A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of the same material of length $2L$ and radius $2r$, is pulled by a force $2f$. Find the increase in length of this wire.

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a f

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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 a material property, so it stays the same for both wires. Using $Y = \frac{f/A}{l/L}$ and substituting the new dimensions and force gives the new extension as $l$ .

The key idea is that Young’s modulus $Y$ is a constant for a given material, independent of the wire’s geometry or the applied force (within the elastic limit). So for both wires, $Y$ is identical. That single fact lets us connect the two situations.

Young’s modulus is defined as the ratio of tensile stress to tensile strain:

Y = \frac{\text{stress}}{\text{strain}} = \frac{f/A}{l/L}

where $A$ is the cross-sectional area, $L$ the original length, and $l$ the extension.

First wire — given $L$ , $r$ , $f$ , and extension $l$ . Its area is $A_1 = \pi r^2$ . So:

Y = \frac{f / (\pi r^2)}{l / L} = \frac{f L}{\pi r^2 l}

Second wire — length $2L$ , radius $2r$ , force $2f$ , unknown extension $l_2$ . Its area is $A_2 = \pi (2r)^2 = 4\pi r^2$ . Write $Y$ for this wire:

Y = \frac{(2f) / (4\pi r^2)}{l_2 / (2L)} = \frac{2f \cdot 2L}{4\pi r^2 l_2} = \frac{4f L}{4\pi r^2 l_2} = \frac{f L}{\pi r^2 l_2}

Equate the two expressions for $Y$ (same material):

\frac{f L}{\pi r^2 l} = \frac{f L}{\pi r^2 l_2}

Cancel the common factor $\frac{f L}{\pi r^2}$ (non-zero), and we get:

$\frac{1}{l} = \frac{1}{l_2} \quad \Rightarrow \quad l_2 = l$ …

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Q.A wire of length LLL and radius rrr is clamped rigidly at one end. When the other end of the wire is pulled by a force fff, its length increases by lll. Another wire of the same material of length 2L2L2L and radius 2r2r2r, is pulled by a force 2f2f2f. Find the increase in length of this wire.

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Q.A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$ , its length increases by $l$ . Another wire of the same material of length $2L$ and radius $2r$ , is pulled by a force $2f$ . Find the increase in length of this wire.

Unlock everything free for 14 days