A iron rod of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurements refer to 20°C. The coefficients of linear expansion of iron and aluminium are 12 x 10⁻⁶ /°C and 24 x 10⁻⁶/°C respectively. The average coefficient of composite system is

The correct option is 20 x 10⁻⁶/°C.

Let us derive the solution step-by-step to find the average coefficient of linear expansion of the composite system.

1. Identify the given parameters:
- Length of the iron rod at 20°C, $L_1=50\text{cm}$
- Coefficient of linear expansion of iron, ${\alpha }_1=12\times {10}^{-6}{\text{/°C}}^{}$
- Length of the aluminium rod at 20°C, $L_2=100\text{cm}$
- Coefficient of linear expansion of aluminium, ${\alpha }_2=24\times {10}^{-6}{\text{/°C}}^{}$

2. Understand the composite system:
The total initial length of the composite rod is the sum of the individual lengths:
$L=L_1+L_2=50\text{cm}+100\text{cm}=150\text{cm}$

When the temperature changes by $\Delta T$, each rod expands. The total expansion of the composite system ($\Delta L$) is the sum of the individual expansions of the iron and aluminium rods:
$\Delta L=\Delta L_1+\Delta L_2$

Using the formula for linear expansion, $\Delta L=L·\alpha ·\Delta T$, we can write:
$L·{\alpha }_{\text{avg}}·\Delta T=L_1·{\alpha }_1·\Delta T+L_2·{\alpha }_2·\Delta T$

Dividing both sides by $\Delta T$, we get:
$(L_1+L_2){\alpha }_{\text{avg}}=L_1{\alpha }_1+L_2{\alpha }_2$

Thus, the average coefficient of linear expansion ${\alpha }_{\text{avg}}$ is given by:
${\alpha }_{\text{avg}}=\frac{L_1{\alpha }_1+L_2{\alpha }_2}{L_1+L_2}$

3. Substitute the values:
${\alpha }_{\text{avg}}=\frac{50\times (12\times {10}^{-6})+100\times (24\times {10}^{-6})}{50+100}$

${\alpha }_{\text{avg}}=\frac{600\times {10}^{-6}+2400\times {10}^{-6}}{150}$

${\alpha }_{\text{avg}}=\frac{3000\times {10}^{-6}}{150}$

${\alpha }_{\text{avg}}=20\times {10}^{-6}{\text{/°C}}^{}$

Therefore, the average coefficient of linear expansion of the composite system is 20 x 10⁻⁶/°C.