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(D) According to Poiseuille's equation,the rate of flow $V$ is given by $V = \frac{P \pi r^4}{8 \eta l}$.From this,the pressure difference $P$ can be

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$\frac{P}{4}$

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(D) According to Poiseuille's equation,the rate of flow $V$ is given by $V = \frac{P \pi r^4}{8 \eta l}$.From this,the pressure difference $P$ can be expressed as $P = \frac{V 8 \eta l}{\pi r^4}$.Let the initial state be $P_1 = P$,$V_1 = V$,$r_1 = r$,and $l_1 = l$.For the final state,we have $V_2 = 2V$,$r_2 = 2r$,and $l_2 = 2l$.Taking the ratio of the pressures: $\frac{P_2}{P_1} = \frac{V_2}{V_1} 	imes \frac{l_2}{l_1} 	imes \left( \frac{r_1}{r_2} ight)^4$.Substituting the values: $\frac{P_2}{P} = 2 	imes 2 	imes \left( \frac{r}{2r} ight)^4 = 4 	imes \left( \frac{1}{2} ight)^4 = 4 	imes \frac{1}{16} = \frac{1}{4}$.Therefore,$P_2 = \frac{P}{4}$.

$A$ liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference $P$. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

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