Question

If dimensions of critical velocity ${\mathrm{V}}_{\mathrm{c}}$ of a liquid flowing through a tube are expressed as $\left[{\mathrm{\eta }}^{\mathrm{x}}{\mathrm{\rho }}^{\mathrm{y}}{\mathrm{r}}^{\mathrm{z}}\right]$, where $\mathrm{\eta },\mathrm{p}$ and $\mathrm{r}$ are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then values of $\mathrm{x},\mathrm{y}$ and $\mathrm{z}$ are given by:

Answer 1

Step 1: Given data

The dimensions of the critical velocity ${\mathrm{V}}_{\mathrm{c}}$ of a liquid flowing through a tube are expressed as: $\left[{\mathrm{\eta }}^{\mathrm{x}}{\mathrm{\rho }}^{\mathrm{y}}{\mathrm{r}}^{\mathrm{z}}\right]$

Where $\mathrm{\eta },\mathrm{p}$ and $\mathrm{r}$ are the coefficient of viscosity of the liquid, the density of liquid and radius of the tube respectively.

Step 2; Determine the value of $\mathrm{x},\mathrm{y}$ and $\mathrm{z}$

According to the principle of homogeneity of dimension, a physical quantity equation will be dimensionally correct, if the dimensions of all the terms occurring on both sides of the equations are the same.

The given critical velocity of liquid flowing through a tube is,

${\mathrm{V}}_{\mathrm{c}}$ is directly proportional to $\left[{\mathrm{\eta }}^{\mathrm{x}}{\mathrm{\rho }}^{\mathrm{y}}{\mathrm{r}}^{\mathrm{z}}\right]$

Coefficient of viscosity of liquid, $\mathrm{\eta }=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]$

Density of liquid, $\mathrm{\rho }=\left[{\mathrm{ML}}^{-3}\right]$

Radius, $\mathrm{r}=\left[{\mathrm{L}}^1\right]$

Critical velocity of liquid ${\mathrm{V}}_{\mathrm{c}}=\left[M^0{L}^1{\mathrm{T}}^{-1}\right]$

$\Rightarrow$ $\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]={\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]}^{\mathrm{x}}{\left[{\mathrm{ML}}^{-3}\right]}^{\mathrm{y}}{\left[\mathrm{L}\right]}^{\mathrm{z}}$

$=\left[{\mathrm{M}}^{\mathrm{x}+\mathrm{y}}{\mathrm{L}}^{-\mathrm{x}-3\mathrm{y}+\mathrm{z}}{\mathrm{T}}^{-\mathrm{x}}\right]$

Step 3: Find the value of $\mathrm{x},\mathrm{y}$ and $\mathrm{z}$

Comparing exponents of $\mathrm{M},\mathrm{L}$ and $\mathrm{T}$, we get,

$\mathrm{x}+\mathrm{y}=0$,

$-\mathrm{x}=-1$

$-\mathrm{x}+-3\mathrm{y}+\mathrm{z}=1$,

$\mathrm{z}=-1$

$\mathrm{x}=1$

$\mathrm{y}=-1$

Therefore, the value of$\mathrm{x},\mathrm{y}$ and $\mathrm{z}$ is $1,-1$ and $-1$ respectively.