First slide

Elastic Modulie

Question

There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is:

Moderate

A

$+ \frac{1}{2}$
B

$- \frac{1}{2}$
C

$+ \frac{1}{4}$
D

$- \frac{1}{4}$

Solution

Volume of cylindrical wire, V = $\frac{{πx}^{2} L}{4}$ , where x is diameter of wire
Differentiating both sides
$\frac{dV}{dx} = \frac{π}{4} [2 xL + x^{2} . \frac{dL}{dx}]$
Also volume remains constant $\frac{dV}{dx} = 0$
2xL + $x^{2} \frac{dL}{dx} = 0 \Rightarrow 2 xL = - x^{2} \frac{dL}{dx}$
$\Rightarrow$ $\frac{\frac{dx}{x}}{\frac{dL}{L}} = - \frac{1}{2}$
$\Rightarrow$ Poisson's ratio = - $\frac{1}{2}$
So the correct choice is (b).

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App