A piece of copper having a rectangular cross-section of $15.2\ \text{mm} \times 19.1\ \text{mm}$ is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

A piece of copper having a rectangular cross-section of 15.2mm × 19.1mm is pulled in tension with 44,500 N for

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15.2\ \text{mm} \times 19.1\ \text{mm}

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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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Using Young's modulus for copper ( $Y = 1.1 \times 10^{11}\ \text{Pa}$ , Table 8.1) and the relation $\sigma = Y \cdot \varepsilon$ , dividing the tensile stress by $Y$ gives the strain. The printed answer for this question is $\boxed{0.127}$ .

Young's modulus is the material's stiffness -- it tells you how much strain (fractional change in length) a given stress produces, as long as the deformation stays elastic. For copper, Table 8.1 gives $Y = 1.1 \times 10^{11}\ \text{Pa}$ . The key relation is:

\sigma = Y \cdot \varepsilon \quad \text{or} \quad \varepsilon = \frac{\sigma}{Y}

where $\sigma$ is tensile stress (force per unit area) and $\varepsilon$ is the strain (dimensionless). So the plan is: compute the cross-sectional area, find the stress, then divide by $Y$ .

Convert dimensions to metres. The cross-section is $15.2\ \text{mm} \times 19.1\ \text{mm}$ .

15.2\ \text{mm} = 0.0152\ \text{m}, \qquad 19.1\ \text{mm} = 0.0191\ \text{m}

Calculate the area.

A = 0.0152 \times 0.0191 = 2.9032 \times 10^{-4}\ \text{m}^2

Find the tensile stress. Force $F = 44{,}500\ \text{N}$ .

\sigma = \frac{F}{A} = \frac{44{,}500}{2.9032 \times 10^{-4}} \approx 1.53 \times 10^{8}\ \text{Pa}

Apply Young's modulus. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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The shear modulus (modulus of rigidity) of an ideal liquid is zero.

Shear modulus G is defined as G = shear stress / shear strain. A liquid (unlike a solid) has no fixed shape; when a tangential (shear) force is applied to a liquid layer, it does not develop a restoring elastic shear stress — it simply flows/deforms continuously. Since a liquid offers essentially zero resistance to shape-changing (shear) deformation whil …

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$Q. Draw the Force-Displacement graph for a body obeying the relation (F = kx), where (k) is a constant and$
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Figure — Force-Displacement graph for F=kx

The Force–Displacement graph for F = kx is a straight line passing through the origin, with slope equal to k.

Given F = kx, F is directly proportional to x (k is a constant, e.g., a spring/elastic constant, as in Hooke's Law).

To draw the graph:

Put displacement x on the horizontal axis and force F on the vertical axis.

Since F = kx is a linear equation in x with zero intercept, the graph is a straight line that starts at the origin (0,0) and rises with a constant slope equal to k, i.e., for every unit increase in x, F increases by k. …
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$Q. State Hooke's law.$
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Within the elastic limit, the stress produced in a body is directly proportional to the strain it undergoes.

When a deforming force is applied to an elastic body, it develops an internal restoring force per unit area called stress, and undergoes a fractional change in dimension called strain. Hooke's law states that as long as the deformation stays within the elastic limit of the material (i.e., the body can still fully recover its original shape when the force is removed), the stress developed is directly proportional to the strain produced:

$\text{Stress} \propto \text{Strain} \quad \Rightarrow \quad \text{Stress} = k \times \text{Strain}$

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Bending a wire repeatedly at the same point strains it beyond its elastic limit again and again, causing progressive internal damage (elastic fatigue) until it fractures, even though one such bend alone would not break it.

When a wire is bent, the material on the outer side of the bend is stretched (tensile strain) while the material on the inner side is compressed. If the bend is sharp, the strain at that point can exceed the elastic limit of the material, so the material does not fully return to its original state when straightened — some permanent (plastic) deformation and internal microscopic damage remains.

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$Q. How are we able to break a wire by repeated bending ?$
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Bending a wire back and forth repeatedly causes elastic fatigue — the metal at the bend loses its elasticity and develops cracks, so it eventually breaks even without a large single force.

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Q.A piece of copper having a rectangular cross-section of 15.2 mm×19.1 mm15.2\ \text{mm} \times 19.1\ \text{mm}15.2 mm×19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

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Q.A piece of copper having a rectangular cross-section of $15.2\ \text{mm} \times 19.1\ \text{mm}$ is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

Unlock everything free for 14 days