Question

A material has Poisson’s ratio $0.5$. If a uniform rod of it suffers a longitudinal strain of $3 \times 10^{-3}$, what will be percentage increase in volume?

• A. 2%
• B. 3%
• C. 5%
• D. 0%

Answer 1

Answer: (D)

$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\eta=0.5$
$\frac{-\Delta r / r}{\Delta / 1}=\frac{1}{2} $
$\frac{-2 \Delta r}{r}=\frac{\Delta l}{l}$
Magnitute wise both are equal but sign's would
be different as both quantities cannot increase
Now volume oc area $\times$ length
$V \propto r^{2} \cdot L $
$\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta L}{L}$
Substituting value of $\frac{\Delta L}{L}$
$\frac{\Delta V}{V}=0$
$\therefore $ No change in volume