JEE Main 2020 — Thermal Properties of Matter Question with Solution

The wire are joined end-to-end, so they are in series combination, then total length of wire is given by
$L=L_1+L_2 ...\left(1\right)$

When the temperature is increase by $∆\theta$, then its final length is given by using concept of linear expansion,
$L^{\text{'}}=L(1+\alpha ∆\theta ) ...\left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$ for the series combination of wires, final length is given by

$L^{\text{'}}={L_1}^{\text{'}}+{L_2}^{\text{'}} ...\left(3\right)$

$\Rightarrow \left(L_1+L_2\right)(1+{\alpha }_{\mathrm{eff}}∆\theta )=L_1(1+{\alpha }_1∆\theta )+L_2(1+{\alpha }_2∆\theta )$

$\Rightarrow {\alpha }_{\mathrm{eff}}=\frac{L_1{\alpha }_1+L_2{\alpha }_2}{\left(L_1+L_2\right)}$