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Chapter 7: Problem 9

Two wires of the same material and length are stretched by the same force. Their masses are in the ratio \(3: 2\). Their elongations are in the ratio (1) \(3: 2\) (2) \(9: 4\) (3) \(2: 3\) (4) \(4: 9\)

Short Answer

Expert verified
The elongation ratio is \(2:3\), so option (3) is correct.

Step by step solution

01

Understanding the Problem

We have two wires made from the same material and with the same initial length stretched by the same force. Their masses are in the ratio of \(3:2\). We need to find the ratio of their elongations.
02

Identify Relevant Formula

The elongation of a wire under a force is given by Hooke's Law: \( \frac{FL}{AE} \), where \( F \) is the force, \( L \) is the original length, \( A \) is the cross-sectional area, and \( E \) is the Young's modulus of the material.
03

Relate Mass to Area

The mass of the wire \( m \) is related to its volume, and for a cylinder (wire shape), volume is \( AL \), thus, \( m = \rho AL \) where \( \rho \) is the density. Given \( m_1 : m_2 = 3 : 2 \), and \( \rho \) and \( L \) are constant, we find \( A_1 : A_2 = 3 : 2 \).
04

Determine Ratio of Elongations

The elongation \( \Delta L \) varies inversely with the cross-sectional area \( A \) since \( \Delta L = \frac{FL}{AE} \). Therefore, \( \Delta L_1 : \Delta L_2 = A_2 : A_1 \). Using \( A_1 : A_2 = 3 : 2 \), we find \( \Delta L_1 : \Delta L_2 = 2 : 3 \).
05

Conclusion

The ratio of the elongations of the wires is \(2:3\), which corresponds to option (3).

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Key Concepts

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

Hooke's Law
Hooke's Law describes the behavior of materials when they are subjected to a force. It states that the extension or compression of an object is directly proportional to the force applied to it, provided the material's elastic limit is not exceeded. This relationship is commonly expressed as:
  • \( F = k \cdot x \)
where:
  • \( F \) is the force applied to the object
  • \( k \) is the stiffness or spring constant of the material
  • \( x \) is the displacement or change in length
This formula is essential for understanding how different materials behave under stress. For the example with the wires, we see how applying the same force to two wires of the same material but different dimensions affects their elongation. Hooke's Law is a foundational concept in mechanics and material science, used extensively to calculate stresses and strains in materials.
Cross-sectional Area
The cross-sectional area of a wire is a critical factor in determining its elongation under force. Essentially, it is the area of the cut face of the wire when it is sliced perpendicular to its length. This area affects how the material behaves under a load because:
  • A larger cross-sectional area can distribute the force over a greater area, leading to less elongation.
  • A smaller cross-sectional area concentrates the force, resulting in greater elongation.
In our exercise, even though both wires have the same material and are subjected to the same force, their differing masses imply different cross-sectional areas. Since mass is proportional to the cross-sectional area (given the same length and density), we find that the wire with the smaller mass has a greater elongation.
Young's Modulus
Young's modulus, often denoted by \( E \), is a measure of the stiffness of a material. It is defined as the ratio of stress to strain in a material in the linear elasticity regime of a uniaxial deformation:
  • \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \)
where:
  • \( F \) is the force applied
  • \( A \) is the cross-sectional area
  • \( \Delta L \) is the change in length
  • \( L \) is the original length of the material
This parameter is crucial because it provides insight into how much a material will deform under a given stress. In the exercise, since both wires are made from the same material, they share the same Young's modulus value. This constant allows us to predict that, all else being equal, changes in elongation are owing to differences in cross-sectional areas and not material stiffness.
Density of Materials
Density is a property of materials that relates mass to volume. It is expressed as \( \rho \):
  • \( \rho = \frac{m}{V} \)
where:
  • \( m \) is the mass
  • \( V \) is the volume
In the context of wires, volume can be expressed as the product of cross-sectional area and length. Hence, for a wire,
  • Mass = Density \( \times \) Cross-sectional Area \( \times \) Length
In our exercise, although the wires have the same density and initial length, their masses are given in a ratio of \( 3:2 \). This indicates different cross-sectional areas, directly affecting how these materials elongate under the same force. Understanding density helps us realize how material distribution within the volume determines its overall mass and thus its response to forces.

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Most popular questions from this chapter

A uniform elastic rod of cross-section area \(A\), natural length \(L\) and Young's modulus \(Y\) is placed on a smooth horizontal surface. Now two horizontal forces (of magnitude \(F\) and \(3 F\) ) directed along the length of rod and in opposite direction act at two of its ends as shown. After the rod has acquired steady state, the extension of the rod will be (1) \(\frac{2 F}{Y A} L\) (2) \(\frac{4 F}{Y A} L\) (3) \(\frac{F}{Y A} L\) (4) \(\frac{3 F}{2 Y A} L\)

A metal cube of side \(10 \mathrm{~cm}\) is subjected to a shearing stress of \(10^{6} \mathrm{~N} / \mathrm{m}^{2}\). Calculate the modulus of rigidity (in \(\left._{0} \times 10^{8} \mathrm{Nm}^{-2}\right)\) if the top of the cube is displaced by \(0.05 \mathrm{~cm}\) with respect to its bottom.

The breaking stress for a metal is \(7.8 \times 10^{9} \mathrm{Nm}^{-2}\). The density of the metal is \(7800 \mathrm{~kg} \mathrm{~m}^{-3}\). If \(g=10 \mathrm{~N} \mathrm{~kg}^{-1}\), Find the maximum length of the wire made of this metal which may be suspended without breaking.

On applying a stress of \(x \mathrm{~N} / \mathrm{m}^{2}\), the length of wire of some material gets doubled. Value of Young's modulus for the material of wire in \(\mathrm{N} / \mathrm{m}^{2}\), is (assume Hooke's law to be valid and go for approx. results) (1) \(x\) (2) \(2 x\) (3) \(x / 2\) (4) Insufficient information

Bulk modulus of water is \(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\). The change in pressure required to increase the density of water by \(0.1 \%\) is (1) \(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (2) \(2 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}\) (3) \(2 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\) (4) \(2 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\)
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