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

Hint: 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
     $\begin{align}
  & {{v}_{c}}=L{{T}^{-}}^{1} \\
 & \eta =M{{L}^{-}}^{1}{{T}^{-}}^{1} \\
 & d=M{{L}^{-}}^{3} \\
 & r=L \\
\end{align}$
Using dimensional analysis with $a,b\text{ and }c$ as integers
$\begin{align}
  & {{v}_{c}}=k{{(\eta )}^{a}}{{(d)}^{b}}{{(r)}^{c}} \\
 & L{{T}^{-}}^{1}={{(M{{L}^{-}}^{1}{{T}^{-}}^{1})}^{a}}{{(M{{L}^{-}}^{3})}^{b}}{{(L)}^{c}} \\
\end{align}$
Comparing the coefficients of $M,L\;\text{and }T$ on both sides
     $\begin{align}
  & a+b=0 \\
 & -a-3b+c=1 \\
 & -a=-1 \\
\end{align}$
Solving for $a,b\text{ and }c$
     $\begin{align}
  & a=+1 \\
 & b=-1 \\
 & c=-1 \\
\end{align}$
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.