The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one w

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Q. The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

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Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

When you pull on a rubber band, it stretches easily. When you pull on a steel rod of the same size, it barely moves. Both are elastic — they return to their original shape when you let go — but they resist stretching very differently. Young’s modulus is the number that tells you exactly how much a material resists being stretched or compressed lengthwise.

The Intuition: Stiffness per Unit Size

Think of a spring. A stiff spring requires a large force to stretch it a little. A soft spring stretches a lot with a small force. Young’s modulus is like the “stiffness” of a material, but it’s cleverly designed to be independent of the object’s shape and size.

If you take a thick steel rod and a thin steel wire of the same length, the rod is harder to stretch. That’s because you’re pulling on more material. Young’s modulus removes this size effect — it tells you the stiffness of the material itself, not the particular piece you’re holding.

The Precise Definition

Young’s modulus ( $E$ or $Y$ ) is defined as the ratio of tensile stress to tensile strain, as long as the material obeys Hooke’s law (the deformation is reversible and proportional to the force).

Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}

Let’s break down the two parts.

Tensile Stress ( $\sigma$ ) is the force per unit area. If you pull with a force $F$ on a rod of cross-sectional area $A$ , the stress is:

\sigma = \frac{F}{A}

Stress has units of pressure — pascals (Pa) or $\text{N/m}^2$ . It tells you how “intense” the pulling is, regardless of the rod’s thickness.

Tensile Strain ( $\varepsilon$ ) is the fractional change in length. If the original length is $L_0$ and it stretches by $\Delta L$ , the strain is:

\varepsilon = \frac{\Delta L}{L_0}

Strain is a pure number — it has no units. A strain of 0.01 means the rod stretched by 1% of its original length.

Putting it together:

Y = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L}

What the Number Tells You

A high Young’s modulus means the material is very stiff — it takes a huge stress to produce even a tiny strain. Steel has $Y \approx 200 \times 10^9$ Pa. A low Young’s modulus means the material is easily stretched. Rubber has $Y \approx 0.01 \times 10^9$ Pa — about 20,000 times smaller than steel.

Note

Young’s modulus is only valid in the elastic region — where the material returns to its original shape after the force is removed. If you stretch too far (past the elastic limit), the material deforms permanently or breaks, and Young’s modulus no longer applies.

A Worked Example

A steel wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ is pulled by a force of 100 N. How much does it stretch? (Young’s modulus of steel = $2.0 \times 10^{11}$ Pa)

From $Y = \frac{F L_0}{A \Delta L}$ , rearrange:

$\Delta L = \frac{F L_0}{A Y} = \frac{100 \times 2.0}{(1.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{200}{2.0 \times 10^5} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}$ …

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Young's modulus is the ratio of stress to strain. Since steel has a much higher Young's modulus than rubber, for the same longitudinal strain, steel will experience a greater tensile stress.

When we talk about how materials respond to forces, two key concepts are stress and strain. Young's modulus connects these two, telling us how stiff a material is. Understanding this relationship is crucial for predicting how materials will deform or break under load.

Understanding Stress and Strain:
- Tensile Stress ( $\sigma$ ) is the internal restoring force per unit cross-sectional area of a material. When you pull on an object, the internal forces resisting that pull are distributed over its cross-section. Mathematically, it's given by $\sigma = F/A$ , where $F$ is the applied force and $A$ is the cross-sectional area. Its SI unit is Pascals (Pa) or N/m $^2$ .
- Longitudinal Strain ( $\epsilon$ ) is the fractional change in length of a material when subjected to a tensile or compressive force. It's a dimensionless quantity, calculated as $\epsilon = \Delta L / L$ , where $\Delta L$ is the change in length and $L$ is the original length.
Introducing Young's Modulus:

Young's modulus ( $Y$ ) is a fundamental material property that quantifies its stiffness or resistance to elastic deformation under tensile or compressive stress. It is defined as the ratio of tensile stress to longitudinal strain within the elastic limit of the material.
$Y = \frac{\text{Tensile Stress}}{\text{Longitudinal Strain}} = \frac{\sigma}{\epsilon}$
A higher Young's modulus means the material is stiffer; it requires a greater stress to produce a given amount of strain. Conversely, a lower Young's modulus indicates a more flexible material.
Applying the Concept to the Problem:

The problem states two key pieces of information:
- Young's modulus for steel ( $Y_{steel}$ ) is much more than that for rubber ( $Y_{rubber}$ ). So, $Y_{steel} > Y_{rubber}$ .
- Both materials are subjected to the same longitudinal strain ( $\epsilon$ ). So, $\epsilon_{steel} = \epsilon_{rubber} = \epsilon$ . …

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Q.The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

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