Viscosity (Fluid statics) | Complete notes | Important questions | Short questions and Numerical problem solutions | Class 12 (Physics) | Physics in Depth

Viscosity

The property of a fluid by virtue of which an internal friction comes in to play when the fluid is in motion and opposes the relative motion of its different layers is called viscosity . If we move through a pool of water we experience a resistance to our motion. This shows that there is a frictional force in liquids. We say this is due to the viscosity of the liquid. If the frictional force is comparatively low, as in water, the viscosity of the liquid is low; if the frictional force is large, as in glue or glycerine, the viscosity of the liquid is high.

Newton's law of viscosity, Coefficient of viscosity

Consider the streamline flow of a liquid as in figure. The layer of liquid in contact with the fixed horizontal surface is at rest and all the other layers move parallel to each other with their velocities increasing uniformly upwards. Consider two uppermost layers at a distance x and x+dx from the horizontal surface with v and v+dv as their respective velocities.

The rate of change of velocity with distance in a direction perpendicular to that of the liquid flow is called the velocity gradient i.e., \(\frac{dv}{dx}\).
According to Newton, the viscous force F depends on the following factors:

(i) It is directly proportional to the area A of layers in contact i.e., \[F \propto A\] (ii) It is directly proportional to the velocity gradient between the layers i.e., \[F \propto \frac{dv}{dx}\] Combining this two factors:
\[F \propto A\frac{dv}{dx}\] \[\therefore F=-\eta A\frac{dv}{dx}\] Here, \(\eta\) is the proportionality constant called the coefficient of viscosity . Its value depends on the nature of the liquid. The negative sign shows that the direction of viscous force, F is opposite to the direction of motion of the liquid.

Coefficient of viscosity of liquid :
If A = 1 unit and \(\frac{dv}{dx}\)=1 unit, then the viscous force is, \[\eta=-F\] Thus, coefficient of viscosity of a liquid is defined as the viscous force acting per unit area of the layer having unit velocity gradient perpendicular to the direction of the flow of the liquid.

The C.G.S unit of the coefficient of viscosity is poise and \[1 \hspace{0.1cm} \text{poise}=1 \hspace{0.1cm}\text{dyne.s.cm}^{-2}\] Equivalently,\[1 \hspace{0.1cm} \text{poise}=0.1 \hspace{0.1cm} \text{N.s.m}^{-2}\]

Laminar and Turbulent flow

A flow of liquid in which different layers glide over one another at a slow and steady velocity (below its critical velocity) without intermixing is called a laminar, streamline or viscous flow . In laminar flow, the velocity of particles of the liquid at every point remains constant.

When the velocity of liquid exceeds a limiting value called its critical velocity , the motion of the particles of the liquid becomes disorderly or irregular (i.e., zigzag).

This type of flow of liquid is called turbulent flow. In turbulent flow, the velocity of the particles of the liquid change continuously and haphazardly.

A large wooden plate of area 10 \(m^2\) flowing on the surface of a river is made to move horizontally with a speed of 2 m\(s^{-1}\) by applying a tangential force. If the river is 1 m deep and the water in contact with the bed is stationary, find the tangential force needed to keep the plate moving. Coefficient of viscosity of water at the temperature of the river = 10\(^{-2}\) poise.

Here, the velocity gradient, \(\frac{dv}{dx}\) = 2 \(s^{-1}\) , \(\eta\) =1 0\(^{-2}\) poise = \(0.1\times 10^{-2}\) Nsm\(^{-2}\) = 10\(^{-3}\) Nsm\(^{-2}\), A = 10 \(m^2\)

From the Newton's law of viscosity, the magnitude of the viscous force or the tangential force required to keep the plate moving is, \[\begin{align*} F&=\eta A \frac{dv}{dx}\\ &=10^{-3} \times 10 \times 2\\ &=0.02\hspace{0.1cm} \text{N} \end{align*}\]

A flat plate of area 0.1 \(m^2\) is placed on a flat surface and is separated from it by a film of 10\(^{-5}\) m thick whose coefficient of viscosity is 1.5 Nsm\(^{-2}\). Calculate the force required to cause the plate to slide on the surface at the constant speed of 1 mm\(s^{-1}\).

