We have two (narrow) capillary tubes T1 and T2 . Their lengths are l1 and l2 and radii of the cross-section are r1 and r2 respectively. The rate of water under a pressure difference P through tube T1 is 8 cm3 s- 1 . If l1 = 2l2 and r1=r2 , what will be the rate of flow when the two tubes are connected in series and the pressure difference across the combination is same as before(= P )

Question

We have two (narrow) capillary tubes T1 and T2 . Their lengths are l1 and l2 and radii of the cross-section are r1 and r2 respectively. The rate of water under a pressure difference P through tube T1 is 8 cm3 s- 1 . If l1 = 2l2 and r1=r2 , what will be the rate of flow when the two tubes are connected in series and the pressure difference across the combination is same as before(= P ) (A) 4 cm3s- 1 (B) (16/3) cm3 s- 1 (C) (8/17) cm3 s- 1 (D) 8 cm3 s- 1

Tardigrade Admin · Accepted Answer

V=(π P r4/8 η l)=8 cm3s- 1 For composite tube V1=(P π r4/8 η (. l + (l)2 .))=(2/3)(π P r4/8 η l)=(2/3)× 8=(1 6/3)(cm)3s- 1[∴ l1 = l = 2 l2 textor l2 = (l/2)]

$4 c m^{3} s^{- 1}$

$\frac{16}{3} c m^{3} s^{- 1}$

$\frac{8}{17} c m^{3} s^{- 1}$

$8 c m^{3} s^{- 1}$

$V = \frac{π P r ^{4}}{8 η l} = 8 c m^{3} s^{- 1}$
For composite tube
$V_{1} = \frac{P π r ^{4}}{8 η ( l + \frac{l}{2} )} = \frac{2}{3} \frac{π P r ^{4}}{8 η l} = \frac{2}{3} \times 8 = \frac{16}{3} (c m)^{3} s^{- 1} [∴ l_{1} = l = 2 l_{2} or l_{2} = \frac{l}{2}]$

4cm3s−1

316​cm3s−1

178​cm3s−1

8cm3s−1

V=8ηlπPr4​=8cm3s−1 For composite tube V1​=8η(l+2l​)Pπr4​=32​8ηlπPr4​=32​×8=316​(cm)3s−1[∴l1​=l=2l2​orl2​=2l​]

$4 c m^{3} s^{- 1}$

$\frac{16}{3} c m^{3} s^{- 1}$

$\frac{8}{17} c m^{3} s^{- 1}$

$8 c m^{3} s^{- 1}$

$V = \frac{π P r ^{4}}{8 η l} = 8 c m^{3} s^{- 1}$
For composite tube
$V_{1} = \frac{P π r ^{4}}{8 η ( l + \frac{l}{2} )} = \frac{2}{3} \frac{π P r ^{4}}{8 η l} = \frac{2}{3} \times 8 = \frac{16}{3} (c m)^{3} s^{- 1} [∴ l_{1} = l = 2 l_{2} or l_{2} = \frac{l}{2}]$