##### The dimension of Physical quantities

The dimension of a physical quantity is simply how that quantity is related to the** 7 fundamental quantities**. Most textbooks will say **3 important fundamental quantities** which are *mass, length and time.*

In the above write up, we were made to know that derived units are gotten from the combination of two or more fundamental units.

However, we must know how to represent the fundamental unit with the correct letters.

**Mass** is represented with **M**

**Length** is represented with **L**

**Time** is represented with** T**

**Temperature** is represented with **θ**

**Current** is represented with **I**

Let’s work with some examples:

##### Example 1: Find the dimension of **Area**.

We have two ways of tackling this:

1. Via the definition of the physical quantity.

2. Via the unit of the physical quantity.

we will apply these two methods to this question

Via method 1;

The area is defined as length x breadth since both quantities are used to measure length.

Area = L x B = L x L = L^{2}

via method 2;

The unit of area is m^{2}

since m indicate meter which is used for measuring length we can replace it with L, making it L^{2}

##### Example 2: We will look at the physical dimension for **density**

Using the definition method

Density = \frac{mass}{volume}

mass can be represented as M

While volume is defined as Length x Breadth x Height. Which can be represented by a dimension of L cubed, L³

Therefore, Density =\frac{M}{L^{3}}

###### Example 3: Let’s repeat this for velocity

using the unit method, the unit of velocity is in m/s

since m indicates Length(L) and seconds indicates time(T)

velocity=\frac{L}{T}

###### Example 4: What does x,y,z represent for **acceleration** in **M^{x}L^{y}T^{z}**

Going with the definition of Acceleration: acceleration is the rate of change of velocity with time.

Acceleration = \frac{Velocity}{time}, but velocity = \frac{L}{T}

Replacing velocity with its dimension gives:

Acceleration = \frac{(L/T)}{T}

simplifying the expression gives: Acceleration = \frac{L}{T^{2}}

writing it out using the laws of indices

Acceleration = LT^{-2}

comparing it with M^{x}L^{y}T^{z};

LT^{-2}=M^{x}L^{y}T^{z}

Since the left-hand side of the equation do not have M, we can say that the power of M in the right-hand side of the equation is 0, Therefore;

x = 0

The power of L in the right-hand side of the equation is 1, Therefore;

y = 1

The power of T in the right-hand side of the equation is -2, Therefore,

z = -2

x = 0, y = 1, z = -2

###### Example 5: Let us find out the dimension of **specific heat capacity**

specific heat = \frac{Heat energy}{mass * temperature change}

c = \frac{Q}{Mθ}

since the dimension of Heat energy is the same as work or energy, therefore;

Let us do the dimension for energy;

energy/work = force x distance

energy/work = \frac{ML}{T^{2}} x L

energy/work = \frac{ML^{2}}{T^{2}}

going back to the specific heat

c = \frac{ML^{2}}{T^{2}} / Mθ

The mass term will cancel out giving us

c = \frac{L^{2}}{θT^{2}} or {L^{2}T^{-2}θ^{-1}}

thanks bigger bro! i find it interesting to educate my self here. thanks.

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Awesome tutorial. Keep up with this, keep giving people good value. Kudos!

22/09/2009