## Dimensions Of Physical Quantities

The dimensions of any physical quantity are those powers which are raised to the basic units to express the unit of that quantity.

To write the dimensions of physical quantities in mechanical and thermodynamics, the basic units for length, time, mass, and temperature are represented by L, T, M, and θ respectively.

If the dimensions of any physical quantity are a in length, b in mass, c in time, and d in temperature, then the dimensions of that quantity will be written as follows [LᵃMᵇTᶜθᵈ].

### Dimensional formula of some physical quantities

**Basic Quantities__Basic Units__Dimensional Formulas**

Area____L.×W.____[L×L]=[L²]

Volume___L×W×H_[L³]

Velocity____D/Time_____[L/T]=[LT⁻¹]

Acceleration___Velocity/Time___[LT⁻¹/T]=[LT⁻²]

Force___Mass×Acceleration__[MLT⁻²]

Deformation=Increase in length/Initial length=[L/L]=[L⁰]

That means there is no dimension in deformation.

Planck’s constant=[ML²T⁻¹]

Angular velocity, frequency, velocity gradient=[T⁻¹]

Viscosity coefficient=[ML⁻¹T⁻¹]

## Some following physical quantities have the same dimensions

### 1. Work, Kinetic Energy, Potential Energy, Torque

=Force×Displacement

=Mass×Acceleration×Displacement

=[MLT⁻²L]

=[ML²T⁻²]

### 2. Momentum, Impulse

=Mass×Velocity

=[MLT⁻¹]

### 3. Young’s elastic coefficient, elastic coefficient, volume elastic coefficient, stress

=Force/Area

=[MLT⁻²/L²]

=[ML⁻¹T⁻²]

** **

### 4. Surface tension and spring force constant

=force/length

=[MLT⁻²/L]

=[MT⁻²]

### 5. Gravitational Potential, latent Heat

L=[L²T⁻²]

## Use Of Dimension

### 1. Dimensional balance test of the equation of motion

**(a)**. Test of v=u+at

Dimension of L.H.S

v=[LT⁻¹]

Dimension of R.H.S

u+at=[LT⁻¹]+[LT⁻²]T

u+at=[LT⁻¹]+[LT⁻¹]

u+at=[LT⁻¹]

Hence v=u+at is balanced by dimensional method.

**(b)**. Test of v²=u²+2as

Dimension of L.H.S

v2=[LT⁻¹]²

v2=[L²T⁻²]

Dimension of R.H.S

u2+2as=[L²T⁻²]+[LT⁻²]L

u2+2as=[L²T⁻²]+[L²T⁻²]

u2+2as=[L²T⁻²]

Hence v²=u²+2as is balanced by dimensional method.

**(c)**. Testing S=ut+1/2at²

Dimension of L.H.S

S=[L]

Dimension of R.H.S

ut+1/2at²=[LT⁻¹]T+[LT⁻²][T²]

ut+1/2at²=[LT⁰]+[LT⁰]

ut+1/2at²=[L]+[L]

ut+1/2at²=[L]

Hence S=ut+1/2at² is balanced by dimensional method.

### 2. Establishing relationships between different physical quantities

**Question– The force acting on an object depends on the mass m and acceleration a of F, then prove by dimensional method that F=m.a.**

**Solution– Since force F depends on mass m and acceleration a.**

**Therefore,**

F∝mᵇaᶜ

F=Kmᵇaᶜ____(1)

where K is a constant of proportionality

Dimension of L.H.S=[MLT⁻²]

Dimension of R.H.S=[Mᵇ][LᶜT⁻²ᶜ]

On comparing the powers of M,L,T,

b=1

-2C=-2

C=1

Putting the values of b and c in equation (1)

F=Km¹a¹

k=1

F=m.a

**Question–**

**The energy E of a body oscillating with simple harmonic motion depends on its mass m, frequency n and oscillation amplitude a. Prove through dimensional analysis that,**

**E=mn²a².**

**Solution– Let the kinetic energy of the body depend on m to the power b, n to the power c, and a to the power d.**

E∝mᵇnᶜaᵈ

E=Kmᵇnᶜaᵈ______(1)

Dimension of L.H.S,

[ML²T⁻²]

Dimension of R.H.S,

[Mᵇ][T⁻ᶜ][Lᵈ]

On comparing the powers of M,T,L,

b=1

d=2

-c=-2

c=2

Putting the values in equation one,

E=Km¹n²a²

k=1

E=mn²a²

### 3. Converting units of one system into units of another system

The product of numerical value of a physical quantity and its corresponding unit is a constant.

