NEET Physics Class 11 Chapter 4 Elasticity And Viscosity Notes

Elasticity And Viscosity Solids

The materials having a definite shape and volume are known as solids. All solids have the property of elasticity by virtue of which solids behave as incompressible substances and exhibit rigidity and mechanical strength. Solids are classified into two categories namely Crystalline solids and amorphous solids (or glassy solids).

Crystalline Solids: A solid in which atoms or molecules are arranged in a regular three-dimensional pattern is known as a crystalline solid shown in figure (1) For example quartz, mica, sugar, copper sulphate, sodium chloride, potassium iodide, cesium chloride, carbon, etc.

Amorphous Solids: A solid in which atoms or molecules are not arranged in a regular manner is known as an amorphous solid shown in figure (2) For example: talc powder, glass, rubber, plastics, etc.

Unit Cell And Crystal Lattice

Unit cell is the building block of a crystal. It is defined as the smallest pattern of atoms in a lattice, the repetition of which in three dimensions forms a crystal lattice.

Crystal lattice: It is defined as a regular arrangement of a large number of points in space, each point representing the position of an atom or a group of atoms in a crystal. The crystal lattice is shown in the figure.

Elasticity And Plasticity

The property of a material body by virtue of which it regains its original configuration (i.e. shape and size) when the external deforming force is removed is called elasticity.

• The property of the material body by virtue of which it does not regain its original configuration when the external force is removed is called plasticity.

Deforming Force: An external force applied to a body that changes its size or shape or both is called a deforming force.

Perfectly Elastic Body: A body is said to be perfectly elastic if it completely regains its original form when the deforming force is removed. Since no material can regain completely its original form the concept of a perfectly elastic body is only an ideal concept. A quartz fiber is the nearest approach to the perfectly elastic body.

Perfectly Plastic Body: A body is said to be perfectly plastic if it does not regain its original form even slightly when the deforming force is removed.

• Since every material partially regains its original form on the removal of the deforming force, the concept of a perfectly plastic body is only an ideal concept. Paraffin wax and wet clay are the nearest approaches to perfectly plastic bodies.

Cause of Elasticity: In a solid, atoms and molecules are arranged in such a way that each atom/molecule is acted upon by the forces due to the neighboring atom/molecules. These forces are known as intermolecular forces.

• When no deforming force is applied to the body, each atom/molecule of the solid (i.e. body) is in its equilibrium position and the intermolecular forces between the molecules of the solid are zero.
• On applying the deforming force on the body, the molecules either come closer or go far apart from each other. As a result of this, the atoms/molecules are atoms displaced from their equilibrium position.
• In other words, equilibrium intermolecular forces get disturbed or changed, and restoring forces are developed on the molecules. When the deforming force is removed, these restoring forces bring the molecules of the solid to their respective equilibrium positions and hence the solid (or the body) regains its original form.

Stress

When deforming force is applied to the body then the upto a certain limit equal restoring force in the opposite direction is developed inside the body. The restoring forces per unit area developed in the body is called stress.

stress = \(=\frac{\text { restoring force }}{\text { Area of the body }}=\frac{F}{A}\)

The unit of stress is N/m² or Nm^-2. There are three types of stress

1. Longitudinal or Normal stress: When an object is one dimensional or linear then force acting per unit area is called longitudinal stress. It is of two types :

1. Compressive stress
2. Tensile stress

Longitudinal or Normal stress Examples:

Consider a block of solid as shown in the figure. Let a force F be applied to the face which has area A. Resolve \(\vec{F}\) into two components:

F_n= F sin θ is called normal force and F_t= F cos θ called tangential force.

∴ Normal (tensile) stress = \(\frac{F_{\mathrm{n}}}{\mathrm{A}}=\frac{\mathrm{F} \sin \theta}{\mathrm{A}}\)

2. Tangential or shear stress: It is defined as the restoring force acting per unit area tangential to the surface of the body.

Tangential (shear) stress = \(\frac{F_t}{A}=\frac{F \cos \theta}{A}\)

The effect of stress is to produce distortion or a change in size, volume, and shape
(i.e. configuration of the body).

3. Bulk stress or All-around stress or Pressure: When force F is acting all along the surface normal (ΔA) to the area, then force acting per unit area is known as pressure. The effect of pressure is to produce volume change. The shape of the body may or may not change depending upon the homogeneity of the body.

