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Stokes' 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 $F=6 \pi \eta a v$.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 value of $a, b$ and $c$ are
$a=1, b=-1, c=4$
$a=-1, b=1, c=4$
$a=2, b=-1, c=3$
$a=1, b=-2, c=-4$
Solution
(a) By Stokes' law,
$F=6 \pi \eta a v$
We have, $\quad \eta=\frac{F}{6 \pi a v}$
Dimensions of viscosity index $\eta$ are
$\Rightarrow \quad[\eta]=\left[\frac{ MLT ^{-2}}{ L \cdot LT ^{-1}}\right]=\left[ ML ^{-1} T ^{-1}\right]$
Now, given relation of volume flow rate is
$\frac{V}{t}=k\left(\frac{p}{l}\right)^a \cdot \eta^b \cdot r^c$
Substituting dimensions of physical quantities and equating dimensions on both sides of equation, we have
$\frac{\left[ L ^3\right]}{[ T ]} =\left[ ML ^{-2} T ^{-2}\right]^a \cdot\left[ ML ^{-1} T ^{-1}\right]^b \cdot[ L ]^c$
$\Rightarrow\left[ M ^0 L ^3 T ^{-1}\right] =\left[ M ^{a+b} L ^{-2 a-b+c} T ^{-2 a-b}\right]$
$a+b=0 \quad \dots(i)$
$-2 a-b+c=3 \quad \dots(ii)$
$-2 a-b=-1 \quad \dots(iii)$
From Eqs. $(ii)$ and $(iii)$, we have
$c=4$
From Eqs. $(i)$ and $(iii)$, we have
$b=-1$
Substituting bin Eq. $(i)$, we have
$a=1$
So, $a=1, b=-1$ and $c=4$
