L T^{- 1}]$ , $[h] = [M L^{2} T^{- 1}]$ .

Then $[L] = [M^{- 1} L^{3} T^{- 2}]^{x} [L T^{- 1}]^{y} [M L^{2} T^{- 1}]^{z}$ .

Comparing powers of $M$ , $L$ , $T$ :

$M$ : $- x + z = 0$
$L$ : $3 x + y + 2 z = 1$
$T$ : $- 2 x - y - z = 0$

Solving: $x = \frac{1}{2}$ , $y = - \frac{3}{2}$ , $z = \frac{1}{2}$ .

$L_{P} = k \frac{G h}{c ^{3}}$

With $k = 1$ , $L_{P} \approx 1 0^{- 35} m$ .

6. Limitations of Dimensional Analysis

Dimensional analysis cannot derive relations that involve sums or differences of terms (e.g. $s = u t + \frac{1}{2} a t^{2}$ ) - only relations of the form of products of powers.
It cannot derive relations involving more than three unknowns, since only three equations (from $M$ , $L$ , $T$ ) are available. It can still be used to check such relations.
It cannot handle dimensionless constants like $π$ , $\frac{1}{2}$ or trigonometric ratios - they must come from experiment or theory.
If a quantity depends on three quantities and two of them have the same dimensions, dimensional analysis fails to give a unique answer.
It cannot distinguish between quantities that share dimensions. For example, both torque and work have dimensions $[M L^{2} T^{- 2}]$ , yet they are physically distinct.

7. Solved Examples on Units and Dimensions

Solved Example 5

Find the dimensional formula of Young's modulus.

Solution:

Young's modulus $Y = \frac{stress}{strain} = \frac{F / A}{Δ L / L}$ .

Strain is dimensionless, so $[Y] = \frac{[ F ]}{[ A ]} = \frac{[ M L T ^{- 2} ]}{[ L ^{2} ]} = [M L^{- 1} T^{- 2}]$ .

Same as pressure. SI unit: pascal (Pa).

Solved Example 6

Find the dimensions of Planck's constant

h

Solution:

From $E = h ν$ , $h = \frac{E}{ν}$ .

$[h] = \frac{[ M L ^{2} T ^{- 2} ]}{[ T ^{- 1} ]} = [M L^{2} T^{- 1}]$ . SI unit: $J s$ .

Solved Example 7

Check dimensionally whether the equation

T = 2 π \frac{l}{g}

(time period of a simple pendulum) is correct.

Solution:

LHS: $[T]$ .

RHS: $\frac{[ L ]}{[ L T ^{- 2} ]} = [T^{2}] = [T]$ . (The $2 π$ is dimensionless.)

LHS = RHS, so the equation is dimensionally correct.

Solved Example 8

The velocity

v

of a wave on a stretched string depends on the tension

T

and the mass per unit length

μ

. Derive the dependence using dimensional analysis.

Solution:

Assume $v = k T^{x} μ^{y}$ .

$[L T^{- 1}] = [M L T^{- 2}]^{x} [M L^{- 1}]^{y} = [M^{x + y} L^{x - y} T^{- 2 x}]$ .

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

$v = k \frac{T}{μ}$

Experiment shows $k = 1$ , giving the well-known result $v = T / μ$ .

Solved Example 9

Which of the following pairs have the same dimensions? (a) Work and torque (b) Angular momentum and Planck's constant (c) Pressure and Young's modulus (d) All of these.

Solution:

Work and torque: both $[M L^{2} T^{- 2}]$ . Same.

Angular momentum $L = I ω$ : $[M L^{2}] [T^{- 1}] = [M L^{2} T^{- 1}]$ . Planck's constant: $[M L^{2} T^{- 1}]$ . Same.

Pressure and Young's modulus: both $[M L^{- 1} T^{- 2}]$ . Same.

Answer: (d) All of these. This is a classic JEE Main/NEET trap - identical dimensions do not imply identical physical quantities.