Question1. Find out longitudinal stress and tangential stress on a fixed block

Answer:

Longitudinal or normal stress ⇒ \(\sigma_{\mathrm{n}}=\frac{100 \sin 30^{\circ}}{5 \times 2}=5 \mathrm{~N} / \mathrm{m}^2\)

Tangential stress ⇒ \(\sigma_{\mathrm{t}}=\frac{100 \cos 30^{\circ}}{5 \times 2}=5 \sqrt{3} \mathrm{~N} / \mathrm{m}^2\)

Question 2. Find out Bulk stress on the spherical object of radius\(\frac{10}{\pi}\) cm if the area and mass of the piston are 50 cm² and 50 kg respectively for a cylinder filled with gas. 

Answer:

⇒ \(p_{\text {gas }}=\frac{m g}{A}+p_a=\frac{50 \times 10}{50 \times 10^{-4}}+1 \times 10^5\)

⇒ \(2 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)

Bulk stress = p_gas= 2 × 10⁵ N/m²

Strain

The ratio of the change in configuration (i.e. shape, length, or volume) to the original configuration of the body is called strain

i.e. Strain,\(\epsilon=\frac{\text { change in configuration }}{\text { original configuration }}\)

Types Of Strain: There are three types of strain

Longitudinal Strain: This type of strain is produced when the deforming force causes a change in the length of the body. It is defined as the ratio of the change in length to the original length of the body.

Consider A Wire Of Length L: When the wire is stretched by a force F, then let the change in length of the wire be ΔL as shown in the Figure.

∴ Longitudinal strain, \(\epsilon_{\ell}=\frac{\text { change in length }}{\text { original length }}\)

or Longitudinal strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\)

Volume Strain: This type of strain is produced when the deforming force produces a change in the volume of the body as shown in the Figure. It is defined as the ratio of the change in volume to the original volume of the body.

If the upper face is displaced through x keeping the lower face fixed at a distance of l then

If ΔV = change in volume

V = original volume

Δ tangentialdisplacement

⇒ \(\epsilon_{\mathrm{v}}=\text { volume strain }\)

⇒ \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\text { tangential displacement }}{\text { transverse distance }}\)

Shear Strain: This type of strain is produced when the deforming force causes a change in the shape of the body. It is defined as the angle (θ) through which a face originally perpendicular to the fixed face is turned as shown in the Figure.

⇒ \(\tan \phi \text { or } \phi=\frac{\mathrm{x}}{\ell}\)

Hooke’s Law And Modulus Of Elasticity

According to this law, within the elastic limit, stress is proportional to the strain.

i.e. stress ∝ strain

or stress = constant × strain or \(\frac{\text { stress }}{\text { strain }}=\text { Modulus of Elasticity. }\)

This Constant Is Called The Modulus Of Elasticity.

Thus, the modulus of elasticity is defined as the ratio of the stress to the strain.

The modulus of elasticity depends on the nature of the material of the body and is independent of its dimensions (i.e. length, volume, etc.).

Unit: The Sl unit of modulus of elasticity is Nm^-2 or Pascal (Pa).

Types Of Modulus Of Elasticity

Corresponding to the three types of strain there are three types of modulus of elasticity.

1. Young’s modulus of elasticity (Y)
2. Bulk modulus of elasticity (B)
3. Modulus of rigidity (η).

Young’s modulus of elasticity

It is defined as the ratio of the normal stress to the longitudinal strain.

i.e. Young’s modulus (Y) = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

Normal stress = F/A,

Longitudinal strain = ΔL/L

Y = \(Y=\frac{F / A}{\Delta L / L}=\frac{F L}{A \Delta L}\)

Question 1. Find out the shift in points B, C, and D in the compound wire shown in Figure.

