Units and Measurements · JEE & NEET Physics

SI Units: notes and previous year questions

Why we need standard units, the seven base units, derived units, prefixes and conversions.

31 JEE Main questions (2002–2026)
3 JEE Advanced questions (1998–2013)
12 NEET questions (2000–2026)

SI Units in short

A measurement is a number times a unit. The number alone means nothing.
Changing the unit changes only the number: a smaller unit gives a bigger number.
SI has seven base units: metre, kilogram, second, ampere, kelvin, mole, candela.
Mass is in kilograms (not grams); temperature is in kelvin (not °C).

1Why we need standard units

A physical quantity is anything we can measure: length, mass, time, speed, force, temperature. To measure a quantity means to compare it with a fixed amount of the same kind of quantity. That fixed amount is called a unit.

So every measurement has two parts: a number that says how many units, and the unit itself. A length of 5 m means “5 times one metre”. The number 5 on its own means nothing, because 5 mm and 5 km are very different.

Q = n\,u

quantity = number × unit

If you measure the same thing with a different unit, the thing itself does not change. Only the number changes. A smaller unit fits in more times, so the number gets bigger:

n_1 u_1 = n_2 u_2 \quad\Rightarrow\quad n \propto \frac{1}{u}

the number is inversely proportional to the size of the unit

A unit is only useful if everyone agrees on it. If I measure a table in “my hand-spans” and you use yours, we get different numbers for the same table. That is why the world agreed on one standard set of units.

2The SI system and its seven base units

SI stands for Système International d'Unités, the International System of Units. It is the modern form of the metric system and is used across the world. It is a decimal system: bigger and smaller units differ by powers of 10, so converting is easy.

SI picks seven base quantities. Their units are the base units. Every other unit is built from these seven.

Base quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature (thermodynamic)	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

Rules for writing units: symbols never take a plural “s” (5 kg, not 5 kgs) and never take a full stop. Units named after people have a capital symbol but a small-letter name: the symbol is N, the unit is written “newton”.

3How the base units are fixed

A unit must never change. Old units were fixed by objects: the kilogram was a metal cylinder kept near Paris. But an object can gain dust or lose atoms. Since 2019, every SI base unit is fixed by giving an exact value to a constant of nature. Constants of nature are the same everywhere and forever, so any good laboratory can rebuild the units.

Unit	Fixed by	In plain words
second (s)	Caesium frequency, $\Delta\nu_{\text{Cs}} = 9\,192\,631\,770$ Hz	One second is the time for 9,192,631,770 waves of radiation from a caesium-133 atom.
metre (m)	Speed of light, $c = 299\,792\,458$ m/s	One metre is the distance light travels in vacuum in 1/299,792,458 of a second.
kilogram (kg)	Planck's constant, $h = 6.626\,070\,15 \times 10^{-34}$ J s	The kilogram is set so that h has exactly this value.
ampere (A)	Elementary charge, $e = 1.602\,176\,634 \times 10^{-19}$ C	One ampere is a flow of $1/(1.602\,176\,634 \times 10^{-19})$ elementary charges per second.
kelvin (K)	Boltzmann's constant, $k = 1.380\,649 \times 10^{-23}$ J/K	The kelvin is set so that k has exactly this value.
mole (mol)	Avogadro's number, $N_A = 6.022\,140\,76 \times 10^{23}$ /mol	One mole is exactly $6.022\,140\,76 \times 10^{23}$ particles.
candela (cd)	Luminous efficacy, $K_{cd} = 683$ lm/W	Fixed using green light of frequency $540 \times 10^{12}$ Hz.

4Derived units

A derived unit is made by multiplying and dividing base units. To find one, write the formula that defines the quantity, then put in the unit of each part.

Quantity	Unit	In base units	From
Area	square metre	m²	length × length
Volume	cubic metre	m³	length³
Velocity	metre per second	m s⁻¹	distance ÷ time
Acceleration	metre per second²	m s⁻²	velocity ÷ time
Force	newton (N)	kg m s⁻²	mass × acceleration
Pressure	pascal (Pa)	kg m⁻¹ s⁻²	force ÷ area
Work, energy	joule (J)	kg m² s⁻²	force × distance
Power	watt (W)	kg m² s⁻³	work ÷ time
Momentum	(no name)	kg m s⁻¹	mass × velocity

Quantity	Unit	In base units
Charge	coulomb (C)	A s
Voltage	volt (V)	kg m² s⁻³ A⁻¹
Resistance	ohm (Ω)	kg m² s⁻³ A⁻²
Capacitance	farad (F)	kg⁻¹ m⁻² s⁴ A²
Magnetic field	tesla (T)	kg s⁻² A⁻¹
Magnetic flux	weber (Wb)	kg m² s⁻² A⁻¹
Frequency	hertz (Hz)	s⁻¹

5SI prefixes

Very big and very small values are written with a prefix in front of the unit. Each prefix stands for a power of 10.

