In an experiment, brass and steel wires of length $1\ \text{m}$ each with areas of cross section $1\ \text{mm}^2$ 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 $0.2\ \text{mm}$ is, [Given, the Young's modulus for steel and brass are, respectively, $120\times10^9\ \text{N/m}^2$ and $60\times10^9\ \text{N/m}^2$]
Answer: (A) $8.0\times10^6\ \text{N/m}^2$
In series both wires carry the same force and, having equal areas, the same stress $\sigma$:
$$\Delta L = \frac{\sigma L}{Y_s} + \frac{\sigma L}{Y_b} = \sigma\left(\frac{1}{120\times10^9} + \frac{1}{60\times10^9}\right) = \frac{\sigma}{40\times10^9}$$
$$\sigma = 0.2\times10^{-3}\times40\times10^9 = 8.0\times10^6\ \text{N/m}^2$$
Solution by Sreeraj P, M.Sc Physics