Here, v = 1 mm\(s^{-1}\) = 10\(^{-3}\) m\(s^{-1}\), x = 10\(^{-5}\) m, \(\frac{dv}{dx}\) = \(\frac{10^{-3}}{10^{-5}}\) = \(10^2\) \(s^{-1}\)

Use the Newton's formula for viscosity and you will get 15 N as the answer.

Poiseuille's formula

Suppose a fluid flows through a narrow tube in steady flow. Because of viscosity, the layer in contact with the wall of the tube remains at rest and the layers away from the wall move fast. Poiseuille found that the volume, V of the liquid flowing per second through a capillary tube as in figure is,

(i) directly proportional to the difference of the pressure, P between the two ends of the tube i.e., \[V \propto P\] (ii) directly proportional to the fourth power of the radius, r of the capilary tube i.e., \[V \propto r^4\] (iii) inversely proportional to the coefficient of viscosity, \(\eta\) of the liquid i.e., \[V \propto \frac{1}{r^4}\] (iv) inversely proportional to the length , l of the capillary tube i.e., \[V \propto \frac{1}{l}\]
Combining these factors, we get \[\begin{align*} V &\propto \frac{Pr^4}{\eta l}\\ V&=\frac{kPr^4}{\eta l} \end{align*}\] Here, \(k=\frac{pi}{8}\) is the constant of proportionality.
\[\therefore V=\frac{\pi P r^4}{8 \eta l}\] This equation is called the poiseuille's equation.

Using dimensional consideration, deduce Poiseuille's formula for the rate of flow of liquid through the tube.
Poiseuille found that the volume of the liquid flowing per unit time through the tube depends on the pressure gradient \(\frac{P}{l}\), the coefficient of viscosity \(\eta\) and the radius r . If V be the volume flowing in time t , then, \[\begin{align*} V &\propto \frac{P}{l}^a \eta^b r^c\\ V&=k\frac{P}{l}^a \eta^b r^c... (i) \end{align*}\] Here, k is the proportionality constant (dimensionless).
Now, taking the dimensions on both sides,

\[\begin{align*} [L^3 T^{-1}]&=\left(\frac{[ML^{-1}T^{-2}]}{[L]}\right)^a [ML^{-1}T^{-1}]^b [L]^c\\ [L^3 T^{-1}]&=[M^{a+b}L^{-2a-b+c}T^{-2a-b}] \end{align*}\] Equating the exponents (i.e., powers ) of M, L and T, we get,
\[\begin{align*} 0&=a+b... (ii)\\ 3&=-2a-b+c... (iii)\\ -1&=-2a-b... (iv) \end{align*}\] Solving this equations,
From (ii),
\(a=-b\)
From (iv),
\(b=-2a+1\)
Then from this two equations, \[b=2b+1\implies b=-1\] So, a = 1 and c = 4

Thus,equation (i) becomes, \[V=k\frac{pr^4}{\eta l}\] Experimentally, it has been found that, \(k=\frac{\pi}{8}\) and thus the poiseuille's formula is, \[V=\frac{\pi P r^4}{8\eta l}\]

Stoke's law

When a small object, such as a small steel ball, is released in a viscous liquid like glycerine it accelerates at first, but its velocity soon reaches a steady value known as the terminal velocity . In this case, the viscous force, F , acting upwards, and the upthrust, U , due to the liquid on the object, are together equal to its weight, mg acting downwards, or F + U = mg . Since the resultant force on the object is zero it now moves with a constant (terminal) velocity. Figure below shows the variation of velocity with time as the terminal velocity v is reached.
Stoke found that the viscous force, F acting on a spherical body of radius r moving with terminal velocity v in a fluid of coefficient of viscosity \(\eta\) is, \[F=6\pi \eta r v\] This is called stoke's law .

Using dimensional analysis, deduce stoke's law.

Stoke found that the viscous force, F depends upon

• coefficient of viscosity, \(\eta\) of the medium,
• the terminal velocity, v of the body, and
• radius, r of the spherical body.

i.e., \[\begin{align*} F &\propto \eta^a v^b r^c\\ F&=k\eta^a v^b r^c ... (i) \end{align*}\] where k is dimensionless constant and a, b, and c are constants to be determined.

