Question

If ${\text{Force}} = \dfrac{\alpha }{{{\beta ^3} + {\text{density}}}}$, then find dimension of $\alpha $ and $\beta $

Answer 1

Hint: For the correctness of an equation the dimension on either side of the equation must be the same. This is known as the principle of homogeneity of dimensions. In equations containing more than two terms, the dimension of each term must be the same.
Dimension of force is $\left[ {ML{T^{ - 2}}} \right]$.
Dimension of density is $\left[ {M{L^{ - 3}}} \right]$

Complete step by step solution:
We can check the relation between various physical quantities by identifying the dimension of physical quantities. This method is known as dimensional analysis. Using dimensional analysis, we can check the correctness of an equation, we can derive the correct relationship between different physical quantities and it can also be used to convert one system of the unit into another.
For the correctness of an equation, the dimension on either side of the equation must be the same. This is known as the principle of homogeneity of dimensions. In equations containing more than two terms, the dimension of each term must be the same.
In the given equation ${\beta ^3}$ and density must have the same dimension.
Density is mass by volume. The unit of density is $kg{m^{ - 3}}$.
Dimension of mass is $\left[ M \right]$ and the dimension of length is $\left[ L \right]$.
Therefore, dimension of density is $\left[ {M{L^{ - 3}}} \right]$
Now, the dimension of force must be equal to the dimension of $\alpha $ divided by the dimension of density.
Force is mass times acceleration. We know that unit of force is $kg-m/s^2$
By analyzing the unit, we can write the dimension of force as $\left[ {ML{T^{ - 2}}} \right]$.
Substituting the dimensions in the given equation, we get
\[
{\text{dimension}}\,{\text{of}}\,{\text{force}} = \dfrac{{{\text{dimension}}\,{\text{of}}\,\alpha }}{{{\text{dimension}}\,{\text{of}}\,{\text{density}}}} \\
  \left[ {ML{T^{ - 2}}} \right] = \dfrac{{{\text{dimension}}\,{\text{of}}\,\alpha }}{{\left[ {M{L^{ - 3}}} \right]}} \\
 \]
\[
  {\text{dimension}}\,{\text{of}}\,\alpha = \left[ {M{L^{ - 3}}} \right] \times \left[ {ML{T^{ - 2}}} \right] \\
   = \left[ {{M^2}{L^{ - 2}}{T^{ - 2}}} \right] \\
 \]
$\therefore $ The dimension of $\alpha $ is $\left[ {{M^2}{L^{ - 2}}{T^{ - 2}}} \right]$ and the dimension of $\beta $ is $\left[ {M{L^{ - 3}}} \right]$

Note:
Here we know that according to the principle of homogeneity, the dimension of each term in the equation is the same. The right-hand side of the equation together forms one term. So, the dimension of the whole term will be equal to that of the dimension of the term on the left-hand side. Don't think that the dimension of $\alpha $, ${\beta ^3}$, and density is separately equal to the dimension of force.