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

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$(r_1^4 + r_2^4)^{1/4}$

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(D) According to Poiseuille's equation,the rate of flow $V$ through a capillary tube is given by $V = \frac{\pi P r^4}{8 \eta l}$.Since the two tubes are connected in parallel,the total rate of flow $V$ is the sum of the rates of flow through each tube: $V = V_1 + V_2$.Substituting the expression for flow rate: $\frac{\pi P r^4}{8 \eta l} = \frac{\pi P r_1^4}{8 \eta l} + \frac{\pi P r_2^4}{8 \eta l}$.Canceling the common terms $\frac{\pi P}{8 \eta l}$ from both sides,we get $r^4 = r_1^4 + r_2^4$.Therefore,the radius of the single tube is $r = (r_1^4 + r_2^4)^{1/4}$.

Two capillary tubes of the same length but different radii $r_1$ and $r_2$ are fitted in parallel to the bottom of a vessel. The pressure head is $P$. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is the same as before?

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