Unit Conversion — SI, CGS, and Everything Between

Why Unit Conversion Matters

Every physical measurement has two parts: a number and a unit. The number means nothing without the unit — “5” could be 5 metres or 5 kilometres or 5 light-years. Unit conversion is the skill of expressing the same physical quantity in different units without changing its physical meaning.

In Indian competitive exams, unit conversion problems appear in two forms: direct conversions (change 5 km/h to m/s) and formula-based conversions (express G in CGS given its SI value). Both test the same underlying understanding.

Key Terms and Definitions

SI System (International System of Units): The modern standard system used in science worldwide. Seven base units: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), candela (cd). All other units are derived from these.

CGS System: An older system using centimetre (cm), gram (g), and second (s) as base units for length, mass, and time respectively. Still used in electromagnetism (Gaussian units) and some older textbooks.

MKS System: Metre-kilogram-second. Essentially the mechanical part of SI.

Dimensional Formula: The expression of a physical quantity in terms of base dimensions — [M], [L], [T]. Used to derive conversion factors systematically.

Conversion Factor: A ratio equal to 1, used to convert between units. For example, $1\ \text{km} = 1000\ \text{m}$ , so $\dfrac{1000\ \text{m}}{1\ \text{km}} = 1$ (multiplying by this changes units, not the quantity).

The Dimensional Analysis Method

The most powerful method for any unit conversion uses the dimensional formula of the physical quantity. If you know how a quantity depends on mass (M), length (L), and time (T), you can convert between any two systems.

If $n_1 u_1 = n_2 u_2$ (same physical quantity in two systems):

n_2 = n_1 \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 $[M^a L^b T^c]$ is the dimensional formula of the quantity, subscript 1 = original system, subscript 2 = new system.

This formula is the engine behind all unit conversions. Once you understand it, you never need to memorise conversion factors for complex quantities.

Common Unit Conversions You Must Know

Length

From	To	Conversion
1 km	m	1000 m
1 m	cm	100 cm
1 cm	mm	10 mm
1 Angstrom (Å)	m	$10^{-10}$ m
1 nanometre (nm)	m	$10^{-9}$ m
1 micrometre (μm)	m	$10^{-6}$ m
1 light-year	km	$9.46 \times 10^{12}$ km
1 parsec	m	$3.086 \times 10^{16}$ m
1 astronomical unit (AU)	m	$1.496 \times 10^{11}$ m

Mass

From	To	Conversion
1 kg	g	1000 g
1 g	mg	1000 mg
1 tonne	kg	1000 kg
1 atomic mass unit (u)	kg	$1.66 \times 10^{-27}$ kg

Time

From	To	Conversion
1 hour	seconds	3600 s
1 day	seconds	86400 s
1 year	seconds	$3.15 \times 10^7$ s

Common Derived Quantities

Quantity	SI	CGS	Conversion
Force	N	dyne	$1\ \text{N} = 10^5\ \text{dyne}$
Pressure	Pa	dyne/cm²	$1\ \text{Pa} = 10\ \text{dyne/cm}^2$
Energy	J	erg	$1\ \text{J} = 10^7\ \text{erg}$
Power	W	erg/s	$1\ \text{W} = 10^7\ \text{erg/s}$
Magnetic field	T	Gauss	$1\ \text{T} = 10^4\ \text{Gauss}$

The ratio $\dfrac{\text{SI value}}{\text{CGS value}}$ for force, energy, and power follow the pattern $10^5$ , $10^7$ , $10^7$ respectively. Remember these three for JEE.

Solved Examples

Example 1 (Easy — CBSE): Convert 72 km/h to m/s

72\ \text{km/h} = 72 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 72 \times \frac{1}{3.6}\ \text{m/s} = 20\ \text{m/s}

Shortcut: Divide km/h by 3.6 to get m/s. This is the most-used conversion in kinematics.

