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Chapter 7: Problem 22

Two identical wires of iron and copper with their Young's modulus in the ratio \(3: 1\) are suspended at same level. They are to be loaded so as to have the same extension and hence level. Ratio of the weight is (1) \(1: 3\) (2) \(2: 1\) (3) \(3: 1\) (4) \(4: 1\)

Short Answer

Expert verified
The weight ratio is 3:1 (Option 3).

Step by step solution

01

Understand the Problem

We're given two wires made of different materials (iron and copper) with identical dimensions. The Young's modulus of iron is three times that of copper, and we need to load these wires so they exhibit the same extension.
02

Recall Young's Modulus Formula

The formula for Young's modulus is \( Y = \frac{FL}{A \Delta L} \), where \( F \) is the force (or weight), \( L \) is the original length, \( A \) is the cross-sectional area, and \( \Delta L \) is the extension. Since the extensions and lengths are the same, the expression simplifies.
03

Set the Extensions Equal

Since the extensions \( \Delta L \) for both wires are the same and using \( Y = \frac{F}{A} \) due to identical dimensions \( L \), we equate the force-weight relationships: \( \frac{W_1}{Y_1} = \frac{W_2}{Y_2} \).
04

Apply Given Young's Modulus Ratio

We know \( \frac{Y_1}{Y_2} = 3 \). So, \( Y_1 = 3Y_2 \). Substitute into the equation: \( \frac{W_1}{3Y_2} = \frac{W_2}{Y_2} \), simplifying to \( W_1 = 3W_2 \).
05

Solve for Weight Ratio

From the relationship \( W_1 = 3W_2 \), solve for the weight ratio: \( \frac{W_1}{W_2} = 3 \). Therefore, the weights must be in the ratio 3:1.

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Key Concepts

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

Elasticity
When we talk about elasticity, we refer to a material's ability to return to its original shape or length after a force is removed.
A key factor in elasticity is Young's Modulus, which measures how stiff a material is.
It is crucial because it helps compare how different materials will respond when forces are applied.

  • The formula for Young's Modulus is given by: \[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
  • Where:
    • \( Y \) is Young's Modulus
    • \( F \) is the force applied (in this case, the weight)
    • \( L \) is the original length of the material
    • \( A \) is the cross-sectional area
    • \( \Delta L \) is the extension (change in length)
The greater the Young's Modulus, the more force is needed to stretch or compress the material, meaning it is less elastic.
In our exercise, despite iron being less elastic than copper (having a greater Young's Modulus), we want both to extend the same length.
Material Properties
Every material has distinct properties that determine its behavior under force.
These include parameters like density, thermal conductivity, and, importantly, Young's Modulus.
Young's Modulus tells us about the stiffness of a material.

In this problem, we are comparing two wires made of iron and copper.
They have the same shape and size, but their internal properties cause them to behave differently when a force is applied.
  • Iron has a higher Young's Modulus compared to copper, making it stiffer.
  • With Young's Modulus ratio \(3:1\), it means iron requires more force to achieve the same extension as copper.
An understanding of these properties will help in identifying how different materials will react in engineering applications.
Mechanics
Mechanics is the branch of physics concerned with the behavior of physical bodies when subjected to forces or displacements.
In this context, it helps us understand how different forces affect the two wires.
To make both wires have the same extension, we have to consider the mechanics behind it.Let's break it down:
  • Both the force \(F\) applied on the wires and the Young's Modulus \(Y\) of the wires play crucial roles,
  • Given that \[\frac{W_1}{Y_1} = \frac{W_2}{Y_2} \] we balance the forces \(W_1\) and \(W_2\) using the modulus,
  • Since \(Y_1 = 3Y_2\), we derive that \[W_1 = 3W_2\]
This means that the force required on iron must be three times greater than that on copper to achieve equivalent extensions.
Understanding these mechanics is key in applied physics and engineering to ensure structures behave correctly under different types of loads.

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Most popular questions from this chapter

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.

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) \(2 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\)

A wire of cross section \(A\) is stretched horizontally between two clamps located \(2 l\) apart. A weight \(W\) is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance \(x<1\) the strain produced is (1) \(\frac{2 x^{2}}{l^{2}}\) (2) \(\frac{x^{2}}{l^{2}}\) (3) \(\frac{x^{2}}{2 l^{2}}\) (4) none of these

When a body of mass \(M\) is attached to lower end of a wire (of length \(L\) ) whose upper end is fixed, then the elongation of the wire is \(l\). In this situation, mark out the correct statement(s). (1) Loss in gravitational potential energy of \(M\) is \(M g l .\) (2) Elastic potential energy stored in the wire is \(\frac{M g l}{2}\). (3) Elastic potential energy stored in the wire is \(\mathrm{Mgl}\). (4) Elastic potential energy stored in the wire is \(\frac{M g l}{3}\).

Two wires of the same length and same material but radil in the ratio of \(1: 2\) are stretched by unequal forces to produce equal elongation. The ratio of the two forces is (1) \(1: 1\) (2) \(1: 2\) (3) \(1: 3\) (4) \(1: 4\)
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