Question

The correct formula of critical velocity $v_c$ is
A. $v_c={\displaystyle \frac{k\eta d}{r}}$
B. $v_c={\displaystyle \frac{k\eta }{dr}}$
C. $v_c={\displaystyle \frac{dr}{k\eta }}$
D. $v_c={\displaystyle \frac{r\eta }{dk}}$

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
     $\begin{array}{rl}& v_c=L{T^{-}}^1\\ & \eta =M{L^{-}}^1{T^{-}}^1\\ & d=M{L^{-}}^3\\ & r=L\end{array}$
Using dimensional analysis with $a,b\text{ and }c$ as integers
$\begin{array}{rl}& v_c=k{(\eta )}^a{(d)}^b{(r)}^c\\ & L{T^{-}}^1={(M{L^{-}}^1{T^{-}}^1)}^a{(M{L^{-}}^3)}^b{(L)}^c\end{array}$
Comparing the coefficients of $M,L\text{and }T$ on both sides
     $\begin{array}{rl}& a+b=0\\ & -a-3b+c=1\\ & -a=-1\end{array}$
Solving for $a,b\text{ and }c$
     $\begin{array}{rl}& a=+1\\ & b=-1\\ & c=-1\end{array}$
Therefore
     $v_c={\displaystyle \frac{k\eta }{dr}}$

The correct answer is option B.

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