a
(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 $c=4$