Why is steel considered more elastic than rubber in physics, contrary to everyday intuition?

In physics, elasticity is defined by a material's resistance to deformation and its ability to regain its original shape. For a given deforming force and area, steel undergoes far less strain than rubber, meaning $Y_{\text{steel}} > Y_{\text{rubber}}$ (since $Y = \frac{\text{Stress}}{\text{Strain}}$). Thus, steel has a significantly higher Young's modulus and is scientifically more elastic.

Why is $50\%$ of the work done by a stretching force lost as heat during wire elongation?

When a constant stretching force $F$ moves through displacement $\Delta L$, work done is $W = F \Delta L$. However, internal restoring force increases linearly from $0$ to $F$, making average restoring force $\frac{F}{2}$. The stored strain energy is $U = \frac{1}{2} F \Delta L$. The difference $W - U = \frac{1}{2} F \Delta L$ is converted into thermal energy due to internal atomic friction.

How do I calculate the elongation of a wire under its own weight?

For a wire of mass $M$, length $L$, area $A$, and density $\rho$, force varies linearly from $0$ at the bottom to $M g$ at the top. Integrating along length gives effective load $\frac{M g}{2}$. Thus, elongation under self-weight is $\Delta L = \frac{M g L}{2 A Y} = \frac{\rho g L^2}{2 Y}$.

Mechanical Properties of Solids Notes, Formulas, Traps & PYQs | ExamiFly

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Key Formulas & High-Yield Equations

Young's Modulus of Elasticity ($Y$)

$Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F L}{A \Delta L}$

Applies strictly to solid wires under uniform axial tensile or compressive load within the proportional limit where Hooke's law holds (

F \propto \Delta L

). Units:

\text{N/m}^2

\text{Pa}

Shear Modulus / Modulus of Rigidity ($\eta$)

$\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F / A}{\theta} = \frac{F L}{A \Delta x}$

Valid exclusively for shape deformation at constant volume within the elastic limit.

\theta

represents shear angle in radians and

\Delta x

is relative lateral displacement.

Bulk Modulus ($B$) and Compressibility ($K$)

$B = - V \frac{\Delta P}{\Delta V}, \quad K = \frac{1}{B} = - \frac{1}{V} \frac{\Delta V}{\Delta P}$

Applies to volume changes under uniform hydrostatic pressure

\Delta P

. Negative sign ensures

B > 0

as volume decreases when pressure increases.

Elastic Potential Energy Density ($u$)

$u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \epsilon^2 = \frac{\text{Stress}^2}{2Y}$

Valid for linear elastic deformations within the proportional limit. Represents elastic energy stored per unit volume (

u = \frac{U}{V}

) expressed in

\text{J/m}^3

Thermal Stress and Force in Rigid Rod

$\text{Stress} = Y \alpha \Delta T, \quad F = Y A \alpha \Delta T$

Valid strictly when both ends of a rod are fixed to unyielding supports so that natural thermal expansion (

\Delta L = L \alpha \Delta T

) is completely constrained.

Poisson's Ratio ($\sigma$)

$\sigma = - \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = - \frac{\Delta d / d}{\Delta L / L}$

Theoretical limit is

-1 < \sigma < 0.5

. For actual real-world isotropic materials, practical values range between

0

and

0.5

. It is a dimensionless quantity.

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2024

A wire of original length $L$ , cross-sectional area $A$ , and Young's modulus $Y$ is stretched by a force $F$ producing an extension $x$ . The elastic potential energy $U$ stored in the wire is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write the expression for elastic strain energy stored in a stretched wire: $U = \frac{1}{2} \times \text{Average Restoring Force} \times \text{Extension}$ .

Step 2

At equilibrium, restoring force equals applied force $F$ , and extension is $x$ . Thus, average restoring force is $\frac{F}{2}$ .

Step 3

Multiply average force by extension to get $U = \frac{1}{2} F x$ , matching the first option.

Select an option & click Check Answer to reveal solution

Question #2JEE Main 2023

Two wires $A$ and $B$ made of the same material have lengths in the ratio $1 : 2$ and radii in the ratio $2 : 1$ . If both wires are stretched by the same stretching force $F$ , the ratio of elastic potential energy stored in wire $A$ to wire $B$ ( $U_A : U_B$ ) is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Express potential energy $U$ in terms of force $F$ , length $L$ , radius $r$ , and Young's modulus $Y$ : $U = \frac{F^2 L}{2 A Y} = \frac{F^2 L}{2 \pi r^2 Y}$ .

Step 2

Since both wires have identical $Y$ and equal stretching force $F$ , energy scales as $U \propto \frac{L}{r^2}$ .

Step 3

Calculate the ratio: $\frac{U_A}{U_B} = \left(\frac{L_A}{L_B}\right) \times \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right)^2 = \frac{1}{8}$ , giving ratio $1 : 8$ .

Select an option & click Check Answer to reveal solution

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Mastering Mechanical Properties of Solids: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Young's Modulus of Elasticity ($Y$)

Shear Modulus / Modulus of Rigidity ($\eta$)

Bulk Modulus ($B$) and Compressibility ($K$)

Elastic Potential Energy Density ($u$)

Thermal Stress and Force in Rigid Rod

Poisson's Ratio ($\sigma$)

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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