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(B) The longitudinal strain is given by $\epsilon_L = \frac{\Delta L}{L} = \frac{F}{AY} = \frac{Mg}{\pi r^2 Y}$.Poisson's ratio $\sigma$ is defined as

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$\frac{Mg \sigma}{\pi rY}$

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(B) The longitudinal strain is given by $\epsilon_L = \frac{\Delta L}{L} = \frac{F}{AY} = \frac{Mg}{\pi r^2 Y}$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{\epsilon_D}{\epsilon_L} = -\frac{\Delta r / r}{\Delta L / L}$.Therefore,the lateral strain is $\frac{\Delta r}{r} = -\sigma \epsilon_L$.The magnitude of the decrease in radius is $\Delta r = r \sigma \epsilon_L$.Substituting the value of $\epsilon_L$: $\Delta r = r \sigma \left( \frac{Mg}{\pi r^2 Y} \right) = \frac{Mg \sigma}{\pi r Y}$.

$A$ wire of length $L$ and radius $r$ is loaded with a weight $Mg$. If $Y$ 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 $(\Delta r)$ is given by:

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