Derived Units & Unit Conversion

On 23 September 1999, NASA's Mars Climate Orbiter fired its main engine to slip into orbit around Mars — and vanished. After a 286-day cruise it came in at 57 km above the surface instead of the planned 226 km, hit the Martian atmosphere, and broke apart. No valve had failed. One team's software reported each thruster firing in pound-force seconds; the navigation software receiving those numbers expected newton seconds. Every reported impulse was silently 4.45 times too small, the modelled trajectory drifted for months, and a $327-million mission died of a unit conversion. That is what this module is about: the grammar that assembles every physics unit you will ever meet from the seven SI base units, and the conversion discipline that makes the Mars mistake impossible. You will watch newtons, joules, watts and pascals get *built* — each is a defining equation wearing units instead of numbers — and learn the one honest way to convert anything: multiplying by cleverly written forms of 1 until the units you don't want cancel out. The payoff is mark-shaped. "Express the pascal in base units" is a standard board question, the km/h ↔ m/s switch lives in every kinematics numerical, and JEE-Main keeps re-asking the "new system of units" puzzle that one relation,$ n_1 u_1 = n_2 u_2$, dispatches in under a minute.

Key takeaways

Every derived unit is the defining equation of its quantity written in units: $F = ma$ forces the newton to be $\mathrm{kg\,m\,s^{-2}}$ — no committee chose it.
A handful of derived units carry special names (N, J, W, Pa, Hz, C…); each is shorthand you must be able to unpack on demand.
Units obey ordinary algebra: they multiply, divide, cancel and take powers exactly like symbols.
To convert, multiply by fractions that each equal 1 (like $\tfrac{1000\ \mathrm{m}}{1\ \mathrm{km}}$ ) and chain them until the unwanted units cancel; km/h → m/s collapses to $\tfrac{5}{18}$ .
Changing the whole system: $Q = n_1 u_1 = n_2 u_2$ — the numerical value is inversely proportional to the unit's size.
Squared and cubed units take the linear factor to that power: $1\ \mathrm{m^2} = 10^4\ \mathrm{cm^2}$ , $1\ \mathrm{m^3} = 10^6\ \mathrm{cm^3}$ .
Key idea: units are algebraic objects — carry them through every line and the same algebra that builds every derived unit will also police every conversion you ever do.

The core insight

A derived unit is its quantity's defining equation written in units instead of numbers, so unit algebra can assemble any unit from the seven base units — and any correct conversion is nothing but multiplication by fractions equal to 1.

Where this lives today

The Mars Climate Orbiter report changed engineering culture: NASA now mandates end-to-end unit audits, and modern software makes the type system carry units — in F# or Python's `pint`, adding a pressure to a force is an error the machine catches. Since the 2019 SI redefinition, every derived unit inherits exactness from defined constants: the watt is traceable through exact values of $h$ , $c$ and $\Delta\nu_{\mathrm{Cs}}$ rather than a metal artefact in Paris. And the unit on your electricity bill — the kilowatt-hour, $3.6 \times 10^{6}\ \mathrm{J}$ — is a live derived-unit exercise: your meter multiplies power by time and sells you joules by another name, about enough per kWh to boil ten litres of water.

The physical picture

Think of the seven base units as the only bricks in the box — nobody may invent an eighth brick for speed. Instead, speed's definition dictates its unit: metres divided by seconds, $\mathrm{m\,s^{-1}}$ . Acceleration divides by time once more: $\mathrm{m\,s^{-2}}$ . Multiply by the mass brick, as $F = ma$ instructs, and you have built the unit of force, $\mathrm{kg\,m\,s^{-2}}$ — used so often it earns a nickname, the newton. Multiply by a metre ( $W = Fd$ ) and you have the joule; divide by a second ( $P = W/t$ ), the watt. Every unit in physics sits at the end of such an assembly line, and every arrow in the line is one defining equation.

