$(a)\;\large\frac{l_{1} \alpha_{1} - l_{2} \alpha_{2}}{l_{1}-l_{2}}\qquad(b)\;\large\frac{l_{1} \alpha_{1} + l_{2} \alpha_{2}}{l_{1}+l_{2}}\qquad(c)\;\alpha_{1}+\alpha_{2}\qquad(d)\;\large\frac{\alpha_{1} + \alpha_{2}}{2}$

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Answer : $\;\large\frac{l_{1} \alpha_{1} + l_{2} \alpha_{2}}{l_{1}+l_{2}}$

Explanation :

$\bigtriangleup l_{1} = l_{1} \alpha_{1} \bigtriangleup t$

$\bigtriangleup l_{2} = l_{2} \alpha_{2} \bigtriangleup t$

Total change in length

$\bigtriangleup l_{1} + \bigtriangleup l_{2} = (l_{1} \alpha_{1}+\alpha_{2} l_{2}) \bigtriangleup t $

Therefore , $\;\alpha_{comp}\; = \large\frac{l_{1} \alpha_{1}+l_{2} \alpha_{2}}{l_{1}+l_{2}}$

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