A flow of viscous fluid is *laminar* if particles of the fluid move in parallel layers (laminas). All particles in a layer move with the same speed. There is a relative motion between particles of adjacent layers. The relative velocity between layers is described by *velocity gradient* $\mathrm{d}v/\mathrm{d}z$. The velocity gradient is the change in velocity as one moves unit distance in a direction perpendicular to the layers.

Each layer applies a viscous force on its adjacent layer. The viscous force on a layer of area $A$ located at a place having velocity gradient $\mathrm{d}v/\mathrm{d}z$ is given by
\begin{align}
F=-\eta A \frac{\mathrm{d}v}{\mathrm{d}z}
\end{align}
where $\eta$ is called viscosity or coefficient of viscosity of the fluid. It is a fluid's property with SI unit is N-s/m^{2} and CGS unit poise (1 poise = 0.1 N-s/m^{2}). The dimensional formula of viscosity is ML^{-1}T^{-1}. The negative sign in the above equation indicates the resistive nature of the viscous force (it opposes relative motion between the layers).

The viscosity of the liquids is very high as compared to that of gases. In liquids, the viscosity is due to cohesive forces among molecules whereas in gases, it is due to diffusion. The viscosity is very sensitive to the temperature. The viscosity of liquids decreasese with an increase in temperature whereas the viscosity of gases increases with an increase in temperature. The viscosity of liquids is measured by a viscometer.

Consider the laminar flow of a fluid of viscosity $\eta$ through a pipe of radius $r$ and length $l$. If the pressure difference between two ends of the pipe is $P$ then volume flow rate (volume V of the fluid flow in time t) through the pipe is given by Poiseuille's law
\begin{align}
\frac{V}{t}=\frac{\pi P r^4}{8\eta l}
\end{align}
Poiseuille's formula can be derived by dimensional analysis. Note that the volume flow rate depends on the 4^{th} power of the tube's radius.

A liquid of density 900 kg/m^{3} is filled in a cylindrical tank of upper radius 0.9 m and lower radius 0.3 m. A capillary tube of length $l$ is attached at the bottom of the tank as shown in the figure. The capillary has outer radius 0.002 m and inner radius $a$. When pressure $p$ is applied at the top of the tank volume flow rate of the liquid is ${8\times{10}^{-6}}\;\mathrm{m^3/s}$ and if capillary tube is detached, the liquid comes out from the tank with a velocity of ${10}\;\mathrm{m/s}$. Determine the coefficient of viscosity of the liquid. (Given : $\pi a^2={{10}^{-6}}\;\mathrm{m^2}$ and $a^2/l={2\times{10}^{-6}}\;\mathrm{m}$.)

**Solution: **
Let $r_1={0.9}\mathrm{m}$ be the radius at the top, $v_1$ be the velocity at the top, $r_2=0.3\;\mathrm{m}$ be the radius at the bottom, and $v_2={10}\;{m/s}$ be the velocity at the bottom (when capillary tube is detached). The continuity equation, $\pi r_1^2 v_1=\pi r_2^2 v_2$, gives $v_1={{10}/{9}}\;{m/s}$. Let $p_0$ be the atmospheric pressure. Apply Bernoulli's equation between the top and bottom,
\begin{align}
p_0+p+\rho g H+\tfrac{1}{2}\rho v_2^2=p_0+0+\tfrac{1}{2}\rho v_1^2,
\end{align}
to get,
\begin{align}
p+\rho g H=\tfrac{1}{2}\rho(v_2^2-v_1^2)={\tfrac{4}{9}\times{10}^{5}}\mathrm{N/m^2}.
\end{align}
**Note:**Bernoulli's theorem is valid for non-viscous fluids. In this problems, the fluid is viscous (we are asked to find its viscosity). We assume that viscous effects are small when fluid flow through a large hole of radius 0.3 m (i.e., when capillary is not connected). We applied Bernoulli's equation with this assumption.

The pressure difference across the capillary is, \begin{align} \Delta p=(p_0+p+\rho gH)-p_0=p+\rho g H. \end{align} By Poiseuille's equation, volume flow rate in a capillary is, \begin{align} \label{glb:eqn:4} Q=\frac{\pi \Delta p a^4}{8\eta l}=\frac{\Delta p (\pi a^2) (a^2/l)}{8\eta}. \end{align} Substitute the values in above equation to get $\eta={\frac{1}{720}}\,\mathrm{N\,s/m^2}$.

Consider a thin square plate floating on a viscous liquid in large tank. The height $h$ of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity $u_0$. Which of the following statements is (are) true?

- The resistive force of liquid on the plate is inversely proportional to $h$.
- The resistive force of liquid on the plate is independent of the area of the plate.
- The tangential (shear) stress on the floor of the tank increases with $u_0$.
- The tangential (shear) stress on the plate varies linearly with the viscosity $\eta$ of the liquid.

**Solution:**
In a viscous liquid, there is no relative motion between the solid surface and the liquid layer in contact with it i.e., the liquid layer moves with the speed of the solid surface. Thus, the liquid layer in contact with the plate moves with a speed $u_0$ and the liquid layer in contact with the floor of the tank is stationary. The velocity gradient is given by
\begin{align}
{\mathrm{d}v}/{\mathrm{d}z}\approx {(u_0-0)}/{h}={u_0}/{h}, \nonumber
\end{align}
where we have used the fact that $h$ is very small in comparison to the tank width. Consider a liquid layer which is in contact with the plate. The viscous drag force (backward) on this layer, due to the layer just below it, is
\begin{align}
F_\text{on top layer}=\eta A ({\mathrm{d} v}/{\mathrm{d}z})=\eta A ({u_0}/{h}). \nonumber
\end{align}

The top layer moves with a constant speed $u_0$. Thus, the plate should apply an equal but opposite force (forward) on the top layer to balance viscous drag on it. In turn, by Newton's third law, the top layer should apply equal but opposite force (backward) force on the plate. Thus, the resistive force of liquid on the plate is \begin{align} F_\text{on plate}=F_\text{on top layer}=\eta A ({u_0}/{h}). \nonumber \end{align} The tangential (shear) stress on the plate is given by \begin{align} \sigma_\text{on plate}=F_\text{on plate}/A=\eta ({u_0}/{h}). \nonumber \end{align} Apply similar argument on liquid layer in contact with the floor of the tank. The resistive force of liquid on the floor of the tank is $F_\text{on floor}=\eta A ({u_0}/{h})$ and the tangential (shear) stress on the floor of the tank is \begin{align} \sigma_\text{on floor}=F_\text{on floor}/A=\eta ({u_0}/{h}). \nonumber \end{align}

Thus, correct options are (A), (C) and (D).

**Question 1:** A layer of oil 1.5 mm thick is placed between two microscope slides. A force of 5.5×10^{−4} N is required to glide one slide over the other at a speed of 1 cm/s when their contact area is 6 cm^{2}. The coefficient of viscosity of the oil is?

**Question 2:** A capillary tube of length l and radius r is joined to another capillary tube of length l/4 and radius r/2. A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is

**Question 3:** Two capillary tubes of the same length but different radii r_{1} and r_{2} are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube of the same length that can replace the two tubes so that the rate of flow is the same as before

**Question 4:** The viscosities of a liquid and a gas varies with increase in temperature as

- IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
- Concepts of Physics Part 1 by HC Verma (Link to Amazon)
- Viscosity and Laminar Flow, Poiseuille's Law (Chapter from College Physics)
- Poiseuille's Law Derivation (without dimensional analysis)
- Fluids Dynamics using Dimensional Analysis (American Journal of Physics)