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(D) According to Poiseuille's equation,the rate of flow $Q$ through a capillary tube is given by:$Q = \frac{\pi P r^4}{8 \eta l}$where $P$ is the pres

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$Q/8$

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(D) According to Poiseuille's equation,the rate of flow $Q$ through a capillary tube is given by:$Q = \frac{\pi P r^4}{8 \eta l}$where $P$ is the pressure difference,$r$ is the radius,$\eta$ is the coefficient of viscosity,and $l$ is the length of the tube.From the formula,we see that $Q \propto P r^4$ (since $\eta$ and $l$ are constants).Given the initial conditions: $Q_1 = Q$,$P_1 = P$,and $r_1 = a$.Given the final conditions: $P_2 = 2P$ and $r_2 = a/2$.Taking the ratio:$\frac{Q_2}{Q_1} = \left( \frac{P_2}{P_1} ight) \left( \frac{r_2}{r_1} ight)^4$$\frac{Q_2}{Q} = \left( \frac{2P}{P} ight) \left( \frac{a/2}{a} ight)^4$$\frac{Q_2}{Q} = 2 	imes \left( \frac{1}{2} ight)^4 = 2 	imes \frac{1}{16} = \frac{1}{8}$Therefore,$Q_2 = Q/8$.

Water flows in a streamlined manner through a capillary tube of radius $a$,the pressure difference being $P$ and the rate of flow $Q$. If the radius is reduced to $a/2$ and the pressure is increased to $2P$,the rate of flow becomes:

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