All questions of Elasticity- Hooke's Law & Young's Modulus for JAMB Exam

Given:
- The flood lamp stretches the wire by 0.18 mm
- The stress is proportional to the strain
To find:
- How much would it have stretched if the wire had the same length but twice the diameter

Let's begin by understanding the given information.

Stress is defined as the force per unit area and is denoted by the symbol σ (sigma). Mathematically, stress is given by:

σ = F / A

where F is the force applied and A is the area over which the force is applied.

Strain is defined as the change in length per unit length and is denoted by the symbol ε (epsilon). Mathematically, strain is given by:

ε = ΔL / L

where ΔL is the change in length and L is the original length.

From the given information, we know that the stress is proportional to the strain. This can be expressed mathematically as:

σ ∝ ε

or

σ = kε

where k is a constant of proportionality.

Now, let's apply this information to the problem at hand.

When the flood lamp is hung from the wire, it exerts a force on the wire which causes it to stretch. Let's assume that the original diameter of the wire is d and the original length is L.

From the given information, we know that the stress is proportional to the strain. Therefore, we can write:

σ = kε

where σ is the stress, k is a constant of proportionality, and ε is the strain.

The stress can be calculated using the formula:

σ = F / A

where F is the force applied and A is the cross-sectional area of the wire.

The force applied is the weight of the flood lamp, which can be calculated using the formula:

F = mg

where m is the mass of the flood lamp and g is the acceleration due to gravity.

The cross-sectional area of the wire can be calculated using the formula:

A = πd^2 / 4

where d is the diameter of the wire.

Therefore, we can write:

σ = (mg) / (πd^2 / 4)

The strain can be calculated using the formula:

ε = ΔL / L

where ΔL is the change in length and L is the original length.

From the given information, we know that the flood lamp stretches the wire by 0.18 mm. Therefore, we can write:

ε = 0.18 / L

Now, let's combine the equations for stress and strain:

σ = kε

σ = (mg) / (πd^2 / 4)

ε = 0.18 / L

Substituting the values of σ and ε, we get:

(mg) / (πd^2 / 4) = k (0.18 / L)

Simplifying, we get:

k = (mgL) / (0.18πd^2)

Now, let's use this value of k to calculate the change in length when the diameter of the wire is doubled.

When the diameter of the wire is doubled, the cross-sectional area of the wire becomes 4 times the original area. Therefore, the new diameter is 2d and the new cross-sectional area is:

A' = π(2d)^2 / 4 = 4πd^2

Using the same formula for stress,

Luggage

When a soild is compressed or elongated, it's P.E increases. Potential energy increases because work has to be done against force of repulsion during compression and against force of attraction during elongation.

From Hooke's Law: F =kx

A

Introduction:
Hooke's Law is a principle in physics that relates the force applied to a spring or elastic material to the resulting deformation or change in length of the material. It states that the force applied to a spring is directly proportional to the displacement or change in length of the spring.

Explanation:
The constant of proportionality in Hooke's Law is known as the modulus of elasticity or Young's modulus. It is represented by the symbol 'E' and is a measure of the stiffness or rigidity of a material. The modulus of elasticity signifies how much a material will deform when a force is applied to it.

Modulus of Elasticity:
The modulus of elasticity is a material property that describes how it responds to stress. It is defined as the ratio of stress to strain within the elastic limit of the material. In other words, it measures how much stress a material can withstand before it starts to deform permanently.

Modulus of Strain:
The modulus of strain is not a property used in Hooke's Law. Strain is the measure of deformation or change in length of a material, and the modulus of strain is not directly related to the constant of proportionality in Hooke's Law.

Elasticity of Wire:
The elasticity of a wire refers to its ability to return to its original shape after being stretched or deformed. It is related to Hooke's Law as the law describes the linear relationship between the force applied to a wire and the resulting deformation or change in length of the wire.

Modulus of Stress:
The modulus of stress is not a term used in Hooke's Law. Stress is defined as the force applied per unit area of a material, and the modulus of stress is not directly related to the constant of proportionality in Hooke's Law.

