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

• A. $+\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $+\frac{1}{4}$
• D. $-\frac{1}{4}$

Answer 1

Answer: (B)

Volume of cylindrical wire, $V=\frac{\pi r^{2} L}{4}$ , where $x$ is the diameter of wire.
Differentiating both sides,
$\frac{dV}{dx}=\frac{\pi }{4}\left[2 xL + x^{2} \cdot \frac{dL}{dx}\right]$
Also, volume remains constant
$\therefore \frac{dV}{dx}=0$
$\therefore \, 2xL+x^{2}\frac{dL}{dx}=0$
$\Longrightarrow 2xL=-x^{2}\frac{dL}{dx}$
$\Longrightarrow \frac{\frac{dx}{x}}{\frac{dL}{L}}=-\frac{1}{2}$
Poisson's ratio $=-\frac{1}{2}$