1. **State the problem:** A steel wire 2 meters long and 4 millimeters in diameter is stretched under a load until its length becomes 200.1 meters. We need to calculate the stress, strain, and modulus of elasticity. 2. **Given data:** - Original length, $L_0 = 2$ m - Final length, $L = 200.1$ m - Diameter, $d = 4$ mm = 0.004 m 3. **Formulas and definitions:** - Strain ($\varepsilon$) is the relative change in length: $$\varepsilon = \frac{\Delta L}{L_0} = \frac{L - L_0}{L_0}$$ - Stress ($\sigma$) is force per unit area: $$\sigma = \frac{F}{A}$$ - Modulus of elasticity ($E$) relates stress and strain: $$E = \frac{\sigma}{\varepsilon}$$ 4. **Calculate strain:** $$\varepsilon = \frac{200.1 - 2}{2} = \frac{198.1}{2} = 99.05$$ 5. **Calculate cross-sectional area:** The cross-sectional area of the wire is a circle: $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.004}{2}\right)^2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} = 1.2566 \times 10^{-5} \text{ m}^2$$ 6. **Calculate stress:** We need the force $F$ to calculate stress, but it is not given. Without the force, stress and modulus of elasticity cannot be numerically determined. 7. **Conclusion:** - Strain is $99.05$ (a very large strain, indicating extreme elongation). - Stress and modulus of elasticity require the applied force $F$ to be known. If the force is provided, we can calculate stress and then modulus of elasticity using the formulas above.