Base Quantity & SI Units

Base Quantity

A base quantity (or basic quantity) is chosen and arbitrarily defined, rather than being derived from a combination of other physical quantities.

The 7 base quantities are:

Physical quantityBase SI unit
Mass (m)Kilogram (Kg)
Length ($l$)Metre (m)
Time (t)Second (s)
Current ($\text{I}$)Ampere (A)
Temperature (T)Kelvin (K)
Amount of sub. (n)Molar (mol)
Luminous Intensity (L)Candela (cd)

SI units are used as standardised units in all measurements in the world. SI is the short form for “International System of Units”.

Additional fundamental physical measures like area, volume, and speed are determined by utilizing mathematical equations derived from these base quantities. As an illustration, speed is characterized as the distance covered divided by the time taken: $\text{speed} = \frac{\text{distance travelled}}{\text{time}}$. Further details on derived quantity are in the next section.

Derived Quantity

A derived quantity is defined based on a combination of base quantities and has a derived unit that is the exponent, product or quotient of these base units.


  • Units, such as the joule, newton, volt and ohm, are SI units, but they are not base SI units.
  • Dimensional analysis: The main idea of “deriving” a derived unit is to treat units like algebraic terms, and manipulate them accordingly to get the right derived unit for the quantity. (See example 3 for a walkthrough)

An example of derived quantity is energy which has a derived unit of Joules which is $\text{kg} \, \text{m}^{2} \, \text{s}^{-2}$ OR $\text{kg} \, \text{m}^{2} / \text{s}^{2}$in base SI units.

Another example of derived quantity is density which has a derived quantity of $\text{kg} \, \text{m}^{-3}$ or $\text{kg}/\text{m}^{3}$ in base SI units.

Examples on Derived Quantities

Derived QuantityUnit NameUnit SymbolBase Units
ForceNewton$\text{N}$$\text{kg m s}^{-2}$
EnergyJoule$\text{J}$$\text{kg m}^{2}\text{ s}^{-2}$
PowerWatt$\text{W}$$\text{kg m}^{2}\text{ s}^{-3}$
ChargeCoulomb$\text{C}$$\text{A s}$
VoltageVolt$\text{V}$$\text{kg m}^{2}\text{s}^{-3}\text{ A}^{-1}$
ResistanceOhm$\Omega$$\text{kg m}^{2}\text{s}^{-3}\text{ A}^{-2}$

Homogeneous Equation

When each of the terms in a physical equation has the same base units, the equation is said to be homogeneous. This means that the units on both sides of the equation must be the same.

An equation which is not homogeneous must be wrong. However, when a physical equation is homogeneous, it does not necessarily imply that the equation is correct.

Consider the following equation:

$$\begin{aligned} s &= ut + \frac{1}{2}at^{2} \\ \text{Units: m} &= \, \text{m s}^{-1} \times \text{s} \,\,\, \text{m s}^{-2} \times \text{s}^{2} \\ \end{aligned}$$

Notice that the unit for all the terms on each side of the equation above is “m”! The equation is a homogeneous equation.

Worked Examples

Example 1

The equation for density is:

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

Now, you know that the unit for mass is $\text{kg}$, while the unit for volume is $\text{m}^{3}$. What is the unit for density in base SI units?

Click here to show/hide answer


$$\begin{aligned} \text{Unit for density} &= \frac{\text{kg}}{\text{m}^{3}} \\ &= \text{kg} / \text{m}^{3} \\ &= \text{kg} \, \text{m}^{-3} \end{aligned}$$

Example 2

The unit of force is Newton. Express this in base SI units.

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The equation for force is:

$$\text{force} = \text{mass} \times \text{acceleration}$$

The units of mass is $\text{kg}$, while the unit of acceleration is $\text{m} \, \text{s}^{-2}$.

$$\begin{aligned} \text{Unit for force} &= \text{kg} \times \text{m} \times \text{s}^{-2} \\ &= \text{kg m s}^{-2} \end{aligned}$$

Example 3

When an object descends in a vacuum, its entire gravitational potential energy transforms into kinetic energy. Demonstrate through unit comparison that the given equation $mgh = \frac{1}{2} m v^{2}$ is a possible solution to this scenario (i.e. it is a homogeneous equation).

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First, look at the units of both sides of the equation.

Left-hand side of the equation:

  • Unit of $m$: $\text{kg}$
  • Unit of $g$: $\text{m s}^{-2}$
  • Unit of $h$: $\text{m}$
  • Combined unit of $mgh$: $\text{kg} \times \text{m s}^{-2} \times \text{m} = \text{kg m}^{2} \, \text{s}^{-2}$

Right-hand side of the equation:

  • Unit of $\frac{1}{2}$: None
  • Unit of $m$: $\text{kg}$
  • Unit of $v$: $\text{m s}^{-1}$
  • Combined unit of $\frac{1}{2} m v^{2}$: $\text{kg} \times (\text{m s}^{-1})^{2} = \text{kg m}^{2} \, \text{s}^{-2}$

Both sides of the equation is the same – it is a homogeneous equation.

