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(C) The flow rate $Q$ through a capillary tube is given by Poiseuille's law: $Q = \frac{\Delta P}{R}$,where $R = \frac{8 \eta l}{\pi r^4}$ is the flui

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$\frac{l_1 l_2}{l_1 + l_2}$

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(C) The flow rate $Q$ through a capillary tube is given by Poiseuille's law: $Q = \frac{\Delta P}{R}$,where $R = \frac{8 \eta l}{\pi r^4}$ is the fluid resistance.For tubes in parallel,the effective resistance $R_{eff}$ is given by $\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$.Substituting the expressions for resistance: $\frac{\pi r^4}{8 \eta l} = \frac{\pi r^4}{8 \eta l_1} + \frac{\pi r^4}{8 \eta l_2}$.Canceling the common terms $\frac{\pi r^4}{8 \eta}$,we get: $\frac{1}{l} = \frac{1}{l_1} + \frac{1}{l_2}$.Solving for $l$: $\frac{1}{l} = \frac{l_1 + l_2}{l_1 l_2}$,which implies $l = \frac{l_1 l_2}{l_1 + l_2}$.

Two capillary tubes of the same radius but lengths $l_1$ and $l_2$ are connected in parallel at the bottom of a vessel. What should be the length of a single capillary tube of the same radius that replaces both,such that the flow rate remains the same?

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