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A wire \(3 \mathrm{~m}\) in length and \(1 \mathrm{~mm}\) in diameter at \(30^{\circ} \mathrm{C}\) is kept in a low temperature at \(-170^{\circ} \mathrm{C}\) and is stretched by hanging a weight of \(10 \mathrm{~kg}\) at one end. The change in length of the wire is \(\left(Y=2 \times 10^{11} \mathrm{Nm}^{-2}, g=10 \mathrm{~ms}^{-2}\right.\) and \(\left.\alpha=1.2 \times 10^{-50} \mathrm{C}^{-1}\right) \quad\) IUP SEE 2006] (a) \(5.2 \mathrm{~mm}\) (b) \(2.5 \mathrm{~mm}\) (c) \(52 \mathrm{~mm}\) (d) \(25 \mathrm{~mm}\)

Short Answer

Expert verified
Resulting change: 6.818 mm.

Step by step solution

01

Calculate the Thermal Contraction

The first step is to find out how the wire contracts due to cooling. The change in length per unit change in temperature is given by the formula \(\Delta L = L \cdot \alpha \cdot \Delta T\), where \(\Delta L\) is the change in length, \(L\) is the original length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature. Here, \(L = 3 \text{ m} = 3000 \text{ mm}\), \(\alpha = 1.2 \times 10^{-5} \text{ }^\circ\text{C}^{-1}\), and \(\Delta T = -170 - 30 = -200 \text{ }^\circ\text{C}\). Thus, substituting, \(\Delta L = 3000 \times 1.2 \times 10^{-5} \times 200 = 7.2 \text{ mm}\). This value is negative because it is a contraction.
02

Calculate the Stretching Due to Weight

Next, we calculate how much the wire stretches due to the weight. The formula for the change in length under force is \(\Delta L = \dfrac{F \cdot L}{A \cdot Y}\), where \(F\) is the force applied, \(A\) is the cross-sectional area, and \(Y\) is Young's modulus. The force \(F = m \cdot g = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}\). The cross-sectional area \(A = \pi \cdot r^2 = \pi \cdot (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6} \text{ m}^2\). Substituting these in, we get \(\Delta L = \dfrac{100 \times 3}{\pi \times 0.25 \times 10^{-6} \times 2 \times 10^{11}}\). Calculating further, we find \(\Delta L = \dfrac{300}{\pi \times 0.5 \times 10^5} \approx 0.382 \text{ mm}\).

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Key Concepts

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

Thermal Expansion
When a material is subjected to a change in temperature, it often experiences a change in dimensions, a phenomenon known as thermal expansion. This happens because the atoms within the material move more vigorously during a temperature increase, or slow down during cooling, affecting the overall dimensions of the object.
The amount that an object expands or contracts depends on its material properties and the degree of temperature change. The linear coefficient of thermal expansion (\( \alpha \)) quantifies how much a given material expands or contracts per degree change in temperature. The formula used to calculate the change in length (\( \Delta L \)) is:
  • \( \Delta L = L \cdot \alpha \cdot \Delta T \)
In this formula, \( L \) is the initial length, \( \alpha \) is the linear expansion coefficient, and \( \Delta T \) is the temperature change. In our exercise, as the wire is cooled from \( 30^{\circ} \text{C} \) to \( -170^{\circ} \text{C} \), the contraction is calculated using this principle, resulting in a contraction of \( 7.2 \text{ mm} \).
Cross-Sectional Area
The cross-sectional area of a wire or any cylindrical object determines how it responds under a particular load. It is essentially the area of the circle that you would see if you cut the object perpendicular to its length. Understanding this concept is crucial because it helps us calculate how stress is distributed across the material.
The cross-sectional area (\( A \)) of a wire is calculated using the formula for the area of a circle:
  • \( A = \pi \cdot r^2 \)
where \( r \) is the radius of the wire. Given a diameter of \( 1 \text{ mm} \), the radius will be half that, \( 0.5 \text{ mm} \). Thus, the area, which is essential for calculating stress and elongation due to applied forces, is determined in our exercise as \(\pi \times 0.25 \times 10^{-6} \text{ m}^2 \).
Force and Pressure
Force and pressure are related but distinct concepts that play major roles in this exercise. Force refers to any interaction that, when unopposed, will change the motion of an object. In our case, it's the gravitational force acting on the mass that stretches the wire. It's calculated as:
  • \( F = m \cdot g \)
where \( m \) is the mass (10 kg in our problem) and \( g \) is the acceleration due to gravity, approximately \( 10 \text{ m/s}^2 \). The force applied is therefore \( 100 \text{ N} \).
Pressure, on the other hand, is the force exerted per unit area. The relation between force, pressure, and area is:
  • \( P = \frac{F}{A} \)
However, in contexts involving wire elongation, like this, we often go further to calculate stress using similar principles. Understanding these definitions allows us to comprehend how different forces and pressures affect materials.
Temperature Change Effects
Temperature changes have significant impacts on material properties, including length, volume, and structural integrity. When an object like a wire undergoes a temperature decrease, it often contracts. This contraction occurs because molecules within the material move less and bunch closer together as they lose energy. However, a simultaneous application of external forces can also cause stretching or deformation.
In the exercise at hand, the wire contracts by \( 7.2 \text{ mm} \) as it cools due to thermal contraction, yet the applied gravitational force causes it to stretch by \( 0.382 \text{ mm} \). Ultimately, the net change in length is a combination of these effects. Such an understanding is essential when examining engineering scenarios where materials are subject to both thermal changes and mechanical forces.

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Most popular questions from this chapter

A thick rope of rubber of density \(1.5 \times 10^{3} \mathrm{kgm}^{-3}\) and Young's modulus \(5 \times 10^{6} \mathrm{Nm}^{-2}, 8 \mathrm{~m}\) in length is hung from the ceiling of a room, the increase in its length due to its own weight is (a) \(9.6 \times 10^{-2} \mathrm{~m}\) (b) \(19.2 \times 10^{-2} \mathrm{~m}\) (c) \(9.6 \times 10^{-3} \mathrm{~m}\) [d) \(9.6 \mathrm{~m}\)

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