The rate of flow of water in a capillary tube | vedclass

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(C) According to Poiseuille's law,the rate of flow of liquid through a capillary tube is given by:$V = \frac{\pi P r^4}{8 \eta \ell}$where $P$ is the

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$8$

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(C) According to Poiseuille's law,the rate of flow of liquid through a capillary tube is given by:$V = \frac{\pi P r^4}{8 \eta \ell}$where $P$ is the pressure difference,$r$ is the radius,$\eta$ is the coefficient of viscosity,and $\ell$ is the length of the tube.Given that $P$ and $\eta$ remain constant,the rate of flow $V$ is proportional to $\frac{r^4}{\ell}$.Therefore,the ratio of the new rate of flow $V^{\prime}$ to the original rate of flow $V$ is:$\frac{V^{\prime}}{V} = \frac{(2r)^4}{2\ell} 	imes \frac{\ell}{r^4}$$\frac{V^{\prime}}{V} = \frac{16r^4}{2\ell} 	imes \frac{\ell}{r^4} = \frac{16}{2} = 8$Thus,$V^{\prime} = 8V$.

The rate of flow of water in a capillary tube of length $\ell$ and radius $r$ is $V.$ The rate of flow in another capillary tube of length $2 \ell$ and radius $2 r$ for the same pressure difference would be $....V$

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