Question

Poisson's ratio of certain material is 0.5. If longitudinal strain of 0.04% is produced in rod, then percentage change in volume will be

• A. $0.12\%$
• B. zero
• C. $0.2\%$
• D. $0.013\%$

Answer 1

Answer: (B)

zero

$\sigma =\frac{\frac{-\Delta r}{r}}{\frac{\Delta l}{l}}⟶\frac{-\Delta r}{r}=\sigma \frac{\Delta l}{l}$

$\frac{-\Delta r}{r}=0.5\left(0.04\right)=2\times {10}^{-2}$

$\frac{\Delta V}{V}=\frac{\Delta l}{l}+\frac{2\Delta r}{r}=4\times {10}^{-2}+(-4\times {10}^{-2})=0$