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

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:

$$r = \frac{d}{2} = \frac{2.50 \times 10^{-3} \text{ m}}{2} = 1.25 \times 10^{-3} \text{ m}$$

The cross-sectional area $A$ of a circular wire is given by the formula $A = \pi r^2$:

$$A = \pi (1.25 \times 10^{-3} \text{ m})^2$$ $$A = \pi (1.5625 \times 10^{-6} \text{ m}^2)$$ $$A \approx 4.9087 \times 10^{-6} \text{ m}^2$$

Stress ($\sigma$) is defined as the force ($F$) applied per unit cross-sectional area ($A$).
The applied force is $F = 12 \text{ N}$.

$$\sigma = \frac{F}{A}$$ $$\sigma = \frac{12 \text{ N}}{4.9087 \times 10^{-6} \text{ m}^2}$$ $$\sigma \approx 2444600 \text{ Pa}$$

We can express this in megapascals (MPa):

$$\sigma \approx 2.44 \text{ MPa}$$

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 ($L_0$):

$$\epsilon = \frac{\Delta L}{L_0}$$ $$\epsilon = \frac{2.53 \times 10^{-3} \text{ m}}{2 \text{ m}}$$ $$\epsilon = 0.001265$$

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