Q.A composite wire is prepared by joining a tungsten wire and a steel wire end to end. Both the wires are of the same length and the same area of cross section. If this composite wire is suspended from a rigid support and a force is applied to its free end, it gets extended by 3.25 mm. Calculate the increase in length of the tungsten wire and the steel wire separately. [Given: Ysteel = 2×10¹¹ N/m², Ytungsten = 3.40×10⁸ N/m²]

Step 1. Since the tungsten and steel segments are joined end to end (in series) with the same length and same cross-sectional area, and the composite wire carries a single tension $F$ throughout, the same force $F$ acts on both segments.

Step 2. For each segment, elongation $= \dfrac{FL}{AY}$; since $F$, $L$ and $A$ are common to both segments, the elongations are in inverse proportion to their respective Young's moduli: $\dfrac{l_{tungsten}}{l_{steel}} = \dfrac{Y_{steel}}{Y_{tungsten}} = \dfrac{2\times10^{11}}{3.40\times10^8} = 588.2$.

Step 3. Total extension: $l_{tungsten} + l_{steel} = 3.25$ mm. Writing $l_{steel} = x$, then $l_{tungsten} = 588.2x$, so $589.2x = 3.25 \Rightarrow x = 0.005515$ mm.

Step 4. So $l_{steel} \approx 0.0055$ mm and $l_{tungsten} = 3.25 - 0.0055 = 3.2445 \approx 3.244$ mm. …

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