Example 2 (Medium — JEE Main): Convert G = 6.67 × 10⁻¹¹ N·m²/kg² from SI to CGS

First, find the dimensional formula of G from $F = \dfrac{Gm_1 m_2}{r^2}$ :

G = \frac{Fr^2}{m_1 m_2} \Rightarrow [G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]

Using the conversion formula with $a = -1$ , $b = 3$ , $c = -2$ :

n_2 = n_1 \left[\frac{M_1}{M_2}\right]^{-1} \left[\frac{L_1}{L_2}\right]^3 \left[\frac{T_1}{T_2}\right]^{-2}

$M_1 = 1\ \text{kg}$ , $M_2 = 1\ \text{g} = 10^{-3}\ \text{kg}$
$L_1 = 1\ \text{m}$ , $L_2 = 1\ \text{cm} = 10^{-2}\ \text{m}$
$T_1 = T_2 = 1\ \text{s}$ (same in both systems)

n_2 = 6.67 \times 10^{-11} \times \left[\frac{1}{10^{-3}}\right]^{-1} \times \left[\frac{1}{10^{-2}}\right]^3 \times 1

= 6.67 \times 10^{-11} \times 10^{-3} \times 10^6

= 6.67 \times 10^{-11} \times 10^3 = 6.67 \times 10^{-8}\ \text{dyne·cm}^2\text{/g}^2

Example 3 (Hard — JEE Advanced): Express 1 kWh in joules and ergs

1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}

Converting to ergs:

3.6 \times 10^6\ \text{J} \times 10^7\ \frac{\text{erg}}{\text{J}} = 3.6 \times 10^{13}\ \text{erg}

This also equals $3.6 \times 10^6$ joules = $3.6$ MJ.

Exam-Specific Tips

In JEE Main, dimensional analysis questions ask you to find the value of a constant in CGS when SI value is given. Master the conversion formula — these are 4-mark questions that take 2 minutes if you know the method.

CBSE Class 11 frequently tests: “Express 1 newton in dynes” and “Express 1 joule in ergs.” These are 2-mark questions — know both answers cold. 1 N = 10⁵ dyne, 1 J = 10⁷ erg.

For speed conversions: km/h ÷ 3.6 = m/s. For the reverse: m/s × 3.6 = km/h. This shortcut appears in at least one kinematics problem in every CBSE board exam.

Common Mistakes to Avoid

Mistake 1: Confusing dyne and newton. 1 N = 10⁵ dyne (not 10³). The extra two zeros come from the cm-to-m conversion squared (since force = mass × acceleration, and acceleration involves length/time²).

Mistake 2: Forgetting that the second (time) is the same in both SI and CGS. The only unit changes are mass (kg→g, factor 10³) and length (m→cm, factor 10²). Students sometimes introduce wrong time conversion factors.

Mistake 3: Using the wrong sign for the exponent in the conversion formula. If converting from SI to CGS and the length exponent is +3 (like in G), then $L_1/L_2 = 1/(10^{-2}) = 10^2$ , so you get $(10^2)^3 = 10^6$ . Students frequently flip the ratio and get $10^{-6}$ instead.

Mistake 4: Treating Angstrom (Å) as a basic SI unit. Å is not an SI unit — it’s a non-SI unit commonly used in atomic/molecular scale. The SI equivalent is $10^{-10}$ m or 0.1 nm.

Mistake 5: Not including units in the answer. A number without a unit is meaningless in physics. Always write the unit — examiners deduct marks for missing units even if the number is correct.

Practice Questions

Q1. Convert 5 m/s to km/h.

5 m/s × 3.6 = 18 km/h

Or: $5 \times \dfrac{3600}{1000} = 18$ km/h

Q2. The density of water is 1 g/cm³. Express this in kg/m³.

$1\ \text{g/cm}^3 = \dfrac{10^{-3}\ \text{kg}}{(10^{-2}\ \text{m})^3} = \dfrac{10^{-3}}{10^{-6}}\ \text{kg/m}^3 = 10^3\ \text{kg/m}^3 = \mathbf{1000\ \text{kg/m}^3}$

Q3. Express 1 pascal (Pa) in CGS units.

Pressure [M¹L⁻¹T⁻²]
$1\ \text{Pa} = \left(\dfrac{M_1}{M_2}\right)^1 \left(\dfrac{L_1}{L_2}\right)^{-1} \left(\dfrac{T_1}{T_2}\right)^{-2}$
$= 10^3 \times (10^2)^{-1} \times 1 = 10^3 \times 10^{-2} = 10\ \text{dyne/cm}^2$

Q4. A car travels at 90 km/h. What is this speed in m/s?

$90 \div 3.6 = \mathbf{25\ \text{m/s}}$

Q5. How many ergs equal 1 joule?

