Question

The rate at which water flows through a capillary of length \[0.5m\] with an internal diameter of \[1mm\], coefficient of viscosity \[1.3 \times {10^{ - 3}}kg{m^{ - 1}}{s^{ - 1}}\] , and pressure head \[20cm\] of water in cc. per second is:
\[(A)7.4 \times {10^{ - 2}}\]
\[(B)3.7 \times {10^{ - 2}}\]
\[(C)74 \times {10^{ - 2}}\]
\[(D)7.4\]

Answer 1

Hint : Use Poiseuille's equation. By Newton’s law of viscous flow, the viscous force is directly proportional to the surface area and also changes in its velocity. The flow of liquid is based on the factors such as viscosity of the fluid, the pressure, length, and therefore the diameter of the tube. Hence by substituting all the values in Poiseuille's equation, we are going to get the answer.

Formula used:
Poiseuille's law for the volume flow rate is
\[Q = \dfrac{{\nabla P\pi {r^4}}}{{8\eta l}}\]
Where, \[Q\] is the volume flow rate, \[\nabla P\] is the pressure difference, \[r\] is the width of the tube for flow, \[l\] is the length of the tube for flow, and \[\eta \] is the coefficient of viscosity.

Complete step-by-step solution:
Given that, pressure difference \[\nabla P = 20cm\]
The width of the tube for flow \[r = 0.05cm\]
The length of the tube for flow \[l = 0.5m = 50cm\]
The coefficient of viscosity \[\eta = 1.3 \times {10^{ - 2}}\]
Substitute all values in Poiseuille's equation,
\[Q = \dfrac{{\nabla P\pi {r^4}}}{{8\eta l}} = \dfrac{{1 \times 980 \times 20 \times \pi \times {{\left( {0.05} \right)}^4}}}{{8 \times 1.3 \times {{10}^{ - 2}} \times 50}}\]
\[Q = 7.4 \times {10^{ - 2}}c{m^3}/s\]
Hence, the correct option is A.
Additional information:
Capillary action - Tubes having very small diameters (narrow cylindrical tubes) are named capillary. If these narrow tubes are dipped during a liquid it’s observed that liquid within the capillary either rises (or) falls relative to the surrounding liquid level. This phenomenon is termed as capillary action and such tubes are named capillary.
For better understanding, we are going to consider the capillary tube as a normal pipe.
The rate of flow through a capillary is given by Poiseuille's equation
Poiseuille's equation
\[Q = \dfrac{{\nabla P\pi {r^4}}}{{8\eta l}}\]

Note: Poiseuille’s law states during a smooth laminar flow of fluid, the rate of flow of the liquid is given because of the ratio of the pressure difference across the tube and therefore the viscous resistance of the fluid. The viscous resistance is always equal to the product of fluid viscosity and the tube length. Poiseuille’s law is only valid for laminar flow. Flow rate is termed as discharge per unit of time or how much liquid is flowing through a pipe within a unit of time.