Now, taking the dimensions of both sides,
\[\begin{align*} [MLT^{-2}]&=[ML^{-1}T^{-1}]^a [LT^{-1}]^b [L]^c\\ [MLT^{-2}]&=[M^a L^{-a+b-c}T^{-a-b}] \end{align*}\] Equating the exponents on both sides (i.e., using principle of homogeneity),

a =1 , -a+b+c = 1 and -a-b = -2

From this three equations,
a = 1 , b = 1 and c = 1
Substituting this values in equation (i), we get, \[F=k\eta r v\] For all spheres, the value of k is found to be \(6\pi\) and the viscous force is, \[F=6\pi \eta r v\] This is stoke's law .

Describe the experimental method to find the coefficient of viscosity of a liquid by Stoke's method.

Fill a tall glass vessel with the liquid and drop a small steel ball P gently into the liquid so that it falls through the liquid as in figure. Towards the middle of the liquid P reaches a constant velocity i.e., terminal velocity v , which is measured by timing its fall through a distance AB or BC.
The downward force acting on the body is weight W and the upward forces are upthrust U and viscous force F . When P attains the terminal velocity, \[F+U=W ... (i)\] From stoke's law, \[F=6\pi \eta r v \] And, \[\begin{align*} W&=mg\\ &=\rho V g\\ &=\rho \frac{4}{3} \pi r^3 g\\ &=\frac{4}{3} \pi r^3 \rho g \end{align*}\] Here, V is the volume of P ,\(\rho\) is the density of P and r is its radius.

Also, \[\begin{align*} U&=m_f g\\ &=\sigma V g\\ &=\sigma \frac{4}{3}\pi r^3 g\\ &=\frac{4}{3}\pi r^3 \sigma g \end{align*}\] Here, \(m_f\) is the mass of the fluid displaced and \(\sigma\) is the density of the liquid.

From all these expressions, equation (i) becomes, \[\begin{align*} 6\pi \eta r v + \frac{4}{3}\pi r^3 \sigma g&=\frac{4}{3} \pi r^3 \rho g\\ 6\pi \eta r v&=\frac{4}{3} \pi r^3 \rho g-\frac{4}{3}\pi r^3 \sigma g\\ &=\frac{4}{3}\pi r^3 (\rho-\sigma)g\\ \eta&=\frac{2r^2(\rho-\sigma)g}{9v} \end{align*}\] This is the expression for the coefficient of viscosity of the liquid or fluid.

Find the terminal velocity of a rain drop of radius 0.01 mm. The coefficient of viscosity of air is 1.8 \(\times 10^{-5}\) Nsm\(^{-2}\) and its density is 1.2 kgm\(^{-3}\). Density of water = 1000 kgm\(^{-3}\). Take g = 10 ms\(^{-2}\).

When the rain drop falls through the air, the forces on it are

1. the weight W = \(\frac{4}{3}\pi r^3 \rho g\) downward,
2. upthrust U = \(\frac{4}{3}\pi r^3\sigma g\) upward,
3. viscous force F = \(6\pi \eta r v\) upward.

Here, \(\rho\) is the density of water and \(\sigma\) is the density of air.

At terminal velocity, \[F+U=W\] Since, the density of air is much smaller than the density of water, upthrust may be neglected. So,
\[\begin{align*} F&=W\\ 6\pi \eta r v &=\frac{4}{3} \pi r^3 \rho g\\ v&=\frac{2r^2 \rho g}{9\eta}\\ &=\frac{2\times (0.01\times 10^{-3})^{2} \times 1000 \times 10}{9 \times 1.8 \times 10^{-5}}\\ &=0.012 \hspace{0.1cm}ms^{-2} \end{align*}\] So, the terminal velocity of a raindrop is \(0.012 \hspace{0.1cm}ms^{-2}\).

A small oil drop falls with a terminal velocity of \(4.0 \times 10^{-4}\) m\(s^{-1}\) through air. Calculate the radius of the drop.
What is the new terminal velocity for an oil drop of half this radius? (Viscosity of air = \(1.8 \times 10^{-5}\) Nsm\(^{-2}\), density of oil = 900 kgm\(m^{-3}\), g = 10m\(s^{-2}\), neglect density of air.)

