We have two (narrow) capillary tubes \[{T_1}\] and \[{T_2}\]. Their lengths are \[{l_1}\], \[{l_2}\] and radii of cross-section are \[{r_1}\],\[{r_2}\] respectively. The rate of flow of water through \[{T_1}\] is \[8c{m^3}{s^{ - 1}}\] when the pressure difference across its ends is p. What will be the rate of flow of water through \[{T_2}\] under the same pressure difference, given that \[{l_1} = {l_2}\] and \[{r_1} = 2{r_2}\]?
A. \[8\,c{m^3}{s^{ - 1}}\]
B. \[4\,c{m^3}{s^{ - 1}}\]
C. \[2\,c{m^3}{s^{ - 1}}\]
D. \[0.5\,c{m^3}{s^{ - 1}}\]
Answer
301.2k+ views
Hint: Poiseuille's theorem states that the velocity of a liquid flowing through a capillary is directly proportional to the pressure of the liquid and the fourth power of the radius of the capillary tube and is inversely proportional to the viscosity of the liquid, and length of the capillary tube.
Formula Used:
The Poiseuille’s formula is,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Where, \[\eta \] is dynamic viscosity, P is pressure, l is length of the pipe and r is radius of the pipe.
Complete step by step solution:
According to the Poiseuille’s formula, the rate of flow of a liquid a in capillary tube is given as,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Here, we have two capillary tubes, then,
\[{Q_1} = \dfrac{{\pi {{\Pr }_1}^4}}{{8\eta {l_1}}}\]
And, \[{Q_2} = \dfrac{{\pi {{\Pr }_2}^4}}{{8\eta {l_2}}}\]……. (1)
Here, \[{l_2} = {l_1}\] and \[{r_2} = \dfrac{{{r_1}}}{2}\]
Substitute the value in equation (1), we obtain,
\[{Q_2} = \dfrac{{\pi {\mathop{\rm P}\nolimits} {{\left( {\dfrac{{{r_1}}}{2}} \right)}^4}}}{{8\eta {l_1}}} \\ \]
\[\Rightarrow {Q_2} = {\left( {\dfrac{1}{2}} \right)^4}\dfrac{{\pi {{\Pr }_1}^4}}{{8\eta {l_1}}} \\ \]
\[\Rightarrow {Q_2} = {\left( {\dfrac{1}{2}} \right)^4}{Q_1} \\ \]
\[\Rightarrow {Q_2} = \dfrac{{{Q_1}}}{{16}}\]
Here, given that, \[{Q_1} = 8c{m^3}{s^{ - 1}}\]
\[{Q_2} = \dfrac{8}{{16}} \\ \]
\[\therefore {Q_2} = 0.5\,c{m^3}{s^{ - 1}}\]
Therefore, the rate of flow of water through \[{T_2}\] under the same pressure difference is \[0.5\,c{m^3}{s^{ - 1}}\].
Hence, option D is the correct answer.
Note: The property of a fluid that opposes the relative motion between its different layers is known as viscosity. The force that opposes the motion between the layers of fluid is known as viscous force. Poiseuille number is a non-dimensional number which characterizes steady, fully-developed, laminar flow of a constant-property fluid through a duct of arbitrary, but constant, cross section.
Formula Used:
The Poiseuille’s formula is,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Where, \[\eta \] is dynamic viscosity, P is pressure, l is length of the pipe and r is radius of the pipe.
Complete step by step solution:
According to the Poiseuille’s formula, the rate of flow of a liquid a in capillary tube is given as,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Here, we have two capillary tubes, then,
\[{Q_1} = \dfrac{{\pi {{\Pr }_1}^4}}{{8\eta {l_1}}}\]
And, \[{Q_2} = \dfrac{{\pi {{\Pr }_2}^4}}{{8\eta {l_2}}}\]……. (1)
Here, \[{l_2} = {l_1}\] and \[{r_2} = \dfrac{{{r_1}}}{2}\]
Substitute the value in equation (1), we obtain,
\[{Q_2} = \dfrac{{\pi {\mathop{\rm P}\nolimits} {{\left( {\dfrac{{{r_1}}}{2}} \right)}^4}}}{{8\eta {l_1}}} \\ \]
\[\Rightarrow {Q_2} = {\left( {\dfrac{1}{2}} \right)^4}\dfrac{{\pi {{\Pr }_1}^4}}{{8\eta {l_1}}} \\ \]
\[\Rightarrow {Q_2} = {\left( {\dfrac{1}{2}} \right)^4}{Q_1} \\ \]
\[\Rightarrow {Q_2} = \dfrac{{{Q_1}}}{{16}}\]
Here, given that, \[{Q_1} = 8c{m^3}{s^{ - 1}}\]
\[{Q_2} = \dfrac{8}{{16}} \\ \]
\[\therefore {Q_2} = 0.5\,c{m^3}{s^{ - 1}}\]
Therefore, the rate of flow of water through \[{T_2}\] under the same pressure difference is \[0.5\,c{m^3}{s^{ - 1}}\].
Hence, option D is the correct answer.
Note: The property of a fluid that opposes the relative motion between its different layers is known as viscosity. The force that opposes the motion between the layers of fluid is known as viscous force. Poiseuille number is a non-dimensional number which characterizes steady, fully-developed, laminar flow of a constant-property fluid through a duct of arbitrary, but constant, cross section.
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