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(D) According to Poiseuille's equation,the rate of flow $Q$ through a capillary tube is given by $Q = \frac{\pi P r^4}{8 \eta L}$.From this,we can see

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$\frac{Q}{64}$

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(D) According to Poiseuille's equation,the rate of flow $Q$ through a capillary tube is given by $Q = \frac{\pi P r^4}{8 \eta L}$.From this,we can see that $Q \propto P 	imes r^4$.Let the initial state be $Q_1 = Q$,$P_1 = P$,and $r_1 = a$.Let the final state be $Q_2$,$P_2 = 4P$,and $r_2 = \frac{a}{4}$.Using the proportionality $Q \propto P 	imes r^4$,we have $\frac{Q_2}{Q_1} = \frac{P_2}{P_1} 	imes \left( \frac{r_2}{r_1} ight)^4$.Substituting the values: $\frac{Q_2}{Q} = \frac{4P}{P} 	imes \left( \frac{a/4}{a} ight)^4$.$\frac{Q_2}{Q} = 4 	imes \left( \frac{1}{4} ight)^4 = 4 	imes \frac{1}{256} = \frac{1}{64}$.Therefore,$Q_2 = \frac{Q}{64}$.

Water flows in a streamline manner through a capillary tube of radius $a$. The pressure difference is $P$ and the rate of flow is $Q$. If the radius is reduced to $\frac{a}{4}$ and the pressure is increased to $4P$,then the rate of flow becomes ................

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