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(D) According to Poiseuille's equation,the volume flow rate $V$ is given by $V = \frac{P \pi r^4}{8 \eta l}$.Rearranging for pressure difference $P$,w

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$\frac{P}{4}$

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(D) According to Poiseuille's equation,the volume flow rate $V$ is given by $V = \frac{P \pi r^4}{8 \eta l}$.Rearranging for pressure difference $P$,we get $P = \frac{V 8 \eta l}{\pi r^4}$.Given the initial state $(P_1, V_1, r_1, l_1)$ and final state $(P_2, V_2, r_2, l_2)$:$V_2 = 2V_1$,$r_2 = 2r_1$,and $l_2 = 2l_1$.Using the ratio: $\frac{P_2}{P_1} = \left( \frac{V_2}{V_1} ight) 	imes \left( \frac{l_2}{l_1} ight) 	imes \left( \frac{r_1}{r_2} ight)^4$.Substituting the values: $\frac{P_2}{P_1} = (2) 	imes (2) 	imes \left( \frac{1}{2} ight)^4 = 4 	imes \frac{1}{16} = \frac{1}{4}$.Therefore,$P_2 = \frac{P_1}{4} = \frac{P}{4}$.

$A$ liquid flows through a pipe with a pressure difference of $P$. What should be the pressure difference required to maintain the same flow rate if the radius is doubled and the length is doubled?

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