Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation, the circular cross section is to remain circular, and the rectangular is to remain rectangular. Their original and deformed dimensions are as follows: $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Circular } \\ (\text { diameter, } \boldsymbol{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \begin{array}{l} \text { Original } \\ \text { dimensions } \end{array} & 18.0 & 20 \times 50 \\ \hline \begin{array}{l} \text { Deformed } \\ \text { dimensions } \end{array} & 15.9 & 13.7 \times 55.1 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?

In the context of materials science and engineering, understanding how to calculate the cross-sectional area of an object is fundamental, particularly when assessing its strength and durability post-deformation. The cross-sectional area is the area of a two-dimensional shape that is exposed when cutting through an object perpendicular to its length.

For example, the circular specimen's cross-sectional area is obtained using the formula: $$A_{circle} = \frac{\pi \times D^2}{4}$$
where D represents the diameter. In the case of the rectangular specimen, the calculation is simpler as it involves multiplication of its length (l) and width (w): $$A_{rectangular} = l \times w$$
Students should visualize and understand that these areas represent the 'footprint' or the face that bears the load or stress during deformation. Knowing how to accurately determine these areas is crucial when analyzing mechanical properties such as deformation, hardness, and ductility of materials.

The percent reduction in area is a significant measure in materials science used to quantify the extent of plastic deformation a material has undergone. It is expressed as a ratio of the change in cross-sectional area to the original cross-sectional area, typically in percentage. This calculation can give us insights about the work hardening of a material, which can result in increased strength and hardness.

Here's the formula: $$\text{Percent reduction} = \left( \frac{\text{Initial area} - \text{Final area}}{\text{Initial area}} \right) \times 100$$
By applying this formula to both the circular and rectangular specimens from our exercise, we can work out which one underwent the most significant dimensional change and therefore, which has possibly become the hardest. It's essential to remember that a larger percent reduction usually correlates with higher hardness due to the greater degree of plastic deformation imposed on the material.

Deformation in materials science refers to the change in shape or size of a material when subjected to external forces. Deformation can be either elastic or plastic. Elastic deformation is temporary — remove the force, and the material returns to its original state. Plastic deformation, on the other hand, is permanent, as the material does not revert back to its initial shape or size when the force is removed.

During plastic deformation, dislocations in the material's crystal structure move, and the movement of these dislocations results in the permanent deformation of the material. Through processes such as work hardening (also known as strain hardening), the repeated movement of dislocations makes a metal harder and stronger yet less ductile.

Understanding the plastic deformation process is critical when predicting how a material will behave in real-world applications, like the two specimens in our exercise. The hardness of a material after plastic deformation is indicative of the stress it has been subjected to, making this concept not only academically interesting but also practically significant in material selection and design.