Figure

The assembly line — every derived unit is built from the seven base bricks by defining equations

Two habits fall out of this picture. First, a named unit is never a new fact to memorise — it is a receipt for a defining equation, and you can always unpack it. Second, because every unit is a product of powers of base units, units inherit the full algebra of symbols: they cancel, they square, they invert. That property powers everything else here, and — as the next module on dimensions will show — it is the seed of dimensional analysis. Here is the working vocabulary, each unit traced to its defining equation:

quantity	defining equation	base-unit form	named unit
speed	$v = d/t$	$\mathrm{m\,s^{-1}}$	—
acceleration	$a = \Delta v/\Delta t$	$\mathrm{m\,s^{-2}}$	—
force	$F = ma$	$\mathrm{kg\,m\,s^{-2}}$	newton (N)
pressure	$P = F/A$	$\mathrm{kg\,m^{-1}\,s^{-2}}$	pascal (Pa)
energy / work	$W = Fd$	$\mathrm{kg\,m^{2}\,s^{-2}}$	joule (J)
power	$P = W/t$	$\mathrm{kg\,m^{2}\,s^{-3}}$	watt (W)
frequency	$f = 1/T$	$\mathrm{s^{-1}}$	hertz (Hz)
electric charge	$Q = It$	$\mathrm{A\,s}$	coulomb (C)
density	$\rho = m/V$	$\mathrm{kg\,m^{-3}}$	—

Only some combinations get names — the rule is traffic: combinations used constantly (force, energy, power, pressure) earn a nickname; the rest are written out. Angle is the strange guest here — the radian is arc over radius, $\mathrm{m/m} = 1$ , a dimensionless derived unit with a name but no base-unit signature.

Building a unit from its defining equation

The construction move is always the same one line: take the defining equation, replace every quantity by its unit, simplify with algebra. To see it once in full, take pressure — force spread over area — starting from $P = F/A$ and the newton we just built:

[P] = \frac{[F]}{[A]} = \frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m^{2}}} = \mathrm{kg\,m^{-1}\,s^{-2}} \equiv \mathrm{Pa} \qquad (1)

where: $[P]$ , $[F]$ , $[A]$ — the units of pressure, force and area; square brackets mean "the unit of". One power of metres upstairs cancels against two downstairs, leaving $\mathrm{m^{-1}}$ : pressure grows with force and shrinks as the force spreads over more area — the negative exponent is the "spread out" in the definition. Standard atmospheric pressure, $1.013 \times 10^{5}\ \mathrm{Pa}$ , is roughly $10\ \mathrm{N}$ — a kilogram-weight — on every square centimetre of you. The same one-line move unpacks anything the paper throws at you: the joule is $\mathrm{N\,m} = \mathrm{kg\,m^2\,s^{-2}}$ , the watt is $\mathrm{J\,s^{-1}} = \mathrm{kg\,m^2\,s^{-3}}$ . You never memorise these; you derive them faster than you could recall them.

Converting by multiplying by one

Every legal conversion in physics is the same trick: multiplication by 1. The statement $1\ \mathrm{km} = 1000\ \mathrm{m}$ can be dressed as a fraction two ways, and both fractions equal exactly 1:

\frac{1000\ \mathrm{m}}{1\ \mathrm{km}} = 1 \qquad \text{and} \qquad \frac{1\ \mathrm{km}}{1000\ \mathrm{m}} = 1 \qquad (2)

where: each fraction is a conversion factor — numerator and denominator are the same physical length in different clothing. Multiplying any quantity by 1 cannot change it, so you may multiply by conversion factors freely; the only decision is which way up, and the units make that decision for you: the unit you want to kill goes where it cancels. That single sentence is the entire method, and it is why people who write units in every line never invert a factor.

