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-1T-2 |
| Elastic Force | MLT-2 |
| Young Modulus | ML-1T-2 |
| Work done in elastic spring/string | ML2T-2 |
| Work | ML2T-2 |
| Work done horizontal | ML2T-2 |
| Work done vertical | ML2T-2 |
| Kinetic energy | ML2T-2 |
| Potential energy | ML2T-2 |
| Power | ML2T-3 |
| Projectile time of flight | T |
| Projectile maximum height | L |
| Projectile range | L |
| Moment of force | ML2T-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-1T-2 |
| Work done by expanding gas | ML2T-2 |
| Wavelength | M |
| Wave velocity | LT-1 |
| Gravitational Force | MLT-2 |
| Gravitational potential | L2T-2 |
| Escape velocity | LT-1 |
| Electric force | MLT-2 |
| Magnetic force | MLT-2 |
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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 m2, we can now write [L2], 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] = [L2]
Now, since volume = Length x Base x Height
The dimension of volume = [L] x [L] x [L] = [L3]
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] = [L2]
Thus, Dimension of pressure = [MLT-2] / [L2]
This implies
Dimension of pressure = [ML-1T-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 = ML2 T-2
b. Velocity = Displacement / Time = L / T = LT-1
c. Acceleration = Velocity / Time = LT-1 / T = LT-2
d. Power = work done / time = ML2 T-2 / T = ML2 T-3
e. Pressure = Force / Area = MLT-2 / L3 = ML-2T-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 = ML2 T-3 / A = [MA-1L2 T-3]
c. Capacitance = charge / potential = [AT] / [MA-1L2 T-3]
Thus, capacitance = [M-1A2L-2 T4]
Reference