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A wire can sustain the weight of \(20 \mathrm{~kg}\) before breaking. If the wire is cut into two equal parts, each part can sustain a weight of (1) \(10 \mathrm{~kg}\) (2) \(20 \mathrm{~kg}\) (3) \(40 \mathrm{~kg}\) (4) \(35 \mathrm{~kg}\)

Short Answer

Expert verified
Option (1) 10 kg.

Step by step solution

01

Understanding the Problem

The original wire can sustain a weight of \(20 \mathrm{~kg}\) before breaking. We need to determine how much weight each half of the wire can sustain when it is cut into two equal parts.
02

Assessing Wire Capacity When Divided

When the wire is divided into two equal parts, the total weight capacity of the entire wire is distributed evenly between the two halves. This means each part will have half the capacity of the original wire.
03

Calculating the Sustenance Capacity for Each Half

Since the original wire has a capacity of \(20 \mathrm{~kg}\), each half will sustain half of this, which is calculated by dividing the original capacity by 2: \( \frac{20}{2} = 10 \mathrm{~kg}\).
04

Selecting the Correct Answer

The capacity of each half of the wire to sustain weight is \(10 \mathrm{~kg}\). Therefore, option (1) \(10 \mathrm{~kg}\) is the correct answer.

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Key Concepts

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

Tensile Strength
Tensile strength is a crucial concept in understanding how materials respond to stress before they fracture. It is the maximum amount of tensile stress that a material can withstand while being stretched or pulled before failing.
When we talk about tensile strength, it specifically relates to how much force a material can bear without breaking.
In the case of the wire in the exercise, it can hold a weight equivalent to a tensile strength of the force needed to support 20 kg.
Once the wire is divided into two equal parts, each part retains only half of this original tensile strength, holding up to 10 kg each before breaking.
Understanding tensile strength helps in predicting the behavior and performance of materials in real-world applications.
  • Useful for construction and engineering to ensure structural integrity.
  • Determines the longevity and durability of products like cables and ropes.
Weight Distribution
Weight distribution is how weight is spread out across a particular length or surface area of a material.
In simple terms, it involves how a material supports weight across its entire length.
Thinking of the wire in the exercise, once it's cut in half, each segment takes on equal responsibility to support the load.
For instance, if the original wire could support 20 kg, slicing it into two equal parts means each new segment can logically only sustain 10 kg.
This happens because each piece bears an equal amount of stress spread evenly over its length.
Balanced weight distribution ensures stability and prevents material fatigue, essential in fields such as:
  • Engineering designs where balancing forces is critical.
  • Product manufacturing that involves load-bearing components.
Material Properties
Material properties refer to the intrinsic characteristics inherent in a material, influencing how it behaves under different conditions.
These properties include tensile strength, elasticity, hardness, and ductility.
Considering our wire example, these properties help determine how it will act when a weight is applied.
The wire's ability to return to its original shape after being stressed, or its ability not to break easily under stress, is part of its elastic and tensile properties.
Material properties impact:
  • The selection of materials based on the needs of specific applications.
  • How a material can be processed or forged into different shapes.
Understanding these properties ensures that the right material is chosen for the task, avoiding material failure and ensuring safety and efficiency.

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Most popular questions from this chapter

A \(0.05 \mathrm{~m}\) cube has its upper face displaced by \(0.2 \mathrm{~cm}\) by a tangential force of \(8 \mathrm{~N}\). Calculate the modulus of rigidity (in \(\left.\times 10^{4} \mathrm{Nm}^{-2}\right)\) of the material of the cube.

According to Hooke's law, within the elastic limit stress/strain = constant. This constant depends on the type of strain or the type of force acting. Tensile stress might result in compressional or elongative strain; however, a tangential stress can only cause a shearing strain. After crossing the elastic limit, the material undergoes elongation and beyond a stage breaks. All modulus of elasticity are basically constants for the materials under stress. Two wires of the same material have length and radius \((l, r)\) and \(\left(2 l, \frac{r}{2}\right)\). The ratio of their Young's modulus is (1) \(1: 2\) (2) \(2: 3\) (3) \(2: 1\) (4) \(1: 1\)

Two parallel and opposite forces, each of magnitude \(4000 \mathrm{~N}\), are applied tangentially to the upper and lower faces of a cubical metal block \(25 \mathrm{~cm}\) on a side. Find the displacement of the upper surface relative to the lower surface (in \(\times 10^{-5} \mathrm{~cm}\) ). The shear modulus for the metal is \(80 \mathrm{GPa}\).

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

A rubber rope of length \(8 \mathrm{~m}\) is hung from the ceiling of a room. What is the increase in length of rope due to its own weight? (Given: Young's modulus of elasticity of rubber \(=5 \times 10^{6} \mathrm{~N} / \mathrm{m}\) and density of rubber \(=1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Take \(\left.g=10 \mathrm{~ms}^{-2}\right)\) (1) \(1.5 \mathrm{~mm}\) (2) \(6 \mathrm{~mm}\) (3) \(24 \mathrm{~mm}\) (4) \(96 \mathrm{~mm}\)

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