In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm² 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…

Question

In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm² 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, 120×10⁹ N/m² and 60×10⁹ N/m²]

Accepted Answer

(4) 4.0×10⁶ N/m². Idea: in series the two wires carry the same force and have the same area, so the stress σ is common; the elongations add. Δ l=(σ L)/(Yₛₜₑₑₗ)+(σ L)/(Y_brass)=σ L(1/(120×10⁹)+1/(60×10⁹)). 1/(120)+1/(60)=1/(120)+2/(120)=3/(120)=1/(40). 0.2×10⁻³=σ×1×1/(40×10⁹) σ=0.2×10⁻³×40×10⁹=8×10⁶ N/m². That is none of the four printed options, and the key marks this question with an asterisk. `answer` is set to (d), the nearest printed value. See DEFECTS.

Step-by-step solution

Why the other options are wrong