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

Derivation: Energy [ML²T⁻²], converting to CGS gives $(10^3)(10^2)^2(1)^{-2} = 10^3 \times 10^4 = 10^7$ .

Q6. A force of 500 dyne acts on a body. Express this in newtons.

$500\ \text{dyne} = \dfrac{500}{10^5}\ \text{N} = 5 \times 10^{-3}\ \text{N} = \mathbf{5\ \text{mN}}$

Additional Worked Examples

Example 4 (JEE Level): Convert viscosity from CGS to SI

The viscosity of water is 0.01 poise (CGS). Express it in SI (Pa·s).

From Newton’s law of viscosity: $\eta = \frac{F/A}{dv/dx}$

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

With $a = 1$ , $b = -1$ , $c = -1$ :

n_2 = 0.01 \times \left(\frac{1}{10^{-3}}\right)^1 \times \left(\frac{1}{10^{-2}}\right)^{-1} \times 1^{-1}

= 0.01 \times 10^3 \times 10^{-2} = 0.01 \times 10 = 0.1

0.01 \text{ poise} = 0.001 \text{ Pa·s} = 10^{-3} \text{ Pa·s}

The conversion factor is: 1 poise = 0.1 Pa·s.

Example 5: Everyday conversion — Power rating

An electric heater is rated 2 kW. Express this in (a) watts, (b) erg/s, (c) horsepower.

(a) $2 \text{ kW} = 2000 \text{ W}$

(b) $2000 \text{ W} = 2000 \times 10^7 \text{ erg/s} = 2 \times 10^{10} \text{ erg/s}$

Quantity	Conversion
1 eV	$1.6 \times 10^{-19}$ J
1 HP	746 W
1 calorie	4.186 J
1 atm	$1.013 \times 10^5$ Pa
1 poise	0.1 Pa·s
1 debye	$3.33 \times 10^{-30}$ C·m

NEET Physics frequently tests the eV to joule conversion in photoelectric effect and nuclear physics problems. Memorise: $1 \text{ eV} = 1.6 \times 10^{-19}$ J. Also remember $hc = 1240$ eV·nm — this shortcut eliminates the need to convert units in photon energy calculations.

Q7. The surface tension of water is 0.072 N/m. Express this in CGS units (dyne/cm).

Surface tension has dimensions $[MT^{-2}]$ . Using the conversion:

$n_2 = 0.072 \times (10^3)^1 \times (1)^{-2} = 72$ dyne/cm.

Alternatively: $1 \text{ N/m} = 10^3 \text{ dyne} / 10^2 \text{ cm} = 10 \text{ dyne/cm}$ , but surface tension is force per length, so $1 \text{ N/m} = 1000 \text{ dyne/cm}$ . Thus $0.072 \text{ N/m} = 72 \text{ dyne/cm}$ .

Q8. Express the gravitational constant $G = 6.67 \times 10^{-11}$ N·m²/kg² in units where mass is in grams, length in cm, and time in seconds.

From Example 2 above: $G = 6.67 \times 10^{-8}$ dyne·cm²/g². The dimensional analysis gives the same factor of $10^3$ multiplied to the SI value.

FAQs

Why do we need multiple unit systems? Different systems developed in different countries and scientific traditions. The CGS system was used by older European science. SI was internationally standardised in 1960 to create a universal language. Older papers and textbooks still use CGS, so conversion skills remain essential.

What is the difference between a unit and a dimension? A dimension is a physical quantity type: [M] for mass, [L] for length, [T] for time. A unit is a specific standard quantity: kg, g, pound are all units of the dimension [M]. Dimensions are universal; units depend on the system chosen.

Can I always use the dimensional analysis method for conversion? Yes, for any physical quantity with a known dimensional formula. The method fails only for dimensionless quantities (like angles in radians vs degrees) or for constants defined by convention (like π).

What is the SI unit of pressure and how is it related to atmosphere? The SI unit of pressure is pascal (Pa). $1\ \text{atm} = 1.013 \times 10^5\ \text{Pa}$ . Another common unit is bar: $1\ \text{bar} = 10^5\ \text{Pa}$ .

Why is the second the same in SI and CGS? The second is defined by atomic transitions (the frequency of radiation from cesium-133 atoms), which is a universal constant unaffected by the choice of length or mass units. So both systems use the same second.