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
A$0.6$
B$0.4$
C$0.2$
D$0$
Medium
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B$0.4$
b (b) $\frac{{dV}}{V} = (1 + 2\sigma )\frac{{dL}}{L}$
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