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

Question

Dr. Abubacker · Accepted Answer

(b) Volume of cylindrical wire, $V = \frac{\pi r^{2}L}{4},$
where $x$ is the diameter of wire
Differentiating both sides
$\frac{dV}{dx} = \frac{\pi}{4}\left\lbrack 2xL + x^{2} \bullet \frac{dL}{dx} ightbrack$
Also, volume remains constant
$	herefore\ \ \frac{dV}{dx} = 0$
$	herefore\ \ \ 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 $= - \frac{1}{2}.$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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