Question

Dimensional analysis of the equation, $\left(\right.$ velocity $=($ pressure difference $) \frac{3}{2} \times($ density gives the value of $x$ as,

Answer 1

First, recall the dimensions of velocity, pressure difference and density,
$\left[v\right]=LT^{- 1}\text{,}\left[P\right]=ML^{- 1}T^{- 2}\text{and}\left[\rho \right]=ML^{- 3}$
Now, put these values in the given expression,
(Velocity) $^{ x }=P^{\frac{3}{2}} \times \rho^{-\frac{3}{2}}=\left(\frac{P}{\rho}\right)^{\frac{3}{2}}=\frac{\left\{M L^{-1} T^{-2}\right\}^{\frac{3}{2}}}{\left( M L ^{-3}\right)^{\frac{3}{2}}}$
$=\left[L^{1} T^{- 1}\right]^{3}$ .
But $\left[L^{1} T^{- 1}\right]$ is the dimension of velocity,
$\Rightarrow x=3$ .