/*! 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 3 An elongation of \(0.1 \%\) in a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 3

An elongation of \(0.1 \%\) in a wire of cross-sectional area \(10^{-6} \mathrm{~m}^{2}\) causes a tension of \(100 \mathrm{~N}\). The Young's modulus is (A) \(10^{12} \mathrm{~N} / \mathrm{m}^{2}\) (B) \(10^{11} \mathrm{~N} / \mathrm{m}^{2}\) (C) \(10^{10} \mathrm{~N} / \mathrm{m}^{2}\) (D) \(10^{2} \mathrm{~N} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The short answer is: The Young's modulus is (B) \(10^{11} N/m^2\).

Step by step solution

01

Understand the given information

We are given the following information: 1. Elongation: \(0.1\% \) 2. Cross-sectional area: \(10^{-6} m^2\) 3. Tension: \(100 N\) Our goal is to find the Young's modulus (Y).
02

Determine the elongation in meters

First, we need to convert the elongation percentage to a ratio and then multiply by the original length (L) to find the elongation in meters. Elongation (螖L) = \(\frac{0.1}{100} \times L = 0.001L\)
03

Use the formula for Young's modulus

Young's modulus (Y) is the measure of the stiffness of a material and is given by the formula: \(Y = \frac{stress}{strain}\) Here, Stress (蟽) = \(\frac{Force}{Area} = \frac{100 N}{10^{-6} m^2}\) Strain (蔚) = \(\frac{\Delta L}{L} = 0.001\) Now, we can find the Young's modulus by substituting the values in the formula: \(Y = \frac{stress}{strain} = \frac{\frac{100 N}{10^{-6} m^2}}{0.001}\)
04

Calculate the Young's modulus

Now let's calculate the Young's modulus: \(Y = \frac{100 N}{10^{-6} m^2 \times 0.001} = \frac{100 N}{10^{-9} m^2} = 10^{11} N/m^2\) Therefore, the correct answer is: (B) \(10^{11} N/m^2\)

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

A ball of mass \(m\) and radius \(r\) is released in viscous liquid. The value of its terminal velocity is proportional to (A) \((1 / r)\) only (B) \(\mathrm{m} / \mathrm{r}\) (C) \((m / r)^{1 / 2}\) (D) \(m\) only

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is \(T\), density of liquid is \(\rho\), and \(L\) is its latent heat of vaporization. (A) \(\sqrt{T / \rho L}\) (B) \(T / \rho L\) (C) \(2 T / \rho L\) (D) \(\rho L / T\)

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities \(d_{1}\) and \(d_{2}\) are filled in the tube. Each liquid subtends \(90^{\circ}\) angle at the centre. Radius joining their interface makes an angle \(\alpha\) with vertical plane. Ratio \(\frac{d_{1}}{d_{2}}\) is [2014](A) \(\frac{1+\sin \alpha}{1-\sin \alpha}\) (B) \(\frac{1+\cos \alpha}{1-\cos \alpha}\) (C) \(\frac{1+\tan \alpha}{1-\tan \alpha}\) (D) \(\frac{1+\sin \alpha}{1-\cos \alpha}\)

Equal volumes of water and alcohol are mixed together. The density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) and the density of alcohol is \(800 \mathrm{~kg} / \mathrm{m}^{3}\). The density of the mixture is (A) \(900 \mathrm{~kg} / \mathrm{m}^{3}\) (B) \(1100 \mathrm{~kg} / \mathrm{m}^{3}\) (C) \(875 \mathrm{~kg} / \mathrm{m}^{3}\) (D) \(950 \mathrm{~kg} / \mathrm{m}^{3}\)

Two capillary tubes of same length \(l\) but radii \(r_{1}\) and \(r_{2}\) are fitted in parallel to the bottom of a vessel. The pressure head is \(P .\) What should be the radius \(r\) of the single tube that can replace the two tubes, so that the rate of flow is same as before? (A) \(r=r_{1}+r_{2}\) (B) \(r=r_{1}^{2}+r_{2}^{2}\) (C) \(r^{4}=r_{1}^{4}+r_{2}^{4}\) (D) \(\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r_{2}}\)
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Physics of Motion

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Force

Read Explanation

Modern Physics

Read Explanation

Classical Mechanics

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.