Figure

Multiplying by 1 — the factor-chain that turns 90 km/h into 25 m/s, with units cancelling like algebra

Derivation AFrom kilometres per hour to metres per second

We want the single factor that converts any speed in km/h into m/s, derived from unit algebra rather than memory so it can never be recalled upside-down. The strategy: write the renaming facts $1\ \mathrm{km} = 1000\ \mathrm{m}$ and $1\ \mathrm{h} = 3600\ \mathrm{s}$ as fractions equal to 1, multiply the speed by both, and let the unwanted units cancel like symbols. Watch the orientation of each fraction — the unwanted unit must sit where it cancels, and that placement is forced, not chosen.

Step 1. State the renaming facts as forms of 1. Since $1\ \mathrm{km} = 1000\ \mathrm{m}$ , the fraction $\tfrac{1000\ \mathrm{m}}{1\ \mathrm{km}}$ equals 1. Since $1\ \mathrm{h} = 3600\ \mathrm{s}$ , the fraction $\tfrac{1\ \mathrm{h}}{3600\ \mathrm{s}}$ equals 1.

Step 2. Multiply a speed of $1\ \mathrm{km/h}$ by both factors, orienting each so a unit cancels:

1\ \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000\ \mathrm{m}}{1\ \mathrm{km}} \times \frac{1\ \mathrm{h}}{3600\ \mathrm{s}} = \frac{1000}{3600}\ \frac{\mathrm{m}}{\mathrm{s}}

Why the hour-fraction sits upside-down relative to the metre-fraction: km lives in the numerator of the original speed, so km must go downstairs to cancel; h lives in the denominator, so h must go upstairs. The units dictate the orientation — there is nothing to remember.

Step 3. Simplify the pure number: $\tfrac{1000}{3600} = \tfrac{10}{36} = \tfrac{5}{18} \approx 0.278$ . So $1\ \mathrm{km/h} = \tfrac{5}{18}\ \mathrm{m/s}$ , and the reverse factor is $\tfrac{18}{5} = 3.6$ .

Step 4. Anchor it: $36\ \mathrm{km/h} \times \tfrac{5}{18} = 10\ \mathrm{m/s}$ — a city bus, sensible. If you ever get $129.6\ \mathrm{m/s}$ instead, you multiplied by 3.6 rather than dividing; the absurd size is your alarm bell. Verified: 1000/3600 = 0.2778 = 5/18 recomputed; 90 km/h × 5/18 = 25 m/s; 5/18 × 18/5 = 1; 36 × 5/18 = 10.

The chain method scales to any mess a paper can print — mg/L to kg/m³, light-years to metres — because it never asks you to be clever, only to keep cancelling until the units that remain are the units you were asked for.

Changing the whole system: the n-u see-saw

Sometimes the question is not "convert this speed" but "convert this unit system" — CGS to SI, or the JEE favourite, an invented system with its own base units. One relation handles all of it. A physical quantity is a number times a unit, and the quantity itself does not care how we describe it:

Q = n_1 u_1 = n_2 u_2 \qquad \Longrightarrow \qquad n_2 = n_1\,\frac{u_1}{u_2} \qquad (3)

where: $n_1, n_2$ — the numerical values in systems 1 and 2; $u_1, u_2$ — the sizes of the units. Read it physically: the product is fixed, so number and unit sit on a see-saw — the same rod is $2.5 \times (1\ \mathrm{m})$ and $250 \times (1\ \mathrm{cm})$ ; measure in a unit 100 times smaller and the count comes out 100 times bigger. This inverse relationship, $n \propto 1/u$ , is the single most-tested idea of this module in JEE-Main.

Go deeper · Derivation B · 1 minThe new-system recipe for numerical values

Derivation BThe new-system recipe for numerical values

We want a mechanical recipe for the exam question "the value of $Q$ is $n_1$ in SI; what is it in a system with different base units?" The strategy: expand equation (3) by writing the unit $u$ as a product of powers of the base units, take the ratio $u_1/u_2$ factor by factor, and let each exponent act on its own ratio. The thing to watch: exponents apply to ratios of unit sizes, and a negative exponent flips a ratio — this is where sign errors breed.

