Lesson 3, Topic 6
In Progress

# Dimensions of Physical Quantities

Lesson Progress
0% Complete

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

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 = $$\frac{M}{L^3} = \scriptsize ML^{-3}$$

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.

Evaluation Questions

Show the following: 