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A material has Poisson's ratio 0.50 . If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
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0
Given, Poisson's ratio,
$$
\sigma=0.5
$$
Longitudinal strain,
$$
\frac{\Delta l}{l}=2 \times 10^{-3}
$$
Volumetric strain $\left(\frac{\Delta V}{V}\right)$ and Iongitudinal strain $\left(\frac{\Delta l}{l}\right)$ are related as
$$
\begin{array}{rlrl}
\frac{\Delta V}{V} & =(1-2 \sigma) \frac{\Delta l}{l} \\
\Rightarrow & & =(1-2 \times 0.5) \times 2 \times 10^{-3} \\
& =(1-1) \times 2 \times 10^{-3}=0 \times 2 \times 10^{-3}=0 \\
\therefore \quad & \frac{\Delta V}{V} \times 100 & =0 \times 100 \%=0
\end{array}
$$
$$
\sigma=0.5
$$
Longitudinal strain,
$$
\frac{\Delta l}{l}=2 \times 10^{-3}
$$
Volumetric strain $\left(\frac{\Delta V}{V}\right)$ and Iongitudinal strain $\left(\frac{\Delta l}{l}\right)$ are related as
$$
\begin{array}{rlrl}
\frac{\Delta V}{V} & =(1-2 \sigma) \frac{\Delta l}{l} \\
\Rightarrow & & =(1-2 \times 0.5) \times 2 \times 10^{-3} \\
& =(1-1) \times 2 \times 10^{-3}=0 \times 2 \times 10^{-3}=0 \\
\therefore \quad & \frac{\Delta V}{V} \times 100 & =0 \times 100 \%=0
\end{array}
$$
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