JEE Main Study Notes for Elasticity: A material body gets deformed for a deforming force. When this deforming force is removed then the material body is able to return to its original shape and size, this ability of the deformed material body is called elasticity. A body with this ability is said to behave elastically. The elastic limit depends markedly on the type of considered solid.

The topic of elasticity has around 3.3% of weightage in JEE Main. Questions from this topic are found to be confusing as there are many formulas that look very similar. But with a proper preparation strategy, students will not face any confusion. Check JEE Main Physics Preparation Tips

JEE Main Study Notes for Elasticity covers some important concepts from the State of Matter Chapter like Stress, Strain, Hook’s Law, and Young’s Modulus of Elasticity. The questions asked in this section are majorly based on calculating the mass after stress is applied, calculating the elastic limit, etc. However, the topic is important but you cannot expect more than one question from this topic. Check JEE Main Physics Syllabus

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Definition of Elasticity

In the case when some external force is applied to a rigid body then there is a change in its length, volume, or shape. When these external forces will be removed then the body tends to regain its original shape and size. This property of a body by virtue of which a body tends to regain its original shape or size when the external forces that are applied to it are removed is called elasticity.

Sample Question

Question: The elastic limit of brass is 379 MPa. What should be the minimum diameter (in mm) of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

1. 1.15

2. 2.15

3. 3.45

4. 4.67

Solution: (1)

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Understanding the Concept of Stress and its types

Stress: The property of the Elastic bodies is that they regain their original shape due to internal restoring forces. This internal restoring force that is acting per unit area of the deformed body is called stress.

Stress = Restoring Force/Unit Area

Types of Stress

There are three types of stress

• Longitudinal stress

• Volume stress

• Tangential stress (or) shear stress

Longitudinal Stress: Longitudinal Stress is the stress which is is normal to the surface area of the body. Longitudinal Stress is classified into two types

• Tensile stress

• Compressive stress.

Tensile stress: Tensile stress occurs when longitudinal stress produced due to an increase in the length of the object.

Compressive stress: It is the Longitudinal stress produced due to the decrease in the length of the object.

Volume Stress: When the equal normal forces are applied everywhere on the surface of a body then change in volume is produced. But there is a force opposing this change in volume per unit area which is called volume stress.

Tangential Stress: When the stress is tangential or parallel to the surface of the body is known as Tangential or Shear stress. Due to this shape of body changes or gets twisted.

Must Read: 

• JEE Main Study Notes for Sound Waves

• JEE Main Study Notes for Center of Mass

• JEE Main Study Notes for Kinematics

Sample Question

Question: Let us consider a certain weight is suspended to a long uniform wire its length increases by one cm. The same weight is suspended to another wire. The material and length of the weight are the same but the diameter half of the first one. Calculate the increase in its length (in cm)?

1. 4

2. 5

3. 6

4. 7

Solution: (1)

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Quick Notes on Strain and its types

Strain is the ratio of charge of in any dimension to its original dimension.

Strain = Change in dimension/ initial dimension

The strain is also classified into three types

• Longitudinal strain

• Volume strain

• Shearing strain

Longitudinal Strain

Longitudinal strain = Change in length of the body/ initial length of the body

= ΔL/ L

Volume Strain

Volume strain = Change in volume of the body/ Original volume of the body

= ΔV/ V

Shearing Strain

When a deforming force is applied to a body parallel to its surface its shape changes but dies remains the same then this is known as shearing strain. The angle of shear ϕ

tan ϕ = ℓ/ L

= Displacement of upper face/ distance between two faces

Sample Question

Question: Let us consider a 2 m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radians. The shear strain developed will be

1. 0.004

2. 0.04

3. 0.4

4. 0.0004

Solution: (1)

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Stress-Strain Graph

1. Proportion limit: It is the limit in which Hook’s law is valid and the stress is directly proportional to strain.

2. Elastic limit: It is that maximum stress which is when removed the deforming force makes the body to recover completely its original state.

3. Yield point: This is the point beyond the elastic limit at which the length of the wire starts increasing with increasing stress.

4. Breaking point: This is the point when the strain becomes so large that the wire breaks down in the last.

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Hooke’s Law

When the deformation is small, the stress in a body is proportional to the corresponding strain this fact is known as Hooke’s law.

Within elastic limit, Stress & strain ⇒ Stress/ Strain = Constant

The constant is known as modulus of elasticity or coefficient of elasticity. It only depends on the type of material used. It is independent of stress and strain that are:

• Young’s modulus of elasticity “y”

• The bulk modulus of elasticity “B”

• Modulus of rigidity

• Poisson’s ratio

Check: JEE Main Study Notes for Current Electricity

Sample Question

Question: The only elastic modulus that applies to fluids is

1. Young's modulus

2. Shear modulus

3. Modulus of rigidity

4. Bulk modulus

Solution: (4)

Question: Let us consider a uniform cube is subjected to volume compression. If each side is decreased by 1%, then calculate the bulk strain?

