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Question: Two tubes of radii \[{r_1}\]​ and \[{r_2}\]​, and lengths \[{l_1}\]and \[{l_2}\]​, respectively, are...

Two tubes of radii r1{r_1}​ and r2{r_2}​, and lengths l1{l_1}and l2{l_2}​, respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1{P_1}​ and P2{P_2}​ are pressure differences across the two tubes. If P2{P_2} is 4P14{P_1}​ and l2{l_2}​ isl14\dfrac{{{l_1}}}{4}​​, then the radius r2{r_2} will be equal to :

a). 2r12{r_1}
b). 4r14{r_1}
c). r1{r_1}
d). r12\dfrac{{{r_1}}}{2}

Explanation

Solution

The tube dimensions for tube A and tube B are given in the question and are said to be connected in series. In a series connection , flow rate in tube A is assumed to be the same as flow rate in tube B. Using the given rule , we can solve for unknown radius.

Complete Step by Step Solution:
We are given 2 tubes A and B that are connected in series to each other. When two tubes are connected in series, we can assume that the flow rate Q in tube A will be equal to that of flow rate in tube B. Flow rate of a fluid depends on factors such as pressure difference between the tubes, viscosity of the fluid, the radii of the tubes and the length of the tube. Greater the tube length, flow rate is said to be less. Mathematically , flow rate can be represented as,
Q=ΔPπr48ηLQ = \dfrac{{\Delta P\pi {r^4}}}{{8\eta L}}(This equation is derived from rearranging poiseuille’s equation)
Now, from the given condition,
Q1=Q2\Rightarrow {Q_1} = {Q_2}
Writing the formula for Q1{Q_1}, we get
Q1=P1πr148ηl1\Rightarrow {Q_1} = \dfrac{{{P_1}\pi {r_1}^4}}{{8\eta {l_1}}}
Writing the formula for the second tube , we get
Q2=P2πr248ηl2\Rightarrow {Q_2} = \dfrac{{{P_2}\pi {r_2}^4}}{{8\eta {l_2}}}
Equating both we get ,
P1πr148ηl1=P2πr248ηl2\Rightarrow \dfrac{{{P_1}\pi {r_1}^4}}{{8\eta {l_1}}} = \dfrac{{{P_2}\pi {r_2}^4}}{{8\eta {l_2}}}
Now, applying the given conditions and cancelling out the common terms, we get
P1r14l1=4P1r24l1/4\Rightarrow \dfrac{{{P_1}{r_1}^4}}{{{l_1}}} = \dfrac{{4{P_1}{r_2}^4}}{{{l_1}/4}}
Cancelling the common terms again, we get
r14=16r24\Rightarrow {r_1}^4 = 16{r_2}^4
r1=2r2\Rightarrow {r_1} = 2{r_2}
r12=r2\Rightarrow \dfrac{{{r_1}}}{2} = {r_2}

Thus, Option (d) is the right answer for the given question.

Note: Poiseuille’s law states in a smooth laminar flow of fluid the flow rate of the liquid is given as the ratio of the pressure difference across the tube and the viscous resistance of the fluid. The viscous resistance is equal to the product of viscosity of the fluid and the length of the tube.