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A copper rod of length \(l_{0}\) at \(0^{\circ} \mathrm{C}\) is placed on smooth surface. The rod is heated up to \(100^{\circ} \mathrm{C}\). The longitudinal strain developed is : \((\alpha=\) coefficient of linear expansion) (a) \(\frac{100 l_{0} \alpha}{l_{0}+100 l_{0} \alpha}\) (b) \(100 \alpha\) (c) zero (d) none of these

Short Answer

Expert verified
The longitudinal strain developed is \(100 \alpha\), which corresponds to option (b).

Step by step solution

01

Understanding the Problem

The problem is asking for the longitudinal strain in a copper rod that is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). The given parameters include the initial length \(l_0\) and the coefficient of linear expansion \(\alpha\). Strain is defined as the change in length divided by the original length.
02

Determine Change in Length

When the rod is heated, its length changes. The change in length \(\Delta l\) due to the temperature change \(\Delta T\) is given by the formula:\[ \Delta l = l_0 \cdot \alpha \cdot \Delta T \]where \(\Delta T = 100^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C} = 100^{\circ} \mathrm{C}\). So, \[ \Delta l = l_0 \cdot \alpha \cdot 100 \]
03

Calculate Longitudinal Strain

Strain is defined as the ratio of the change in length to the original length:\[ \text{Strain} = \frac{\Delta l}{l_0} = \frac{l_0 \cdot \alpha \cdot 100}{l_0} \]Simplifying this gives:\[ \text{Strain} = 100 \cdot \alpha \]
04

Identify the Correct Option

Compare the calculated strain with the given choices. The calculated longitudinal strain is \(100 \alpha\), which matches option (b) in the given choices.

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Key Concepts

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

Coefficient of Linear Expansion
The Coefficient of Linear Expansion is a fundamental concept in understanding how materials expand or contract with temperature changes. It is denoted by the symbol \( \alpha \). Essentially, this coefficient tells us how much a unit length of a material will expand per degree change in temperature. This value is unique for different materials, and knowing it helps predict dimensional changes in objects due to temperature variation.

Here's what happens when a material is heated:
  • The atoms gain energy and start vibrating more.
  • This increased vibration pushes the atoms slightly further apart.
  • As a result, the material expands, increasing its length.
Understanding the Coefficient of Linear Expansion is crucial in many engineering applications. By knowing \( \alpha \), engineers can design structures that can accommodate or counteract thermal expansion, ensuring structural integrity under varying temperatures.
Longitudinal Strain
Longitudinal Strain is a measure of deformation representing how much a material stretches or compresses along its length. It is calculated as the ratio of the change in length (\( \Delta l \)) to the original length (\( l_0 \)). Mathematically, it is expressed as:

\[ \text{Strain} = \frac{\Delta l}{l_0} \]This concept helps us quantify how thermal expansion affects the overall dimensions of an object. When a material, like a copper rod, is subjected to heat, it tends to increase in size. The longitudinal strain tells us by what fraction of its original length it underwent expansion.

In real-world applications, measuring longitudinal strain is vital for ensuring that materials perform safely under thermal stresses. This is crucial in construction, manufacturing, and aerospace, where materials undergo varying temperatures.
Change in Length Formula
The Change in Length Formula links the physical change experienced by a material to the temperature change it undergoes. This formula is given by:

\[ \Delta l = l_0 \cdot \alpha \cdot \Delta T \]

Where:
  • \( \Delta l \) is the change in length.
  • \( l_0 \) is the initial length of the material.
  • \( \alpha \) is the coefficient of linear expansion.
  • \( \Delta T \) is the change in temperature.
By using this formula, one can predict how much a material will change in size with a given temperature change. This is particularly useful in situations such as fitting railway tracks, designing bridges, or creating components for electronic devices, where precise dimensional tolerances are critical.

With this understanding, engineers can better control thermal expansion effects, ensuring that components fit together properly even across a wide range of temperatures.

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Most popular questions from this chapter

If same amount of heat is supplied to two identical spheres (one is hollow and other is solid), then: (a) the expansion in hollow is greater than the solid (b) the expansion in hollow is same as that in solid (c) the expansion in hollow is lesser than the solid (d) the temperature of both must be same to each other.

At temperature \(T_{0}\), two metal strips of length \(l_{0}\) and thickness \(d\), is bolted so that their ends coincide. The upper strip is made up of metal \(A\) and have coefficient of expansion \(\alpha_{A}\) and lower strip is made up of metal \(B\) with coefficient of expansion \(\alpha_{B} \cdot\left(\alpha_{A}>\alpha_{B}\right)\). When temperature of their blastic strip is in-seased from \(T_{0}\) to \(\left(T_{0}+\Delta T\right)\), on strip become longer than the other and blastic strip is bend in the form of a circle as shown in fig. Calculate the radius of furvature \(R\) of the strip: (a) \(R:=\frac{\left[2+\left(\alpha_{A}+\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T}\) (b) \(R=\frac{\left[2-\left(\alpha_{A}+\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T}\) (c) \(R=\frac{\left[2+\left(\alpha_{A}-\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T_{i}}\) (d) \(R=\frac{\left[2-\left(\alpha_{A}-\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta \mathcal{k}}\)

Using the following, data, at what temperature will the wood just sink in benzene? Density of wood at \(0^{\circ} \mathrm{C}=8.8 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\) Density of benzene at \(0^{\circ} \mathrm{C}=9 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\) Cubical expansivity of wood \(=1.5 \times 10^{-4} \mathrm{~K}^{-1}\) Cubical expansivity of benzene \(=1.2 \times 10^{-3} \mathrm{~K}^{-1}\) (a) \(27^{\circ} \mathrm{C}\) (b) \(21.7^{\circ} \mathrm{C}\) (c) \(31^{\circ} \mathrm{C}\) (d) \(31.7^{\circ} \mathrm{C}\)

The molar heat capacity of rock salt at low temperatures varies with temperature according to Debye's \(T^{3}\) law". Thus, \(C=k \frac{T^{3}}{\theta^{3}}\) where, \(k=1940 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \theta=281 \mathrm{~K}\). How much heat is required to raise the temperature of 2 moles of rock salt from \(10 \mathrm{~K}\) to \(50 \mathrm{~K}\) ? (a) \(800 \mathrm{~J}\) (b) \(373 \mathrm{~J}\) (c) \(273 \mathrm{~J}\) (d) None of these

What is the change in the temperature on Fahrenheit scale and on Kelvin scale, if a iron piece is heated from \(30^{\circ}\) to \(90^{\circ} \mathrm{C} ?\) (a) \(108^{\circ} \mathrm{F}, 60 \mathrm{~K}\) (b) \(100^{\circ} \mathrm{F}, 55 \mathrm{~K}\) (c) \(100^{\circ} \mathrm{F}, 65 \mathrm{~K}\) (d) \(60^{\circ} \mathrm{F}, 108 \mathrm{~K}\)

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