A steel wire of length 2 m and Young's modulus $$2.0 \times 10^{11}$$ Nm$$^{-2}$$ 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).

Young’s modulus of the steel wire is $$E = 2.0 \times 10^{11} \,\text{N m}^{-2}$$.
Poisson ratio is $$\nu = 0.2$$.
Given transverse (lateral) strain, $$\varepsilon_t = 10^{-3}$$.

For a rod under uniaxial tension, Poisson ratio is defined as
$$\nu = - \dfrac{\varepsilon_t}{\varepsilon_l}$$, where $$\varepsilon_l$$ is the longitudinal (axial) strain.

Therefore,
$$\varepsilon_l = -\,\dfrac{\varepsilon_t}{\nu} = -\,\dfrac{10^{-3}}{0.2} = -\,5 \times 10^{-3}$$.
(The negative sign only indicates that the lateral and longitudinal strains are opposite in sense; for energy we need the magnitude.)
So, $$|\varepsilon_l| = 5 \times 10^{-3}$$.

The corresponding longitudinal stress is obtained from Hooke’s law:
$$\sigma = E\,\varepsilon_l = 2.0 \times 10^{11} \times 5 \times 10^{-3}\; \text{N m}^{-2}$$.
$$\sigma = 1.0 \times 10^{9}\; \text{N m}^{-2}$$.

Strain-energy density (elastic potential energy per unit volume) for uniaxial loading is
$$u = \dfrac{1}{2}\,\sigma\,\varepsilon_l$$.

Substituting the values:
$$u = \dfrac{1}{2}\,(1.0 \times 10^{9})\,(5 \times 10^{-3})$$
$$u = 0.5 \times 5 \times 10^{6}$$
$$u = 2.5 \times 10^{6}\;\text{J m}^{-3}$$.

Expressing the result as $$\times 10^{5}$$ J m$$^{-3}$$:
$$u = 25 \times 10^{5}\;\text{J m}^{-3}$$.

Hence, the elastic potential energy density of the wire is 25 × 105 J m-3.