Question

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

a). \[2{r_1}\]
b). \[4{r_1}\]
c). \[{r_1}\]
d). \[\dfrac{{{r_1}}}{2}\]

Answer 1

Hint: 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,
\[Q = \dfrac{{\Delta P\pi {r^4}}}{{8\eta L}}\](This equation is derived from rearranging poiseuille’s equation)
Now, from the given condition,
\[ \Rightarrow {Q_1} = {Q_2}\]
Writing the formula for \[{Q_1}\], we get
\[ \Rightarrow {Q_1} = \dfrac{{{P_1}\pi {r_1}^4}}{{8\eta {l_1}}}\]
Writing the formula for the second tube , we get
\[ \Rightarrow {Q_2} = \dfrac{{{P_2}\pi {r_2}^4}}{{8\eta {l_2}}}\]
Equating both we get ,
\[ \Rightarrow \dfrac{{{P_1}\pi {r_1}^4}}{{8\eta {l_1}}} = \dfrac{{{P_2}\pi {r_2}^4}}{{8\eta {l_2}}}\]
Now, applying the given conditions and cancelling out the common terms, we get
\[ \Rightarrow \dfrac{{{P_1}{r_1}^4}}{{{l_1}}} = \dfrac{{4{P_1}{r_2}^4}}{{{l_1}/4}}\]
Cancelling the common terms again, we get
\[ \Rightarrow {r_1}^4 = 16{r_2}^4\]
\[ \Rightarrow {r_1} = 2{r_2}\]
\[ \Rightarrow \dfrac{{{r_1}}}{2} = {r_2}\]

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.