Stokes' law states that the viscous drag | vedclass

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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $c=4$

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