Dimensions Of Physical Quantities

Dimensions of Physical Quantities

Physical quantities are the measurable aspects of our world, describing everything from the speed of a car to the brightness of a star. Each physical quantity is associated with a dimension, such as length, mass, or time, which helps us understand and quantify the universe around us. For instance, consider the dimensions of force, which involve mass, length, and time—this is what allows engineers to design safe bridges or vehicles that can withstand different forces.

Let's now discuss Important points one by one

Dimension

The dimension of physical quantity may be defined as the power to which fundamental quantities must be raised in order to express the given physical quantities.

For representing dimensions of different quantities, we use the following symbols:

Mass - M

Length - L

Time - T

Electric current - A

Temperature - K

Amount of substance- mol

Luminous intensity - cd

Frequency, Angular Frequency, Angular Velocity, Velocity Gradient

All these quantities will have the same dimensional formula which is equal to $M^0{L}^0{T}^{-1}$

While the SI unit of Frequency and velocity gradient is ${\sec }^{-1}$,

The SI unit of angular frequency and angular velocity is radians per sec

Note:- Angle is a dimensionless quantity

Work, Potential Energy, Kinetic Energy, Torque

All these quantities will have the same dimensional formula which is equal to $L^2{T}^{-2}$

All these quantities will have the same unit in the SI system which is equal to N-m or Joule.

Momentum, Impulse, Angular Momentum, Angular Impulse

Momentum and Impulse both have the same dimensional formula which is equal to $ML{T}^{-1}$

Both have the same SI unit which is equal to ${\mathrm{kgms}}^{-1}$

Angular Momentum and Angular Impulse have the same dimensional formula which is equal to $M{L}^2{T}^{-1}$ and have the same SI unit which is equal to $kg(m{)}^2(\sec {)}^{-1}$

Heat, Latent heat, Specific Heat capacity and Temperature

Temperature

It is a fundamental quantity.

• Dimensional formula- $M^0{L}^0{T}^0{K}^1$ (where K represents Kelvin)
• SI unit- Kelvin

Heat

• Dimensional formula- $M{L}^2{T}^{-2}$

• SI unit- Joule

Latent Heat

Its dimensional formula is equal to $M^0{L}^2{T}^{-2}$

And its SI unit is equal to $m^2{s}^{-2}$ or $J/\mathrm{kg}$

Specific Heat Capacity

Dimensional formula $M^0{L}^2{T}^{-2}{K}^{-1}$

SI unit- $\frac{J}{kgK}$

Surface Tension, Surface Energy

Surface tension- Dimensional formula- $M^1{L}^0{T}^{-2}$

SI unit- ${\mathrm{kgs}}^{-2}$

Surface energy(per unit area) - Dimensional formula- $M^1{L}^0{T}^{-2}$

SI unit- ${\mathrm{kgs}}^{-2}$

But (Surface tension) and (surface energy per unit area) will have the same dimensional formula and SI units.

Vander Waals Constant (a and b)

The real gas equation is given as

$(P+\frac{n^{2}a}{V^2})(V-nb)=nRT$

Where a and b are called Vander Waal's constant.

1) Vander Waal's constant (a)

Dimension- $M{L}^5{T}^{-2}$
Unit- Newton $-m^4$

2) Vander waal 's constant (b)

Dimension- $M^0{L}^3{T}^0$
Unit- $m^3$

Voltage, Resistance and Resistivity

Voltage (V)

Dimension- $M{L}^2{T}^{-3}{A}^{-1}$
Unit- Volt

Resistance (R)

Dimension- $M{L}^2{T}^{-3}{A}^{-2}$
Unit- Ohm

Resistivity $(\rho )$

Dimension- $M{L}^3{T}^{-3}{A}^{-2}$

Unit- Ohm - meter

Permittivity of Free Space and Dielectric Constant (k)

The permittivity of free space $\left({ϵ}_o\right)$

Dimension- $M^{-1}{L}^{-3}{T}^4{A}^2$
Unit- $C^{-2}{N}^1{m}^{-2}$ or farad/metre

Dielectric constant (k)

Dimension- $M^0{L}^0{T}^0$

Unit- Unitless

Magnetic Field, Permeability of Free Space, Magnetic Flux and Self-Inductance

Magnetic Field (B)

Dimension- $M^1{L}^0{T}^{-2}{A}^{-1}$
Unit- $\frac{\text{newton}}{\text{ampere}-\text{metre}}$ or $\frac{\text{volt}-\text{second}}{{\text{metre}}^2}$

Permeability of free space

The dimension of permeability of free space $({\mu }_{o)-}{M}^1{L}^1{T}^{-2}{A}^{-2}$

SI unit- Newton/ampere²

Magnetic flux (φ)

Dimension- $M{L}^2{T}^{-2}{A}^{-1}$

Unit- Weber or Volt-second

Coefficient of self-induction (L)

Dimension- $M{L}^2{T}^{-2}{A}^{-2}$

Unit- Henry

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Solved Examples Based on Dimensions of Physical Quantities

Example 1: Which one of the following represents the correct dimensions of the coefficient of viscosity?

