There is no change in the volume of a wire due to change in its length on stretching. What is the possion's ratio of the material of the wire $\ldots \ldots \ldots$ (A) \(+0.5\) (B) \(-0.50\) (C) \(0.25\) (D) \(-0.25\)

Short Answer

Expert verified
The correct Poisson's ratio for the case of no change in volume of a wire due to change in its length on stretching is \(\sigma = 0\). However, this option is not available in the given choices. There might be an error in the provided choices or the question itself.

Step by step solution

01

Understand the given information

We are given that the volume of the wire does not change when its length changes. So, the product of the length, cross-sectional area, and Poisson's ratio should not change.
02

Represent the variables

Let the initial length of the wire be \(L_0\), the initial diameter be \(D_0\), and the volume be \(V_0\). Also, let the final length of the wire be \(L_1\), the final diameter be \(D_1\), and the Poisson's ratio be \(\sigma\).
03

Relate the volume to length and diameter

We know that the volume of the wire is the product of its length and cross-sectional area. Lets assume that the wire has a cylindrical shape, then the cross-sectional area could be represented as \(A_0 = \frac{\pi D_0^2}{4}\). Thus, we have: \[V_0 = L_0 A_0\] Since the volume of the wire does not change, its final volume is also \(V_0\). Therefore: \[V_0 = L_1 A_1\] Where \(A_1 = \frac{\pi D_1^2}{4}\). Now, we have: \[L_0 A_0 = L_1 A_1\]
04

Calculate longitudinal and transverse strains

Longitudinal strain is defined as the ratio of change in length to the original length: \[\epsilon_L = \frac{L_1 - L_0}{L_0}\] Transverse strain is defined as the ratio of change in diameter to the original diameter: \[\epsilon_T = \frac{D_1 - D_0}{D_0}\]
05

Write the equation for Poisson's ratio

Poisson's ratio is defined as the ratio of transverse strain to longitudinal strain: \[\sigma = -\frac{\epsilon_T}{\epsilon_L}\]
06

Express Poisson's ratio in terms of volume change

From steps 3 and 4, we have: \[\frac{L_1 - L_0}{L_0} = -\frac{D_1 - D_0}{D_0} \times \sigma\] Using the result from Step 3, we can write: \[\frac{L_1 - L_0}{L_0} = -\frac{L_0}{L_1} \times \sigma\]
07

Solve for Poisson's ratio

Now, we can rearrange the equation from Step 6 to find \(\sigma\): \[\sigma = -\frac{(L_1 - L_0)L_1}{L_0^2}\] Since there is no change in volume, we have: \[L_1 = L_0\] Therefore, Poisson's ratio is: \[\sigma = -\frac{(L_1 - L_0)L_0}{L_0^2} = -\frac{0}{L_0} = 0\] From the given choices, none of them match the correct value of Poisson's ratio. It's possible that there's a mistake in the given choices or the question is flawed.

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 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

A soap bubble of radius \(r\) is blown up to form a bubble of radius $2 \mathrm{r}\( under isothermal conditions if the \)\mathrm{T}$ is the surface tension of soap solution the energy spent in the slowing is. (A) \(3 \pi \mathrm{Tr}^{2}\) (B) \(6 \pi \mathrm{Tr}^{2}\) (C) \(12 \pi \mathrm{Tr}^{2}\) (D) \(24 \pi \mathrm{Tr}^{2}\)

A rubber ball when taken to the bottom of a \(100 \mathrm{~m}\) deep take decrease in volume by \(1 \%\) Hence, the bulk modulus of rubber is $\ldots \ldots \ldots . . .\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]$ (A) \(10^{6} \mathrm{~Pa}\) (B) \(10^{8} \mathrm{~Pa}\) (C) \(10^{7} \mathrm{~Pa}\) (D) \(10^{9} \mathrm{~Pa}\)

An incompressible fluid flows steadily through a cylindrical pipe which has radius \(2 \mathrm{r}\) at point \(\mathrm{A}\) and radius \(\mathrm{r}\) at \(\mathrm{B}\) further along the flow direction. It the velocity at point \(\mathrm{A}\) is \(\mathrm{V}\), its velocity at point \(\mathrm{B}\). (A) \(2 \mathrm{~V}\) (B) V (C) \((\mathrm{V} / 2)\) (D) \(4 \mathrm{~V}\)

Melting point of ice (A) Increases with increasing pressure (B) Decreases with increasing pressure (C) Is independent of pressure (D) is proportional of pressure

The amount of work done in blowing a soap bubble such that its diameter increases from \(d\) to \(D\) is \((T=\) Surface tension of solution) (A) \(4 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (B) \(8 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (C) \(\pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (D) \(2 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\)

See all solutions

Recommended explanations on English Textbooks

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