1. 0.01

2. 0.06

3. 0.02

4. 0.03

Solution: (4)

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Young’s Modulus of Elasticity “y”

The ratio of longitudinal stress and longitudinal strain within the elastic limit is called Young’s modulus of elasticity (y).

y = Longitudinal stress/ Longitudinal strain = (F/A)/ (ℓ/L) = FL/ Aℓ

The force acting upon a unit area of a wire within the elastic limit by which the length of wire becomes double is equivalent to Young’s modulus of elasticity of the material of the wire. L is the length of wire, r is the radius and the increase in the length of wire by suspending a weight (mg) at its one end then young’s modulus of elasticity of the wire becomes,

y = (F/A)/ (ℓ/L) = FL/ Aℓ = mgL/ πr2ℓ

The increment of the length of an object by its own weight

Let us consider a rope of mass M and length (L) is hanged vertically. As the tension of different points on the rope is different, similarly stress as well as the strain will be different at different points.

• Maximum stress at hanging point

• Minimum stress at a lower point

Consider a dx element of rope at x distance from the lower end then tension.

T = (M/L) X g

So stress = T/A = (M/L) xg/A

Let us consider the increase in the length of element dx is dy then

Strain = Change in length/ Original Length = Δy/Δx = dy/dx

After calculating stress, Young’s Modulus of Elasticity “y” will be 

Y = Stress/strain = [(M/L) xg/A]/ (dy/dx) ⇒ (M/L) xg/A dx = 1/dy

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Work done in stretching a wire

Elastic potential energy U is stored in the wire, so U = 1/2F×l

= ½ (stress) × (strain) × volume of the wire

So, elastic potential energy per unit volume of the wire

U = ½ (stress) × (strain)

= ½ (Young’s modulus × strain) × strain (as Young’s modulus = stress/strain)

So, U = ½ (young’s modulus) × (strain)2

Analogy of rod as a spring

F = kx

k = yA/L = constant

yA/L = constant which only functions its material property.

Sample Question

Question: Young's moduli of two wires A and B are in the ratio 7:4. Wire A is 2 m long with a radius R. Wire B is 1.5 m long with a radius of 2 mm. If the two wires stretch by the same length for a given load, then the value of R (in mm) is close to:

1. 1.75

2. 2.75

3. 3.75

4. 4.75

Solution: (1)

Bulk Modulus, Rigidity Modulus & Poisson’s Ratio

Bulk Modulus

Within the elastic limit, the ratio of the volume stress and the volume strain is called the bulk modulus of elasticity.

B = Volume Stress/ Volume Strain = (F/A)/ (-ΔV/V) = ΔP/ (-ΔV/V)

Rigidity Modulus

Within the elastic limit, the ratio of shearing stress and shearing strain is called the modulus of rigidity.

η = Shearing stress/ Shearing strain = (F tangential/ A)/ ϕ = F tangential/ ϕ

Φ = angle of shear

Poisson’s Ratio

Within the elastic limit, the ratio of lateral strain and longitudinal strain is called the poison’s ratio. 

Poisson’s Ration σ = Lateral Starin/ Longitudinal Strain = β/ α

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Tips to Solve Questions on Elasticity in JEE Main

1. Question: A wire whose cross-section area is A1 is stretched by L1 by a certain weight. Calculate that how far will a wire of same material and the same length and cross-section area A2 stretch if same weight is applied to it

Tip to Solve the Question: Here if you know this formula then you can easily solve this question ΔL=FL/AY. Just remember that AΔL is a constant.

Solution: A1L1/L2

2. Question: A wire is stretched by x mm when a load F is hanged on it.If the same wire goes over a pulley and two-weight F each is hung at the two ends the what will be the elongation in the wire

Tip to Solve the Question: Use formula Y = (F/A)/ (x/L). Consider the elongation to be y then the value of x will be y and of L will be 5L. Equate the equations and you will get the answer. 

Solution: x

3. Question: One end of a uniform wire of length L and μ mass per unit length is attached rigidly to a point in a ceiling. Let us consider that a Mass M is suspended from its lower end. The area of cross-section of the wire will be A. Find the stress at a distance x from the ceiling point?

Tip to Solve the Question: Weight of wire below the point (L-x)μg +Mg

Stress = F/A

Enter the values in the formula and calculator the answer

Solution: [(L-x)μg +Mg]/ A

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Tricks to Solve the Questions on Elasticity

• You can start your question from this section by making a rough diagram. This tip is very useful to solve questions of this section and also the questions from other sections. 

• The diagram created by you will give you an idea about the point where the stress is applied, mention mass, and other given variables. 

• Try to solve your questions by using the formulas of dimensions. This way you will not get confused with equations.

• But while calculating you will find that value of the options is equal by using dimensions then you have to solve the question by creating diagrams and following equations only. 

JEE Main Elasticity FAQs