## What is Dimension in Physics

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. To understand dimension, we need to understand the method of measurements, fundamental quantities, derived quantities, and even their corresponding units in physics. This dimension helps us in finding the relationship between quantities. As we mentioned earlier, we apply Mass in the form of M, Length in the form of L, and Time in the form of T. Additionally, units like temperature will be in the form of K or θ, electric current in the form of (A) which is an expression for ampere, and finally luminous intensity as (II). Below are 2 videos explaining dimensions and their calculations.

**Video 1**

**Video 2**

## Table of Dimensions in Physics

QUANTITY | DIMENSION |

Speed | LT^{-1} |

Velocity | LT^{-1} |

Acceleration | LT^{-2} |

Force | MLT^{-2} |

Tensile stress | ML^{-1}T^{-2} |

Elastic Force | MLT^{-2} |

Young Modulus | ML^{-1}T^{-2} |

Work done in elastic spring/string | ML^{2}T^{-2} |

Work | ML^{2}T^{-2} |

Work done horizontal | ML^{2}T^{-2} |

Work done vertical | ML^{2}T^{-2} |

Kinetic energy | ML^{2}T^{-2} |

Potential energy | ML^{2}T^{-2} |

Power | ML^{2}T^{-3} |

Projectile time of flight | T |

Projectile maximum height | L |

Projectile range | L |

Moment of force | ML^{2}T^{-2} |

Density | ML^{-3} |

Upthrust | MLT^{-2} |

Centripetal acceleration | LT^{-2} |

Centripetal force | MLT^{-2} |

Linear velocity | LT^{-1} |

Linear acceleration | LT^{-2} |

Period | T |

Frequency | T^{-1} |

Period of simple pendulum | T |

Momentum | MLT^{-1} |

Impulse | |

Pressure | ML^{-1}T^{-2} |

Work done by expanding gas | ML^{2}T^{-2} |

Wavelength | M |

Wave velocity | LT^{-1} |

Gravitational Force | MLT^{-2} |

Gravitational potential | L^{2}T^{-2} |

Escape velocity | LT^{-1} |

Electric force | MLT^{-2} |

Magnetic force | MLT^{-2} |

*You may also like to read:*

SS1 Lesson Note: Introduction to Physics For First Term

## How to Calculate Dimensions of Physical Quantity

To calculate the dimensions of a physical quantity, we need to be guided on the measurement of quantities in units of Length, Mass, and Time. These involve the dimensions 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**

We know that the unit for Distance = meter (m) and meter is a length, which can be, Dimension of **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}]**.

## 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 the dimension of 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

This 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 the 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**