A wire of length 'L' and radius ' $\mathrm{r}$ ' is loaded with a weight ' $\mathrm{Mg}$ '. If '…

A wire of length 'L' and radius ' $\mathrm{r}$ ' is loaded with a weight ' $\mathrm{Mg}$ '. If ' $\mathrm{Y}$ ' and ' $\sigma$ ' denote the Young's modulus and poisson's ratio of the material of the wire respectively, then the decrease in the radius of the wire $(\Delta \mathrm{r})$ is given by
  1. $\frac{\mathrm{MgY}}{\pi \mathrm{r} \sigma}$
  2. $\frac{\mathrm{Mg} \sigma}{\pi \mathrm{rY}}$
  3. $\frac{\sigma \pi \mathrm{r}}{\mathrm{MgY}}$
  4. $\frac{\mathrm{Mgr}}{\sigma \pi \mathrm{Y}}$

Solution

$\begin{aligned} & Y=\frac{M g L}{\pi r^2 \Delta L} \therefore \frac{\Delta L}{L}=\frac{M g}{\pi r^2 Y} \ldots(1) \\ & \text { and } \sigma-\frac{\frac{\Delta D}{D}}{\frac{\Delta L}{L}}=\frac{\frac{\Delta r}{r}}{\frac{\Delta L}{L}} \\ & \therefore \frac{\Delta r}{r}=\sigma\left(\frac{\Delta L}{L}\right)=\sigma \cdot \frac{M g}{\pi r^2 Y} \ldots \text { from (1) } \\ & \therefore \Delta r=\sigma \cdot \frac{M g}{\pi r^2 Y} \times r=\frac{\sigma M g}{\pi r Y}\end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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