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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 \frac{\alpha _{1}L_{1}+\alpha _{2}L_{2}}{L_{1}+L_{2}}
Option: 2 2\sqrt{\alpha _{1}\alpha _{2}}
Option: 3 \frac{4\alpha _{1}\alpha _{2}}{(\alpha _{1}+\alpha _{2})}\times \frac{L_{2}L_{1}}{(L_{1}+L_{2})^{2}}
Option: 4 \frac{\alpha _{1}+\alpha _{2}}{2}

Answers (1)

best_answer

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}}

 

Posted by

Deependra Verma

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