/*! 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 19 The deformation of a wire under ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 19

The deformation of a wire under its own weight compared to the deformation of same wire subjected to a load equal to the weight of the wire is (A) Same (B) One-third (C) Half (D) One-fourth

Short Answer

Expert verified
The deformation of a wire under its own weight is equal to the deformation of the same wire subjected to a load equal to the weight of the wire. Therefore, the correct answer is (A) Same.

Step by step solution

01

Determine the formula for elongation

The formula for elongation (螖L) of a wire subjected to force (F) can be determined by the following formula: 螖L = \(\frac{FL}{EA}\) Where: - 螖L is the elongation (deformation), - F is the force applied, - L is the original length of the wire, - E is the modulus of elasticity of the material, - A is the cross-sectional area of the wire.
02

Calculate the elongation due to the wire's own weight

To find the elongation when the wire is only subjected to its own weight, we will need to account for the weight force acting on the entire wire. The weight force (W) can be calculated using the formula: W = 蟻gL Where: - W is the weight force, - 蟻 is the linear mass density of the wire (mass per unit length), - g is the acceleration due to gravity (approximately 9.81 m/s^2), - L is the original length of the wire. Using the formula for elongation from step 1, we can find the elongation due to the wire's own weight (螖L1): 螖L1 = \(\frac{(蟻gL)L}{EA}\)
03

Calculate the elongation when subjected to a load equal to the weight of the wire

In this case, we will consider the wire experiencing the load (F) equal to its own weight (W), so F = 蟻gL. Plugging this load value into our elongation formula, we can find the elongation when subjected to a load equal to the weight of the wire (螖L2): 螖L2 = \(\frac{(蟻gL)L}{EA}\) Notice that this equation is the same as the one used for calculating 螖L1 in step 2.
04

Determine the ratio of the two elongations

Now, we need to find the ratio of the elongation due to the wire's own weight (螖L1) to the elongation when subjected to a load equal to the weight of the wire (螖L2). This can be found using the following formula: Ratio = \(\frac{螖L1}{螖L2}\) As calculated in steps 2 and 3, 螖L1 and 螖L2 have the same formula, which gives us: Ratio = \(\frac{\frac{(蟻gL)L}{EA}}{\frac{(蟻gL)L}{EA}}\) The terms on the numerator and the denominator will cancel out, giving us a ratio of 1: Ratio = 1 This means that the deformation of a wire under its own weight is equal to the deformation of the same wire subjected to a load equal to the weight of the wire. The correct answer is (A) Same.

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!

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

Terminal velocity of water drops depends upon the (A) Radius of drop (B) Charge of drop (C) Temperature of drop (D) Velocity of light

If a small sphere is let to fall vertically in a large quantity of a still liquid of density smaller than that of the material of the sphere (A) At first its velocity increases, but soon approaches a constant value. (B) It falls with constant velocity all along from the very beginning. (C) At first it falls with a constant velocity which after some time goes on decreasing. (D) Nothing can be said about its motion.

An open vessel containing water is given a constant acceleration \(a\) in the horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle \(\theta\) given by (A) \(\theta=\tan ^{-1}\left(\frac{a}{g}\right)\) (B) \(\theta=\tan ^{-1}\left(\frac{g}{a}\right)\) (C) \(\theta=\sin ^{-1}\left(\frac{a}{g}\right)\) (D) \(\theta=\cos ^{-1}\left(\frac{g}{a}\right)\)

The viscous force on a small sphere of radius \(R\) moving in a fluid varies as \((\mathrm{A}) \propto R^{2}\) (B) \(\propto R\) (C) \(\propto(1 / R)\) \((\mathrm{D}) \propto(1 / R)^{2}\)

An incompressible, non-viscous fluid flows steadily through a cylindrical pipe, which has radius \(2 R\) at point \(A\) and radius \(R\) at point \(B\) farther along the flow direction. If the velocity of flow at point \(A\) is \(v\), the velocity of flow at point \(B\) will be (A) \(2 v\) (B) \(v\) (C) \(v / 2\) (D) \(4 \mathrm{v}\)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Particle Model of Matter

Read Explanation

Electromagnetism

Read Explanation

Physics of Motion

Read Explanation

Kinematics Physics

Read Explanation

Magnetism and Electromagnetic Induction

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.