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:Â

1. Vernier Callipers reading

(i) 3.42cm (ii) 4.38cm (iii) 6.72cm

2. Micrometer screw gauge reading

i. 4.65mm (ii) 5.48mm (iii) 7.33mm

3. Distinguish between mass and weight

4. What are the instruments for measuring mass of an object?

#### Responses

1. Nice leaaon. Thanks a lot