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Chapter 11: Problem 12

A load of \(4.0 \mathrm{~kg}\) is suspended from a ceiling through a steel wire of length \(2.0 \mathrm{~m}\) and radius \(2.0 \mathrm{~mm}\). It is found that the length of the wire increases by \(0.031 \mathrm{~mm}\) as equilibrium is achieved. Taking, \(g=3.1 \pi \mathrm{ms}^{-2}\), the Young's modulus of steel is (a) \(2.0 \times 10^{8} \mathrm{Nm}^{-2}\) (b) \(2.0 \times 10^{9} \mathrm{Nm}^{-2}\) (c) \(2.0 \times 10^{11} \mathrm{Nm}^{-2}\) (d) \(2.0 \times 10^{13} \mathrm{Nm}^{-2}\)

Short Answer

Expert verified
The Young's modulus of steel is \(2.0 \times 10^{11} \mathrm{Nm}^{-2}\).

Step by step solution

01

Determine the force applied

The force applied on the wire by the load is the weight of the load. We calculate it using the formula: \[ F = mg \]where \( m = 4.0 \text{ kg} \) and \( g = 3.1\pi \text{ m/s}^2 \).Substituting the given values, we get:\[ F = 4.0 \times 3.1\pi = 12.4\pi \text{ N} \].
02

Calculate the cross-sectional area of the wire

The wire is cylindrical, so its cross-section is a circle. The area \( A \) is calculated using the formula:\[ A = \pi r^2 \]where \( r \) is the radius of the wire, given as \(2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}\).Thus,\[ A = \pi (2.0 \times 10^{-3})^2 = \pi \times 4.0 \times 10^{-6} = 4.0 \times 10^{-6} \pi \text{ m}^2 \].
03

Define Young's Modulus formula

Young's Modulus (\( E \)) is defined as the ratio of the stress to the strain in the material. We can express it by the formula:\[ E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/L_0}{\Delta L/L_0} = \frac{F \times L_0}{A \times \Delta L} \]where \( L_0 \) is the original length of the wire \(= 2.0 \text{ m} \), \( \Delta L \) is the change in length \(= 0.031 \text{ mm} = 0.031 \times 10^{-3} \text{ m} \), and \( A \) is the cross-sectional area that we calculated.
04

Substitute values and solve for Young's Modulus

Substituting all known values into the formula for Young's Modulus, we get:\[ E = \frac{12.4\pi \times 2.0}{4.0 \times 10^{-6} \pi \times 0.031 \times 10^{-3}} \]This simplifies to:\[ E \approx \frac{24.8}{1.24 \times 10^{-10}} \approx 2.0 \times 10^{11} \text{ N/m}^2 \].
05

Identify the correct answer choice

Compare the computed value of Young's Modulus with the given options:(a) \(2.0 \times 10^{8} \text{ Nm}^{-2}\)(b) \(2.0 \times 10^{9} \text{ Nm}^{-2}\)(c) \(2.0 \times 10^{11} \text{ Nm}^{-2}\)(d) \(2.0 \times 10^{13} \text{ Nm}^{-2}\)The correct option is (c) \(2.0 \times 10^{11} \text{ Nm}^{-2}\).