Follow the similar process as above to get the original radius \(r=1.9\times 10^{-6}\) m.
Now, when the oil drop is made half the original radius, i.e., \(r'=\frac{r}{2}\)
We know that, the terminal velocity\(\propto\) (radius)\(^2\). So,
\[\begin{align*} \frac{v'}{v}&=\left(\frac{r'}{r}\right)^2\\ v'&=v\times \frac{(r')^2}{(r)^2}\\ v'&=v \times \frac{1}{4}\\ &=4\times 10^{-4}\times \frac{1}{4}\\ \therefore v'&=1.0\times 10^{-4}\hspace{0.1cm} m/s \end{align*}\]

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed 5 cm/s. If they coalesce to form one big drop, what will be its terminal speed?

Let r be the radius of the small spherical drops with terminal velocity v and R be the radius of the big drop formed when the small drops coalesce with terminal velocity V . Let \(\rho\) be the density of rain drops.
For small drops,
At terminal velocity, \[\begin{align*} F&=W\\ 6\pi \eta r v&=\frac{4}{3}\pi r^3 \rho g ... (i) \end{align*}\] Since, eight spherical rain drops coalesce to form a single. We write, \[\begin{align*} \text{volume of big drop}&=8 \times \text{volume of small drops}\\ \frac{4}{3}\pi R^3&=8 \times \frac{4}{3} \pi r^3\\ R&=2r \end{align*}\] So, \[6\pi \eta R V=\frac{4}{3}\pi R^3 \rho g ... (ii)\] Dividing equation (ii) by (i), we get, \[\begin{align*} \frac{R}{r} \frac{V}{v}&=\left(\frac{R}{r}\right)^3\\ \frac{V}{v}&=\left(\frac{R}{r}\right)^2\\ V&=\left(\frac{2r}{r}\right)^2 \times v\\ \therefore V&=0.2 \hspace{0.1cm} m/s \end{align*}\]

An air bubble of diameter 2 mm rises steadily through a solution of the density 1750 kg \(m^{-3}\) at the rate of 0.35 cm/s. Calculate the coefficient of viscosity of the solution. The density of air is negligible.

Since, the air bubble is rising upward, the viscous force and weight is acting downward and the upthrust is upward.

As the density of air is negligible, we neglect the weight of the air.
So, At terminal velocity, \[\begin{align*} F&=U\\ 6\pi \eta r v&=\frac{4}{3}\pi r^3 \sigma g\\ \eta&=\frac{2r^2\sigma g}{9v}\\ &=\frac{2 \times (1\times 10^{-3})^2 1750 \times 10}{9 \times 0.35\times 10^{-2}}\\ \therefore \eta&= 1.11 \hspace{0.1cm} \text{Nsm}^{-2} \end{align*}\]

Equation of continuity

Consider the streamline flow of a non-viscous liquid on a pipe of varying cross sectional area \(a_1\) and \(a_2\) as in figure. Let \(v_1\) and \(\rho_1\) be the velocity of low of the liquid and density of the liquid respectively at point A of the pipe and, \(v_2\) and \(\rho_2\) be the corresponding values at point B of the pipe.
Volume of the liquid entering per second at A = \(a_1v_1\)
Remember! volume = area \(\times\) length and volume per second is \(\frac{\text{area} \times \text{length}}{\text{time}}\) and as \(\frac{\text{length}}{\text{time}}\) is velocity and thus we wrote \(a_1v_1\) in above expression.
Mass of liquid entering per second at A , \(m_A\) = \(a_1v_1\rho_1\)
Similarly, mass of liquid leaving per second at B , \(m_B\) = \(a_2v_2\rho_2\)
From the conservation of mass,
\[\begin{align*} m_A&=m_B\\ a_1v_1\rho_1&=a_2v_2\rho_2\\ a_1v_1&=a_2v_2 \end{align*}\] Here, we have considered the incompressible liquid i.e., \(\rho_1=\rho_2\).
We thus get, \[av=\text{constant}\] This is called the equation of continuity .

