Question

A metal wire of length L is loaded and an elongation of $\Delta L$ is produced. If the area of cross section of the wire is A, then the change in volume of the wire, when elongated is . Take Poisson's ratio as 0.25

• A. $\Delta V=(\Delta L{)}^{2}A/L$
• B. $\Delta V=(\Delta L{)}^{2}A/4L$
• C. $\Delta V=(\Delta L{)}^{2}A/2L$
• D. $\Delta V=(\Delta L{)}^{2}A/3L$

Answer 1

The correct option is B $\Delta V=(\Delta L{)}^{2}A/4L$

We know that $\sigma =\left(\Delta A/A\right)/\left(\Delta L/L\right)=\left(\Delta V/V\right)/(\Delta L/L{)}^2⟹\Delta V=\sigma (\Delta L/L{)}^{2}V=\sigma \left(\Delta L/L{)}^2\right(LA)=\sigma (\Delta L{)}^{2}A/L$
Substituting $\sigma =0.25$, we get, $\Delta V=(\Delta L{)}^{2}A/4L$
The correct option is (b)