Answer:

⇒ \(\Delta L_B=\Delta L_{A B}=\frac{F L}{A Y}=\frac{M g L}{A Y}\)

⇒ \(\frac{10 \times 10 \times 0.1}{10^{-7} \times 2.5 \times 10^{10}}=4 \times 10^{-3} \mathrm{~m}=4 \mathrm{~mm}\)

⇒ \(\Delta \mathrm{L}_{\mathrm{c}}=\Delta \mathrm{L}_{\mathrm{B}}+\Delta \mathrm{L}_{\mathrm{BC}}=4 \times 10^{-3}+\frac{100 \times 0.2}{10^{-7} \times 4 \times 10^{10}}\)

⇒ \(4 \times 10^{-3}+5 \times 10^{-3}=9 \mathrm{~mm}\)

⇒ \(\Delta \mathrm{L}_{\mathrm{D}}=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{CD}}=9 \times 10^{-3}+\frac{100 \times 0.15}{10^{-7} \times 1 \times 10^{10}}\)

⇒ \(9 \times 10^{-3}+15 \times 10^{-3}=24 \mathrm{~mm}\)

Elongation Of Rod Under Its Self Weight

Let rod has a self-weight ‘W’, area of cross-section ‘A” and length ‘L’. Considering on element of length dx at a distance ‘x’ from the bottom. then \(\mathrm{T}=\frac{\mathrm{W}}{\mathrm{L}} \mathrm{x}\) is down word force on dx elongation in ‘dx’ element = \(\frac{\text { T.dx }}{\mathrm{AY}}\)

Total elongation s = \(\int_0^L \frac{T d x}{A Y}=\int_0^L \frac{W x d x}{L A Y}=\frac{W L}{2 A Y}\)

Note: One can do this directly by considering total weight at C.M. and using effective length L/2.

Question 1. Find out the elongation in the block shown in Figure. If mass, area of cross-section, and Young modulus of a block are m, A, and Y respectively.

Answer:

Acceleration, a = \(\frac{T d x}{A Y}\) then T = m′a

where ⇒ m′ = \(\frac{m}{L} x\)

T = \(\frac{m}{L} \times \frac{F}{m}=\frac{F x}{L}\)

Elongation in element ‘dx’ = \(\frac{T d x}{A Y}\) total elongation, δ = \(\int_0^L \frac{T d x}{A Y}=\int_0^L \frac{F x d x}{A L Y}=\frac{F L}{2 A Y}\)

Note: Try this problem, if friction is given between block and surface (µ = friction coefficient), and Case :

F < µmg

F > µmg

In both cases answer will be \(\frac{F L}{2 A Y}\)

Bulk Modulus:

It is defined as the ratio of the normal stress to the volume strain i.e. B = \(\frac{\text { Pressure }}{\text { Volume strain }}\)

The stress is the normal force applied per unit area and is equal to the pressure applied (p). p

⇒ \(B=\frac{p}{\frac{-\Delta V}{V}}=-\frac{p V}{\Delta V}\)

A negative sign shows that an increase in pressure (p) causes a decrease in volume (ΔV).

Compressibility: The reciprocal of bulk modulus of elasticity is called compressibility K. The Unit of compressibility in Sl is N^-1 m² or pascal^-1(Pa^-1).

The bulk modulus of solids is about fifty times that of liquids, and for gases it is 10–8 times of solids.

B_Solids > B_liquids > B_gases

Isothermal modulus of elasticity of gas B = P (pressure of gas)

Adiabatic modulus of elasticity of gas B = γ × P where γ = \(\) = ratio of specific beats.

Modulus of Rigidity:

It is defined as the ratio of the tangential stress to the shear strain. Let us consider a cube whose lower face is fixed and a tangential force F acts on the upper face whose area is A.

∴ Tangential stress = F/A.

Let the vertical sides of the cube shift through an angle θ, called shear strain

∴ The modulus of rigidity is given by

η = \(\frac{\text { Tangential stress }}{\text { Shear strain }}\)

or \(\eta=\frac{F / A}{\phi}=\frac{F}{A \phi}\)

Question 2. A rubber cube of side 5 cm has one side fixed while a tangential force equal to 1800 N is applied to the opposite face to find the shearing strain and the lateral displacement of the strained face. The modulus of rigidity for rubber is 2.4 × 10⁶ N/m².
Answer:

L = 5 × 10^-2 m ⇒ \(\frac{F}{A}=\eta \frac{x}{L}\)

Strain θ = \(\frac{F}{A \eta}\)

⇒ \(\frac{1800}{25 \times 10^{-4} \times 2.4 \times 10^6}\)

⇒ \(\frac{180}{25 \times 24}\)

⇒ \(\frac{3}{10}\)

=0.3 radian

⇒ \(\frac{x}{L}=0.3\)

x = 0.3 × 5 × 10^-2

= 1.5 × 10^-2 m

= 1.5 mm

Variation Of Strain With Stress

When a wire is stretched by a load, it is seen that for a small value of the load, the extension produced in the wire is proportional to the load. On removing the load, the wire returns to its original length.

