Question

A metal wire is stretched by a load. Then the fractional change in its volume $\frac{\Delta V}{V}$ is proportional to

• A. $\frac{\Delta L}{L}$
• B. ${\left(\frac{\Delta L}{2}\right)}^2$
• C. $\sqrt{\frac{\Delta L}{L}}$
• D. ${\left(\frac{\Delta L}{L}\right)}^3$

Answer 1

The correct option is A $\frac{\Delta L}{L}$

From the data given in the question, a load $F$ acting on the metallic wire is as shown in figure.


It is given that, the wire will stretch.
$\therefore$ The length of wire will increase
We know, Poisson's ratio, $e=-\frac{\frac{\Delta d}{d}}{\frac{\Delta L}{L}}--\left(1\right)$
Volume of wire, $V=A\times L=\frac{\pi d^{2}L}{4}$ where $d$ is the diameter of the wire.
$\Rightarrow \Delta V=\frac{\pi d^2}{4}\Delta L+\frac{\pi L}{4}\times 2d\Delta \text{d}$
$\Rightarrow \frac{\Delta V}{V}=\frac{\Delta L}{L}+\frac{2\Delta d}{d}$
$\Rightarrow \frac{\Delta V}{V}=(1-2e)\frac{\Delta L}{L}$
[$\therefore \text{using (1)}$]
Therefore, $\frac{\Delta V}{V}\propto \frac{\Delta L}{L}$