Calculate the stress and strain in a metal wire given its dimensions, applied force, and elongation.

Step 1: Calculate the cross-sectional area of the wire. The diameter of the wire is $d = 2.50 \text{ mm}$. We convert this to meters: $d = 2.50 \times 10^{-3} \text{ m}$ The radius $r$ is half the diameter: $r = \frac{d}{2} = \frac{2.50 \times 10^{-3} m}{2} = 1.25 \times 10^{-3} \text{ m}$ The cross-sectional area $A$ of a circular wire is given by the formula $A = \pi r^2$: $A = \pi (1.25 \times 10^{-3} m)^2$ $A = \pi (1.5625 \times 10^{-6} m^2)$ $A \approx 4.9087 \times 10^{-6} m^2$

Step 2: Calculate the stress in the wire. Stress ($\sigma$) is defined as the force ($F$) applied per unit cross-sectional area ($A$). The applied force is $F = 12 \text{ N}$. $\sigma = \frac{F}{A}$ $\sigma = \frac{12 N}{4.9087 \times 10^{-6} m^2}$ $\sigma \approx 2444600 \text{ Pa}$ We can express this in megapascals (MPa): $\sigma \approx 2.44 \text{ MPa}$

Step 3: Calculate the strain in the wire. The original length of the wire is $L_0 = 2 \text{ m}$. The problem states "it stretches to a new length of 2.53 mm". Given that the original length is 2 m, it is physically impossible for the new length to be 2.53 mm after stretching. Therefore, we assume that "2.53 mm" refers to the elongation or change in length ($\Delta L$). So, $\Delta L = 2.53 \text{ mm}$. We convert this to meters: $\Delta L = 2.53 \times 10^{-3} \text{ m}$ Strain ($\epsilon$) is defined as the change in length ($\Delta L$) divided by the original length ($L_0$): $\epsilon = \frac{\Delta L}{L_0}$ $\epsilon = \frac{2.53 \times 10^{-3} m}{2 m}$ $\epsilon = 0.001265$

The stress in the wire is $\boxed{2.44 MPa}$. The strain in the wire is $\boxed{0.001265}$.