• The wire regains its original dimensions only when the load applied is less or equal to a certain limit. This limit is called the elastic limit.
• Thus, the elastic limit is the maximum stress on whose removal, the bodies regain their original dimensions.
• In the shown figure, this type of behavior is represented by the OB portion of the graph. Till A the stress is proportional to strain and from A to B if deforming forces are removed then the wire comes to its original length but here stress is not proportional to strain.

OA → Limit of Proportionality

OB → Elastic limit

C → Yield Point

CD → Plastic behavior

D → Ultimate point

DE → Fracture

As we go beyond point B, then even for a very small increase in stress, the strain produced is very large. This type of behaviour is observed around point C and at this stage the wire begins to flow like a viscous fluid.

• The point C is called yield point. If the stress is further increased, then the wire breaks off at point D called the breaking point.
• The stress corresponding to this point is called breaking stress or tensile strength of the material of the wire. A material for which the plastic range CD is relatively high is called ductile material.
• These materials get permanently deformed before breaking. The materials for which the plastic range is relatively small are called brittle materials. These materials break as soon as the elastic limit is crossed.

Important Points

• Breaking stress = Breaking force/area of cross-section.
• Breaking stress is constant for a material
• Breaking force depends upon the area of the section of the wire of a given material.
• The working stress is always kept lower than that of breaking stress so the safety factor = breaking stress/working stress may have a large value.
• Breaking strain = elongation or compression/original dimension.
• Breaking strain is constant for the material.

Elastic After Effect

We know that some material bodies take some time to regain their original configuration when the deforming force is removed. The delay in regaining the original configuration by the bodies on the removal of deforming force is called elastic after effect.

• The elastic after-effect is negligibly small for quartz fiber and phosphor bronze. For this reason, the suspensions made from quartz and phosphor bronze are used in galvanometers and electrometers.
• For glass fibre elastic after effect is very large. It takes hours for glass fiber to return to its original state on removal of a deforming force.

Elastic Fatigue

The loss of strength of the material due to repeated strains on the material is called elastic fatigue. That is why bridges are declared unsafe after a long time of their use.

Analogy Of Rod As A Spring

⇒ \(\mathrm{Y}=\frac{\text { stress }}{\text { strain }}\)

⇒ \(\mathrm{Y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}\)

or \(\mathrm{F}=\frac{\mathrm{AY}}{\ell} \Delta \ell\)

⇒ \(\frac{\mathrm{AY}}{\ell}\) = constant, depending on the type of material and geometry of the rod. F = kΔl

where k = \(\frac{\mathrm{AY}}{\ell}\) = equivalent spring constant.

for the system of rods shown in the figure

1. The replaced spring system is shown in the figure
2. Two spring in series]. Figure
3. Represents an equivalent spring system.

Figure (4) represents another combination of rods and their replaced spring system.

Question 1. A mass ‘m’ is attached with rods as shown in the figure. This mass is slightly stretched and released whether the motion of the mass is S.H.M., if yes then find out the time period.

Answer:

⇒ \(k_{e q}=\frac{k_1 k_2}{k_1+k_2}\)

where \(\mathrm{k}_1=\frac{\mathrm{A}_1 \mathrm{Y}_1}{\ell_1}\)

⇒ \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{\mathrm{eq}}}}\)

and \(\mathrm{k}_2=\frac{\mathrm{A}_2 \mathrm{Y}_2}{\ell_2}\)

Elastic Potential Energy Stored In A Stretched Wire Or In A Rod

Strain energy stored in equivalent spring

U = \(\frac{1}{2} k x^2\)

where x = \(\frac{\mathrm{F} \ell}{\mathrm{AY}}\)

k = \(\frac{\mathrm{AY}}{\ell}\)

U = \(\frac{1}{2} \frac{A Y}{\ell} \frac{F^2 \ell^2}{A^2 Y^2}\)