Prefix	Symbol	Factor	Example
tera	T	10¹²	1 THz = 10¹² Hz
giga	G	10⁹	1 GW = 10⁹ W
mega	M	10⁶	1 MHz = 10⁶ Hz
kilo	k	10³	1 km = 1000 m
hecto	h	10²	1 hm = 100 m
deca	da	10¹	1 dam = 10 m
deci	d	10⁻¹	1 dm = 0.1 m
centi	c	10⁻²	1 cm = 0.01 m
milli	m	10⁻³	1 mm = 0.001 m
micro	μ	10⁻⁶	1 μm = 10⁻⁶ m
nano	n	10⁻⁹	1 nm = 10⁻⁹ m
pico	p	10⁻¹²	1 pm = 10⁻¹² m
femto	f	10⁻¹⁵	1 fm = 10⁻¹⁵ m

Letter case matters: capital M is mega ( $10^6$ ), small m is milli ( $10^{-3}$ ). Use one prefix at a time: write nm, not mμm.

6Useful non-SI units

Some older or special units are still common. You must be able to change them into SI.

Quantity	Unit	In SI
Length	1 angstrom (Å)	10⁻¹⁰ m
Length	1 light year (ly)	9.46 × 10¹⁵ m
Length	1 parsec (pc)	3.08 × 10¹⁶ m
Mass	1 atomic mass unit (u)	1.66 × 10⁻²⁷ kg
Time	1 year	3.156 × 10⁷ s ≈ π × 10⁷ s
Energy	1 electron volt (eV)	1.6 × 10⁻¹⁹ J
Energy	1 calorie (cal)	4.186 J
Energy	1 kilowatt-hour (kWh)	3.6 × 10⁶ J
Pressure	1 atmosphere (atm)	1.013 × 10⁵ Pa
Pressure	1 bar	10⁵ Pa
Power	1 horsepower (hp)	746 W
Force	1 dyne	10⁻⁵ N

7Converting units

A conversion factor such as $\frac{1000\ \text{m}}{1\ \text{km}}$ is equal to 1, because the top and bottom are the same length. Multiplying by 1 never changes a quantity. It only changes the unit it is written in.

Write the value with its unit.
Decide which unit you want.
Multiply by conversion factors equal to 1.
Cancel units that appear on the top and the bottom.
Calculate the number.

Because $1000/3600 = 5/18$ : to change km/h into m/s, multiply by 5/18. To change m/s into km/h, multiply by 18/5. So 90 km/h = 25 m/s and 10 m/s = 36 km/h.

Summary

Key ideas

A measurement is a number times a unit. The number alone means nothing.
Changing the unit changes only the number: a smaller unit gives a bigger number.
SI has seven base units: metre, kilogram, second, ampere, kelvin, mole, candela.
Mass is in kilograms (not grams); temperature is in kelvin (not °C).
Since 2019 every base unit is fixed by an exact constant of nature ( $c, h, e, k, N_A, \Delta\nu_{\text{Cs}}, K_{cd}$ ).
Derived units are products and quotients of base units: put each part's unit into the defining formula.
Prefixes are powers of 10. A power applies to the prefix too: 1 cm³ = 10⁻⁶ m³.
Radian and steradian are ratios, so angles are pure numbers.
A light year is a distance, not a time.
To convert, multiply by factors equal to 1 and cancel units. Never mix units in one calculation.

Every equation

Measurement: $Q = n\,u$
Changing units: $n_1 u_1 = n_2 u_2$
Newton: $1\ \text{N} = 1\ \text{kg m s}^{-2}$
Joule: $1\ \text{J} = 1\ \text{N m} = 1\ \text{kg m}^2\,\text{s}^{-2}$
Watt: $1\ \text{W} = 1\ \text{J s}^{-1} = 1\ \text{kg m}^2\,\text{s}^{-3}$
Pascal: $1\ \text{Pa} = 1\ \text{N m}^{-2} = 1\ \text{kg m}^{-1}\,\text{s}^{-2}$
Coulomb: $1\ \text{C} = 1\ \text{A s}$
Volt: $1\ \text{V} = 1\ \text{kg m}^2\,\text{s}^{-3}\,\text{A}^{-1}$
Unit of G: $\text{N m}^2\,\text{kg}^{-2} = \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$
Speed of light (exact): $c = 299\,792\,458\ \text{m/s}$
Plane and solid angle: $\theta = \frac{\text{arc}}{r}\ \text{rad}, \qquad \Omega = \frac{\text{area}}{r^2}\ \text{sr}$
Powers of prefixes: $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3, \qquad 1\ \text{L} = 10^{-3}\ \text{m}^3 = 1000\ \text{cm}^3$
Speed: $1\ \text{km/h} = \tfrac{5}{18}\ \text{m/s}, \qquad 1\ \text{m/s} = \tfrac{18}{5}\ \text{km/h}$
Density: $1\ \text{g/cm}^3 = 10^3\ \text{kg/m}^3$
Kilowatt-hour: $1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$
Electron volt: $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$
Calorie: $1\ \text{cal} = 4.186\ \text{J}$
Angstrom: $1\ \text{Å} = 10^{-10}\ \text{m}$
Light year: $1\ \text{ly} = 9.46 \times 10^{15}\ \text{m}$
Parsec: $1\ \text{pc} = 3.08 \times 10^{16}\ \text{m}$
Atomic mass unit: $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$
Atmosphere, bar: $1\ \text{atm} = 1.013 \times 10^5\ \text{Pa}, \quad 1\ \text{bar} = 10^5\ \text{Pa}$
Horsepower: $1\ \text{hp} = 746\ \text{W}$
Dyne: $1\ \text{dyne} = 10^{-5}\ \text{N}$
Year: $1\ \text{year} = 3.156 \times 10^7\ \text{s} \approx \pi \times 10^7\ \text{s}$

