Question

A wire of length \(2 \mathrm{~m}\) is made from \(10 \mathrm{~cm}^{3}\) of copper. A force \(\mathrm{F}\) is applied so that its length increases by \(2 \mathrm{~mm}\). Another wire of length \(8 \mathrm{~m}\) is made from the same volume of copper. If the force \(\mathrm{F}\) is applied to it, its length will increase by......... (A) \(0.8 \mathrm{~cm}\) (B) \(1.6 \mathrm{~cm}\) (C) \(2.4 \mathrm{~cm}\) (D) \(3.2 \mathrm{~cm}\)

Short answer

The increase in length of the second wire is \(0.8 \mathrm{~cm}\).

Step-by-step solution

Find the cross-sectional area of the first wire.

The volume of the first wire is given as 10 cm³, and its length is 2 m or 200 cm. To find the area A, we can use the formula: \[ Volume = Area \cdot Length \] \[ A = \frac{Volume}{Length} \] \[ A = \frac{10}{200} = 0.05 cm^{2} \]

Find the cross-sectional area of the second wire.

Since the volume of copper used for both wires is the same and the second wire's length is 8 m or 800 cm, we can find its cross-sectional area A' using the same formula: \[ A' = \frac{Volume}{Length} \] \[ A' = \frac{10}{800} = 0.0125 cm^{2}\]

Determine the ratio of the cross-sectional areas.

Now, let's find the ratio of the cross-sectional areas of the two wires, which we will call R: \[R = \frac{A'}{A} \] \[R = \frac{0.0125}{0.05} \] \[R = 0.25\]

Calculate the increase in length of the second wire.

We know that the length of the first wire increases by 2 mm or 0.2 cm. We can now use the ratio R to find the increase in length of the second wire (ΔL): \[ \Delta L = \frac{\Delta L_{1}}{R}\] \[ \Delta L = \frac{0.2}{0.25} = 0.8 cm\] The increase in length of the second wire is 0.8 cm. Therefore, the correct answer is (A) \(0.8 \mathrm{~cm}\).

Key concepts

Young's Modulus

Young's Modulus is an essential concept in the field of materials science and engineering. It is a measure of the stiffness of a material, defined as the ratio of stress to strain in the linear portion of a material's deformation curve. Young's Modulus is often represented by the letter E, and it tells us how much a material will deform under a given load.

Mathematically, Young's Modulus is calculated using: \[ E = \frac{\text{Stress}}{\text{Strain}} \] Where stress is the force applied per unit area, and strain is the change in length divided by the original length. This modulus helps to predict how materials will behave under stress, which is crucial in design and structural engineering.

Materials with a high Young's Modulus are typically rigid and deform very little under load, such as metals and ceramics, whereas materials with a low Young's Modulus, like rubber, are more flexible. Understanding Young's Modulus helps engineers to choose the correct material for specific applications.

Stress and Strain

Stress and strain are fundamental concepts in the study of elasticity, dealing with the behavior of materials when subjected to forces. Stress is defined as the internal force per unit area within materials that arises from externally applied forces. It is expressed in units like Pascals (Pa) and can be calculated using the formula: \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \] Strain, on the other hand, is a measure of the deformation of the material. It is a dimensionless quantity that represents the change in length relative to the original length of an object. Mathematically, strain can be expressed as: \[ \text{Strain} = \frac{\Delta L}{L_0} \] where \(\Delta L\) is the change in length and \(L_0\) is the original length.

In our example, when the copper wire is stretched, it experiences both stress and strain. Understanding these concepts helps in determining how much a material will elongate or compress under a given force, which is critical in applications that require specific precision and safety.

Hooke's Law

Hooke's Law is a principle of physics that defines the relationship between the force applied to a spring or material and its resulting displacement. It is expressed in the linear elasticity regime of the force-displacement curve before the material yields or breaks. Hooke's Law is given by: \[ F = k \times x \] where \(F\) is the force applied, \(k\) is the spring constant or stiffness of the material, and \(x\) is the displacement or change in length.

Hooke's Law is crucial in understanding how materials behave under small deformations. For solids, this law illustrates that the extension (deformation) is proportional to the force applied, up to the elastic limit of that material.

In our exercise, Hooke's Law plays a role in understanding why the copper wire length changes when a force is applied. The law helps explain the initial linear relationship seen in the exercise, where an increase in force leads to a proportional increase in deformation.