⇒ \(\frac{1}{2} \frac{F^2 \ell}{A Y}\)

equation can be re-arranged

⇒ \(\mathrm{U}=\frac{1}{2} \frac{\mathrm{F}^2}{\mathrm{~A}^2} \times \frac{\ell \mathrm{A}}{\mathrm{Y}}\) [lA = volume of rod, F/A = stress]

⇒ \(\mathrm{U}=\frac{1}{2} \frac{(\text { stress) })^2}{\mathrm{Y}} \times \text { volume }\) again, U = \(\frac{1}{2} \frac{\mathrm{F}}{\mathrm{A}} \times \frac{\mathrm{F}}{\mathrm{AY}} \times \mathrm{A} \ell\)

[Stain = \(\frac{\mathrm{F}}{\mathrm{AY}}\)]

⇒ \(\mathrm{U}=\frac{1}{2} \text { stress } \times \text { strain } \times \text { volume }\) again, U = \(\frac{1}{2} \frac{F^2}{A^2 Y^2} A \ell Y\)

⇒ \(U=\frac{1}{2} Y(\text { strain })^2 \times \text { volume }\)

strain energy density ⇒ \(\frac{\text { strain energy }}{\text { volume }}\)

⇒ \(\frac{1}{2} \frac{\text { (stress) }^2}{Y}\)

⇒ \(\frac{1}{2} Y(\text { strain })^2\)

⇒ \(\frac{1}{2} \text { stress } \times \text { strain }\)

Posson’s Ratio (σ)

Within the elastic limit, the ratio between the lateral strain and the linear strain is a constant. This constant is called Poisson’s ratio.

σ = \(\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Lateral Strain: Change in length of the body from its initial length perpendicular to deforming force lateral strain = \(-\frac{\Delta D}{D}=\frac{D-(D-\Delta D)}{D}=\beta\)

Longitudinal Strain: Change in length of the body from its initial length in the direction of deforming force

Longitudinal strain \(\varepsilon_{\ell}=\frac{\text { change in length }}{\text { original length }}=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha\)

⇒ \(\sigma=\frac{\beta}{\alpha}\)

theoretical limit of σ –1 < σ < 0.5

The experimental value of σ lies between 0.2 and 0.4

Four important Relations between Y, B, η and σ

1. \(\eta=\frac{Y}{2(1+\sigma)}\)
2. \(\frac{9}{Y}=\frac{3}{\eta}+\frac{1}{B}\)
3. \(\sigma=\frac{3 B-2 \eta}{6 B+2 \eta}\)
4. \(B=\frac{Y}{3(1-2 \sigma)}\)

Thermal Stress :

If the temperature of the rod is increased by ΔT, then the change in length

Δl = l α ΔT strain = \(\frac{\text { stress }}{\text { strain }}\)

But due to rigid support, there is no strain. Supports provide force on stresses to keep the length of the rod the same

Y = \(\frac{\text { stress }}{\text { strain }}\)

thermal stress = Y strain = Y α ΔT

⇒ \(\frac{F}{A}=Y \alpha \Delta T\)

F = AY α ΔT

Applications Of Elasticity

Some of the important applications of the elasticity of the materials are discussed below:

1. The material used in bridges loses its elastic strength with time, and bridges are declared unsafe after long use.
2. To estimate the maximum height of a mountain :

The pressure at the base of the mountain = hρg = stress. The elastic limit of a typical rock is 3 × 10⁸ N m^-2

The stress must be less than the elastic limits, otherwise the rock begins to flow.

⇒ \(h<\frac{3 \times 10^8}{\rho g}<\frac{3 \times 10^8}{3 \times 10^3 \times 10}<10^4 \mathrm{~m}\)

∵ \(\left(\rho=3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} ; g=10 \mathrm{~ms}^{-2}\right)\)

h = 10km

It may be noted that the height of Mount Everest is nearly 9 km.

Torsion Constant Of A Wire

C = \(\frac{\pi \eta r^4}{2 \ell}\) Where η is the modulus of rigidity r and l are the radius and length of the wire respectively.

1. Toque required for twisting by angle θ, τ = Cθ.
2. Work done in twisting by angle θ, W = \(\frac{1}{2} C \theta^2\)

Viscosity

When a solid body slides over another solid body, a frictional force begins to act between them. This force opposes the relative motion of the bodies.