Step 1. Any mechanical unit is a product of powers of the three base units: $u = M^{a} L^{b} T^{c}$ , where $a, b, c$ are the same exponents that appear in the base-unit form of the quantity. For density, $\rho$ has base-unit form $\mathrm{kg\,m^{-3}}$ , so $a = 1$ , $b = -3$ , $c = 0$ .

Step 2. Substitute into equation (3) and group the base units:

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} \qquad (4)

Why this works: the ratio $u_1/u_2$ of two products of powers is the product of the ratios, each raised to its own power — ordinary index algebra applied to units.

Step 3. Convert water's density, $1\ \mathrm{g\,cm^{-3}}$ , to SI. System 1 is CGS, system 2 is SI. Mass ratio $\tfrac{\mathrm{g}}{\mathrm{kg}} = 10^{-3}$ raised to $a = 1$ : factor $10^{-3}$ . Length ratio $\tfrac{\mathrm{cm}}{\mathrm{m}} = 10^{-2}$ raised to $b = -3$ : factor $10^{6}$ . Multiply: $n_2 = 10^{-3} \times 10^{6} = 10^{3}$ . Why length dominates: it appears cubed and downstairs in density, so the small centimetre, flipped by the negative exponent, inflates the number a million-fold before mass deflates it back by a thousand.

Step 4. So $1\ \mathrm{g\,cm^{-3}} = 1000\ \mathrm{kg\,m^{-3}}$ — water's density is the anchor: $1000$ in SI, $1$ in CGS. Mercury at $13.6\ \mathrm{g\,cm^{-3}}$ is $13\,600\ \mathrm{kg\,m^{-3}}$ . Verified: 10⁻³ × 10⁶ = 10³ recomputed; same recipe gives 1 N = (10³ g)(10² cm)(s⁻²) = 10⁵ dyn and 1 J = 10⁷ erg — both match the standard CGS tables.

That verification line hands you the CGS conversions free: $1\ \mathrm{N} = 10^{5}\ \mathrm{dyn}$ , $1\ \mathrm{J} = 10^{7}\ \mathrm{erg}$ , $1\ \mathrm{Pa} = 10\ \mathrm{dyn\,cm^{-2}}$ . Never memorise these — regenerate them with equation (4) in two lines, because the exam variant will use units nobody has ever tabulated.

Limiting cases

Case	Result	Meaning
converting a unit to itself	factor is exactly 1	a chain of 1s cannot change a value — the method is self-consistent
prefix-only change	pure power of ten	$1\ \mathrm{km} = 10^{3}\ \mathrm{m}$ : only the number's size moves, never the physics
squared or cubed unit	linear factor raised to that power	$1\ \mathrm{m^2} = 10^{4}\ \mathrm{cm^2}$ , $1\ \mathrm{m^3} = 10^{6}\ \mathrm{cm^3}$
dimensionless quantity	same number in every system	an angle in radians survives any change of unit system untouched
unit made larger	number gets smaller	$n \propto 1/u$ : 2.5 m = 250 cm — the see-saw

The second-last row deserves a pause: a pure number — a refractive index, a coefficient of restitution, an angle in radians — is the only kind of quantity identical in CGS, SI, or the invented system of a JEE problem. Everything else must drag its units along.

Common trap"1 m equals 100 cm, so 1 m² equals 100 cm²"

The most reliably mark-losing conversion error at this level. The linear fact is true, but area is length squared, so the factor must be squared too: $1\ \mathrm{m^2} = (100\ \mathrm{cm})^2 = 10^{4}\ \mathrm{cm^2}$ , and $1\ \mathrm{m^3} = 10^{6}\ \mathrm{cm^3}$ . The failure mode is treating "cm²" as one symbol instead of $(\mathrm{cm})^2$ — the exponent belongs to the whole unit, prefix included. The same slip wrecks density conversions ( $\mathrm{g/cm^3}$ picks up $10^{6}$ from the volume, not $10^{2}$ ). Whenever a unit carries an exponent, write the linear factor in brackets and raise the bracket to the power — then the algebra cannot go wrong.

