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Chapter 11: Problem 62

A wire of Young's modulus \(1.5 \times 10^{12} \mathrm{Nm}^{-2}\) is stretched by a force so as to produce a strain of \(2 \times 10^{4}\). The energy stored per unit volume is (a) \(3 \times 10^{\mathrm{fi}} \mathrm{Jm}^{-1}\) (b) \(3 \times 10^{3} \mathrm{Jm}^{-1}\) (c) \(6 \times 10^{3} \mathrm{Jm}^{-3}\) (d) \(3 \times 10^{4} \mathrm{Jm}^{-1}\)

Short Answer

Expert verified
Energy stored per unit volume is approximately \(3 \times 10^{4} \mathrm{Jm}^{-3}\).

Step by step solution

01

Understand the formula for energy stored per unit volume

The energy stored per unit volume (also known as energy density) in a stretched wire can be calculated using the formula \( U = \frac{1}{2} \times \text{stress} \times \text{strain} \). This is derived from the potential energy in elastic deformation.
02

Calculate stress from Young's modulus and strain

Young's modulus \( Y \) is defined as the stress divided by the strain. Therefore, stress can be calculated as \( \text{stress} = Y \times \text{strain} \). Substituting the given values, \( Y = 1.5 \times 10^{12} \mathrm{Nm}^{-2} \) and \( \text{strain} = 2 \times 10^{-4} \), we get \( \text{stress} = 1.5 \times 10^{12} \times 2 \times 10^{-4} = 3 \times 10^{8} \mathrm{Nm}^{-2} \).
03

Substitute values into the energy density formula

Substitute the calculated stress and given strain back into the energy density formula: \( U = \frac{1}{2} \times 3 \times 10^{8} \times 2 \times 10^{-4} \).
04

Simplify and calculate the answer

Simplify the expression: \( U = \frac{1}{2} \times 6 \times 10^{4} = 3 \times 10^{4} \). Since \( U \) is measured in \( \mathrm{Jm}^{-3} \), we check the options and find that \( 3 \times 10^{4} \mathrm{Jm}^{-3} \) matches none, so there is a likely error in given options.

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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 is an important property of materials describing their ability to withstand changes in length under tension or compression. It is a measure of the material's stiffness and is denoted by the symbol \( Y \). Mathematically, Young's modulus is defined as the ratio of stress to strain:
  • \( Y = \frac{\text{Stress}}{\text{Strain}} \)
It has units of pressure, usually measured in \( \mathrm{Nm}^{-2} \).
In the case of a wire, Young's modulus helps us understand how much a wire will stretch when a force is applied. High Young's modulus means that the material is stiff and does not stretch easily, while a lower value indicates the material is more elastic and stretches more under the same force.
Stress
Stress is a concept that describes the internal force exerted by the material per unit area. When a material is subjected to an external force, stress is the measure of the force distribution across that material's cross-sectional area. It is given by the formula:
  • \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \)
The unit of stress is the same as pressure, \( \mathrm{Nm}^{-2} \).
Calculating stress is vital when determining how a material will react when a force is applied, and it helps predict structural integrity in various applications.
Strain
Strain is a measure of how much an object deforms in response to stress, expressed as a relative change in shape or size. It is a dimensionless quantity given by the ratio of change in length to the original length:
  • \( \text{Strain} = \frac{\text{Change in Length}}{\text{Original Length}} \)
Since strain is a ratio, it has no units. This concept helps quantify the extent of deformation experienced by a material when subjected to stress.
Understanding strain is essential in engineering and manufacturing to design components that can safely withstand operational stresses.
Elastic Deformation
Elastic deformation refers to a temporary change in the shape or size of an object due to stress. The object returns to its original shape once the force is removed. It occurs when the material's internal cohesive forces withstand an applied stress without permanent alteration.
In practical terms, elastic deformation is desirable because it allows materials to handle stresses without long-term damage. Many designs engineer materials to operate within their elastic limit for better durability.
The transition from elastic to plastic deformation marks the yield point, where permanent deformation begins.
Energy Density
Energy density, often used interchangeably with energy stored per unit volume, refers to the amount of energy stored in a stretched or compressed material. It is calculated using the formula:
  • \( U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \)
Energy density is expressed in \( \mathrm{Jm}^{-3} \), indicating energy stored per cubic meter. This value is critical in designing materials and structures because it predicts how much potential energy can be absorbed and released by elastic deformation.
Understanding energy density helps engineers create components that can safely perform with energy storage without reaching breaking points.

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Most popular questions from this chapter

A copper bar of length \(L\) and area of cross-section \(A\) is placed in a chamber at atmospheric pressure. If the chamber is evacuated, the percentage change in its volume will be (compressibility of copper is \(8 \times 10^{12} \mathrm{~m}^{2} \mathrm{~N}^{-1}\) and 1 atm \(\left.=10^{5} \mathrm{~N} \mathrm{~m}^{-2}\right)\) (a) \(8 \times 10^{-7}\) (b) \(8 \times 10^{-5}\) (c) \(1.25 \times 10^{-4}\) (d) \(1.25 \times 10^{-5}\)

Determine the volume contraction of a solid copper cube, \(10 \mathrm{~cm}\) on an edge, when subjected to a hydraulic pressure of \(7 \times 10^{6} \mathrm{~Pa} . K\) for copper \(=140 \times 10^{9} \mathrm{~Pa}\). [NCERT] (a) \(5 \times 10^{-7} \mathrm{~m}^{3}\) (b) \(4 \times 10^{-6} \mathrm{~m}^{3}\) [c) \(5 \times 10^{-8} \mathrm{~m}^{3}\) (d) \(6 \times 10^{-4} \mathrm{~m}^{2}\)

A height spring extends \(40 \mathrm{~mm}\) when stretched by a force of \(10 \mathrm{~N}\), and for tensions upto this value the extension is proportional to the stretching force. Two such springs are joined end-to-end and the double-length spring is stretched \(40 \mathrm{~mm}\) beyond its natural length. The total strain energy in (joule), stored in the double spring is (a) \(0.05\) (b) \(0.10\) (c) \(0.80\) (d) \(0.40\)

How much should the pressure on a litre of water be changed to compress it by \(0.10 \% .\) Bulk modulus of elasticity of water \(=2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\). (a) \(2.2 \times 10^{-6} \mathrm{~N}-\mathrm{m}^{-2}\) (b) \(4.2 \times 10^{-6} \mathrm{~N}-\mathrm{m}^{-2}\) (c) \(2.2 \times 10^{6} \mathrm{~N}-\mathrm{m}^{2}\) (d) \(4.2 \times 10^{4} \mathrm{~N}-\mathrm{m}^{2}\)

A rectangular bar \(2 \mathrm{~cm}\) in breadth and \(1 \mathrm{~cm}\) in depth and \(100 \mathrm{~cm}\) in length is supported at its ends and a load of \(2 \mathrm{~kg}\) is applied at its middle. If Young's modulus of the material of the bar is \(20 \times 10^{11}\) dyne \(\mathrm{cm}^{-2}\), the depression in the bar is (a) \(0.2450 \mathrm{~cm}\) (b) \(0.3675 \mathrm{~cm}\) (c) \(0.1225 \mathrm{~cm}\) (d) \(0.9800 \mathrm{~cm}\)
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