1) ML^-1T^-2

2) MLT^-1

3) ML^-1T ^-1

4) ML^-2T^-2

Solution:

Viscous force η = F * L / (A * V)
η = F / (L* V)

where F is the force acting on the fluid, η is the coefficient of viscosity, L is the length of the fluid layer, A is the area of the fluid layer, and V is the velocity of the fluid.

$\begin{array}{rl}& \eta =\mathrm{F}/\left({\mathrm{L}}^{\ast }\mathrm{V}\right)\\ & [\eta ]=\frac{\left[ML{T}^{-2}\right]}{[L]\left[L{T}^{-1}\right]}\\ & \Rightarrow [\eta ]=\left[M{L}^{-1}{T}^{-1}\right]\end{array}$

Hence, the correct answer is the option (3).

Example 2: Out of the following pairs which one does not have identical dimensions is

1) moment of inertia and moment of force

2)work and torque

3)angular momentum and Planck’s constant

4)impulse and momentum

Solution:

Dimension of Work, Potential Energy, Kinetic Energy, Torque is $M{L}^2{T}^{-2}$ and that of Momentum and Impulse - $ML{T}^{-1}$ and that of angular momentum \& Plank's Constant ( h ) is - $M^1{L}^2{T}^{-1}$

Moment of inertia is defined as $(I)=$ Mass $(M)x{\mathrm{radius}}^2(r)$
The dimensional formula is $={\mathrm{ML}}^2$
Know that the moment of force, $T=$ radius $(r)x$ force $(F)$
The dimensional formula for the moment of a force is- ${\mathrm{ML}}^2{\mathrm{T}}^{-2}$

Therefore the dimension of a moment of inertia and a moment of force does not have an identical dimension and torque is also called a moment of force.

Example 3: Given below are two statements :
Statements (I): Planck’s constant and angular momentum have the same dimensions
Statements (II): Linear momentum and moment of force have the same dimensions
In light of the above statements choose the correct answer from the option given below

1)Statement (I) is true but Statement II is false

2) Statement (I) is false but Statement II is true

3)Both statement I and statement II are true
4)Both statement I and statement II are false

Solution:

(1) $L=\frac{n}{2\pi }h$

Statement I is true.

$$\begin{array}{rl}& \text{(2)}p=mv\Rightarrow [p]=\left[ML{T}^{-1}\right]\\ & \tau =rF\Rightarrow [\tau ]=\left[M{L}^2{T}^{-2}\right]\end{array}$$

Statement II is false.

Hence, the answer is the option (1).

Example 4: The dimensional formula of heat (Q) is

1) $\left[M^0{L}^2{T}^{-2}\right]$
2) $\left[M^1{L}^2{T}^{-2}\right]$
3) $\left[M^2{L}^1{T}^{-2}\right]$
4) $\left[M^1{L}^2{T}^{-1}\right]$

Solution:

Heat is a form of energy. Hence, The dimensions of heat must be equal to that of energy or work done.

Work done
$W=Fd$
Dimensions of work done or heat is

$\begin{array}{rl}& =\left[M^1{L}^1{T}^{-2}\right]\left[L^1\right]\\ & \Rightarrow \left[M{L}^2{T}^{-2}\right]\end{array}$

Hence, the answer is the option (2).

Example 5: The dimensional formula $\left[M^0{L}^2{T}^{-2}\right]$ is equal to the dimensional formula of

1) Electric Potential

2) Gravitational potential

3) Latent heat

4) Both B and C

Solution:

Latent heat and gravitational potential

$\begin{array}{rl}& M^0{L}^2{T}^{-2}\\ & \text{wherein}\\ & \text{dimension-}{m}^2{s}^{-2}\\ & \text{Electrical Potential}=\frac{J}{c}\\ & [V]=\frac{\left[M{L}^2{T}^{-2}\right]}{AT}=\left[M{L}^2{T}^{-3}A\right]\\ & \text{Gravitational potential}=\frac{\text{Potential Energy}}{\text{mass}}\\ & =\frac{\left[M{L}^2{T}^{-2}\right]}{M}=\left[M^0{L}^2{T}^{-2}\right]\\ & \text{Latent Heat (L)}\frac{Q}{m}\\ & [L]=\frac{\left[M{L}^2{T}^{-2}\right]}{M}=\left[L^2{T}^{-2}\right]\end{array}$

Hence, the answer is the option (4).

Summary

Dimensions of physical quantities The nature of a PHYSICAL QUANTITY is expressed in terms of fundamental quantities such as length L, mass M, and time T. For example, the dimensions for speed are length per time, L/T, whereas for force, it is mass times acceleration, ML/T². These types of dimensions help us make sense of physical relationships between various variables and keep the equations consistent. With this analysis, we could check the rightness of equations, change units, and even solve very complicated problems. Understanding the dimensions of a physical quantity is an exceedingly important area of acquiring accurate measurements, prosperous scientific experimentation, and practical utility in all life pursuits.