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]; so P₁ [M₁^a L₁^b T₁^c ]= P₂[M₂^a L₂^bT₂^c]
We have P₂ = P₁ \([{M_1\over M_2}]^a[{L_1\over 2}]^b[{T_1\over T_2} ]^c\)
M₁ = 1g, M₂ =1kg
L₁ = 1 cm, L₂ = 1m
T₁ = 1 s, T₂ = 1s
So, a = 1, b = -1, and c =-2
Then, P₁ = 76\(\times\)13.6\(\times\)980
\([{1g\over 1kg}]^1[{1cm\over 1m}]^{-1}[{1s\over 1s}]^{-1}=76\times13.6\times980[{10^{-3}kg\over 1kg}]^1[{10^{-2}m\over1m}]^{-1}[{1s\over 1s}]^{-2}\)
= 76\(\times\)13.6\(\times\)980\(\times\)[10^-3]\(\times\)10²
P₂ = 1.01\(\times\)10⁵ Nm^-2

Dimension formula for 
 \(\frac { 1 }{ 2 } \)mv² = [M][LT^-1]² = [ML²T^-2]
Dimension formula for 
mgh = [M][LT^-2][L] = [ML²T^-2]
[ML²T^-2] = [ML²T^-2]
Both sides are dimensionally the same, hence the equations \(\frac { 1 }{ 2 } \)mv² = mgh is dimensionally correct

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 \(T=2\pi \sqrt{l\over g}\)

\((C+\triangle C)=(3.0\pm0.1)\mu F\)
\((V+\triangle V)=(18\pm 0.4)V\)
Q = CV
\(Q=3.0\times10^{-6}\times18=54\times10^{-6}\) coulomb
\(\text {Error in } C=\frac{\Delta C}{C} \times 100=\frac{0.1}{3} \times 100=3.3 \%\)
Error in \(V=\frac{\triangle V}{V}\times100=\frac{0.4}{18}\times100=2.2\%\)
Error in Q = Error in C + Error in V
= 3.3% + 2.2% = 5.5%
\(\therefore\) Charge Q = (54\(\times\)10⁶ 土 5.5%)C

F ∝ m^a v^b r^c ;
F = k m^a v^b r^c
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k= 1, we have
[MLT^-2] = [M]^a [LT^-1]^b [L]^c
= [M^aL^bT^-bL^c]
[MLT^-2] = [M^aL^b+cT^-b]
Comparing the powers of M, L and T on both sides
 a = 1 ; b + c = 1; -b =-2
2 + c = 1 ; b = 2;
a = 1 b = 2 and c = -1
From the above equation we get
F = m^av^br^c
F =m¹v²r^-1
or F = \(\frac { m{ v }^{ 2 } }{ r } \)