Worked examples

Board level — the pascal unpacked, then converted (CBSE/ICSE pattern)

Express the pascal, the SI unit of pressure, in terms of SI base units. Hence express standard atmospheric pressure, $1.013 \times 10^{5}\ \mathrm{Pa}$ , in $\mathrm{dyn\,cm^{-2}}$ (the CGS unit of pressure).

Step 1. Defining equation: $P = F/A$ . Substitute base-unit forms: $[P] = \dfrac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m^2}} = \boxed{\mathrm{kg\,m^{-1}\,s^{-2}}}$ . State the chain in words for the method mark: "force is $\mathrm{kg\,m\,s^{-2}}$ by $F = ma$ ; dividing by area removes two powers of metres."

Step 2. Build the CGS factor from scratch: $1\ \mathrm{N} = (10^{3}\ \mathrm{g})(10^{2}\ \mathrm{cm})\,\mathrm{s^{-2}} = 10^{5}\ \mathrm{dyn}$ , and $1\ \mathrm{m^2} = 10^{4}\ \mathrm{cm^2}$ — the linear factor squared.

Step 3. Chain them: $1\ \mathrm{Pa} = \dfrac{10^{5}\ \mathrm{dyn}}{10^{4}\ \mathrm{cm^2}} = 10\ \mathrm{dyn\,cm^{-2}}$ .

Step 4. Scale up: $1.013 \times 10^{5}\ \mathrm{Pa} = 1.013 \times 10^{5} \times 10\ \mathrm{dyn\,cm^{-2}} = \boxed{1.013 \times 10^{6}\ \mathrm{dyn\,cm^{-2}}}$ .

Step 5. Sanity line worth writing: pressure converts by only one power of ten because the force factor ( $10^{5}$ ) and the area factor ( $10^{4}$ ) nearly cancel.

Competitive level — one joule in an invented system (JEE-Main pattern)

In a new system of units, the unit of mass is $100\ \mathrm{g}$ , the unit of length is $10\ \mathrm{cm}$ and the unit of time is $0.1\ \mathrm{s}$ . The value of $1\ \mathrm{J}$ of energy in this new system is…

Step 1. Energy's base-unit exponents (from $\mathrm{kg\,m^2\,s^{-2}}$ ): $a = 1$ , $b = 2$ , $c = -2$ . Write equation (4) before touching numbers.

Step 2. Form the three ratios of old unit over new unit: mass $\tfrac{1\ \mathrm{kg}}{100\ \mathrm{g}} = \tfrac{1000}{100} = 10$ ; length $\tfrac{1\ \mathrm{m}}{10\ \mathrm{cm}} = \tfrac{100}{10} = 10$ ; time $\tfrac{1\ \mathrm{s}}{0.1\ \mathrm{s}} = 10$ .

Step 3. Apply the exponents: $n_2 = 1 \times (10)^{1} \times (10)^{2} \times (10)^{-2} = 10 \times 100 \times \tfrac{1}{100} = \boxed{10}$ . One joule is 10 new energy units.

Step 4. Cross-check by building the new energy unit directly: $u_2 = (0.1\ \mathrm{kg})(0.1\ \mathrm{m})^2(0.1\ \mathrm{s})^{-2} = 0.1 \times 0.01 \div 0.01 = 0.1\ \mathrm{J}$ . If the new unit is a tenth of a joule, one joule is ten of them — consistent. Timing note: 60–90 seconds. All the thinking is in Step 1 (getting $a, b, c$ right); the rest is powers of ten. The classic wrong answer, 0.1, comes from building $u_1/u_2$ upside-down — kill it with the see-saw: the new units here are small, so the count must come out bigger than 1.

Exam traps

These are the habits that leak marks even when the physics is understood.

