Stokes' law states that the viscous drag forc | vedclass

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(A) By Stokes' law,$F=6 \pi \eta a v$. We have,$\eta = \frac{F}{6 \pi a v}$. Dimensions of viscosity $\eta$ are $[\eta] = [ML^{-1}T^{-1}]$. The given

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

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(A) By Stokes' law,$F=6 \pi \eta a v$. We have,$\eta = \frac{F}{6 \pi a v}$. Dimensions of viscosity $\eta$ are $[\eta] = [ML^{-1}T^{-1}]$. The given relation for volume flow rate is $\frac{V}{t} = k \left(\frac{p}{l}\right)^a \eta^b r^c$. Substituting dimensions: $[L^3 T^{-1}] = [ML^{-2}T^{-2}]^a [ML^{-1}T^{-1}]^b [L]^c$. $[M^0 L^3 T^{-1}] = [M^{a+b} L^{-2a-b+c} T^{-2a-b}]$. Equating powers: $a+b = 0$ $(i)$ $-2a-b+c = 3$ (ii) $-2a-b = -1$ (iii) From (iii),$2a+b = 1$. Subtracting $(i)$ from this,$a = 1$. Substituting $a=1$ in $(i)$,$b = -1$. Substituting $a=1, b=-1$ in (ii),$-2(1) - (-1) + c = 3 \Rightarrow -2 + 1 + c = 3 \Rightarrow c = 4$. Thus,$a=1, b=-1, c=4$.

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