A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1$$\pi$$ m s$$^{-2}$$, what will be the tensile stress that would be developed in the wire?

We have a vertical wire that is supporting a stationary load, so the force producing the stress is simply the weight of the load.

First, recall the definition of tensile stress.
$$\text{Tensile stress} \; (\sigma) \;=\; \dfrac{\text{Force}}{\text{Cross-sectional area}}$$

The force is the weight $$W$$ of the 4 kg mass. By Newton’s second law, weight is mass times acceleration due to gravity:

$$W \;=\; m\,g$$ where $$m = 4\ \text{kg}$$ and $$g = 3.1\pi\ \text{m s}^{-2}$$.

Substituting the numbers,

$$W \;=\; 4 \times (3.1\pi) \;=\; 12.4\pi\ \text{N}.$$

Next we need the cross-sectional area $$A$$ of the wire. The radius is given as 2.0 mm, so

$$r \;=\; 2.0\ \text{mm} \;=\; 2.0 \times 10^{-3}\ \text{m}.$$

For a circle, the area formula is

$$A \;=\; \pi r^{2}.$$

Substituting the value of $$r$$,

$$A \;=\; \pi \left(2.0 \times 10^{-3}\right)^{2} \;=\; \pi \left(4.0 \times 10^{-6}\right) \;=\; 4.0\pi \times 10^{-6}\ \text{m}^{2}.$$

Now we divide the force by the area to obtain the stress:

$$\sigma \;=\; \dfrac{W}{A} \;=\; \dfrac{12.4\pi}{4.0\pi \times 10^{-6}}.$$

The factor $$\pi$$ appears in both numerator and denominator, so it cancels out:

$$\sigma \;=\; \dfrac{12.4}{4.0} \times 10^{6} \;=\; 3.1 \times 10^{6}\ \text{N m}^{-2}.$$

Hence, the correct answer is Option D.