Example 4

The mass M of the largest stone which can be moved by water of a flowing river is assumed to be

$$M = kv^{r} \rho g^{s}$$

, where

$k$ = dimensionless constant

$v$ = velocity of water flow

$\rho$ = density of water

$g$ = acceleration due to gravity

and $r$, $s$ are dimensionless exponents.

Find value of $r$.

Click here to show/hide answer

Comparing the base units of LHS (left hand side) and RHS (right hand side) of the equation:

$$\text{Units of LHS} = \text{kg}$$

$$\begin{aligned} \text{Units of RHS} &= \left( \text{m s}^{-1} \right)^{r} \times \left( \text{kg m}^{-3} \right) \times \left( \text{m s}^{-2} \right)^{s} \\ &= \text{m}^{\left( r-3+s \right)} \times \text{s}^{\left( -r -2s \right)} \times \text{kg} \end{aligned}$$

Since LHS only has $\text{kg}$ as the only unit left, this means that:

$$r – 3 + s = 0$$

$$-r -2s =0$$

Solving for both equation, you will get:

$$r = 6$$

Example 5

The relationship between four physical quantities $z$, $p$, $q$, and $t$ is given by

$$z^2 = p + qt$$

,where $t$ is the time in seconds.

If q has the units $\text{m s}^{-1}$, then p must have the units

  1. $\text{m s}^{-1}$
  2. $\text{m s}^{-3}$
  3. $\text{m s}$
  4. $\text{m}$
Click here to show/hide answer

$$z^{2} = p + qt$$

As q has the units $\text{m s}^{-1}$, this means that:

$$\text{Units of } qt = \text{m s}^{-1} \text{ s} = \text{m}$$

Hence, units of $p$ is also $m$ as the units of each term in a homogenous equation must be the same.

Answer: 4

Example 6

(i) By Newton’s Second law of motion, the resultant force F acting on a constant mass m is given by F = ma, where a is the acceleration of the mass.

A student expresses the unit for mass as $\text{N s}^{2}\text{m}^{-1}$. Explain why its inappropriate to express mass in this unit.

ii) A student defines speed as “distance travelled per second”. Explain clearly why this statement is technically wrong. Hence, write down the correct definition for this speed.

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i) Mass is a base quantity and has the base unit, $kg$. It is not a derived quantity and so it is inappropriate to express the unit for mass as $\text{N s}^{2}\text{m}^{-1}$. Furthermore, the defining equation for mass is not force per unit acceleration.

ii) Distance is a physical quantity while second is a unit. The physical quantity speed should be defined in terms of quantities, and not a mixture of a quantity and a unit. The correct definition for speed is the distance travelled per unit time.

Example 7

A spinning ball moving with a speed v in air of density $\rho$ experiences a difference in pressure $\Delta p$ on its sides. This causes its path of motion to e deflected.

i) By comparing the powers of the base units, suggest a possible relationship between the difference in pressure $\Delta p$ and the quantities $v$ and $\rho$.

ii) Suggest a reason why the relationship that you have found in (c)(i) may not be physically correct.

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Let $\Delta p = k \rho^{x} v^{y}$, where $k$, $x$ and $y$ are dimensionless constants.

$$\begin{aligned} \text{Units of } \Delta p &= \text{Units of } k \rho^{x} v^{y} \\ \left( \text{kg} \right) \left( \text{m s}^{-2} \right) \left( \text{m}^{-2} \right) &= \left( \text{kg m^{-3}} \right)^{x} \left( \text{m s}^{-1} \right)^{y} \\ \text{kg m}^{-1} \text{s}^{-2} &= \text{kg}^{x} \text{m}^{-3x +y} \text{s}^{-y} \end{aligned}$$

By comparing powers,

$$x = 1$$

$$y = 2$$

ii) An equation that is dimensionally correct may be physically incorrect because there might be missing coefficients in the equation, or there might be terms which are missing.

Example 8

Express the pascal, $\text{Pa}$, in SI base units.

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The pascal is a unit for pressure.

$$p = \frac{F}{A}$$

In SI base units,

$$\begin{aligned} \text{Pa} &= \frac{\text{kg m s}^{-2}}{\text{m}^{2}} \\ &= \text{kg m}^{-1}\text{s}^{-2} \end{aligned}$$

Example 9

Suppose two quantities A and B have different dimensions. Determine which of the following arithmetic operations could be physically meaningful:

(a) $A + B$

(b) $\frac{A}{B}$

(c) $B-A$

(d) $AB$

Click here to show/hide answer

(b) and (d). You cannot add or subtract quantities of different dimension.

Example 10

If an equation is dimensionally correct, does that mean that the equation must be true? If an equation is not dimensionally correct, does that mean that the equation cannot be true?

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A dimensionally correct equation need not be true. Example: 1 monkey = 2 monkey is dimensionally correct. However, if an equation is not dimensionally correct, it cannot be correct.

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