• Similarly, when a layer of a liquid slides over another layer of the same liquid, a frictional force acts between them which opposes the relative motion between the layers.
• This force is called ‘internal frictional force’. This is due to intermolecular forces. Suppose a liquid is flowing in streamlined motion on a fixed horizontal surface AB.
• The layer of the liquid that is in contact with the surface is at rest, while the velocity of other layers increases with distance from the fixed surface. In the Figure, the lengths of the arrows represent the increasing velocity of the layers.
• Thus there is a relative motion between adjacent layers of the liquid. Let us consider three parallel layers a, b, and c. Their velocities are in increasing order. The layer a tends to retard the layer b, while b tends to retard c.
• Thus each layer tends to decrease the velocity of the layer above it. Similarly, each layer tends to increase the velocity of the layer below it.
• This means that in between any two layers of the liquid, internal tangential forces act which try to destroy the relative motion between the layers. These forces are called ‘viscous forces’.
• If the flow of the liquid is to be maintained, an external force must be applied to overcome the dragging viscous forces. In the absence of the external force, the viscous forces would soon bring the liquid to rest.
• The property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is known as ‘viscosity’.

The property of viscosity is seen in the following examples :

A stirred liquid, when left, comes to rest on account of viscosity. Thicker liquids like honey, coaltar, glycerine, etc. have a larger viscosity than thinner ones like water.

• If we pour coaltar and water on a table, the coaltar will stop flowing soon while the water will flow upto quite a large distance.
• If we pour water and honey in separate funnels, water comes out readily from the hole in the funnel while honey takes enough time to do so.
• This is because honey is much more viscous than water. As honey tends to flow down under gravity, the relative motion between its layers is opposed strongly.
• We can walk fast in the air, but not in water. The reason is again viscosity which is very small for air but comparatively much larger for water.
• The cloud particles fall down very slowly because of the viscosity of air and hence appear floating in the sky.
• Viscosity comes into play only when there is a relative motion between the layers of the same material. This is why it does not act in solids.

Flow Of Liquid In A Tube Critical Velocity

When a liquid flows ‘in a tube, the viscous forces oppose the flow of the liquid, Hence a pressure difference is applied between the ends of the tube which maintains the flow of the liquid.

• If all particles of the liquid passing through a particular point in the tube move along the same path, the flow” of the liquid is called ‘stream-lined flow’.
• This occurs only when the velocity of flow of the liquid is below a certain limiting value called ‘critical velocity’. When the velocity of flow exceeds the critical velocity, the flow is no longer stream-lined but becomes turbulent.
• In this type of flow, the motion of the liquid becomes zig-zag, and eddy-currents are developed in it as shown in Figure.

• Reynold observed that the critical velocity for a liquid flowing in a tube is vc= Rη/ρa. where ρ is density η is the viscosity of the liquid, a is the radius of the tube and R is ‘Reynold’s number’ (whose value for a narrow tube and for water is about 1000).
• When the velocity of the flow of the liquid is less than the critical velocity, then the flow of the liquid is controlled by the viscosity, the density having no effect on it.
• But when the velocity of flow is larger than the critical velocity, then the flow is mainly governed by the density, and the effect of viscosity becomes less important.
• It is because of this reason that when a volcano erupts, then the lava coming out of it flows speedily in spite of being very thick (of large viscosity).

Velocity Gradient And Coefficient Of Viscosity

The property of a liquid by virtue of which an opposing force (internal friction) comes into play whenever there is a relative motion between the different layers of the liquid is called viscosity.

Consider a flow of a liquid over the horizontal solid surface as shown in Figure. Let us consider two layers AB and CD moving with velocities \(\vec{v}\)and \([\vec{v}+d \vec{v}/latex] a distance z and (z + dz) respectively from the fixed solid surface.

According to Newton, the viscous drag or backward force (F) between these layers happens to be.

1. Directly proportional to the area (A) of the layer and
2. Directly proportional to the velocity gradient [latex]\left(\frac{d v}{d x}\right)\) between the layers. Defined as a change in velocity per unit perpendicular separation between the layers. Thus

⇒ \(F \propto A \frac{d v}{d z} \text { or } F=-\eta A \frac{d v}{d z}\)……..(1)

η is called the Coefficient of viscosity. A negative sign shows that the direction of viscous drag (F) is just opposite to the direction of the motion of the liquid.

