Question
The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is 20/11. When the joined wire is kept under certain tension the elongations in the wires A and B are equal. If the length of wire A is 2.2 m , then the length of wire $B$ is $\_\_\_\_$ m.
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Solution
Sol. [Key Concept: Equal elongations with $\Delta\ l = \frac{Tl}{AY}$ and equal area/tension give $l \propto \ Y$, so $l_{B} = l_{A}\frac{Y_{B}}{Y_{A}} = 2.2 \times \frac{11}{20}$.]
$${\Delta\mathcal{l} = \frac{T\mathcal{l}}{YA} }{\mathcal{l} \propto Y }{\frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} = \frac{Y_{1}}{Y_{2}} = \frac{20}{11} }{\because\mathcal{l}_{1} = 2.2\text{ }m }{\therefore\mathcal{l}_{1} = 1.21\text{ }m }$$
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