Question
10kg mass is attached to one end of a copper wire $3m$ long and $1mm$ in diameter. Calculate the lateral compression produced in it (Poisson's ratio is $0.25$ and Young's modulus of the material of the wire is $12.5 \times 10^{10}N/ m^2$).
✓
Answer
Let Dl be the increase in length of the wire, $\therefore\text{Y}=\frac{\frac{\text{F}}{\text{A}}}{\frac{\Delta\text{l}}{\text{l}}}$
$\text{Y}=\frac{\text{F}.\text{l}}{\text{A.}\Delta\text{l}}$
$\text{Y}=\frac{\text{F}.\text{l}}{\pi\text{r}^2.\Delta\text{l}}$
$\therefore\Delta\text{l}=\frac{\text{F}.\text{l}}{\pi\text{r}^2.\text{Y}}=\frac{10\times9.8\times3}{3.14\times(0.5\times10^{-3})\times12.5\times10^{10}}$
$=0.2993\times10^{-2}\text{m}$ Now the Poisson's ratio $\sigma=-\frac{\Delta\text{D}}{\text{D}}\times\frac{\text{l}}{\Delta\text{l}}$
$\therefore\Delta\text{D}=\frac{\sigma.\text{D}.\Delta\text{l}}{\text{l}}=\frac{-0.2993\times10^{-3}\times0.25}{3}$
$=-2.5\times10^{-7}=-0.25\times10^{-6}=-0.25\mu\text{m}$ Here the lateral compression $=0.25\mu\text{m}$ (approx).
$\text{Y}=\frac{\text{F}.\text{l}}{\text{A.}\Delta\text{l}}$
$\text{Y}=\frac{\text{F}.\text{l}}{\pi\text{r}^2.\Delta\text{l}}$
$\therefore\Delta\text{l}=\frac{\text{F}.\text{l}}{\pi\text{r}^2.\text{Y}}=\frac{10\times9.8\times3}{3.14\times(0.5\times10^{-3})\times12.5\times10^{10}}$
$=0.2993\times10^{-2}\text{m}$ Now the Poisson's ratio $\sigma=-\frac{\Delta\text{D}}{\text{D}}\times\frac{\text{l}}{\Delta\text{l}}$
$\therefore\Delta\text{D}=\frac{\sigma.\text{D}.\Delta\text{l}}{\text{l}}=\frac{-0.2993\times10^{-3}\times0.25}{3}$
$=-2.5\times10^{-7}=-0.25\times10^{-6}=-0.25\mu\text{m}$ Here the lateral compression $=0.25\mu\text{m}$ (approx).
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