Question

The correct formula of critical velocity ${{v}_{c}}$ is
A. ${{v}_{c}}=\dfrac{k\eta d}{r}$
B. ${{v}_{c}}=\dfrac{k\eta }{dr}$
C. ${{v}_{c}}=\dfrac{dr}{k\eta }$
D. ${{v}_{c}}=\dfrac{r\eta }{dk}$

Answer 1

Derive the formula of critical velocity using dimensional analysis.

Complete step by step solution:
Critical velocity depends on coefficient of viscosity $\eta$, density of fluid $d$ and radius of tube $r$ . The dimension of each is
$\begin{aligned} & {{v}_{c}}=L{{T}^{-}}^{1} \\\ & \eta =M{{L}^{-}}^{1}{{T}^{-}}^{1} \\\ & d=M{{L}^{-}}^{3} \\\ & r=L \\\ \end{aligned}$
Using dimensional analysis with $a,b\text{ and }c$ as integers
$\begin{aligned} & {{v}_{c}}=k{{(\eta )}^{a}}{{(d)}^{b}}{{(r)}^{c}} \\\ & L{{T}^{-}}^{1}={{(M{{L}^{-}}^{1}{{T}^{-}}^{1})}^{a}}{{(M{{L}^{-}}^{3})}^{b}}{{(L)}^{c}} \\\ \end{aligned}$
Comparing the coefficients of $M,L\;\text{and }T$ on both sides
$\begin{aligned} & a+b=0 \\\ & -a-3b+c=1 \\\ & -a=-1 \\\ \end{aligned}$
Solving for $a,b\text{ and }c$
$\begin{aligned} & a=+1 \\\ & b=-1 \\\ & c=-1 \\\ \end{aligned}$
Therefore
${{v}_{c}}=\dfrac{k\eta }{dr}$

The correct answer is option B.

Note: Critical velocity is the velocity at which a liquid transitions from subcritical flow to supercritical flow.