## Concept of Dimension in Physics

To understand dimension in physics we need to know its definition

**Definition:** The dimension of a physical quantity shows the application of the method of fundamental quantities like Mass (M), Length (L), and Time (T) to relate the fundamental units of a physical quantity.

In dimension, we derive quantities by applying physical quantities. Understanding dimension, demands understanding the method of measurements, fundamental quantities, derived quantities, and even their corresponding units in physics.

Dimension gives us opportunity to find the relationship between quantities. As we mentioned earlier, we apply Mass in form of M, Length in form of L, and Time in form of T.

Additionally, units like temperature will be in form of K or θ, electric current in form of (A) which is an expression for ampere, and finally luminous intensity as (II).

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## How to Calculate Dimensions of Physical Quantity

To calculate dimensions of physical quantity, we need to understand that the measurement of quantities in units of Lengths, Mass, and Time involves the dimension Length (L), Mass (M), and Time (T) respectively.

To find the dimension of any quantity, we substitute the units of the quantity into L, M, and T. You insert the symbols into a bracket in this way: [L], [M], and [T]. For example, to find the dimension of speed, we first

write the formula as

Speed = Distance / Time

and we know that the unit for

Distance = meter (m) and meter is a length which can be

Distance = [L]

Note: We only have m which is [L]. If we have m^{2}, we can now write [L^{2}]

and Time is in seconds which can also be T

Thus, the dimension of Time = T

To derive the dimension of speed, we say:

Speed = Distance / Time

This implies that

Speed = [L] / [T]

Therefore, the dimension of Speed = [LT^{-1}] or [L][T^{-1}]

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## Worked Examples on Dimension in Physics

Here are examples to make you understand dimension in physics: how to calculate dimensions of physical quantities

### Example 1

- Find the dimension of an Area and Volume.

**Solution:**

For Area = Length x Breadth

Which shows that Area = [L] x [L] = [L^{2}]

Now, since volume = Length x Base x Height

The dimension of volume = [L] x [L] x [L] = [L^{3}]

### Example 2

2. Deduce the dimension of (a) force and (b) Pressure

**Solution:**

(a) force = mass x acceleration

which shows that

Dimension of force = dimension of mass x dimension of acceleration

and the Dimension of mass = [M]

**while** the Dimension of acceleration = Dimension of Velocity / Dimension of Time

and Dimension of Velocity = Dimension of Displacement [L] / Dimension of Time [T]

Thus, Dimension of Velocity = [L] / [T] = [L] [T^{-1}] or [LT^{-1}]

Therefore, by substituting the dimensions of velocity and time into acceleration,

we have

dimension of acceleration = [LT^{-1}] / [T] = [LT^{-2}]

Now, to finally derive the acceleration of force,

we say

Dimension of force = [M] x [LT^{-2}] = [M] [LT^{-2}] or [MLT^{-2}]

(b) To find the dimension of pressure

We apply the formula pressure = force / area

which means Dimension of pressure = Dimension of force / Dimension of area

and Dimension of force = Dimension mass x Dimension of acceleration

you can remember that we earlier calculated the dimension of force as [MLT^{-2}]

Also, Dimension of area = [L] x [L] = [L^{2}]

Thus, Dimension of pressure = [MLT^{-2}] / [L^{2}]

This implies

Dimension of pressure = [ML^{-1}T^{-2}]

### Example 3

Find the dimension of the following:

a. work

b. Velocity

c. Linear acceleration

d. Power

e. Pressure

f. Momentum

g. Impulse

**Solution:**

**a.** Work = Force x Distance

Dimension of Work = Dimension of Force x Dimension of Distance

Therefore, Dimension of Work = MLT^{-2} x L

Which implies that, Dimension of work = ML^{2} T^{-2}

**b.** Velocity = Displacement / Time = L / T = LT^{-1}

**c.** Acceleration = Velocity / Time = LT^{-1} / T = LT^{-2}

**d.** Power = work done / time = ML^{2} T^{-2} / T = ML^{2} T^{-3}

**e.** Pressure = Force / Area = MLT^{-2} / L^{3} = ML^{-2}T^{-2}

**f.** Momentum = mass x velocity = M x LT^{-1} = MLT^{-1}

**g.** Impulse = Force x Time = MLT^{-2} x T = MLT^{-1}

### Example 4

Find the dimension in electrical system of the following:

**a.** Electric charge

**b. **Electrical Potential (voltage)

**c.** Capacitance (c)

**Solution:**

**a.** To find electric charge, we use the formula

electric charge = current (A) x time (T)

Thus, electric charge = [A] x [T] = [AT]

**b.** Electric potential (voltage) = power / current

Dimension voltage = ML^{2} T^{-3} / A = [MA^{-1}L^{2} T^{-3}]

**c.** Capacitance = charge / potential = [AT] / [MA^{-1}L^{2} T^{-3}]

Thus, capacitance = [M^{-1}A^{2}L^{-2} T^{4}]

Reference