A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{-3}$ respectively, then the elastic potential energy density of the wire is __________ $\times 10^5$ (in SI units).

<p>To find the elastic potential energy density of the steel wire, we need to use the given information and formulae for strain and energy density.</p> <p>Given:</p> <p><p>The length of the wire, $ \ell = 2 \, \text{m} $</p></p> <p><p>Young's modulus, $ Y = 2.0 \times 10^{11} \, \text{N/m}^2 $</p></p> <p><p>Poisson's ratio, $ \mu = 0.2 $</p></p> <p><p>Transverse strain, $ \frac{\Delta r}{r} = 10^{-3} $</p></p> <p>The formula for Poisson's ratio is:</p> <p>$ \mu = -\frac{\left(\frac{\Delta r}{r}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} $</p> <p>From this, we solve for the longitudinal strain $\frac{\Delta \ell}{\ell}$:</p> <p>$ \frac{\Delta \ell}{\ell} = \frac{1}{\mu} \times \left(\frac{\Delta r}{r}\right) $</p> <p>Substitute the given values:</p> <p>$ \frac{\Delta \ell}{\ell} = \frac{1}{0.2} \times 10^{-3} = 5 \times 10^{-3} $</p> <p>The elastic potential energy density $ u $ is given by:</p> <p>$ u = \frac{1}{2} Y \varepsilon_{\ell}^2 $</p> <p>where $ \varepsilon_{\ell} = \frac{\Delta \ell}{\ell} $. Plug in the values:</p> <p>$ u = \frac{1}{2} \times 2 \times 10^{11} \times \left(5 \times 10^{-3}\right)^2 $</p> <p>Simplify further:</p> <p>$ u = \frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{-6} $</p> <p>$ u = 25 \times 10^5 \, \text{(in SI units)} $ </p> <p>Thus, the elastic potential energy density of the wire is $ 25 \times 10^5 $ SI units.</p>