/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 When a certain weight is suspend... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 17

When a certain weight is suspended from a long uniform wire, its length increases by \(1 \mathrm{~cm}\). If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one, the increase in length will be (1) \(0.5 \mathrm{~cm}\) (2) \(2 \mathrm{~cm}\) (3) \(4 \mathrm{~cm}\) (4) \(8 \mathrm{~cm}\)

Short Answer

Expert verified
4 cm

Step by step solution

01

Understanding the Problem

We need to determine how the extension (increase in length) of a wire changes when the diameter of the wire is halved, while keeping material and length constant. The extension depends on several physical properties.
02

Using the Formula for Extension

The extension of a wire is described by the formula: \( \Delta L = \frac{F L}{A Y} \), where \( F \) is the force (weight) applied, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus. Since the material and length are constant, \( F \) and \( Y \) do not change.
03

Calculating Cross-sectional Area for the First Wire

The cross-sectional area \( A \) for a wire of diameter \( d \) is \( A = \frac{\pi d^2}{4} \). The first wire has an extension of \( 1 \text{ cm} \); let's denote its diameter as \( d_1 \).
04

Understanding the Effect of Halving Diameter

When the diameter is halved, the cross-sectional area of the second wire becomes \( A_2 = \frac{\pi (d_1/2)^2}{4} = \frac{\pi d_1^2}{16}\). This means the area of the second wire is \( \frac{1}{4} \) of the first wire's area.
05

Predicting the New Wire Extension

Because the area is reduced to \( \frac{1}{4} \), the extension \( \Delta L \) for the same force will be 4 times greater (since \( \Delta L \propto \frac{1}{A} \)). Hence, the increase in length for the wire with half the diameter will be \( 1 \text{ cm} \times 4 = 4 \text{ cm}.\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cross-Sectional Area
The cross-sectional area of a wire greatly influences its mechanical behavior, especially when a force is applied. Essentially, the cross-sectional area is the size of the cut made through the center of an object, perpendicular to its length. For a cylindrical wire, calculating this area requires knowing the diameter. The formula is:
\[ A = \frac{\pi d^2}{4} \]
  • The diameter \(d\) acts as a critical factor, as the area is proportional to the square of the diameter.
  • Halving the diameter reduces the area to one-fourth.
  • Such changes directly influence how materials extend when subjected to force.
Understanding how to calculate cross-sectional area helps in predicting extensions when forces are applied. This concept is vital in fields like materials science and engineering.
Mechanical Properties of Materials
Mechanical properties describe how materials behave under various forces. Young's modulus, a key property, measures a material's stiffness. It defines how much a material will deform under a given force.
  • Young's Modulus \(Y\) is given by the formula:
    \[ Y = \frac{stress}{strain} = \frac{F/A}{\Delta L/L} \]
  • It remains constant for a given material, making it a reliable indicator when altering dimensions, like halving the diameter.
  • These properties are crucial for predicting behaviors like the elongation of a wire when a weight is suspended.
By understanding mechanical properties, we can anticipate how materials behave under stress, ensuring they perform optimally in practical applications.
Force and Extension
When a force is applied to a material, it extends proportionally based on several factors. This relationship is critical when examining how materials stretch or compress.
  • The formula for extension \(\Delta L\) of a wire is:
    \[ \Delta L = \frac{F L}{A Y} \]
  • Since force \(F\) and Young's modulus \(Y\) are constant, changes in cross-sectional area \(A\) have a direct impact on extension.
  • A reduced cross-sectional area means increased extension; thus, halving the diameter results in quadruple the extension for the same force.
This relationship guides the engineering of materials to ensure they meet desired performance under specific loads.
JEE Advanced Mechanics
The concept of Young's modulus and related mechanical properties is essential in advanced mechanics, especially for competitive exams like JEE Advanced. Understanding how materials react under different forces and dimensions is crucial for solving complex problems.
  • Questions often involve manipulating formulas to find variables like extension or stress.
  • Knowledge of how dimensions affect mechanical behavior can lead to quick and accurate solutions.
  • This requires a grasp of how fundamental principles like cross-sectional area and force relate.
Mastery of these concepts not only aids in exams but also lays the groundwork for future studies in engineering and material science.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sphere of radius \(0.1 \mathrm{~m}\) and mass \(8 \pi \mathrm{kg}\) is attached to the lower end of a steel wire of length \(5 \mathrm{~m}\) and diameter \(10^{-5} \mathrm{~m}\). The wire is suspended from \(5.22 \mathrm{~m}\) high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Young's modulus of steel is \(2 \times 10^{11} \mathrm{Nm}^{-2}\). Find the velocity of the sphere at the lowest position in \(\mathrm{m} / \mathrm{s}\). (Given: \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )

The length of a wire is increased by \(1 \mathrm{~mm}\) on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is (1) \(0.25 \mathrm{~mm}\) (2) \(0.5 \mathrm{~mm}\) (3) \(2 \mathrm{~mm}\) (4) \(4 \mathrm{~mm}\)

Two wires of the same material have lengths in the ratio \(1: 2\) and their radii are in the ratio \(1: \sqrt{2}\). If they are stretched by applying equal forces, the increase in their lengths will be in the ratio (1) \(\sqrt{2}: 2\) (2) \(2: \sqrt{2}\) (3) \(1: 1\) (4) \(1: 2\)

A small but heavy block of mass \(10 \mathrm{~kg}\) is attached to a wire \(0.3 \mathrm{~m}\) long. Its breaking stress is \(4.8 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\). The area of the cross section of the wire is \(10^{-6} \mathrm{~m}^{2}\). The maximum angular velocity with which the block can be rotated in the horizontal circle is (1) \(4 \mathrm{rad} / \mathrm{s}\) (2) \(8 \mathrm{rad} / \mathrm{s}\) (3) \(10 \mathrm{rad} / \mathrm{s}\) (4) \(32 \mathrm{rad} / \mathrm{s}\)

The breaking stress for a substance is \(10^{6} \mathrm{~N} / \mathrm{m}^{2}\). What length of the wire of this substance should be suspended vertically so that the wire breaks under its own weight? (Given: density of material of the wire \(=4 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and \(g=10 \mathrm{~ms}^{-2}\) ) (1) \(10 \mathrm{~m}\) (2) \(15 \mathrm{~m}\) (3) \(25 \mathrm{~m}\) (4) \(34 \mathrm{~m}\)
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Nuclear Physics

Read Explanation

Rotational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.