Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 898

A \(5 \mathrm{~m}\) long aluminum wire $\left[\mathrm{Y}=7 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\right]\( of diameter \)3 \mathrm{~mm}\( supports a \)40 \mathrm{~kg}$ mass. In order to have the same elongation in a copper wire $\mathrm{Y}=12 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ of the same length under the same weight, the diameter should now be in \(m m \ldots \ldots \ldots \ldots\) (A) \(1.75\) (B) \(1.5\) (C) \(2.5\) (D) \(5.0\)

Short Answer

Expert verified
The required diameter of the copper wire for the same elongation is approximately \(1.5 \ mm\), so the answer is (B) 1.5 mm.

Step by step solution

01

List the known information

We know that: - Aluminum wire length (L1) = 5m - Aluminum wire diameter (D1) = 3mm = 0.003m - Aluminum Wire Young's modulus (Y1) = 7 x 10^10 N/m^2 - Copper Wire length (L2) = 5m - Copper Wire Young's modulus (Y2) = 12 x 10^10 N/m^2 - Mass supported (m) = 40kg - Gravity (g) = 9.81 m/s^2
02

Calculate the elongation in the aluminum wire

To find the elongation in the aluminum wire, we will use the equation: Elongation (ΔL) = (F * L) / (A * Y) First, we need to calculate the force (F) exerted by the weight: F = m * g = 40 * 9.81 = 392.4 N Next, let's calculate the area (A) of the aluminum wire cross-section: A = (π * D1^2) / 4 = (π * (0.003)^2) / 4 ≈ 7.069 x 10^-6 m^2 Now, we can find the elongation (ΔL1) in the aluminum wire: ΔL1 = (392.4 * 5) / (7.069 x 10^-6 * 7 x 10^10) ≈ 2.002 x 10^-5 m
03

Find the required diameter of the copper wire for the same elongation

We should use the same equation of elongation for the copper wire but rearranging it to solve for the area (A2): A2 = (F * L2) / (ΔL1 * Y2) Substituting the known values, we get: A2 = (392.4 * 5) / (2.002 x 10^-5 * 12 x 10^10) ≈ 1.628 x 10^-6 m^2 Now that we have the area, we will solve for the diameter (D2): D2 = √(4 * A2 / π) ≈ 0.001442 m Convert the diameter to millimeters: D2 ≈ 1.442 mm
04

Determine the answer

From the given options: (A) 1.75 mm (B) 1.5 mm (C) 2.5 mm (D) 5.0 mm The closest option to our calculated diameter of 1.442 mm is (B) 1.5 mm. So the answer is (B) 1.5 mm.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Read the assertion and reason carefully to mark the correct option out of the option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the reason. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. (e) If assertion is false but reason is true. Assertion: Specific heat capacity is the cause of formation of land and sea breeze. Reason: The specific heat of water is more then land. (A) a (B) \(b\) (C) \(c\) (D) \(\mathrm{d}\) \((\mathrm{E}) \mathrm{e}\)

Two small drops mercury, each of radius \(\mathrm{r}\), coalesces the form a single large drop. The ratio of the total surface energies before and after the change is. (A) \(1: 2^{(1 / 2)}\) (B) \(2^{(1 / 3)}: 1\) (C) 2: 1 (D) \(1: 2\)

When more than \(20 \mathrm{~kg}\) mass is tied to the end of wire it breaks what is maximum mass that can be tied to the end of a wire of same material with half the radius? (A) \(20 \mathrm{~kg}\) (B) \(5 \mathrm{~kg}\) (C) \(80 \mathrm{~kg}\) (D) \(160 \mathrm{~kg}\)

Two wires \(A \& \mathrm{~B}\) of same length and of the same material have the respective radius \(\mathrm{r}_{1} \& \mathrm{r}_{2}\) their one end is fixed with a rigid support and at the other end equal twisting couple is applied. Then what will we be the ratio of the angle of twist at the end of \(\mathrm{A}\) and the angle of twist at the end of \(\mathrm{B}\). (A) \(\left(\mathrm{r}_{1}^{2} / \mathrm{r}_{2}^{2}\right)\) (B) \(\left(\mathrm{r}_{2}^{2} / \mathrm{r}_{1}^{2}\right)\) (C) \(\left(\mathrm{r}_{2}^{4} / \mathrm{r}_{1}^{4}\right)\) (D) \(\left(\mathrm{r}_{1}^{4} / \mathrm{r}_{2}^{4}\right)\)

A square frame of side \(\mathrm{L}\) is dipped in a liquid on taking out a membrance is formed if the surface tension of the liquid is \(\mathrm{T}\), the force acting on the frame will be. (A) \(2 \mathrm{TL}\) (B) \(4 \mathrm{TL}\) (C) \(8 \mathrm{TL}\) (D) \(10 \mathrm{TL}\)
See all solutions

Recommended explanations on English Textbooks

English Grammar

Read Explanation

Lexis and Semantics Summary

Read Explanation

Listening and Speaking

Read Explanation

Prosody

Read Explanation

Email

Read Explanation

Lexis and Semantics

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.

Sign-up for free