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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) \(\frac{1}{2} Y x^{2}\)

Short Answer

Expert verified
The correct answer is (d) \( \frac{1}{2} Y x^{2} \).

Step by step solution

01

Identify the Formula for Strain Energy

The strain energy per unit volume or energy density in a material under simple tension or compression is given by the formula \( u = \frac{1}{2} \sigma x \), where \( \sigma \) is the stress and \( x \) is the longitudinal strain.
02

Relate Stress to Young's Modulus

The stress \( \sigma \) is related to Young’s modulus \( Y \) and strain \( x \) by the equation \( \sigma = Y \times x \). This relationship comes from the definition of Young's modulus.
03

Substitute Stress into the Energy Formula

Replace the stress \( \sigma \) in the energy density formula with \( Y \times x \). This gives us: \[ u = \frac{1}{2} (Y \times x) \times x \].
04

Simplify the Expression

Simplify the expression \( \frac{1}{2} (Y \times x) \times x \) to find \( u = \frac{1}{2} Y x^2 \).
05

Select the Correct Option

Based on the simplified formula, the correct choice is (d) \( \frac{1}{2} Y x^{2} \).

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Key Concepts

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

Strain Energy
Strain energy is the energy stored in a material due to deformation under applied stress. Think of it like the potential energy stored in a rubber band when stretched. When an object is subjected to tension or compression, its internal structure undergoes a change, storing energy in this form.
In the context of materials and engineering, understanding strain energy is crucial. It represents how much energy per unit volume is stored in a material and is an indicator of that material's capacity to resist deformation.
The formula for strain energy density, which is the energy stored per unit volume, is given by:
  • \( u = \frac{1}{2} \sigma x \)
Here, \( u \) stands for strain energy per unit volume, \( \sigma \) represents stress, and \( x \) denotes the longitudinal strain. By calculating this, engineers and physicists can predict and analyze how a material will perform under various loads, ensuring safety and efficiency in design.
Longitudinal Strain
Longitudinal Strain is an important concept key to understanding how materials behave when forces are applied. It refers to the deformation or change in length of an object in the direction of the applied force.
When you stretch or compress a material, you are affecting its dimensions. Longitudinal strain, represented by the symbol \( x \), quantifies this change as a ratio. Specifically, it's the change in length divided by the original length:
  • Longitudinal Strain \( x = \frac{\text{Change in Length}}{\text{Original Length}} \)
This dimensionless number gives a clear idea of how much deformation has occurred. For example, if a metal rod of original length 1 meter is stretched to 1.02 meters, the longitudinal strain is 0.02.
Understanding this concept is crucial because it tells us how responsive or resistant a material is to changes in its shape, which is vital in applications ranging from construction to manufacturing.
Stress-Strain Relationship
The stress-strain relationship is foundational to understanding how materials respond to forces and pressure. Stress, denoted by \( \sigma \), is the internal force per unit area, while strain, defined as earlier, is the deformation per unit length.
Young's Modulus \( Y \) is a central player in this relationship, representing a material's stiffness. It's derived from the ratio of stress to strain:
  • \( Y = \frac{\sigma}{x} \)
This glaring equation tells us that for a given material, the ratio of applied stress to resulting strain remains constant, at least within the elastic region.
Simplifying, we see stress expressed as: \( \sigma = Y \times x \). In essence, Young's Modulus helps us predict how much a material will stretch or compress under a given load, letting designers and engineers choose appropriate materials for different applications.
This relationship helps ensure that structures and components operate safely within their elastic limits, preventing permanent deformation or failure.

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Most popular questions from this chapter

Assertion Identical springs of steel and copper are equally stretched. More work will be done on the steel spring. Reason Steel is more elastic than copper.

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