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

### Related Formula beta = frac-Delta PfracDelta VV$$\beta = \frac{-\Delta P}{\frac{\Delta V}{V}}$$ Delta P = rho g h$$\Delta P = \rho g h$$ ### Core Logic The change in pressure Delta P$\Delta P$ is the hydrostatic pressure at depth h$h$. Delta P = -beta fracDelta VV$$\Delta P = -\beta \frac{\Delta V}{V}$$ rho g h = -beta fracDelta VV$$\rho g h = -\beta \frac{\Delta V}{V}$$ ### Step 2: Calculation Given values: rho = 10^3mathrm\,kg/m^3$\rho = 10^3\mathrm{\,kg/m^3}$ g = 10mathrm\,m/s^2$g = 10\mathrm{\,m/s^2}$ beta = 9 times 10^8mathrm\,N/m^2$\beta = 9 \times 10^8\mathrm{\,N/m^2}$ fracDelta VV = -0.02\% = -frac0.02100$\frac{\Delta V}{V} = -0.02\% = -\frac{0.02}{100}$ Substitute into the equation: 10^3 times 10 times h = - (9 times 10^8) times left(-frac0.02100right)$$10^3 \times 10 \times h = - (9 \times 10^8) \times \left(-\frac{0.02}{100}\right)$$ 10^4 times h = 9 times 10^8 times 2 times 10^-4$$10^4 \times h = 9 \times 10^8 \times 2 \times 10^{-4}$$ 10^4 h = 18 times 10^4$$10^4 h = 18 \times 10^4$$ h = 18mathrm\,m$$h = 18\mathrm{\,m}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids