Elasticity Class 11 Physics | Notes

Elasticity 

Elasticity is the property of the body by virtue of which a deformed body regains its original shape, size and position after the removal of deforming force. 

▪ The elasticity of the object generally decreases with increase in temperature.

▪ The elasticity of molten cadmium (Z = 48) and molten copper increases with increase in temperature.

▪ The elasticity of invar is unaffected with the change in temperature.

Deforming force 

A force is said to be a deforming force if it can change shape, size and position of the object.

Restoring force 

A force is said to be restoring force if it brings the object back to its original shape, size and position after the removal of deforming force.

Types of object on the basis of elasticity

1. Elastic object:

An object is said to be elastic if it regains its original shape, size and position after the removal of deforming force.

For example; quartz, rubber, steel, iron etc.

2. Plastic object:

An object is said to be plastic if it doesn’t regain its shape, size and collision after the removal of deforming force.

For example: polythene, wax, etc.

3. Rigid object:

An object is said to be rigid if it doesn’t get reformed by an external force

For example: bricks, stones, etc.

Stress:

Stress is defined as the deforming force per unit area.

Mathematically, 

Stress = $\frac{Deforming\,force}{Area}$

It’s unit is N/m2.

Types of stress

1. Normal stress:

A stress is said to be normal stress if the deforming force is perpendicular to the area of cross section. Normal stress changes the shape and size of the object.

Types of normal stress:

a. Tensile stress: 

Tensile stress is a type of normal stress which can increase the length of the object.

elasticity notes class 11

b. Compression stress: 

Compression stress is a type of normal stress which can decrease the length of the object.

elasticity notes class 11

2. Tangential stress:

A stress is said to be tangential if the deforming force is acting in the direction of the area of cross-section (tangentially). It cannot change shape and size but changes the position of the object.

elasticity notes class 11

Strain:

Strain is defined as the change in configuration per unit original configuration. (Configuration means length, volume, angle, etc.). Mathematically,

Strain = $\frac{Change\,\,in\,\,configuration}{Original\,\,configuration}$

Since it is the ratio of the same physical quantity so it has no unit.

Types of strain

1. Longitudinal strain:

Longitudinal stress is defined as the change in length per unit original length.

Mathematically,

Longitudinal strain = $\frac{Change\,\,in\,\,length}{Original\,\,length}$

elasticity notes class 11

Longitudinal strain = $\frac{\Delta l}{l}$

Longitudinal strain = $\frac{e}{l}$

Where ‘e’ stands for elongation

2. Volumetric strain:

Volumetric strain is defined as the change in volume per unit original volume.

Volumetric strain = $\frac{Change\,\,in\,\,volume}{Original\,\,volume}$

Volumetric strain = $\frac{\Delta V}{V}$

elasticity notes class 11

3. Shear strain:

Shear strain is defined as the angle through which an object is sheared with the application of tangential stress.

elasticity notes class 11

From figure, tan$\theta $ = $\frac{x}{l}$

For very small angle ($\theta $), tan$\theta $ is nearly equal to$\theta $

Therefore, shear strain ($\theta $) = $\frac{x}{l}$

Stress-strain curve:

elasticity notes class 11

Let us consider a uniform wire suspended at it’s one end on the other end attached to a scale pan. The weights on the pan are gradually increased and corresponding elongations are noted. A graph of stress and strain is plotted as shown in the figure above. Here OA is a straight line and AB is a curve, ‘A’ is called proportional limit, within the proportional limit stress is directly proportional to strain and B is called elastic limit, within the elastic limit the object retains elastic property. Beyond the elastic limit (point B), the object loses its elasticity i.e. the object cannot regain its original shape, size and position after the removal of deforming force.

Hooke’s law:

Hooke’s law states that within the proportional limit, stress is directly proportional to strain. i.e.

  Stress $\propto $ Strain

Or, Stress = E strain (where ‘E’ is proportionality constant and known as modulus of elasticity.)

∴ Modulus of elasticity (E) = $\frac{Stress}{Strain}$

It’s unit is Nm–2.

Experimental verification of Hooke’s Law:

Elasticity Class 11 Physics | Notes
Fig: Verification of Hooke’s law

The experimental setup for the verification of Hooke’s law is shown above. It consists of an experimental wire, vernier-scale, a hanger with slotted weights. The weights on the hanger are gradually increased and corresponding elongations of wire are noted.

elasticity notes class 11

When a graph of different stress versus corresponding strain is plotted, a straight line passing through origin is obtained. This shows that stress is directly proportional to the strain which verifies Hooke’s law.

Types of modulus of elasticity

(i) Young’s modulus of electricity (Y):

Young’s modulus of electricity is defined as the ratio of normal stress to the longitudinal strain. It is denoted by ‘Y’ i.e.

Y = $\frac{Normal\,\,stress}{Longitudinal\,\,strain}$

elasticity notes class 11

Y  = $\frac{F/A}{e/l}$

Y = $\frac{Fl}{eA}$

It’s unit is Nm–2.

Since Young’s modulus of elasticity ‘Y’ is the property of material and is independent of external dimensions of the wire. 

Note: 

Young’s modulus of elasticity is given by

Y = $\frac{Fl}{eA}$

For same F, l and A, we have

Y $\propto $ $\frac{1}{e}$

i.e., Young’s modulus is inversely proportional to elongation.

For example; rubber is less elastic than Steel because elongation produced in the rubber is more than that in steel. So, Young’s modulus of rubber is less. Hence rubber is less elastic. Similarly, elongation produced in steel is more than that in rubber. So, Young’s modulus of steel is more. Hence steel is more elastic than rubber.

Determination of young’s modulus of elasticity ‘Y’

elasticity notes class 11

Let’s us consider a wire of length ‘l’ and cross-sectional area ‘A’ is suspended from a rigid support. A force is applied to the free end of the wire to produce an elongation ‘e’ as shown in figure above. Let ‘Y’ be the Young’s modulus of elasticity of the given wire then,

Normal stress = $\frac{Force}{Area}$ = $\frac{F}{A}$

And longitudinal strain = $\frac{Change\,\,in\,\,length}{Original\,\,length}$ = $\frac{e}{l}$

Young’s modulus of elasticity ‘Y’ =  $\frac{Normal\,\,stress}{Longitudinal\,\,strain}$

Or, Y = $\frac{F/A}{e/l}$

$\therefore $ Y = $\frac{\mathbf{Fl}}{\mathbf{eA}}$

Measuring the values of Force (F), original length (l), elongation (e) and area of cross-section (A), Young’s modulus of elasticity is determined.

(ii) Bulk modulus of elasticity (K) or (B):

It is defined as the ratio of normal stress to the volumetric strain i.e.,

K = $\frac{Normal\,\,stress}{Volumetric\,\,strain}$

elasticity notes class 11

Let us consider a spherical object of volume V is compressed by the forces and its volume is decreased by Δv.

Now,

Bulk modulus of elasticity (K) = $\frac{Normal\,\,stress}{Volumetric\,\,strain}$

Or, K = $\frac{F/A}{\Delta V/V}$

Or, K = $\frac{F\,\,V}{\Delta V\,A}$

Note: Compressibility (C) is the reciprocal of bulk modulus of elasticity i.e., C = $\frac{1}{K}$

(iii) Modulus of rigidity ($\eta $):

Modulus of rigidity is defined as the ratio of tangential stress to shear strain i.e.

Modulus of rigidity ($\eta $) = $\frac{Tangential\,\,stress}{Shear\,\,strain}$

elasticity notes class 11

From figure tangential stress = $\frac{Force}{Area}$ = $\frac{F}{A}$

And, shear strain ($\theta $) = $\frac{x}{l}$

Now,

Modulus of rigidity ($\eta $) = $\frac{Tangential\,\,stress}{Shear\,\,strain}$

$\eta $ = $\frac{F/A}{\theta }$

$\therefore $ $\eta $ = $\frac{F}{\theta A}$

Energy stored in a stretched wire:

elasticity notes class 11

Let us consider a uniform wire of length ‘l’ and area of cross section ‘A’ is fixed to a rigid support XY. A force ‘F’ is applied to the lower end of the wire to produce elongation from x = 0 to x = e as shown in figure above. Let ‘Y’ be the Young’s modulus of the wire then from the definition, 

Young’s modulus (Y) = $\frac{Normal\,\,stress}{Longitudinal\,\,strain}$

Y = $\frac{F/A}{x/l}$

Y = $\frac{Fl}{xA}$

$\therefore $ F = $\frac{YAx}{l}$ ………(i)

Let dx be the small elongation at any instant of time then small amount of work done in stretching the wire through ‘dx’ distance is 

dW = F.dx 

Using equation (i)

dW = $\frac{YAx}{l}$.dx ……(ii)

Now, the total amount of work done in stretching the wire through the distance (e) can be obtained by integrating equation (ii) from x = 0 to x = e, we get,

Or, $\int\limits_{0}^{e}{dW}$ = $\int\limits_{0}^{e}{{}}$$\frac{YAx}{l}$.dx

Or, W = $\frac{YA}{l}$$\int\limits_{0}^{e}{{}}$x dx

Or, W = $\frac{YA}{l}$ $\left[ \frac{{{x}^{1+1}}}{1+1} \right]_{0}^{e}$ [$\because $ $\int\limits_{}^{}{{{x}^{n}}}$dx = $\frac{{{x}^{n+1}}}{n+1}$ ]

Or, W = $\frac{YA}{l}$$\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{e}$

Or, W = $\frac{YA}{l}$ $\frac{{{e}^{2}}}{2}$

Or, W = $\frac{YAe}{l}$ $\frac{e}{2}$

Or, W =  $\frac{1}{2}$F e  [$\because $ F = $\frac{YAx}{l}$, x = e]

This amount of work done is stored as elastic potential energy in the wire.

So the energy stored in stretched wire is

E =  $\frac{\mathbf{1}}{\mathbf{2}}$F e  

Energy Density:

Energy density is defined as energy stored in a stretched wire per unit volume.

We know,

$\frac{E}{V}$ = $\frac{1/2F.e}{V}$

${{\rho }_{E}}$= $\frac{1/2F.e}{A\times l}$ $\because $ V = A × l

${{\rho }_{E}}$= $\frac{1}{2}$ $\frac{F}{A}$× $\frac{e}{l}$

$\therefore $ Energy density, ${{\rho }_{E}}$ = $\frac{1}{2}$ stress × strain

Poisson’s ratio ($\sigma $):

Poisson’s ratio is defined as the ratio of lateral strain to longitudinal strain. It is denoted by ($\sigma $) and mathematically given by Poisson’s ratio ($\sigma $) = $\frac{Lateral\,\,strain}{Longitudinal\,\,strain}$

elasticity notes class 11

Let us consider a uniform wire of length ‘l’ and diameter ‘D’ is fixed to a rigid support XY. A force is applied to the lower end of the wire, so that it’s length is increased by ‘e’ and diameter is decreased by ‘$\Delta $D’ as shown in figure above.

From Figure,

Lateral strain = $\frac{\Delta D}{D}$

And, longitudinal strain = $\frac{e}{l}$

Now, 

Poisson’s ratio ($\sigma $) = $\frac{Lateral\,\,strain}{Longitudinal\,\,strain}$

$\sigma $ = $\frac{-\Delta D/D}{e/l}$

$\sigma $ = $\frac{-\Delta Dl}{eD}$ = $\frac{-\Delta Rl}{eR}$

The theoretical value of Poisson ratio lies between -1 to $\frac{1}{2}$ but it’s practical value lies between 0 to $\frac{1}{2}$.

The Poisson’s ratio has maximum value for rubber since there is no change in volume of rubber when it is elongated or compressed.

Elastic After Effect:

The delaying in the time to regain its original shape and size after the removal of deforming force is known as elastic after effect. The glass has a maximum value of elastic after effect.

Elastic Fatigue:

With the application of alternate cycles of stress and strain, an elastic object loses its elastic property which is known as elastic fatigue. Due to elastic fatigue bridges are declared unsafe after long use.

Also Read: Gravitation Notes Class 11

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