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A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 \%$. If the Poisson's ratio for the material of the rod be $0.2,$ which of the following statements is correct? "The volume approximately ...........
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The correct answer is:
decreases by $0.6 \%$
Given, Decrement in the length $=1 \%$ Poisson's ratio for material of the rod, $\sigma=0.2$
As we know that, Volume, $v=\pi r^{2} l,[$ where $, l$ is the length of the rod $]$
$$
\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}...(i)
$$
since, Poisson's ratio, $\sigma=\frac{-\Delta D / D}{\Delta L / L}$
$$
=\frac{-\Delta r / r}{\Delta L / L}
$$
So, Eq. (i) can be written as,
$$
\begin{aligned}
\frac{\Delta V}{V}=\frac{\Delta l}{l}(1-2 \sigma) \\
\frac{\Delta V}{V} \times 100 &=\left(\frac{\Delta l}{l} \times 100\right) \times(1-2 \sigma) \\
&=-1 \times[1-2 \times(0.2)] \\
&=-1 \times[1-0.4] \\
&=-0.6 \%
\end{aligned}
$$
Here, negative sign shows the decrement in the volume.
As we know that, Volume, $v=\pi r^{2} l,[$ where $, l$ is the length of the rod $]$
$$
\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}...(i)
$$
since, Poisson's ratio, $\sigma=\frac{-\Delta D / D}{\Delta L / L}$
$$
=\frac{-\Delta r / r}{\Delta L / L}
$$
So, Eq. (i) can be written as,
$$
\begin{aligned}
\frac{\Delta V}{V}=\frac{\Delta l}{l}(1-2 \sigma) \\
\frac{\Delta V}{V} \times 100 &=\left(\frac{\Delta l}{l} \times 100\right) \times(1-2 \sigma) \\
&=-1 \times[1-2 \times(0.2)] \\
&=-1 \times[1-0.4] \\
&=-0.6 \%
\end{aligned}
$$
Here, negative sign shows the decrement in the volume.
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