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Chapter 12: Problem 12

Two wires one of copper and other of steel having same cross-sectional area and lengths \(1.0 \mathrm{~m}\) and \(0.5 \mathrm{~m}\) respectively, are fastened end to end and stretched by a load \(M\). If copper wire is stretched by \(1 \mathrm{~mm}\), the total extension of the combined wire is: (Given: Young's modulii are \(Y_{\text {copper }}=1 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\), and \(\left.Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\) (a) \(0.125 \mathrm{~cm}\) (b) \(0.2 \mathrm{~cm}\) (c) \(0.120 \mathrm{~cm}\) (d) \(0.25 \mathrm{~cm}\)

Short Answer

Expert verified
The total extension is 0.2 cm, choice (b).

Step by step solution

01

Understand the Problem Context

We have two wires, one made of copper and the other of steel, both with the same cross-sectional area but different lengths. They are connected end-to-end and subjected to identical tensile forces due to a load M.
02

Use the Young's Modulus Formula

The stress-strain relationship for a material can be expressed using Young's modulus: \( \text{Strain} = \frac{\text{Stress}}{Y} \). For the copper wire, if the extension is \(1 \text{ mm} = 0.001 \text{ m}\), its strain is given by \(\frac{\Delta L_{c}}{L_{c}} = \frac{0.001}{1.0}\).
03

Calculate the Extension in the Steel Wire

The same force acts on both wires since they are in series. Use Young's modulus for the steel wire: \( \frac{\Delta L_{s}}{L_{s}} = \frac{\Delta L_{c} \cdot Y_{\text{copper}}}{L_{c} \cdot Y_{\text{steel}}} \). Substitute \( L_{s} = 0.5 \text{ m} \), \( \Delta L_{c} = 0.001 \text{ m} \), \( Y_{\text{copper}} = 1 \times 10^{11} \text{ N/m}^2 \), and \( Y_{\text{steel}} = 2 \times 10^{11} \text{ N/m}^2 \).
04

Calculate the Total Extension

The total extension of the wire system is the sum of extensions of both copper and steel wires: \( \Delta L_{\text{total}} = \Delta L_{c} + \Delta L_{s} \). Determine \( \Delta L_{s} \) from the previous step and add the two extensions.
05

Convert Units and Choose the Right Answer

Convert the total extension from meters to centimeters as all the given options are in centimeters. Verify which option matches the computed value.

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Key Concepts

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

Stress-Strain Relationship
When a material is subjected to a force, it undergoes a deformation, which is quantified by the terms stress and strain.
Stress is defined as the force per unit area applied to the material, while strain is the measure of the deformation or extension experienced by the material relative to its original length.
Young's modulus, denoted as \(Y\), is a material constant that describes the relationship between stress and strain, formulated as:
  • \( ext{Stress} = Y imes ext{Strain} \)

This formula can be rearranged to define strain as \( ext{Strain} = rac{ ext{Stress}}{Y} \).
This relationship indicates that the amount a material stretches or contracts is dependent on its Young's modulus value, with lower values indicating more flexibility.
Understanding this relationship is crucial when calculating the extension of materials under load, such as in the given problem with copper and steel wires.
Material Extension
The extension of a material is calculated using its initial length, the amount of force applied, and its Young鈥檚 modulus.
In the problem, two wires of different materials (copper and steel) undergo extension when subjected to the same force.
For the copper wire, we have:
  • Given extension: \(1 ext{ mm} = 0.001 ext{ m}\)
  • Initial length: \(L_c = 1.0 ext{ m}\)
  • Young's modulus: \(Y_{ ext{copper}} = 1 imes 10^{11} ext{ N/m}^2\)

The strain for the copper wire is \( rac{ ext{Extension}}{L_c} = rac{0.001}{1.0} \).
Using the stress-strain relationship, the strain in the steel wire can be calculated, and subsequently, the extension can be determined:
  • \( rac{ ext{Extension in steel}}{L_s} = rac{ ext{Extension in copper} imes Y_{ ext{copper}}}{L_c imes Y_{ ext{steel}}} \)
  • Here, \(L_s = 0.5 ext{ m}\) and \(Y_{ ext{steel}} = 2 imes 10^{11} ext{ N/m}^2\).

This method ensures each wire鈥檚 extension is considered based on its material properties and original length.
Series Connection of Wires
In a series connection of wires, the same tensile force acts throughout, ensuring that each material experiences the same stress.
This configuration allows us to analyze the extensions separately for each material and then combine them.
The total extension of the system is a simple sum of the individual extensions:
  • \( ext{Total Extension} = ext{Extension of copper wire} + ext{Extension of steel wire} \)

This step is crucial as it accounts for the contributions of both wires to the overall deformation.
After calculating the extensions for each wire, the values are added together to obtain the total elongation, which is then converted from meters to centimeters.
This conversion is necessary because the answer options are provided in centimeters, ensuring consistency in the final result presentation.
Understanding this concept is essential for solving problems involving multiple materials joined in series, particularly in structural and materials engineering.

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Most popular questions from this chapter

One end of a steel wire is fixed to ceiling of an elevator moving up with an acceleration \(2 \mathrm{~m} / \mathrm{s}^{2}\) and a load of \(10 \mathrm{~kg}\) hangs from other end. Area of cross-section of the wire is 2 \(\mathrm{cm}^{2}\). The longitudinal strain in the wire is: (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) and \(\left.Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\) (a) \(4 \times 10^{11}\) (b) \(3 \times 10^{-6}\) (c) \(8 \times 10^{-6}\) (d) \(2 \times 10^{-6}\)

The common radius of curvature ' \(r^{\prime}\), when two soap bubbles with radii \(r_{1}\) and \(r_{2}\left(r_{1}>r_{2}\right)\) come in contact, is: (a) \(r=\frac{r_{1}+r_{2}}{2}\) (b) \(r=\frac{r_{1} r_{2}}{r_{1}-r_{2}}\) (c) \(r=\frac{r_{1} r_{2}}{r_{1}+r_{2}}\) (d) \(r=\sqrt{r_{1} r_{2}}\)

When a sphere is taken to bottom of sea \(1 \mathrm{~km}\) deep, it contracts by \(0.01 \%\). The bulk modulus of elasticity of the material of sphere is: (Given: Density of water \(\left.=1 \mathrm{~g} / \mathrm{cm}^{3}\right)\) (a) \(9.8 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (b) \(10.2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (c) \(0.98 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.4 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\)

Young's modulus is defined for: (a) solid (b) liquid (c) gas (d) all of these

The work done against surface tension in formation of a drop of mercury of radius \(4 \mathrm{~cm}\) is : (surface tension for mercury \(=465\) dyne \(/ \mathrm{cm}\) ) (a) \(9.34 \times 10^{-3} \mathrm{~J}\) (b) \(10 \times 10^{-2} \mathrm{~J}\) (c) \(4 \times 10^{-3} \mathrm{~J}\) (d) \(466 \mathrm{~J}\)
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