A material has Poisson's ratio $0.50.$ If a uniform rod of it suffers a longitudinal strain of $2 \times {10^{ - 3}}$, then the percentage change in volume is
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(b) $\frac{{dV}}{V} = (1 + 2\sigma )\frac{{dL}}{L}$
$\frac{{dV}}{V} = 2 \times 2 \times {10^{ - 3}} = 4 \times {10^{ - 3}}$
Percentage change in volume =$4 \times {10^{ - 1}} = 0.4\% $
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