Question

If there is no change in the volume of wire on stretching, then poison's ratio for the material of wire is -

• A. –1
• B. 0
• C. 0.5
• D. 0.25

Answer 1

Answer: (C)

0.5

Answer 2

Change in volume

$\frac{\Delta V}{V}$ = 2$\left\{ \frac{\Delta D}{D} \right\}$+ $\left\{ \frac{\Delta\mathcal{l}}{\mathcal{l}} \right\}$

so $\frac{\Delta V}{V}$= 0$\Rightarrow$ $\frac{\Delta\mathcal{l}}{\mathcal{l}}$= –2$\left\{ \frac{\Delta D}{D} \right\}$

so Poisson's ratio

s = – $\left\{ \frac{\Delta D/D}{\Delta\mathcal{l/l}} \right\}$ = $\frac{1}{2}$ = 0.5