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

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$V / 32$

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(D) According to Poiseuille's equation,the rate of flow of liquid $V$ through a capillary tube is given by:$V = \frac{\pi P r^4}{8 \eta l}$where $P$ is the pressure difference,$r$ is the radius of the bore,$\eta$ is the coefficient of viscosity,and $l$ is the length of the tube.From the formula,we see that $V \propto \frac{r^4}{l}$.Given that the length is doubled $(l_2 = 2l_1)$ and the diameter is halved,the radius is also halved $(r_2 = r_1 / 2)$.Substituting these values into the ratio:$\frac{V_2}{V_1} = \left( \frac{r_2}{r_1} ight)^4 	imes \left( \frac{l_1}{l_2} ight)$$\frac{V_2}{V_1} = \left( \frac{1}{2} ight)^4 	imes \left( \frac{1}{2} ight) = \frac{1}{16} 	imes \frac{1}{2} = \frac{1}{32}$Therefore,the new rate of flow is $V_2 = V / 32$.

Under a constant pressure head,the rate of flow of liquid through a capillary tube is $V$. If the length of the capillary is doubled and the diameter of the bore is halved,the rate of flow would become

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