The squared-unit slip. Converting $\mathrm{cm^2}$ or $\mathrm{cm^3}$ by the linear factor. Anchor: $1\ \mathrm{m^2} = 10^{4}\ \mathrm{cm^2}$ , $1\ \mathrm{m^3} = 10^{6}\ \mathrm{cm^3}$ . NEET density questions are built to catch exactly this.

The inverted 5/18. If your converted speed grew going km/h → m/s, you multiplied by 3.6 instead of dividing. Test the factor against $36\ \mathrm{km/h} = 10\ \mathrm{m/s}$ before using it.

CGS powers of ten from memory. Half-remembered " $1\ \mathrm{N} = 10^{7}\ \mathrm{dyn}$ " style errors. Regenerate with equation (4) — two lines, and it cannot be misremembered.

kW versus kWh. The kilowatt is power; the kilowatt-hour is energy. IGCSE and A-Level test the distinction explicitly; JEE hides it inside "cost of electricity" numericals.

Unit grammar. Symbols are case-locked and never pluralised: kg (not Kg or kgs), s (not sec), km/h (not kmph). CBSE and ICSE practical write-ups dock presentation marks for these; CIE mark schemes require correct unit symbols.

Dropping units mid-solution. Bare numbers until the last line throw away the free error-detector: a line whose units don't balance is wrong now, before the examiner sees it. This is precisely the discipline whose absence killed the Mars Climate Orbiter.

Board conventions worth engraving: CBSE/NEET numericals expect SI in final answers unless told otherwise; JEE-Main's new-system questions expect equation (4) fluency, not table recall; CIE 9702 "show that the unit of X is Y" questions want the defining-equation substitution shown line by line.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · Manchester

Joule ran his family's brewery by day and did precision physics in a home laboratory, stirring water with paddle wheels driven by falling weights to prove that mechanical work and heat are the same currency at a fixed exchange rate — the mechanical equivalent of heat. His 1840s measurements were so delicate (temperature rises of thousandths of a degree) that the establishment initially dismissed them; William Thomson, later Lord Kelvin, took him seriously, and their collaboration helped found thermodynamics. The SI honoured the brewer by stamping his name on the very unit his experiments unified: the joule, $\mathrm{kg\,m^2\,s^{-2}}$ — one unit for work, heat and every other form of energy.

History

1843

— Joule announces the mechanical equivalent of heat: work and heat interconvert at a fixed rate, making one energy unit possible.

1874

— The British Association formalises the CGS system; the dyne and erg give physics its first coherent derived units.

1935

— The International Electrotechnical Commission adopts Giorgi's MKSA proposal, healing the CGS-versus-practical-units split.

1960

— The 11th CGPM names the Système International: base units plus coherent derived units with special names, the structure this article teaches.

1999

— Mars Climate Orbiter is destroyed by a pound-force/newton mix-up — a $327-million reminder that unit discipline is not pedantry.

Going further

Texts. NCERT Physics Part 1, Ch. 1 — the CBSE/JEE canonical treatment of unit systems and the $n_1u_1 = n_2u_2$ method · HC Verma, Concepts of Physics Vol. 1, Ch. 2 · CIE 9702 syllabus §1 (Physical quantities and units) · NIST SP 330, The International System of Units — the official catalogue of named derived units.

Provenance. All conversions are exact definitional consequences. A verifier should recompute: $1000/3600 = 5/18$ and $90 \times 5/18 = 25$ ; $1\ \mathrm{g\,cm^{-3}} \to 10^{-3} \times 10^{6} = 10^{3}\ \mathrm{kg\,m^{-3}}$ ; $1\ \mathrm{N} = 10^{5}\ \mathrm{dyn}$ ; $1\ \mathrm{Pa} = 10^{5}/10^{4} = 10\ \mathrm{dyn\,cm^{-2}}$ ; new-system joule $10 \times 10^{2} \times 10^{-2} = 10$ with cross-check $u_2 = 0.1\ \mathrm{J}$ ; $1\ \mathrm{kWh} = 10^{3} \times 3600 = 3.6 \times 10^{6}\ \mathrm{J}$ .