Common Mistakes to Avoid

Watch out

Confusing charge $[Q]$ with current $[A]$ as fundamental. Modern SI takes current as fundamental. Some older books use charge - both work as long as you are consistent.
Treating Celsius as a fundamental unit. Only kelvin is fundamental in SI. Celsius is a derived scale ( $t °C = T K - 273.15$ ).
Assuming that "same dimensions" means "same quantity". Torque, work and energy all have $[M L^{2} T^{- 2}]$ but are physically distinct.
Using dimensional analysis to derive equations with additive terms. It works only for products of powers, not for equations like $s = u t + \frac{1}{2} a t^{2}$ .
Forgetting that trigonometric arguments, exponents and logarithms are dimensionless. Anything inside $sin$ , $cos$ , $e^{x}$ or $lo g$ must have zero dimensions.
Missing the $2 π$ or numeric factor. Dimensional analysis gives the correct powers but never the pure numerical constant - always cross-check with the standard formula.

Frequently Asked Questions

Q1. What is the difference between fundamental and derived units?

Fundamental units are independent base units that do not depend on other units, like the metre, kilogram and second. Derived units are formed by combining fundamental units, for example the newton ( $kg m/s^{2}$ ) or the joule ( $kg m^{2} / s^{2}$ ). The SI system has 7 fundamental units and countless derived units.

Q2. How important is units and dimensions for JEE Main and JEE Advanced?

Units and dimensions is a scoring topic in JEE. Typically 1-2 direct questions appear in JEE Main, and it forms the foundation for solving problems in mechanics, electromagnetism and modern physics. Common patterns include finding dimensional formulas, checking equations and identifying quantities with the same dimensions.

Q3. Is units and dimensions asked in NEET?

Yes. NEET regularly asks 1 question directly on dimensions or unit conversion, and the topic underlies every numerical problem in physics. Common NEET questions focus on identifying dimensional formulas and matching pairs of quantities with the same dimensions.

Q4. What are the seven SI base units?

The seven SI base units are the kilogram (mass), metre (length), second (time), ampere (electric current), kelvin (temperature), candela (luminous intensity) and mole (amount of substance). Every other SI unit is derived from these seven.

Q5. What is the principle of homogeneity of dimensions?

The principle states that in any valid physical equation, every term must have the same dimensional formula. This principle is the basis for checking whether an equation is dimensionally correct and for deriving relations by dimensional analysis.

Q6. Can two physical quantities have the same dimensions?

Yes. Many physical quantities share dimensions. Examples: work, energy and torque all have $[M L^{2} T^{- 2}]$ ; pressure, stress and Young's modulus all have $[M L^{- 1} T^{- 2}]$ ; angular momentum and Planck's constant both have $[M L^{2} T^{- 1}]$ . Same dimensions does not mean same physical meaning.

Q7. What are the limitations of dimensional analysis?

Dimensional analysis cannot determine dimensionless constants like $π$ or $\frac{1}{2}$ , cannot handle equations with sums of terms, cannot distinguish between quantities of the same dimension, and fails when more than three unknowns are involved. It is a checking and guiding tool, not a complete substitute for physics.

Q8. Why was the kilogram redefined in 2019?

Before 2019 the kilogram was defined as the mass of a specific platinum-iridium cylinder kept in Paris (the IPK). Over decades the IPK's mass drifted slightly compared to its official copies. In 2019 the SI redefined the kilogram in terms of a fixed value of Planck's constant, giving a permanent, universal standard.

Q9. What is a dimensional formula?

A dimensional formula expresses a physical quantity as a product of powers of the base dimensions - typically written as $[M^{a} L^{b} T^{c}]$ . For example, force has dimensional formula $[M L T^{- 2}]$ , energy has $[M L^{2} T^{- 2}]$ and pressure has $[M L^{- 1} T^{- 2}]$ .

Previous year questions on Units And Dimensions

41 questions from past papers, each with a step-by-step solution.

Show all 41 questions

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