Question

A copper wire of length \(4 \mathrm{~m}\) and area of cross-section \(1.2 \mathrm{~cm}^{2}\) is stretched with a force of \(4.8 \times 10^{3} \mathrm{~N}\). If young's modulus for copper is \(1.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\). What will be the length increase of the wire? (A) \(1.33 \mathrm{~mm}\) (B) \(1.33 \mathrm{~cm}\) (C) \(2.66 \mathrm{~mm}\) (D) \(2.66 \mathrm{~cm}\)

Short answer

The length increase of the copper wire is \(\Delta L = 1.33\,\text{mm}\), which corresponds to answer (A).

Step-by-step solution

Write down the given values

First, we need to write down all the given information: - Length of the copper wire, \(L = 4\,\text{m}\) - Area of cross-section, \(A = 1.2\,\text{cm}^2 = 1.2 \times 10^{-4}\,\text{m}^2\) (converted to square meters) - Stretching force, \(F = 4.8 \times 10^3\,\text{N}\) - Young's modulus for copper, \(Y = 1.2 \times 10^{11}\,\text{N/m}^2\) - Length increase of the wire, \(\Delta L = ?\)

Rearrange the Young's modulus formula to solve for the length increase

We will rearrange the Young's modulus formula to solve for the length increase, \(\Delta L\). \(Y = \frac{F \cdot L}{A \cdot \Delta L}\) Rearrange for \(\Delta L\): \(\Delta L = \frac{F \cdot L}{A \cdot Y}\)

Substitute the given values into the formula

Now, we can substitute all the given values into the rearranged formula: \(\Delta L = \frac{(4.8 \times 10^3\,\text{N}) \cdot (4\,\text{m})}{(1.2 \times 10^{-4}\,\text{m}^2) \cdot (1.2 \times 10^{11}\,\text{N/m}^2)}\)

Calculate the length increase

Calculate the length increase, \(\Delta L\): \(\Delta L = \frac{1.92 \times 10^4\,\text{N} \cdot \text{m}}{1.44 \times 10^7\,\text{N} \cdot \text{m}^2} = 1.33 \times 10^{-3}\, \text{m}\) Convert the result into millimeters: \(\Delta L = 1.33 \times 10^{-3}\, \text{m} \cdot \frac{1000\,\text{mm}}{1\,\text{m}} = 1.33\,\text{mm}\) So the length increase of the copper wire is \(\Delta L = 1.33\,\text{mm}\), which corresponds to answer (A).

Key concepts

Deformation of materials

When we talk about deformation of materials, we're describing how a object's shape changes when a force is applied. Imagine you are pulling on a rubber band; as you pull, the band stretches. Here, the change in length, or the degree to which something is stretched, is what we call deformation.
In the case of solid materials like metals, understanding deformation is crucial for knowing how they behave under forces. There are different types of deformation: elastic and plastic. Elastic deformation is reversible, like the rubber band, while plastic deformation means the material doesn't return to its original shape once the force is removed.
For metals such as copper, when subjected to force, they undergo elastic deformation up to a certain point, called the "elastic limit." Beyond that, they deform plastically, meaning they cannot go back to their original form.

Stress and strain

Stress and strain go hand in hand when discussing material deformation. Stress is essentially the force exerted per unit area on a material. In a formula, it's represented as \( \sigma = \frac{F}{A} \), where \( F \) is the force and \( A \) is the area of cross-section.
Strain, on the other hand, is a measure of deformation, and it tells us how much a material stretches per unit length. This is defined as \( \epsilon = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length and \( L \) is the original length.
To connect these concepts, we use Young's modulus \( Y \), which describes a material's ability to withstand length changes under tension or pressure. The formula is \( Y = \frac{\text{stress}}{\text{strain}} = \frac{F \cdot L}{A \cdot \Delta L} \). By knowing \( Y \), we can predict how much an object will stretch for a given force.

Units conversion in physics

Units conversion is a fundamental skill in physics that enables us to express physical quantities in consistent units. For example, in our problem, we had to convert the area of the copper wire's cross-section from square centimeters to square meters. This is done using the conversion factor, where \( 1 \text{~cm}^2 = 10^{-4} \text{~m}^2 \).
Another conversion involved in this calculation is converting the result from meters to millimeters. We know 1 meter equals 1000 millimeters, so multiplying by 1000 gives the desired millimeter unit.
Learning to convert units correctly is essential not just for solving physics problems accurately but also for communicating with scientists and engineers around the world who might use different systems of measurement.

Copper wire characteristics

Copper is a notable metal widely used in electrical wiring due to its excellent conductivity and flexibility. Its characteristics make it an ideal choice for applications where ductility and toughness are required.
Copper's Young's modulus, approximately \( 1.2 \times 10^{11} \text{~N/m}^2 \), indicates its stiffness. This modulus value shows that copper can endure a considerable amount of stress before deforming, but still stretches more easily compared to some other metals like steel.
These properties allow copper wires to maintain functionality across long distances, even when subjected to the stretching forces during installation or load-bearing utility. Understanding these intrinsic characteristics helps in selecting the right materials for various engineering and construction tasks.