Question

Stoke's law states that the viscous drag force F experienced by a sphere of radius a, moving with a speed v through a fluid with coefficient of viscosity $\eta$, is given by $F=6\pi \eta av$. If this fluid is flowing through a cylindrical pipe of radius r, length l and a pressure difference of P across its two ends, then the volume of water V which flows through the pipe in time t can be written as $\frac{V}{t}=k{\left(\frac{p}{l}\right)}^a{\eta }^b{r}^c$. where k is a dimensionless constant. Correct values of a, b and c are.

• A. $a=1,b=-1,c=4$
• B. $a=-1,b=1,c=4$
• C. $a=2,b=-1,c=3$
• D. $a=1,b=-2,c=-4$

Answer 1

Answer: (A)

$a=1,b=-1,c=4$

We get $\eta =\frac{F}{6\pi av}$
Dimensions of $\eta =\frac{\left[ML{T}^{-2}\right]}{\left[L\right]\left[L{T}^{-1}\right]}=\left[M{L}^{-1}{T}^{-1}\right]$
Dimensions of $\frac{p}{l}=\frac{\left[M{L}^{-1}{T}^{-2}\right]}{\left[L\right]}=\left[M{L}^{-2}{T}^{-2}\right]$
Dimensions of $\frac{V}{t}=\left[L^3{T}^{-1}\right]$
Dimensions of $r=\left[L\right]$
Using dimensional analysis,
$\left[L^3{T}^{-1}\right]$ $=\left[M{L}^{-2}{T}^{-2}{]}^a\right[M{L}^{-1}{T}^{-1}{]}^b[L{]}^c$
Or $\left[L^3{T}^{-1}\right]=\left[M^{a+b}{L}^{-2a-b+c}{T}^{-2a-b}\right]$
So, $a+b=0$ $⟹b=-a$
Also, $-2a-b=-1$
or $-2a+a=-1$ $⟹a=1$
$⟹b=-a=-1$
Also, $-2a-b+c=3$
Or $-2\left(1\right)-(-1)+c=3$ $⟹c=4$