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 `v^(2)=u^(2) + 2as`, the dimensions of `v^(2),u^(2)` and 2 as are the same and equal to `[L^(2)T^(-2)]`.
(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, `n_(1),[u_(1)]` = constant (or) `n_(1)[u_(1)]= n_(2)[u_(2)]`.
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 `M_(1), L_(1), "and" T_(1)` and the other system are `M_(2), L_(2) "and" T_(2)` respectively, then we can write, `n_(1), [M_(1)^(a)L_(1)^(b)T_(1)^(c)]=n_(2)[M_(2)^(a)L_(2)^(b)T_(2)^(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 xx 13.6 xx 980` dyne `cm^(-2)`
The dimensional formula of pressure P is `[ML^(-1)T^(-2)]`
(ii) To check the dimensional correctness of a given physical equation:
Example: The equation `1/2 mv^(2)` = mgh can be checked by using this method as follows.
Solution: Dimensional formula for
`1/2 mv^(2) = [M][LT^(-1)]^(2) = [ML^(2)T^(-2)]`
Dimensional formula for
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 equation `1/2 mv^(2)` = mgh is dimensinally correct.
(iii) To establish the relation among various physical quantities:
Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon (i) miss 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. Let the constant involved is K = `2pi`.
Solution:
`Talpha m^(a) l^(b) g^(c)`
`T = k. m^(a) l^(b) g^(c)`
Here is the dimensionless constant. Rewriting the above equation with dimensions.
`[T^(1)]=[M^(a)][L^(b)][LT^(-2)]^(c)`
`[M^(0)L^(0)T^(1)] = [M^(a)][L^(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^(0) l^(1//2) g^(-1//2)`
`T = k(l/g)^(1//2) = k sqrt(l/g)`
Experimentally k = `2pi`, hence
`T = 2pi sqrt(i=l/g)`
