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Chapter 11: Problem 23

When a weight of \(5 \mathrm{~kg}\) is suspended from a copper wire of length \(30 \mathrm{~m}\) and diameter \(0.5 \mathrm{~mm}\), the length of the wire increases by \(2.4 \mathrm{~cm}\). If the diameter is doubled, the extension produced is (a) \(1.2 \mathrm{~cm}\) (b) \(0.6 \mathrm{~cm}\) (c) \(0.3 \mathrm{~cm}\) (d) \(0.15 \mathrm{~cm}\)

Short Answer

Expert verified
The extension produced with a doubled diameter is \(0.6 \, \text{cm}\), choice (b).

Step by step solution

01

Identify the formula for extension

To find the extension produced in a wire under a load, we use the formula for linear strain: \( \text{Strain} = \frac{\text{Extension} (\Delta L)}{\text{Original Length} (L)} = \frac{F}{A \cdot Y} \), where \( F \) is the force applied, \( A \) is the cross-sectional area of the wire, and \( Y \) is Young's modulus of the material.
02

Calculate initial extension and area

For the initial scenario, the force \( F = mg = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N} \). The cross-sectional area \( A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.5 \times 10^{-3}}{2}\right)^2 \, \text{m}^2 \). Extension is given as \( 2.4 \, \text{cm} = 0.024 \, \text{m} \).
03

Determine Young's modulus

Using the typical scenario with a known extension, we can solve for Young's modulus \( Y \) using \( \Delta L = \frac{FL}{A \cdot Y} \), substituting the known values: \( Y = \frac{FL}{\Delta L \cdot A} \). This calculation confirms the proportionality in the model between load, original length, elongation, and modulus.
04

Analyze the effect with doubled diameter

When the diameter of the wire is doubled, the cross-sectional area becomes \( 4 \times \text{original area} \) because \( A \propto d^2 \). Therefore, the new cross-sectional area \( A' = 4A \).
05

Calculate new extension

With the new area, use \( \Delta L' = \frac{FL'}{A' \cdot Y} = \frac{FL}{4A \cdot Y} \). Given that all other factors including length and load remain constant, the new extension \( \Delta L' = \frac{1}{4} \cdot 0.024 \, \text{m} = 0.006 \, \text{m} \) or \( 0.6 \, \text{cm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Linear Strain
Linear strain is a measure of deformation representing the elongation of a material when subjected to an external load. To calculate linear strain, we use the formula: \( \text{Strain} = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length and \( L \) is the original length of the material. It essentially describes how much a material stretches when a force is applied.
In this exercise, a copper wire experiences a linear strain due to the weight suspended from it. By understanding linear strain, students can predict how materials behave under different loads. Materials with a large linear strain tend to be more ductile, while those with smaller strain values are often more brittle.
The Significance of Cross-Sectional Area
Cross-sectional area plays a critical role in determining how a material responds to a force. It is the area of the cross-section of a cable or wire, usually perpendicular to its length, and is denoted by \( A \). The formula for the cross-sectional area of a circular wire with diameter \( d \) is \( A = \pi \left(\frac{d}{2}\right)^2 \).
When the diameter of the wire is doubled, as in this problem, the cross-sectional area increases by a factor of four. An increased cross-sectional area means the wire can support more stress without elongating as much, hence reducing the extension produced under the same load. In essence, a thicker wire can carry more load and undergo less deformation due to the increased area resisting the applied force.
Exploring the Stress-Strain Relationship
The stress-strain relationship is fundamental to understanding how materials deform under stress. It is often described by the formula \( \frac{F}{A} = E \times \frac{\Delta L}{L} \), where \( F \) is the external force applied, \( A \) is the cross-sectional area, \( E \) is Young's modulus, \( \Delta L \) is the extension, and \( L \) is the original length. Young's modulus \( E \) is a constant that indicates the stiffness of a material.
In this particular exercise, the relationship is assessed by altering the diameter of the wire, which inherently changes the cross-sectional area and hence the stress and strain. An increase in the wire's diameter quadruples the area, lowering the stress and effectively reducing the strain (extension) for the same force. Thus, understanding this relationship helps in designing materials and structures that withstand forces without failing.

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Most popular questions from this chapter

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass \(M\) is attached to each of the free ends at the centre of the rods. (a) Both the rods will clongate but there shall be no perceptible change in shape (b) The stecl rod will elongate and change shape but the rubber rod will only elongate (c) The steel rod will elongate without ary perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse. (d) The steel rod will clongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre

Two wires of the same material and length are stretched by the same force. Their masses are in the ratio 3:2. Their elongations are in the ratio (a) \(3: 2\) (b) \(9: 4\) (c) \(2: 3\) (d) \(4: 9\)

On increasing the length by \(0.5 \mathrm{~mm}\) in a steel wire of length \(2 \mathrm{~m}\) and area of cross-section \(2 \mathrm{~mm}^{2}\), the force required is \(\left[Y\right.\) for steel \(\left.=2.2 \times 10^{11} \mathrm{Nm}^{-2}\right]\) (a) \(1.1 \times 10^{5} \mathrm{~N}\) (b) \(1.1 \times 10^{4} \mathrm{~N}\) (c) \(1.1 \times 10^{3} \mathrm{~N}\) (d) \(1.1 \times 10^{2} \mathrm{~N}\)

Two wires of the same material have lengths in the ratio \(1: 2\) and their radii are in the ratio \(1: \sqrt{2}\). If they are stretched by applying equal forces, the increase in their lengths will be in the ratio of (a) \(\sqrt{2}: 2\) (b) \(2: \sqrt{2}\) (c) \(1: 1\) (d) \(1: 2\)

The upper end of a wire of radius \(4 \mathrm{~mm}\) and length \(100 \mathrm{~cm}\) is clamped and its other end is twisted through an angle of \(30^{\circ}\). Then angle of shear is (a) \(12^{*}\) (b) \(0.12^{\circ}\) (c) \(1.2^{*}\) (d) \(0.012^{*}\)
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