# Two different wires having length L1 and L2 and respective temperature coefficient of linear expression$\alpha _{1}and \alpha _{2}$ are joined end to end. then the effective temperature coefficent of linear expansin is: Option: 1 Option: 2 Option: 3 Option: 4

Correct Answer: a) $\frac{\alpha _{1}L_{1}+\alpha _{2}L_{2}}{L_{1}+L_{2}}$

As,

The coefficient of linear expansion of first wire of length L1? is

$\Delta L_{1}=\alpha, L_{1} \Delta t$

and,

The coefficient of linear expansion of first wire of length L2? is

$\Delta L_{2}=\alpha_{2} L_{2} \Delta t$

Now, if a single wire of linear expansion α is taken instead of two wires,

$\alpha=\frac{\Delta \mathrm{L}_{1}+\Delta \mathrm{L}_{2}}{\left(L_{1}+\mathrm{L}_{2}\right) \times \Delta \mathrm{t}}$

Putting the value of $\Delta L_{1} \text{and} \Delta L_{2}$

We get,

$\alpha=\frac{L_{1} \alpha_{1}+L_{2} \alpha_{2}}{L_{1}+L_{2}}$

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