\(\text {Dimension of permittivity } \varepsilon=A^{2} C^{2} N^{-1} m^{-2}\)
\(\text {Dimension of Boltzmann constant } k_{B}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]\)
\(\text {Dimension of Absolute Temperature }=T\)
No. of charged particles per unit
Volume = No. of dimensions = a
Dimension of charge = AT
Dimension of length = L
\(\text { Dimension of }=\frac{\sqrt{\varepsilon k_{B} T}}{n q^{2}}\)
\(=\frac{\sqrt{\left[\mathrm{A}^{2} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right]}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{T}]}{[\mathrm{AT}]^{2}}\)
\(=\sqrt{L^{2}}=L\)
\(\therefore \text { Dimension of length }=\mathrm{L}\)

Dimension of Planck's constant is \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)
Dimension of Gravitational constant is \(\left[\mathrm{M}^{-1} \mathrm{~L}^{+3} \mathrm{~T}^{-1}\right]\)
Dimension of Velocity constant is LT^-1
Dimension of Length is L
\(\therefore \text { Dimension of } \frac{\sqrt{h G}}{C^{\frac{3}{2}}}\)
\(=\frac{\sqrt{\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)\left(\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right)}}{\left(\mathrm{LT}^{-1}\right)^{3 / 2}} \)
\(=\frac{\sqrt{\mathrm{L}^{5} \mathrm{~T}^{-3}}}{\mathrm{~L}^{3 / 2} \mathrm{~T}^{-3 / 2}} \)
\(=\frac{\mathrm{L}^{5 / 2} \mathrm{~T}^{-3 / 2}}{\mathrm{~L}^{3 / 2} \mathrm{~T}^{-3 / 2}} \)
\(=\mathrm{L}^{5 / 2-3 / 2} \mathrm{~T}^{3 / 2+3 / 2}=\mathrm{L}^{1} \mathrm{~T}^{0}=\mathrm{L}\)
Dimension of length = L

\(\text { Velocity of light } c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(c=\left(\mu_{0} \varepsilon_{0}\right)^{-\frac{1}{2}}\)
\(\text { Hence dimension }\left(\mu_{0} \varepsilon_{0}\right)^{-\frac{1}{2}} \text { is that of velocity. }\)

F = kv²
Dimensional of k
\(=\frac{\text { Dimension of } \mathrm{F}}{\text { Dimension of }(v)^{2}}\)
\(=\frac{\mathrm{MLT}^{-2}}{\left(\mathrm{LT}^{-1}\right)^{2}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2} \mathrm{~T}^{-2}} \)
\(=\left[\mathrm{ML}^{-1-2} \mathrm{~T}^{-2+2}\right] \)
Dimension of proportionality constant \(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]\)

Let us assume that the Energy E depends on mass m and velocity of light c.
\(E\alpha m^ac^b\)
\(E=km^ac^b\) where K a constant
Dimensions of E = [ML²T^-2]
Dimensions of m = [M]
Dimensions ofc = [LT^-1]
Substituting the values in the above equation
[ML²T^-2] = K[M]^a [LT^-1]^b
By equating the dimensions
a = 1
b = 2
-b = -2
E = k.mc²
The value of constant k = 1
E = mc². This is Einstein's mass energy relation.

Voltage V = (100 ± 5)V
Current I = (10 ± 0.2)A
Resistance R = ?
Then resistance R is given by Ohm's law,
\(R={V\over I}={100\over 10}=10\Omega\)
\({\triangle R\over R}=({\triangle V\over V}+{\triangle I\over I})\)
\({\triangle R\over R}=({\triangle V\over V}+{\triangle I\over I})R=({5\over 100}+{0.2\over 10})10\)
= (0.05 + 0.02)10 = 0.07\(\times\)10 = 0.7
The resistance R = (10 ± 0.7)\(\Omega\)

t₁ = (75.4 ± 0.5)⁰C
t₂ = (56.8 ± 0.2)⁰C
Fall in temperature = (75.4 ± 0.5°C) - (56.8 ± 0.2°C)
t = (18.6 ± 0.7)⁰C

