Question

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

Answer 1

Answer: 3

First, recall the dimensions of velocity, pressure difference and density,
$\left[v\right]=L{T}^{-1}\text{,}\left[P\right]=M{L}^{-1}{T}^{-2}\text{and}\left[\rho \right]=M{L}^{-3}$
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,
$\Rightarrow x=3$ .