Two different wires having lengths L1 and L2, and respective temperature coefficient of linear expansion α 1 and α 2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is: (a) 2√(α₁ α₂) (b) 4(α₁ α₂)/(α₁ + α₂) (L₂ L₁)/((L₂ + L₁)²) (c) (α₁ + α₂)/(2) (d) (α₁ L₁ + α₂ L₂)/(L₁ + L₂)

Answer: (d). Total expansion: Δ L = (L₁α₁ + L₂α₂)Δ T. Also Δ L = (L₁+L₂)αeffΔ T. αeff = (α₁ L₁ + α₂ L₂)/(L₁ + L₂), so the answer is (d).

Two different wires having lengths L1 and L2, and respective… | JEE Main 2020 (5 Sep, Shift 2)

Q55JEE Main 2020 · 5 SEP · SHIFT 2Medium

Thermometry & Thermal Expansion

Two different wires having lengths L1 and L2, 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:

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Answer: (d)

Total expansion: $\Delta L = (L_1\alpha_1 + L_2\alpha_2)\Delta T$ .
Also $\Delta L = (L_1+L_2)\alpha_{eff}\Delta T$ .
$\alpha_{eff} = \dfrac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}$ , so the answer is (d).

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