Previous year questions with solutions

Real JEE and NEET questions on si units. Try each one before you open the solution.

Q1NEET 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.

Q2JEE Main 2026One correct option

A new unit ( $\alpha$ ) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes 6 min. 40 s to cover this distance?

A $200\alpha$
B $400\alpha$
C $300\alpha$
D $500\alpha$

Show answer and solution

Answer: Option B

The time first: $6\ \mathrm{min}\ 40\ \mathrm{s}$ is $360 + 40 = 400\ \mathrm{s}$ . One $\alpha$ is the distance light travels in one second, so $400\ \mathrm{s}$ of travel is $400\alpha$ , option B.

Q3JEE Advanced 2013One correct option

Match List I with List II and select the correct answer using the codes given below the lists:

List I

P. Boltzmann Constant

Q. Coefficient of viscosity

R. Plank Constant

S. Thermal conductivity

List II

1. [ML²T⁻¹]

2. [ML⁻¹T⁻¹]

3. [MLT⁻³K⁻¹]

4. [ML²T⁻²K⁻¹]

A $\begin{matrix} & P Q R S \\ & 3 1 2 4\end{matrix}$
B $\begin{matrix} & P Q R S \\ & 3 2 1 4\end{matrix}$
C $\begin{matrix} & P Q R S \\ & 4 2 1 3\end{matrix}$
D $\begin{matrix} & P Q R S \\ & 4 1 2 3\end{matrix}$

Show answer and solution

Answer: Option C

The same four constants as the question before, relabelled. Boltzmann's constant is energy over temperature, $[\mathrm{ML^{2}T^{-2}K^{-1}}]$ — 4. The coefficient of viscosity is $[\mathrm{ML^{-1}T^{-1}}]$ — 2. Planck's constant is energy over frequency, $[\mathrm{ML^{2}T^{-1}}]$ — 1. Thermal conductivity is $[\mathrm{MLT^{-3}K^{-1}}]$ — 3. So P-4, Q-2, R-1, S-3, which is option C.

Practice questions, easy to hard

Three questions from the si units 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

Energy density is the energy packed into each cubic metre: energy divided by volume. Energy is $\mathrm{kg\,m^{2}\,s^{-2}}$ and volume is $\mathrm{m^{3}}$ .

What is the SI unit of energy density?

A $\mathrm{kg\,m^{-2}\,s^{-2}}$
B $\mathrm{kg\,m^{5}\,s^{-2}}$
C $\mathrm{kg\,m^{-1}\,s^{-2}}$
D $\mathrm{kg\,m\,s^{-2}}$

Show answer and solution

Answer: Option C

$\mathrm{m^{2}} \div \mathrm{m^{3}} = \mathrm{m^{-1}}$ , so the unit is $\mathrm{kg\,m^{-1}\,s^{-2}}$ . Pressure written in base units comes out the same, since a pressure is also an energy shared over a volume — whatever kind of energy it happens to be does not change the derivation.

Q6One correct option

Now divide that by an acceleration: $\dfrac{v^{2}}{a} = \dfrac{[\mathrm{L^{2}T^{-2}}]}{[\mathrm{LT^{-2}}]}$ .

The $\mathrm{T^{-2}}$ above and below cancel exactly, and $\mathrm{L^{2}}$ divided by $\mathrm{L}$ leaves a single $\mathrm{L}$ .

So a velocity squared divided by an acceleration has the dimensions of what?

AA time
BA velocity
CA length
DAn acceleration

Show answer and solution

Answer: Option C

Everything cancels except one $[\mathrm{L}]$ , so $\dfrac{v^{2}}{a}$ is a length. This is the first half of the trick: if you know how velocity and acceleration behave between two systems, this combination hands you the length.