**Example-**

**(a)**If the numerical values of a physical quantity are n₁ and n₂ in two different systems and its units are u₁ and u₂ respectively, then

K=n₁u₁=n₂u₂

Force K=F,

F=1Newton=10⁵ Dyne

K=n₁u₁=n₂u₂

Where?

1Newton=n₁u₁

10⁵ dyne=n₂u₂

**(b)**If the dimensions of a physical quantity are a in mass, b in length, and c in time, then the dimensional formula of that quantity is,

K=n₁[M₁ᵃL₁ᵇT₁ᶜ]______(1)

Similarly, in the second method,

K=n₂[M₂ᵃL₂ᵇT₂ᶜ]______(2)

From equations (1) and (2),

n₁[M₁ᵃL₁ᵇT₁ᶜ]=n₂[M₂ᵃL₂ᵇT₂ᶜ]

n₁=n₂[M₁/M₂]ᵃ[L₁/L₂]ᵇ[T₁/T₂]ᶜ

Or,

n₂=n₁[M₂/M₁]ᵃ[L₂/L₁]ᵇ[T₂/T₁]ᶜ

**Question– Convert 1Newton to Dyne.**

**Solution– 1Newton(n₁)=Dyne(n₂)**

**Dimension of force,**

F=[MLT⁻²]

n₂=n₁[M₂/M₁]ᵃ[L₂/L₁]ᵇ[T₂/T₁]ᶜ

n₂=n₁[M₂/M₁]¹[L₂/L₁]¹[T₂/T₁]⁻²

n₂=1[kg/gram]¹[meter/cm]¹[second/second]⁻²

n₂=[1000g/gram][100g/gram][1/1]⁻²

n₂=[1000gram/gram][100gram/gram]

n₂=10⁵ Dyne

1 Newton=10⁵Dyne

**Question– Convert 1Joule to Arg.**

**Solution–**

1Joule(n₁)=Arg(n₂)

n₂=n₁[M₂/M₁]ᵃ[L₂/L₁]ᵇ[T₂/T₁]ᶜ

Dimension of energy

E=[ML²T⁻²]

n₂=1[kg/gram]¹[meter/cm]²[second/second]⁻²

n₂=[1000g/g][100g/g]²[1/1]⁻²

n₂=[1000gram/gram][10000gram/gram]

n₂=10⁷ arg

1Joule=10⁷arg

### 4. Example of oscillation period of simple pendulum

Suppose the oscillation period can depend on the following quantities.

At mass m of the pendulum, length of string l and acceleration due to gravity g,

Let its powers be a, b and c respectively,

T∝mᵃlᵇgᶜ

T=Kmᵃlᵇgᶜ____(1)

Dimension of L.H.S,

[T]=[M⁰L⁰T]

Dimension of R.H.S,

[M]ᵃ[L]ᵇ[LT⁻²]ᶜ

[MᵃLᵇLᶜT⁻²ᶜ]

[MᵃLᵇ⁺ᶜT⁻²ᶜ]

By comparing powers,

a=0

b+c=0____(2)

-2c=1

c=-1/2

Putting the value of c in equ. (2),

b+c=0

b-1/2=0

b=1/2

Putting the above values in equation (1),

T=Km⁰√(l)√(1/g)

T=K√(l)√(1/g)

T=K√(l/g)____(3)

T∝√(l/g)

**Note-**

The value of K cannot be determined from the dimensional equation.

so,

Let us assume K=2π experimentally.

Then,

From equation (3),

T=2π√(l/g)