Question

Using dimensional analysis show that $\dfrac{1N}{m^2}=\dfrac{10dyne}{cm^2}$

Answer 1

Hint: We know that dimensions are physical quantities that are expressed as the powers of the fundamental units. The dimensions are generally expressed in terms of the basic or the supplementary. We can convert the dimensions from one unit to another as follows.

Complete step-by-step solution:
We know that dimensional analysis is the process of checking relations between physical quantities and their dimensions. There are seven most basic or the fundamental dimensions , namely, the mass, length, time, electric current, temperature, intensity of light and quantity of substance. And two supplementary quantities, plane angle and solid angle. Other physical quantities are derived from the basic dimensions.
There exist dimension variables, with dimensions and no fixed value. These examples include acceleration and force. There also exists dimensionless variables, these as the name suggests neither have dimensions or a fixed value. Some examples include specific gravity , coefficient of friction and refractive index to name a few.
Consider the dimension of$\dfrac{1N}{m^2}$
$\implies \dfrac{1N}{m^2}=\dfrac{[MLT^{-2}]}{L^{2}}$

Now since, newton is expressed as kilograms, meters and seconds, we have
$\implies \dfrac{1N}{m^2}=\dfrac{1kg\times 1m\times s^{-2}}{1m^{2}}$
Converting mass into grams, distance into cm we have
$\implies \dfrac{1N}{m^2}=\dfrac{1000g\times 100cm\times s^{-2}}{100cm^{2}}$
$\implies\dfrac{1N}{m^2}=\dfrac{1000gcm\times s^{-2}}{100cm^{2}}$
$\implies\dfrac{1N}{m^2}=\dfrac{10gcm\times s^{-2}}{1cm^{2}}$
Since, dynes is expressed in terms of grams, centimetres and seconds, we get
$\therefore\dfrac{1N}{m^2}=\dfrac{10dynes}{cm^{2}}$
Hence proved.

Note: Dimensional analysis is used to check the validity of the equations, helps in the derivation of relation between different terms and cover the units from one system to another. Dimensional analysis is also called the factor label method or unit factor method. This is frequently used to check the homogeneity of the term, which are expressed in an equation.