Question

If ${\left(velocity\right)}^x={\left(pressure\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\left(density\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, then x would be ______.

Answer 1

1. Dimensional analysis can be used to evaluate both sides.
2. The dimensions of terms on both sides of an equation should be the same.
3. The dimensions of the given quantities are:
   $$\begin{gathered} Velocity=\left[L{T}^{-1}\right] \\ Pressure=\left[M{L}^{-1}{T}^{-2}\right] \\ Density=\left[M{L}^{-3}\right] \end{gathered}$$
4. Using dimensions on both sides, we get:
   $$\begin{gathered} {\left[L{T}^{-1}\right]}^x={\left[M{L}^{-1}{T}^{-2}\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\left[M{L}^{-3}\right]}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =M^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{L}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{T}^{-3}\times M^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{L}^{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =M^{\left(\frac{3}{2}+\frac{(-3)}{2}\right)}{L}^{\left(\frac{-3}{2}+\frac{9}{2}\right)}{T}^{-3} \\ =M^0{L}^{\left(\frac{-3+9}{2}\right)}{T}^{-3} \\ =L^{\left(\frac{6}{2}\right)}{T}^{-3} \\ =L^3{T}^{-3} \\ {\left[L{T}^{-1}\right]}^x={\left[L{T}^{-1}\right]}^3 \end{gathered}$$
5. On comparing, we get $x=3$

Thus, if ${\left(velocity\right)}^x={\left(pressure\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times {\left(density\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, then x would be $\mathbf{3}$.