Similarities And Differences Between Viscosity And Solid Friction

Similarities: Viscosity and solid friction are similar as

1. Both oppose the relative motion. Whereas viscosity opposes the relative motion between two adjacent liquid layers, solid friction opposes the relative motion between two solid layers.
2. Both come into play, whenever there is relative motion between layers of liquid or solid surfaces as the case may be.
3. Both are due to molecular attractions.

Differences between them are given below

Some Applications Of Viscosity

Knowledge of the viscosity of various liquids and gases has been put to use in daily life. Some applications of its knowledge are discussed as under:

1. As the viscosity of liquids varies with temperature, the proper choice of lubricant is made depending on season.
2. Liquids of high viscosity are used in shock absorbers and buffers at railway stations.
3. The phenomenon of viscosity of air and liquid is used to dampen the motion of some instruments.
4. The knowledge of the coefficient of viscosity of organic liquids is used in determining the molecular weight and shape of the organic molecules.
5. It finds an important use in the circulation of blood through arteries and veins of the human body.

Units Of Coefficient Of Viscosity

From the above formula, we have \(\eta=\frac{F}{A\left(\Delta v_x / \Delta z\right)}\)

∴ dimensions of \(\eta=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]\left[\mathrm{LT}^{-1} / \mathrm{L}\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2 \mathrm{~T}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\)

Its unit is kg/(meter-second)

In the C.G.S. system, the unit of coefficient of viscosity is dyne s cm–2 and is called poise. In SI the unit of coefficient of viscosity is N sm–2 and is called decompose.

1 decapoise = 1 N sm^-2 = (10⁵ dyne) × s × (10² cm)^-2 = 10 dyne s cm^-2 = 10 poise

Question 1. A man is rowing a boat with a constant velocity ‘v₀’ in a river the contact area of the boat is ‘A’ and the coefficient of viscosity is η. The depth of the river is ‘D’. Find the force required to row the boat.
Answer:

F – F_T= m a_res

As the boat moves with constant velocity a_res = 0

F = F_T

But \(F_T=\eta A \frac{d v}{d z}\)

but \(\frac{d v}{d z}=\frac{v_0-0}{D}=\frac{v_0}{D}\)

then F = F_T = \(\frac{\eta A v_0}{D}\)

Question 2. A cubical block (of side 2m) of mass 20 kg slides on an inclined plane lubricated with the oil of viscosity η = 10^-1 poise with a constant velocity of 10 m/sec. (g = 10 m/sec²) find out the thickness of a layer of liquid.

Answer:

F = F’\(\eta A \frac{d v}{d z}=m g \sin \theta ; \frac{d v}{d z}=\frac{v}{h}\)

∴ 20 × 10 × sin 30° = η × 4 × \(\frac{10}{h}\)

⇒ \(\frac{40 \times 10^{-2}}{100}\) – [η = 10^-1 poise = 10^-2 N-sec-m^-2] = 4 × 10^-3 m = 4 mm

Effect Of Temperature On The Viscosity

The viscosity of liquids decreases with an increase in temperature and increases with a decrease in temperature. That is, \(\eta \propto \frac{1}{\sqrt{T}}\) T. On the other hand, the value of viscosity of gases increases with the increase in temperature and vice-versa. That is, η ∝ T.

Stoke’S Law

Stokes proved that the viscous drag (F) on a spherical body of radius r moving with velocity v in a fluid of viscosity η is given by F = 6 π η r v. This is called Stokes’ law.

Terminal Velocity

When a body is dropped in a viscous fluid, it is first accelerated. As its velocity increases, viscous force also increases so ultimately its acceleration becomes zero and it attains a constant velocity called terminal velocity.

Calculation Of Terminal Velocity

Let us consider a small ball, whose radius is r and density is ρ, is falling freely in a liquid (or gas), whose density is σ and coefficient of viscosity η. When it attains a terminal velocity v. It is subjected to two forces:

1. Effective weight force acting downward

⇒ \(V(\rho-\sigma) g=\frac{4}{3} \pi r^3(\rho-\sigma) g\)

2. A viscous force acting upward

= 6 π η rv.

