In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm$$^2$$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is,
[Given, the Young's Modulus for steel and brass are, respectively, $$120 \times 10^9$$ N/m$$^2$$ and $$60 \times 10^9$$ N/m$$^2$$]

We have two wires, one of brass and one of steel, each of length $$L = 1 \text{ m}$$ and cross-sectional area $$A = 1 \text{ mm}^2 = 1 \times 10^{-6}\,\text{m}^2$$. The wires are joined in series, so when a tensile force is applied the same force $$F$$ acts on both wires. Because the area is the same for both, the stress $$\sigma$$ in each wire is also the same, where

$$\sigma = \dfrac{F}{A}.$$

The elongation of a wire obeys Hooke’s law for small strains:

$$\Delta L = \dfrac{\sigma L}{Y},$$

where $$Y$$ is Young’s modulus. First we write down the given values:

$$Y_{\text{steel}} = 120 \times 10^{9}\,\text{N m}^{-2},\qquad Y_{\text{brass}} = 60 \times 10^{9}\,\text{N m}^{-2}.$$

For the brass wire the extension is

$$\Delta L_{\text{brass}} = \dfrac{\sigma L}{Y_{\text{brass}}},$$

and for the steel wire

$$\Delta L_{\text{steel}} = \dfrac{\sigma L}{Y_{\text{steel}}}.$$

The two wires are in series, so the total elongation is the sum of the individual elongations:

$$\Delta L_{\text{total}} = \Delta L_{\text{brass}} + \Delta L_{\text{steel}} = \dfrac{\sigma L}{Y_{\text{brass}}} + \dfrac{\sigma L}{Y_{\text{steel}}} = \sigma L\left(\dfrac{1}{Y_{\text{brass}}} + \dfrac{1}{Y_{\text{steel}}}\right).$$

The experiment specifies a required total elongation of

$$\Delta L_{\text{total}} = 0.2 \text{ mm} = 0.2 \times 10^{-3}\,\text{m}.$$

Substituting $$L = 1 \text{ m}$$ and solving for $$\sigma$$ gives

$$\sigma = \dfrac{\Delta L_{\text{total}}} {L\left(\dfrac{1}{Y_{\text{brass}}} + \dfrac{1}{Y_{\text{steel}}}\right)}.$$

Now we evaluate the bracketed term:

$$\dfrac{1}{Y_{\text{brass}}} + \dfrac{1}{Y_{\text{steel}}} = \dfrac{1}{60 \times 10^{9}} + \dfrac{1}{120 \times 10^{9}} = \dfrac{2}{120 \times 10^{9}} + \dfrac{1}{120 \times 10^{9}} = \dfrac{3}{120 \times 10^{9}} = \dfrac{1}{40 \times 10^{9}} = 2.5 \times 10^{-11}\,\text{m}^2\text{N}^{-1}.$$

Putting everything together, we obtain

$$\sigma = \dfrac{0.2 \times 10^{-3}} {1 \times 2.5 \times 10^{-11}} = \dfrac{0.2}{2.5} \times 10^{8} = 0.08 \times 10^{8} = 8.0 \times 10^{6}\,\text{N m}^{-2}.$$

Hence, the correct answer is Option A.