We have two (narrow) capillary tubes T1 and T2. Their lengths are l1 and l2 and radii of cross-section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1 = 2l2 and r1 =r2, 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)

V=πPr48ηl=8cm3secFor composite tube V1=Pπr48ηl+l2=23πPr48ηl=23×8=163cm3sec∵l1=l=2l2 or l2=l2

We have two (narrow) capillary tubes T1 and T2. Their lengths are l1 and l2 and radii of cross-section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1 = 2l2 and r1 =r2, 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)

V=πPr48ηl=8cm3secFor composite tube V1=Pπr48ηl+l2=23πPr48ηl=23×8=163cm3sec∵l1=l=2l2 or l2=l2

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We have two (narrow) capillary tubes T₁and T₂. Their lengths are l₁ and l₂ and radii of cross-section are r₁ and r₂ respectively. The rate of flow of water under a pressure difference P through tube T₁ is 8cm³/sec. If l₁ = 2l₂ and r₁ =r₂, 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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4 cm³/sec

(16/3) cm³/sec

(8/17) cm³/sec

None of these

answer is B.

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Detailed Solution

$V = \frac{{πPr}^{4}}{8 ηl} = \frac{8 {cm}^{3}}{\sec}$

For composite tube

$\begin{array}{l} V_{1} = \frac{{Pπr}^{4}}{8 η (l + \frac{l}{2})} = \frac{2}{3} \frac{{πPr}^{4}}{8 ηl} = \frac{2}{3} \times 8 = \frac{16}{3} \frac{{cm}^{3}}{\sec} \\ [∵ l_{1} = l = 2 l_{2} or l_{2} = \frac{l}{2}] \end{array}$

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