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.
  • 0.50
• B. – 0.50
• C.
  • 0.25
• D. – 0.25

Answer 1

Answer: (B)

– 0.50

Answer 2

$= ( 1 + 2 \sigma ) \frac { d L } { L }$= 0 [As there is no change in the volume of the wire]

$\therefore 1 + 2 \sigma = 0$ ⇒ $\sigma = - \frac { 1 } { 2 }$