Question

There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of the wire is

• A. 0.5
• B. -0.5
• C. 0.25
• D. -0.25

Answer 1

The Correct Option is A

Let the material of length 1 and side s If a material maintains constant volume during stretching $V = 1 \times s ^{2}$
Differentiate wrt dl $d V = s ^{2} \cdot d l + 1 . 2 s \cdot d s$
dl.s $=$ 2l.ds $\frac{d s}{d l}=-\frac{1}{2} \frac{s}{1}$
$\eta=-\frac{\frac{d s}{s}}{\frac{d l}{1}}=\frac{1}{2}$