Dimensions of Physical Quantities

Any quantity that can be measured in units also has dimension, hence the dimension for length, mass, and time is as follows:

Quantity	Symbol
Mass	M
Length	L
Time	T

Some examples of dimension for derived quantities are:

Velocity = \( \frac {length}{time} \\ = \frac {L}{T} \)

Dimension of Velocity is \(\scriptsize [L][T^{-1}] \)

Momentum = Mass x Velocity

= \(\scriptsize M \times LT^{-1} \\ \scriptsize = M LT^{-1} \)

Acceleration = \( \frac{velocity}{time} \\ = \frac{LT^{-1}}{T} \\ = \scriptsize LT^{-2}\)

Force = Mass x Acceleration

= \( \scriptsize M \times LT^{-2} \\ \scriptsize = M LT^{-2} \)

Work = Force x Distance

W = \( \scriptsize MLT^{-2} \: \times \: L \\ \scriptsize = ML^2T^{-2} \)

Pressure = \(\frac {force}{area}\)

= \(\frac {M LT^{-2}}{L^2}\\ = \frac {MT^{-2}}{L} \\ \scriptsize ML^{-1}T^{-2} \)

Dimension of Area, A = length x breadth

A = \( \scriptsize L \: \times \: L \\ \scriptsize L^2 \)

Domension of Volume, V = \( \scriptsize L \: \times \: L \: \times \: L \\ = \scriptsize L^3 \)

Worked Example:

What is the dimension of density?

Step I : formula of density

Step II : write down the formula of density,

Density = \( \frac {mass}{volume} = \frac {mass}{l \; \times \; b \; \times \; h} \)

Step III : what are the quantities that are present in the formula of density

The quantities that are present are l, b and h can all be called length

Step IV: Symbol of length is L, Symbol of mass is M

Step V : Replace the quantities with their symbols

We will have, Dimension of density = \( \frac{M}{L^3} = \scriptsize ML^{-3}\)

Derived Quantity	Dimension
Velocity	\(\scriptsize [L]\:[T^{-1}] \)
Acceleration	\(\scriptsize [L]\:[T^{-2}]\)
Force	\(\scriptsize [M]\: [L]\:[T^{-2}] \)
Area	\(\scriptsize [L^2] \)
Volume	\(\scriptsize [L^3] \)
Density	\( \scriptsize [M]\:[L^{-3}]\)
Pressure	\(\scriptsize [M]\:[L^{-1}]\:[T^{-2}] \)
Work	\( \scriptsize [M] \: [L^2]\:[T^{-2}] \)

Dimensional analysis can be used to verify whether a physical equation is correct or not.

For example, consider the formula v = u + at

The dimension of each term on each side of the equation must be the same.

u, initial velocity = \(\scriptsize [L]\:[T^{-1}] \)

a, acceleration = \(\scriptsize [L]\:[T^{-2}]\)

t = T

Therefore,

v = u + at

\(\scriptsize L\:T^{-1} = L\:T^{-1} \: + \: LT^{-2} \: \times \: T \)

=\(\scriptsize L\:T^{-1} = L\:T^{-1} \: + \: LT^{-1} \)

The equation is dimensionally correct.