Since the ball is moving with a constant velocity v i.e., there is no acceleration in it, the net force acting on it must be zero. That is

6πηrv = \(\frac{4}{3} p r^3(\rho-\sigma) g\)

or \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

Thus, the terminal velocity of the ball is directly proportional to the square of its radius Important point

The air bubble in water always goes up. This is because the density of air (ρ) is less than the density of water (σ).

So the terminal velocity for the air bubble is Negative, which implies that the air bubble will go up. Positive terminal velocity means the body will fall down.

Question 1. A spherical ball is moving with terminal velocity inside a liquid. Determine the relationship between the rate of heat loss with the radius of the ball.
Answer:

Rate of heat loss = power = F × v = 6 π η r v × v = 6 π η r v2 = 6p η r \(\left[\frac{2}{9} \frac{g r^2\left(\rho_0-\rho_{\ell}\right)}{\eta}\right]^2\)

Rate of heat loss α r⁵

Question 2. A drop of water of radius 0.0015 mm is falling in the air. If the coefficient of viscosity of air is 1.8 × 10^-5 kg /(m-s), what will be the terminal velocity of the drop? (density of water = 1.0 × 10³ kg/m² and g = 9.8 N/kg.) The density of air can be neglected.
Answer:

By Stoke’s law, the terminal velocity of a water drop of radius r is given by 2

⇒ \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

where ρ is the density of water, σ is the density of air, and η the coefficient of viscosity of air. Here σ is negligible and r = 0.0015 mm = 1.5 × 10^-3 mm = 1.5 × 10^-6 m. Substituting the values:

⇒ \(v=\frac{2}{9} \times \frac{\left(1.5 \times 10^{-6}\right)^2 \times\left(1.0 \times 10^3\right) \times 9.8}{1.8 \times 10^{-5}}\)

= 2.72×10^-4

Question 3. A metallic sphere of radius 1.0 × 10^-3 m and density 1.0 × 10⁴ kg/m³ enters a tank of water, after a free fall through a distance of h in the earth’s gravitational field. If its velocity remains unchanged after entering water, determine the value of h. Given : coefficient of viscosity of water = 1.0 × 10^-3 N-s/m², g = 10 m/s² and density of water = 1.0 × 10³ kg/m³.
Answer:

The velocity attained by the sphere in falling freely from a height h is

ν = \(\sqrt{2 \mathrm{gh}}\)….(1)

This is the terminal velocity of the sphere in water. Hence by Stoke’s law, we have 2

ν = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)…….(2)

where r is the radius of the sphere, ρ is the density of the material of the sphere σ (= 1.0 × 10³ kg/m³) is the density of water and η is the coefficient of viscosity of water.

∴ ν = \(\frac{2 \times\left(1.0 \times 10^{-3}\right)^2\left(1.0 \times 10^4-1.0 \times 10^3\right) \times 10}{9 \times 1.0 \times 10^{-3}}\)

= 20 m/s

from equation (1), we have h = \(\frac{v^2}{2 g}=\frac{20 \times 20}{2 \times 10}\)

= 20 m

Applications of Stokes’ Formula

In determining the Electronic Charge by Millikan’s Oil Drop Experiment: Stokes’ formula is used in Millikan’s method for determining the electronic charge. In this method, the formula is applied to find out the radii of small oil drops by measuring their terminal velocity in the air.

The velocity of Raindrops: Raindrops are formed by the condensation of water vapor on dust particles. When they fall under gravity, their motion is opposed by the viscous drag in the air.

• As the velocity of their fall increases, the viscous drag also increases and finally becomes equal to the effective force of gravity.
• The drops then attain a (constant) terminal velocity which is directly proportional to the square of the radius of the drops.
• In the beginning, the raindrops are very small in size and so they fall with such a small velocity that they appear floating in the sky as clouds. As they grow in size by further condensation, then they reach the earth with appreciable velocity,

Parachute: When a soldier with a parachute jumps from a flying airplane, he descends very slowly in the air.

• In the beginning, the soldier falls with gravity acceleration g, but soon the acceleration goes on decreasing rapidly when a parachute is fully opened.
• Therefore, in the beginning, the speed of the falling soldier increases somewhat rapidly but then very slowly.
• Due to the viscosity of air, the acceleration of the soldier becomes ultimately zero and the soldier then falls with a constant terminal speed. The figure shows the speed of the falling soldier with time.