JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
Two different wires having lengths \(L _{1}\) and \(L _{2}\) and respective temperature coefficient of linear expansion \(\alpha_{1}\) and \(\alpha_{2},\) are joined end-to-end. Then the effective temperature coefficient of linear expansion is
- A \(4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{ L _{2} L _{1}}{\left( L _{2}+ L _{1}\right)^{2}}\)
- B \(2 \sqrt{\alpha_{1} \alpha_{2}}\)
- C \(\frac{\alpha_{1}+\alpha_{2}}{2}\)
- D \(\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}\)
Step-by-step Solution
Detailed explanation
At \(T^{\circ} C \quad L = L _{1}+ L _{2}\) At \(T +\Delta T \quad L _{ eq }= L _{1}+ L _{2}\) where \(L _{1}= L _{1}\left(1+\alpha_{1} \Delta T \right)\) \(L _{2}^{\prime}= L _{2}\left(1+\alpha_{2} \Delta T \right)\)…
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