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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.

Short Answer

Expert verified
The maximum length of the wire is approximately 100,000 meters.

Step by step solution

01

Understanding the Problem

We are given the breaking stress of a metal, its density, and the acceleration due to gravity. We need to find the maximum length of a wire of this metal that can be suspended without breaking.
02

Defining Breaking Stress

Breaking stress () is the maximum stress that a material can withstand while being stretched or pulled before failing or breaking. It is given by the formula \( \sigma = \frac{F}{A} \), where \(F\) is the force and \(A\) is the cross-sectional area.
03

Relating Force, Mass, and Weight

The force on the wire due to its weight (weight = mass \( \times \) gravity) can be expressed as \( F = mg = \rho A L \cdot g \), where \( \rho \) is the density, \( A \) is the area, \( L \) is the length, and \( g \) is the gravitational acceleration.
04

Setting the Breaking Condition

The wire will break when the force due to its weight equals the breaking stress. Therefore, set the breaking stress equal to the weight per unit area: \( \sigma = \rho g L \).
05

Solving for Maximum Length

Rearrange the equation to solve for \( L \):\[ L = \frac{\sigma}{\rho g} \]Substitute the given values: \( \sigma = 7.8 \times 10^9 \ \mathrm{Nm}^{-2} \), \( \rho = 7800 \ \mathrm{kg/m^3} \), and \( g = 10 \ \mathrm{N/kg} \):\[ L = \frac{7.8 \times 10^9}{7800 \times 10} \approx 100 000 \mathrm{~m} \]
06

Confirming the Result

Review the calculated length to ensure it makes sense with the given parameters and check for any computational errors.

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Key Concepts

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

Density of a Material
Density is a fundamental property of materials that measures how much mass is contained in a given volume. It is expressed in units of mass per unit volume, such as kilograms per cubic meter (kg/m³). To calculate density, we use the formula:

\[ \text{Density} (\rho) = \frac{\text{Mass (m)}}{\text{Volume (V)}} \]

Understanding density is crucial because it helps us predict how materials will behave. In the context of a wire, the density helps determine the wire's weight since the mass of any section is the product of density, cross-sectional area, and length. When you suspend a wire, its density contributes to the overall force due to gravity that it experiences, which is crucial in calculating the maximum length of the wire before it breaks. Imagine if the density was higher, the wire would be heavier per unit length, thus requiring more strength to stay suspended without snapping.
Maximum Length of a Wire
When determining how long a wire can be before it risks breaking under its own weight, we must consider the breaking stress of the material and its density. The breaking stress, in this problem, is the maximum stress a material can endure while being stretched or pulled before failing. To find out the maximum length, we start by understanding that the wire's weight increases with its length.

Using the formula:
\[ L = \frac{\sigma}{\rho g} \]
where \( L \) is the maximum length, \( \sigma \) is the breaking stress, \( \rho \) is the density, and \( g \) is gravitational acceleration. All these factors determine how long the wire can be without collapsing under its own weight. In our example, substituting the given values calculated the maximum length to be about 100,000 meters. This formula highlights how the wire's potential length decreases with heavier materials, as high density contributes more weight per unit.
Force and Stress Relationship
In physics, force and stress have a closely knit relationship, especially in the context of materials science. Stress refers to the force applied per area, represented by the formula:
\[ \sigma = \frac{F}{A} \]
where \( \sigma \) is the stress, \( F \) is the force, and \( A \) is the cross-sectional area. When a force is applied to a wire, the stress depends on both the magnitude of the force and the area over which it is distributed.

This relationship is crucial in understanding when a material will fail. For a wire hanging vertically, the force is due to the weight of the wire itself, which can be further described as the product of its mass and the gravitational force (\( mg \)), with mass expressed through density as \( \rho \cdot A \cdot L \). The wire will break when the stress exceeds its breaking stress, \( \sigma \). This interplay between force and the wire's cross-sectional area directly impacts how much stress it experiences and can withstand, thus influencing its structural integrity.

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Most popular questions from this chapter

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

The edges of an aluminium cube are \(10 \mathrm{~cm}\) long. One face of the cube is firmly fixed to a vertical wall. A mass of \(100 \mathrm{~kg}\) is then attached to the opposite face of the cube. Shear modulus of aluminium is \(25 \times 10^{9} \mathrm{~Pa}\), the vertical deflection in the face to which mass is attached is (1) \(4 \times 10^{-4} \mathrm{~m}\) (2) \(4 \times 10^{-7} \mathrm{~m}\) (3) \(25 \times 10^{-6} \mathrm{~m}\) (4) \(6 \times 10^{-7} \mathrm{~m}\)

The density of water at the surface of ocean is \(\rho\). If the bulk modulus of water is \(B\), then the density of ocean water at depth, when the pressure is \(\alpha p_{0}\) and \(p_{0}\) is the atmospheric pressure, is (1) \(\frac{\rho B}{B-(\alpha-1) p_{0}}\) (2) \(\frac{\rho B}{B+(\alpha-1) p_{0}}\) (3) \(\frac{\rho B}{B-\alpha_{0}}\) (4) \(\frac{\rho B}{B+\alpha p_{0}}\)

A massive stone pillar \(20 \mathrm{~m}\) high and of uniform cross section rests on a rigid base and supports a vertical load of \(5.0 \times 10^{5} \mathrm{~N}\) at its upper end. If the compressive stress in the pillar is not exceed \(16 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\), what is the minimum cross-sectional area of the pillar? (Density of the stone \(=2.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) take \(\left.g=10 \mathrm{~N} / \mathrm{kg} .\right)\) (1) \(0.15 \mathrm{~m}^{2}\) (2) \(0.25 \mathrm{~m}^{2}\) (3) \(0.35 \mathrm{~m}^{2}\) (4) \(0.45 \mathrm{~m}^{2}\)

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

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