12.7

248000

101.6\(\times\)10⁶

The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v²= u² + 2as, the dimensions of v², u² and 2 as are the same and equal to [L²T^-2].
This method is used to
(i) Convert a physical quantity from one system of units to another.
(ii) Check the dimensional correctness of a given physical equation.
(iii) Establish relations among various physical quantities.
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant. i.e, n₁[u₁] = constant (or) n, n₁[u₁ ] = n₂[u₂].
Consider a physical quantity which has dimension 'a' in mass, 'b' in length and 'c' in time.
If the fundamental units in one system are M₁, L₁ and T₁ and the other system are M₂, L_2, and T₂ respectively, then we can write, n₁ [M₁^a L₁^b T₁^c] = n₂ [ M ₂^a  L₂^b T₂^c]
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm^-2 using the method of dimensions.
Solution: In cgs system 76 cm of mercury pressure =76\(\times\)13.6\(\times\)980 dyne cm^-2 
The dimensional formula of pressure P is [ML^-1T^-2]
\(P_{1}\left[M_{1}^{a} L_{1}^{b} T_{1}^{c}\right]=P_{2}\left[M_{2}^{a} L_{2}^{b} T_{2}^{c}\right]\)
We have
\(P_{2} =\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]^{a}\left[\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}\right]^{b}\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{c} \)
\(M_{1} =1 \mathrm{~g}, \mathrm{M}_{2}=1 \mathrm{~kg}\)
\(L_{1}=1 \mathrm{~cm}, \mathrm{~L}_{2}=1 \mathrm{~m} \)
\(T_{1}=1 \mathrm{~s}, T_{2}=1 \mathrm{~s}\)
So a=1, b=1 and c=-2
Then
\(P_{2} =76 \times 13.6 \times 980\left[\frac{\mathrm{g}}{1 \mathrm{~kg}}\right]^{1}\left[\frac{\mathrm{cm}}{1 \mathrm{~m}}\right]^{-1}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2} \)
\(=76 \times 13.6 \times 980\left[\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~kg}}\right]^{1}\left[\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right]^{-1}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2} \)
\(=76 \times 13.6 \times 980 \times\left[10^{-3}\right] \times 10^{2} \)
\(P_{2} =1.01 \times 10^{5} \mathrm{Nm}^{-2}\)
(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(1\over 2\) mv² = mgh can be checked by using this method as follows.
Solution: Dimensional formula for
\(\boxed{{1\over 2}mv^2=[M][LT^{-1}]^2=[ML^2T^{-2}]}\)
Dimensional formula for
\(\boxed {mgh=[M][LT^{-2}][L]=[ML^{2}T^{-2}] \\ [ML^{2}T^{2}]=[ML^{2}T^{-2}]}\)
Both sides are dimensionally the same, hence the equations\(1\over 2\) mv² = mgh is dimensionally correct.
(iii) To establish the relation among various physical quantities:
 If the physical quantity Q depends upon the quantities Q₁, Q₂ and Q₃ ie. Q is proportional to Q₁, Q₂ and Q₃.
Then,
\(Q \alpha Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}  \)
\(Q=k Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}\)
where k is a dimensionless constant. When the dimensional formula of Q₁, Q₂ and Q₃ are substituted, then according to the principle of homogeneity, the powers of M, L, T are made equal on both sides of the equation. From this, we get the values of a, b, c.
Example:
Obtain an expression for the time period T of a simple pendulum. The time period T depend upon (i) mass 'm' of the bob (ii) length 'l' of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended. (Constant k=2π ) i.e
Solution:
\(\boxed{T \alpha m^a l^b g^c \\ T=k.m^al^bg^c}\)
Here k is the dimensionless constant. Rewriting the above equation with dimensions.
\(\boxed{[T^1]=[M^a][L^b][LT^{-2}]^c\\ [M^oL^oT^1]=[M^aL^{b+c}T^{-2c}]}\)
Comparing the powers of M, L and T on both sides, a = 0, b + C = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation
T = k. m^o l^1/2 g-^1/2
T=\(k{1\over g}^{1\over 2}=k\sqrt{1\over g}\)
Experimentally k = 2\(\pi\) , hence \(T=2\pi \sqrt{l\over g}\)

Let the height of the pole be h
Solution \(\frac { x+45 }{ h } \)  = cot 30^o ⇒ h =\(\frac { x+45 }{ cot\quad { 30 }^{ o } } \)
\(\frac { x }{ h } \) = cot 30^o ⇒ x = h cot 60^o
Substituting the values of x in the above equation
h = \(\frac { h\quad cot \ { 60 }^{ o }+45 }{ cot \ { 30 }^{ o } } \)
\(h \cot 30^{\circ} =h \cot 60^{\circ}+45  \)
\(h\left(\cot 30^{\circ}-\cot 60^{\circ}\right) =45  \)
\(h =\frac{45}{\cot 30^{\circ}-\cot 60^{\circ}}=\frac{45}{\sqrt{3}-\frac{1}{\sqrt{3}}}=38.97 \mathrm{~m}\)