/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 The temperature of a wire of len... [FREE SOLUTION] | 91Ó°ÊÓ

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The temperature of a wire of length \(1 \mathrm{~m}\) and area of cross-section \(1 \mathrm{~cm}^{2}\) is increased from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). If the rod is not allowed to increased in length, the force required will be \(\left(\alpha=10^{-5 / \circ} \mathrm{C}\right.\) and \(\left.Y=10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\) (a) \(10^{3} \mathrm{~N}\) (b) \(10^{4} \mathrm{~N}\) (c) \(10^{5} \mathrm{~N}\) (d) \(10^{9} \mathrm{~N}\)

Short Answer

Expert verified
The calculated force is 100 N, not matching any given options, likely due to an error.

Step by step solution

01

Identify Given Parameters

Let's start by identifying all the given parameters in the problem. The length of the wire is \( L = 1 \text{ m} \), the area of cross-section is \( A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \) (since \( 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \)), the temperature change is from \( 0^{\circ} \text{C} \) to \( 100^{\circ} \text{C} \) which implies a temperature change of \( \Delta T = 100^{\circ} \text{C} \), the coefficient of linear expansion \( \alpha = 10^{-5} /^{\circ} \text{C} \), and Young's modulus of elasticity \( Y = 10^{11} \text{ N/m}^2 \).
02

Formula for Thermal Stress

The force required to prevent the length from increasing can be found using the thermal stress formula, which is derived from the linear expansion and Young's modulus. The expression for thermal stress \( \sigma \) is \( \sigma = Y \alpha \Delta T \).
03

Calculate Thermal Stress

Substitute the given values into the thermal stress formula: \( \sigma = Y \alpha \Delta T = 10^{11} \times 10^{-5} \times 100 = 10^6 \text{ N/m}^2 \).
04

Calculate Force Using Stress

The formula for stress is given by \( \sigma = \frac{F}{A} \), where \( F \) is the force and \( A \) is the area. We can rearrange this to find the force \( F = \sigma A \). Substitute the calculated stress and area into the equation: \( F = 10^6 \times 10^{-4} = 100 \text{ N} \).
05

Determine Correct Answer

Compare the calculated force with the given options. None of the options provided matches \( 100 \text{ N} \). There seems to be a discrepancy between the given values and the provided multiple choices. However, based on calculations, \( F = 100 \text{ N} \) is the correct force required.

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Key Concepts

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

Young’s Modulus
Young’s Modulus, denoted by the symbol \( Y \), is a fundamental concept in physics and engineering. It measures a material's ability to withstand length changes under tensile (stretching) or compressive (squeezing) forces. In simpler terms, it tells us how "stiff" or "flexible" a material is. For example, steel with a high Young’s Modulus is very stiff.
Young's Modulus is expressed mathematically as:
  • \( Y = \frac{\text{stress}}{\text{strain}} \)
Here, stress is the force per unit area (\( \sigma = \frac{F}{A} \)) and strain is the fractional change in length (\( \epsilon = \frac{\Delta L}{L} \)).
This dimensionless ratio highlights how much force a material can endure before it deforms.
In our exercise, Young’s Modulus is given as \( 10^{11} \text{ N/m}^2 \), indicating the wire’s strong resistance to deformation under stress.
Coefficient of Linear Expansion
The Coefficient of Linear Expansion, often represented by \( \alpha \), describes how the size of an object changes with temperature. Specifically, it measures the fractional change in length per degree change in temperature. Materials expand when heated and contract when cooled, and \( \alpha \) quantifies this behavior.
The formula for calculating the change in length is:
  • \( \Delta L = \alpha \cdot L \cdot \Delta T \)
Here, \( L \) is the original length, and \( \Delta T \) is the temperature change.
In simpler terms, \( \alpha \) tells us how much longer or shorter an object becomes when the temperature changes. For example, materials with a high \( \alpha \) would expand significantly with a small temperature rise.
In the given exercise, \( \alpha \) is \( 10^{-5} /^{\circ}\text{C} \), indicating that the wire expands slightly for each degree Celsius increase in temperature.
Temperature Change Effect on Solids
Temperature changes have a substantial impact on the physical dimensions of solids, which can cause stress if the expansion or contraction is restricted. When solids are heated, their atoms vibrate more vigorously, resulting in expansion and an increase in length.
In our scenario, to prevent the wire from expanding when heated from \( 0^{\circ} \text{C} \) to \( 100^{\circ} \text{C} \), an internal force is required. This results in thermal stress:
  • Calculating this force involves the formula: \( \sigma = Y \alpha \Delta T \)
  • Where \( \Delta T \) is the change in temperature, \( Y \) and \( \alpha \) are material properties.
This thermal stress is, in essence, the stress developed when the material doesn't expand due to external constraints. It's crucial in engineering, particularly in designing structures that sustain temperature fluctuation, to prevent damage or failure.
Understanding the effect of temperature changes on solids assists in predicting behavior and planning for material selection and design processes.

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Most popular questions from this chapter

A rigid bar of mass \(15 \mathrm{~kg}\) is supported symmetrically by three wires each \(2 \mathrm{~m}\) long. These at each end are of copper and middle one is of iron. Determine the ratio of their diameters if each is to have the same tension. Young's modulus of elasticity for copper and steel are \(110 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) and \(190 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) respectively. [NCERT] (a) \(1: 1.3\) (b) \(1.3: 1\) (c) \(2.3: 1.3\) (d) \(2.3: 1\)

If \(x\) is longitudinal strain produced in a wire of Young's modulus \(Y\), then energy stored in the material of the wire per unit volume is (a) \(Y x^{2}\) (b) \(2 Y \underline{x^{2}}\) (c) \(\frac{1}{2} \gamma^{2} x\) (d) \(\frac{1}{2} Y x^{2}\)

One end of steel wire is fixed to ceiling of an elevator moving up with an acceleration \(2 \mathrm{~ms}^{-2}\) 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 \(g=10 \mathrm{~ms}^{-2}\) and \(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\) ) (a) \(4 \times 10^{11}\) (b) \(3 \times 10^{-6}\) (c) \(8 \times 10^{-6}\) (d) \(2 \times 10^{-6}\)

Two wires of the same material and length but diameters in the ratio \(1: 2\) are stretched by the same force. The potential energy per unit volume for the two wires when stretched will be in the ratio (a) \(16: 1\) (b) \(4: 1\) (c) \(2: 1\) (d) \(1: 1\)

The pressure of a medium is changed from \(1.010 \times 10^{5}\) to \(1.165 \times 10^{5} \mathrm{~Pa}\) and change in volume is 10\% keeping temperature constant. The Bulk modulus of the medium is \(\quad\) [III Screening 2005] (a) \(204.8 \times 10^{5} \mathrm{~Pa}\) (b) \(102.4 \times 10^{5} \mathrm{~Pa}\) (c) \(51.2 \times 10^{5} \mathrm{~Pa}\) (d) \(1.55 \times 10^{5} \mathrm{~Pa}\)

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