Bernoulli's theorem

It states that for the streamline flow of an ideal liquid, the total energy (i.e., the sum of the pressure energy, potential energy and kinetic energy) per unit mass remains constant at every cross-section throughout the flow. i.e., \[\frac{P}{\rho}+gh+\frac{v^2}{2}=\text{constant}\]

Let us take the entire volume of the ideal fluid as in figure. From the work-energy theorem,\[W=(K.E.)_f-(K.E.)_i ... (i)\] The change in kinetic energy results from the change in speed between the ends of the tube and is, \[(K.E.)_f-(K.E.)_i=\frac{1}{2}m v_2^2-\frac{1}{2}m v_1^2=\frac{1}{2}\rho V(v_2^2-v_1^2) ... (ii)\] Where \(m=\rho V\) is the mass of fluid transferred (i.e. mass of the fluid that enters at the input end and leaves at the output end) during a small time interval \(\Delta t\).

During the vertical lift of the fluid of mass m , the work done by the gravitational force is , \[\begin{align*} W_g&=-mg(h_2-h_1)\\ &=-\rho g V (h_2-h_1) \end{align*} ... (iii)\] This work is negative because the upward displacement and the downward gravitational force have opposite directions.

Work must be done on the system (at the input end) to push the entering fluid into the tube and by the system (at the output end) to push forward the fluid that is located ahead of the emerging fluid.
In general, the work done by the force of magnitude F , acting on a fluid sample contained in a tube of area A to move the liquid through a distance \(x\) is, \[Fx=(PA)(x)=P(Ax)=PV\] So, the work done on the system is \(P_1V\) and the work done by the system is -\(P_2V\). Their sum is, \[W_p=-P_2 V + P_1 V=-(P_2-P_1)V ... (iv)\] Then from (i), (ii), (iii) and (iv), we get, \[\begin{align*} W_g+W_p&=\frac{1}{2}\rho V(v_2^2-v_1^2)\\ -\rho g V (h_2-h_1)-(P_2-P_1)V&=\frac{1}{2}\rho V(v_2^2-v_1^2)\\ P_1+\frac{1}{2} \rho v_1^2+\rho g h_1&=P_2+\frac{1}{2}\rho v_2^2+\rho g h_2\\ P+\frac{1}{2}\rho v^2+\rho g h&=\text{constant} ... (v) \end{align*}\] This proves the Bernoulli's theorem .

For the liquid flowing through the horizontal pipe, i.e., \(h_1=h_2\), equation (v) becomes, \[P+\rho \frac{v^2}{2}=\text{constant}\]

Airports at high elevations have longer runways to take offs and landings than do airports at sea level, why?
The Bernoulli's equation is, \[\frac{P}{\rho}+\rho \frac{v^2}{2}+\rho g h=\text{constant}\]

With the increase in height, the air density becomes less. Due to this reason, the pressure difference between the convex and concave part of the wings is not quite sufficient to produce the lift of an aeroplane if the runways are same as that of sea level. To increase the thrust (i.e., pressure difference) plane necessitates the higher speed and thus require longer runway path in higher altitudes. However, at the sea level on a small run, the pressure difference is sufficient to lift an aeroplane.

Water flows steadily through a horizontal pipe of non-uniform cross section. If the pressure of water is \(4\times 10^4\) N \(m^{-2}\) at a point where the velocity of flow is 2 m/s and cross section is 0.02 \(m^2\), what is the pressure at a point where cross section reduces to 0.01 \(m^2\)?

Here, \(P_1= 4\times 10^4\) N \(m^{-2}\), \(v_1\) = 2 m/s , \(a_1\) = 0.02 \(m^2\), \(P_2\) = ?, \(v_2\) = ?, \(a_2\) = 0.01 \(m^2\)

From the equation of continuity, \[\begin{align*} a_1v_1&=a_2v_2\\ 0.02 \times 2&=0.01 \times v_2\\ v_2&=\frac{0.02 \times 2}{0.01}\\ v_2&=4 \hspace{0.1cm} \text{m/s} \end{align*}\]
From the bernoulli's equation, \[\begin{align*} P_1+\frac{\rho v_1^2}{2}&=P_2+\frac{\rho v_2^2}{2}\\ 4\times 10^4+\frac{1000 \times 2^2}{2}&=P_2+\frac{1000 \times 4^2}{2}\\ P_2&=4\times 10^4+2000-8000\\ &=34000\\ \therefore P_2&=3.4 \times 10^4 \hspace{0.1cm} \text{N/m}^2 \end{align*}\]

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