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We note that same physical quantity is expressed in different units. Clearly, conversion of unit is associated with single physical quantity. We know that different units for physical quantities have evolved through human progress in different parts of the world and have prevailed despite efforts towards uniform system. Secondly, a physical quantity has different scales of measurement - small, large etc. Whatever be the reason of different units being used, we need to have a “conversion factor” which coverts one unit to another. The conversion related to magnitude of unit is defined by the system of measurement itself. For example, 1 kilogram is equal to 1000 gm. Here, we shall derive a process to affect the conversion of one unit of one kind to another kind. In another words, we shall derive a “dimensionless constant” as conversion factor.
The underlying principle of process involved here is that a given physical quantity has same dimensional constitution – whatever be the system or unit. This fact is underlined by the fact that a given quantity of a given physical entity remains same. For, example, whether we call 1 inch or 2.54 cm, the length remains same. Now, physical quantity is product of “measurement” and “unit. Hence, We know that :
$$Q={n}_{1}{u}_{1}={n}_{2}{u}_{2}$$
Since unit represents a standard division of the quantity, the dimensional formula of the unit quantity is same as that of the quantity in question. Let dimensions of units in two systems are :
$${u}_{1}=\left[{M}_{1}^{a}{L}_{1}^{b}{T}_{1}^{c}\right]$$
$${u}_{2}=\left[{M}_{2}^{a}{L}_{2}^{b}{T}_{2}^{c}\right]$$
Combining equations, we have :
$$\Rightarrow {n}_{1}\left[{M}_{1}^{a}{L}_{1}^{b}{T}_{1}^{c}\right]={n}_{2}\left[{M}_{2}^{a}{L}_{2}^{b}{T}_{2}^{c}\right]$$
$$\Rightarrow {n}_{2}=\frac{\left[{M}_{1}^{a}{L}_{1}^{b}{T}_{1}^{c}\right]}{\left[{M}_{2}^{a}{L}_{2}^{b}{T}_{2}^{c}\right]}{n}_{1}$$
This is the formula used to convert a quantity from one system of units to another. In case, only 1 unit of the quantity in first system is involved, then we can put ${n}_{1}=1$ in the above equation.
Problem : Convert 1 Joule in SI system into ergs in “cgs” system
Solution : Putting ${n}_{1}=1$ and applying formula, we have :
$$\Rightarrow {n}_{2}=\frac{\left[{M}_{1}^{a}{L}_{1}^{b}{T}_{1}^{c}\right]}{\left[{M}_{2}^{a}{L}_{2}^{b}{T}_{2}^{c}\right]}$$
Now, the dimensional formula of energy is $\left[M{L}^{2}{T}^{-2}\right]$ . Hence, a = 1, b = 2 and c = -2. Putting values in the equation and rearranging,
$$\Rightarrow {n}_{2}=\left[\frac{1\phantom{\rule{2pt}{0ex}}kg}{1\phantom{\rule{2pt}{0ex}}gm}{]}^{1}X\right[\frac{1\phantom{\rule{2pt}{0ex}}m}{1\phantom{\rule{2pt}{0ex}}cm}{]}^{2}X[\frac{1\phantom{\rule{2pt}{0ex}}s}{1\phantom{\rule{2pt}{0ex}}s}{]}^{-2}$$
The basic units of SI system is related to corresponding basic units in “cgs” system as :
$$1\phantom{\rule{2pt}{0ex}}Kg={10}^{3}\phantom{\rule{2pt}{0ex}}gm$$
$$1\phantom{\rule{1em}{0ex}}m={10}^{2}\phantom{\rule{1em}{0ex}}cm$$
$$1\phantom{\rule{2pt}{0ex}}s=1\phantom{\rule{2pt}{0ex}}s$$
Putting in the conversion formula, we have :
$$\Rightarrow {n}_{2}=\left[\frac{{10}^{3}\phantom{\rule{2pt}{0ex}}gm}{1\phantom{\rule{2pt}{0ex}}gm}{]}^{1}X\right[\frac{{10}^{2}\phantom{\rule{2pt}{0ex}}cm}{1\phantom{\rule{2pt}{0ex}}cm}{]}^{2}X[\frac{1\phantom{\rule{2pt}{0ex}}s}{1\phantom{\rule{2pt}{0ex}}s}{]}^{-2}$$
$$\Rightarrow {n}_{2}={10}^{3}X{10}^{4}={10}^{7}$$
Hence,
$$1\phantom{\rule{2pt}{0ex}}J={10}^{7}\phantom{\rule{1em}{0ex}}ergs$$
Dimensional analysis is based on the simple understanding of the fact that only similar quantities can be added, subtracted and equated. Can we add 2 mangoes and 3 oranges? This requirement of similar terms is the underlying principle of dimensional analysis. This principle is known as principle of homogeneity.
Dimensional analysis gives an insight into the composition of a physical quantity. Take the case of frequency. Its SI unit is Hertz. What is its true relation with basic quantities? Notwithstanding the exact definition, its dimensional formula tells us that it is actually inverse of time (T). It has the dimension of “-1” in time.
In this section, we shall discuss three important applications of dimensional analysis involving : (i) Transcendental functions (ii) checking of dimensional formula and (iii) derivation of a formula.
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