Conclusion:
In conclusion, the constant of proportionality in Hooke's Law signifies the modulus of elasticity. It is a measure of the stiffness or rigidity of a material and describes how much a material will deform when a force is applied to it. The modulus of elasticity is a fundamental property used to understand the behavior of elastic materials and is essential in various fields such as engineering and materials science.

Tensile stress (or tension) is the stress state leading to expansion; that is, the length of a material tends to increase in the tensile direction. The volume of the material stays constant. When equal and opposite forces are applied on a body, then the stress due to this force is called tensile stress.

I section is generally used as a beam because of its high section modulus as it's most of the area is situated away from it's neutral axis hence it has high moment of inertia i.e high section modulus i.e high moment carrying capacity which is the major requirement for a good beam section.

Understanding the Problem
To find the shear strain produced in the rubber cube when a tangential force is applied, we can use the relationship between shear stress, shear modulus, and shear strain.
Given Data
- Side of the cube = 10 cm = 0.1 m
- Tangential Force (F) = 1800 N
- Shear Modulus (η) = 2 x 10^6 N/m²
Calculating Shear Stress
Shear stress (τ) can be calculated using the formula:
τ = F / A
Where A is the area of the face being sheared.
- Area (A) = side × side = 0.1 m × 0.1 m = 0.01 m²
Now, substituting the values:
τ = 1800 N / 0.01 m² = 180000 N/m²
Calculating Shear Strain
Shear strain (γ) can be calculated using the relationship:
γ = τ / η
Substituting the values of shear stress and shear modulus:
γ = 180000 N/m² / (2 x 10^6 N/m²)
Calculating this gives:
γ = 0.09
Conclusion
The shear strain produced in the rubber cube is 0.09. Hence, the correct answer is option 'C'.

Alternation of generations (also known as metagenesis) is the type of life cycle that occurs in those plants and algae in the Archaeplastida and the Heterokontophyta that have distinct sexual haploid and asexual diploid stages.

Brittle materials have shorter plastic region, i.e., the proportional and fracture points are close.

An elastomer is a polymer with viscoelasticity (i. e., both viscosity and elasticity) and very weak intermolecular forces, and generally low Young's modulus and high failure strain compared with other materials. Elastomer rubber compounds are made from five to ten ingredients, each ingredient playing a specific role. Polymer is the main component, and determines heat and chemical resistance, as well as low- temperature performance. Reinforcing filler is used, typically carbon black, for strength properties.

Understanding the Problem
To determine the minimum diameter required for a circular steel wire, we must consider the relationship between tensile stress, strain, and Young's modulus.
Key Given Data:
- Length of wire (L) = 2.00 m
- Maximum stretch (ΔL) = 0.25 cm = 0.0025 m
- Tensile force (F) = 400 N
- Young's modulus for steel (E) = 2.00 × 10^11 N/m²
Formulas to Use:
1. Strain (ε): ε = ΔL / L
2. Stress (σ): σ = F / A
3. Young's Modulus (E): E = σ / ε
Calculating Strain:
- Strain (ε) = ΔL / L
- ε = 0.0025 m / 2.00 m = 0.00125
Calculating Stress:
Using Young's modulus, we can rearrange the formula to find stress:
- σ = E × ε
- σ = (2.00 × 10^11 N/m²) × (0.00125) = 2.5 × 10^8 N/m²
Finding the Required Area:
From the stress formula, we can find the cross-sectional area (A):
- A = F / σ
- A = 400 N / (2.5 × 10^8 N/m²) = 1.6 × 10^-6 m²
Calculating Diameter:
The area of a circle is given by A = (π/4) × d². Rearranging gives:
- d² = (4A) / π
- d = √((4 × 1.6 × 10^-6 m²) / π)
- d ≈ 0.0014 m = 1.4 mm
Conclusion:
The minimum diameter required for the wire is 1.4 mm, confirming option 'A' as the correct answer.

This part of stress strain curve is called hardening region.When the load reaches a maximum value, the engineering stress at this point is called the tensile strength or ultimate tensile strength of the material.