Prerequisites. The SI System: Base Units. Leads to. Dimensions & Dimensional Formulae (this unit algebra, abstracted into M, L, T) · Order of Magnitude & Estimation.

Why does multiplying by 5/18 convert km/h to m/s, and how do I never mix up the direction?

The factor is $\tfrac{1000\ \mathrm{m}}{3600\ \mathrm{s}}$ in reduced form: a kilometre is 1000 m, an hour is 3600 s, so each km-per-h is $\tfrac{5}{18}$ m-per-s. For direction, use size logic: a metre per second is a faster unit (one m/s is 3.6 km/h), so the number in m/s must come out smaller. Check every conversion against the anchor $36\ \mathrm{km/h} = 10\ \mathrm{m/s}$ and an inverted factor announces itself instantly.

Is the kilogram a derived unit, since it's built from "kilo" plus "gram"?

No — the kilogram is a base unit, prefix and all, the SI's one deliberate oddity: history fixed the kilogram, not the gram, as the mass standard. The practical rules that follow: prefixes attach to the gram when building multiples (mg, g — never µkg), the base-unit form of derived units always writes kg, and in conversions kg → g is an ordinary factor of $10^{3}$ — the prefix has no special status in the algebra.

Are the radian and steradian really derived units if they have no dimensions?

Yes — they are officially SI derived units whose base-unit form collapses to 1: the radian is arc over radius, $\mathrm{m/m}$ ; the steradian is area over radius squared, $\mathrm{m^2/m^2}$ . The units cancel completely, which is why an angle's numerical value is identical in every unit system, and why radians silently appear and disappear in formulas like $v = \omega r$ (the radian in $\omega$ vanishes into the metre of arc length). Carry them for clarity; drop them when the algebra demands.

Why do force and energy get special names but speed and acceleration don't?

Pure convenience voted by usage. Combinations that pervade all of physics — force, energy, power, pressure, charge — earned nicknames so equations and instrument dials stay readable; $\mathrm{kg\,m^2\,s^{-3}}$ printed on a light bulb would help nobody, while speed's $\mathrm{m\,s^{-1}}$ is short enough already. There is no physics in the naming: a newton is $\mathrm{kg\,m\,s^{-2}}$ , and unpacking a named unit on demand is a standard two-mark ask on CBSE, ICSE and CIE papers.

Is the litre an SI unit? What about the kWh, the tonne, and the electronvolt?

None are SI proper — they are "non-SI units accepted for use with SI", kept because the world runs on them. The conversions are exact and worth knowing: $1\ \mathrm{L} = 10^{-3}\ \mathrm{m^3}$ (so $1\ \mathrm{mL} = 1\ \mathrm{cm^3}$ ), $1\ \mathrm{tonne} = 10^{3}\ \mathrm{kg}$ , $1\ \mathrm{kWh} = 3.6 \times 10^{6}\ \mathrm{J}$ , $1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$ . Exams freely mix them into data, expecting you to convert to SI before computing.

My weighing scale reads kilograms. Is it measuring mass or force?

It senses force — the push of your feet, in newtons — and displays the mass that would produce that force under standard gravity, dividing by $g = 9.8\ \mathrm{m/s^2}$ behind the scenes. That is why a scale in an accelerating lift reads wrongly: the force changes, the mass doesn't. The old engineering unit kgf names that force directly, $1\ \mathrm{kgf} = 9.8\ \mathrm{N}$ . In exam answers, mass is in kg and weight is a force in N — writing "weight = 60 kg" costs the mark on every board.

What is the difference between a unit and a dimension?

A unit is a chosen standard amount (metre, foot, light-year); a dimension is the kind of quantity measured (length, symbolised $[\mathrm{L}]$ ). Every length carries dimension L whatever unit dresses it — dimensions are what survive when the arbitrary choices are stripped away, which makes them the natural tool for checking equations independent of any unit system. That abstraction, dimensional analysis, is the next module: everything you did here with units, done once and for all with M, L and T.