1.1 Dimensional analysis  (Page 6/7)

Problem : Find the value of “x” in the equation :

$$\int \frac{dx}{\sqrt{\left(2ax-x^2\right)}}=a^x\sin \left(\frac{x}{a}-1\right)$$

Solution : It appears that this question is out of the context of dimensional analysis. Actually, however, we can find the solution of "x" for the unique situation by applying principle of homogeneity. We know that the argument of a trigonometric function is an angle, which does not have dimension. In other words, the argument of trigonometric function is a dimensionless variable. Hence, “ $\frac{x}{a}-1$ ” is dimensionless. For this,

$$\left[a\right]=\left[x\right]=\left[L\right]$$

With this information, let us, now, find the dimension of the expression enclosed in the integral on the left hand side (LHS).

$$\Rightarrow \left[LHS\right]=\frac{\left[dx\right]}{\sqrt{\left([2ax-x^2]\right)}}$$

We know that dimension of difference is same as that of the quantity. Hence, $\left[dx\right]=\left[L\right]$ . Putting dimensions,

$$\Rightarrow \left[LHS\right]=\frac{\left[L\right]}{\sqrt{\left(\left[L^2\right]-\left[L^2\right]\right)}}=\frac{L}{L}=L^0$$

We see that LHS is dimensionless. Hence, RHS should also be dimensionless. The trigonometric function on the right is also dimensionless. It means that factor “ $a^x$ ” should also be dimensionless. But, the base “a” has a dimension that of length. The factor can be dimensionless, only if it evaluates to a constant. An exponential with variable power like “ $a^x$ ” is constant if it evaluates to 1 for x=0. Hence,

$$\Rightarrow x=0$$

Problem : The Vander Wall’s equation is given as :

$$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT$$

where “P” is pressure, “V” is volume, “T” is temperature and “R” is gas constant. What are the units of constants “a” and “b”?

Solution : Applying principle of homogeneity of dimensions, we realize that dimensions of the terms connected with “plus” or “minus” signs are equal. Hence,

$$\Rightarrow \left[\frac{a}{V^2}\right]=\left[P\right]=\left[\frac{F}{A}\right]=\left[\frac{ML{T}^{-2}}{L^2}\right]=\left[M{L}^{-1}{T}^{-2}\right]$$

$$\Rightarrow \left[a\right]=\left[M{L}^{-1}{T}^{-2}\right]\left[V^2\right]=\left[M{L}^{-1}{T}^{-2}\right][L^3{]}^2$$

$$\Rightarrow \left[a\right]=\left[M{L}^5{T}^{-2}\right]$$

The unit of “a” is “ $\mathrm{kg}-m^5/s^2$ ”.

Also,

$$\Rightarrow \left[b\right]=\left[V\right]=\left[L^3\right]$$

The unit of “b” is same as that of volume i.e. " $m^3$ " .