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 same as before
Added by Arch K.
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Step 1: Start with the given equation for the rate of flow through a capillary tube: \[V = \frac{\pi r^4 \Delta P}{8 \eta L}\] Show more…
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Aakash G.
Two capillary tubes of same length $l$ but 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 $r$ of the single tube that can replace the two tubes, so that the rate of flow is same as before? (A) $r=r_{1}+r_{2}$ (B) $r=r_{1}^{2}+r_{2}^{2}$ (C) $r^{4}=r_{1}^{4}+r_{2}^{4}$ (D) $\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r_{2}}$
We have two (narrow) capillary tubes T1 and T2 . Their lengths are l 1 and l 2 and radii of cross-section are r 1 and r 2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3 /sec. If l 1 = 2l 2 and r 1 =r 2 , what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)
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