Under a pressure head,the rate of orderly vol | vedclass

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Question

(A) According to Poiseuille's law,the rate of volume flow $Q$ is given by $Q = \frac{\pi P r^4}{8 \eta \ell}$,where $P$ is the pressure difference,$r$

Vedclass · Accepted Answer

$\frac{Q}{32}$

Vedclass · Answer

(A) According to Poiseuille's law,the rate of volume flow $Q$ is given by $Q = \frac{\pi P r^4}{8 \eta \ell}$,where $P$ is the pressure difference,$r$ is the radius,$\eta$ is the viscosity,and $\ell$ is the length of the tube.From this formula,$Q \propto \frac{r^4}{\ell}$.Let the initial length be $\ell_1 = \ell$ and the initial radius be $r_1 = r$. The new length is $\ell_2 = 2\ell$ and the new diameter is halved,so the new radius is $r_2 = \frac{r}{2}$.The new rate of flow $Q^{\prime}$ is given by:$Q^{\prime} = Q 	imes \left(\frac{r_2}{r_1}ight)^4 	imes \left(\frac{\ell_1}{\ell_2}ight)$$Q^{\prime} = Q 	imes \left(\frac{r/2}{r}ight)^4 	imes \left(\frac{\ell}{2\ell}ight)$$Q^{\prime} = Q 	imes \left(\frac{1}{2}ight)^4 	imes \left(\frac{1}{2}ight)$$Q^{\prime} = Q 	imes \frac{1}{16} 	imes \frac{1}{2} = \frac{Q}{32}$.

Under a pressure head,the rate of orderly volume flow of a liquid through a capillary tube is $Q$. If the length of the capillary tube is doubled and the diameter of the tube is halved,the rate of flow would become:

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