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Key Concepts

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

Stress
Stress is a fundamental concept in materials science and engineering, representing how much force an object experiences per unit area. It is essential in determining how materials deform under various forces. Stress is defined mathematically as:\[ \text{Stress} \sigma = \frac{F}{A} \]Here, \( F \) represents the force applied to the material, and \( A \) symbolizes the cross-sectional area over which the force is distributed.
  • Normal Stress: Occurs when the force is perpendicular to the surface, as seen in stretching or compressing.
  • Shear Stress: Develops when forces are parallel to the surface area, leading to a shearing effect.
In our exercise, the normal stress is calculated with the weight force of the load, making it crucial for determining how the steel wire would elongate once the weight is applied.
Strain
Strain is the measure of deformation that the material undergoes due to stress. It quantifies the degree of deformation and is derived from the change in the material's dimensions relative to its original size. Mathematically, it's expressed as:\[ \text{Strain} \epsilon = \frac{\Delta L}{L_0} \]Where \( \Delta L \) represents the change in length, and \( L_0 \) is the original length of the material.
  • Elastic Strain: This is the reversible deformation that a material undergoes when the applied load is removed.
  • Plastic Strain: This is the irreversible deformation that remains even when the load is removed.
In the context of the solution provided, the strain of the steel wire was calculated based on its elongation from the suspension of the load. Understanding strain helps us predict how materials would perform under various conditions.
Force calculation
The calculation of force exerted on an object is crucial in physics and engineering. It helps in predicting how objects will react in different scenarios. The force can be calculated using the formula for weight, which is:\[ F = mg \]In this equation, \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
  • This force calculation helps determine the stress on materials, essential for understanding their strength and durability.
  • In the problem, the force exerted on the steel wire is computed using the mass of the load and a given gravitational constant.
Grasping the concept of force calculation is vital for proceeding with further analysis regarding stress and other related attributes of materials.
Cross-sectional area
The cross-sectional area is a critical factor in understanding how materials behave under various forces. It is the area of a material's section perpendicular to the axis of force application. For cylindrical objects like wires, it's calculated using the formula for the area of a circle:\[ A = \pi r^2 \]Where \( r \) is the radius of the cylinder. The cross-sectional area influences how stress distributes across a material.
  • The thinner the material, the greater the stress for a given force.
  • In our problem, the wire's cross-sectional area was calculated using its given radius, which is crucial for finding the stress and, subsequently, the Young's Modulus.
Understanding cross-sectional area assists in the design and selection of materials based on load-bearing requirements and expected stress levels.

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Most popular questions from this chapter

Find the extension produced in a copper of length \(2 \mathrm{~m}\) and diameter \(3 \mathrm{~mm}\), when a force of \(30 \mathrm{~N}\) is applied. Young's modulus for copper \(=1.1 \times 10^{11} \mathrm{Nm}^{-2}\) (a) \(0.2 \mathrm{~mm}\) (b) \(0.04 \mathrm{~mm}\) (c) \(0.08 \mathrm{~mm}\) (d) \(0.68 \mathrm{~mm}\)

The twisting couple per unit twist for a solid cylinder of radius \(3 \mathrm{~cm}\) is \(0.1 \mathrm{~N}-\mathrm{m} .\) The twisting couple per unit twist, for a hollow cylinder of same material with outer and inner radius \(5 \mathrm{~cm}\) and \(4 \mathrm{~cm}\) respectively, will be (a) \(0.1 \mathrm{~N}-\mathrm{m}\) (b) \(0.455 \mathrm{~N}-\mathrm{m}\) (c) \(0.91 \overline{N-m}\) (d) \(1.82 \mathrm{~N}-\mathrm{m}\)

A wire \(\left(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\right)\) has length \(1 \mathrm{~m}\) and area of cross-section \(1 \mathrm{~mm}^{2}\). The work required to increase its length by \(2 \mathrm{~mm}\) is (a) \(400 \mathrm{~J}\) (b) \(40 \mathrm{~J}\) (c) \(4 \mathrm{~J}\) (d) \(0.4 \mathrm{~J}\)

A \(100 \mathrm{~N}\) force stretches the length of a hanging wire by \(0.5 \mathrm{~mm}\). The force required to stretch a wire, of the same material and length but having four times the diameter, by \(0.5 \mathrm{~mm}\) is (a) \(100 \mathrm{~N}\) (b) \(400 \mathrm{~N}\) (c) \(1200 \mathrm{~N}\) (d) \(1600 \mathrm{~N}\)

The following data were obtained when a wire was stretched within the elastic region Force applied to wire \(100 \mathrm{~N}\) Area of cross-section of wire \(10^{-6} \mathrm{~m}^{2}\) Extension of wire \(2 \times 10^{-9} \mathrm{~m}\) Which of the following deductions can be correctly made from this data? 1\. The value of Young's modulus is \(10^{11} \mathrm{Nm}^{-2}\) 2\. The strain is \(10^{-3}\) 3\. The energy stored in the wire when the load is applied is \(10 \mathrm{~J}\) (a) \(1,2,3\) are correct (b) 1,2 are correct (c) 1 only (d) 3 only
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