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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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0.4
$\frac{d V}{V}=(1+2 \sigma) \frac{d L}{L}$
$\frac{d V}{V}=2 \times 2 \times 10^{-3}=4 \times 10^{-3}\left[\because \sigma=0.5=\frac{1}{2}\right]$
$\therefore$ Percentage change in volume $=4 \times 10^{-1}=0.4 \%$
$\frac{d V}{V}=2 \times 2 \times 10^{-3}=4 \times 10^{-3}\left[\because \sigma=0.5=\frac{1}{2}\right]$
$\therefore$ Percentage change in volume $=4 \times 10^{-1}=0.4 \%$
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