Physics

# The rate of flow of a liquid through a capillary tube of length $L$ and radius $r$ under a pressure difference $p$ is given by the poiseuille's formula $V = \pi \,p\,{r^4}/8\eta L$. Determine the dimensions of the coefficient of viscosity $\eta$ of the liquid.

##### SOLUTION

It is given that

$V=\dfrac{\pi {{\Pr }^{4}}}{8\eta L}$

$\eta =\dfrac{\pi {{\Pr }^{4}}}{8VL}$

Where V is volume rate $\left[ {{L}^{3}}{{T}^{-1}} \right]$

P is pressure $\left[ M{{L}^{-1}}{{T}^{-2}} \right]$

$\eta$  is viscosity

L is length $\left[ L \right]$

So the dimensional formula of viscosity is

$\dfrac{\left[ M{{L}^{-1}}{{T}^{-2}} \right]\left[ {{L}^{4}} \right]}{\left[ {{L}^{3}}{{T}^{-1}} \right]\left[ L \right]}$

$\left[ M{{L}^{-1}}{{T}^{-1}} \right]$

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Subjective Medium Published on 18th 08, 2020
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