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Heat and Thermodynamics question

2020 · 5 Sep · Shift 2 · Q48
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  5. /2020 · 5 Sep · Shift 2 · Q48

Heat and Thermodynamics question

2020 · 5 Sep · Shift 2 · Q48

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Two different wires having lengths L1 and L2, and respective temperature coefficient of linear expansion α\alphaα 1 and α\alphaα 2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :
  1. A
    2α1α22\sqrt {{\alpha _1}{\alpha _2}}2α1​α2​​
  2. B
    4α1α2α1+α2L2L1(L2+L1)24{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}{{{L_2}{L_1}} \over {{{\left( {{L_2} + {L_1}} \right)}^2}}}4α1​+α2​α1​α2​​(L2​+L1​)2L2​L1​​
  3. C
    α1+α22{{{\alpha _1} + {\alpha _2}} \over 2}2α1​+α2​​
  4. D
    α1L1+α2L2L1+L2{{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}L1​+L2​α1​L1​+α2​L2​​
View written solutionFree

Correct answer: D

  1. Thermal expansion of each wire

If a wire of length LLL has coefficient of linear expansion α\alphaα, then for a temperature rise ΔT\Delta TΔT:

ΔL=αLΔT\Delta L = \alpha L \Delta TΔL=αLΔT

So for the two wires:

  • First wire:

    ΔL1=α1L1ΔT\Delta L_1 = \alpha_1 L_1 \Delta TΔL1​=α1​L1​ΔT
  • Second wire:

    ΔL2=α2L2ΔT\Delta L_2 = \alpha_2 L_2 \Delta TΔL2​=α2​L2​ΔT
  1. Total initial length and total expansion

Since the wires are joined end-to-end, the total initial length is:

L=L1+L2L = L_1 + L_2L=L1​+L2​

The total increase in length is the sum of the individual expansions:

ΔL=ΔL1+ΔL2=α1L1ΔT+α2L2ΔT\Delta L = \Delta L_1 + \Delta L_2 = \alpha_1 L_1 \Delta T + \alpha_2 L_2 \Delta TΔL=ΔL1​+ΔL2​=α1​L1​ΔT+α2​L2​ΔT

Thus,

ΔL=(α1L1+α2L2)ΔT\Delta L = (\alpha_1 L_1 + \alpha_2 L_2)\Delta TΔL=(α1​L1​+α2​L2​)ΔT
  1. Define effective coefficient

Let the effective coefficient of linear expansion be αeff\alpha_{\text{eff}}αeff​. Then for the combined wire:

ΔL=αeff(L1+L2)ΔT\Delta L = \alpha_{\text{eff}} (L_1 + L_2) \Delta TΔL=αeff​(L1​+L2​)ΔT

Substitute the total expansion:

αeff(L1+L2)ΔT=(α1L1+α2L2)ΔT\alpha_{\text{eff}} (L_1 + L_2) \Delta T = (\alpha_1 L_1 + \alpha_2 L_2)\Delta Tαeff​(L1​+L2​)ΔT=(α1​L1​+α2​L2​)ΔT

Cancel ΔT\Delta TΔT:

αeff(L1+L2)=α1L1+α2L2\alpha_{\text{eff}} (L_1 + L_2) = \alpha_1 L_1 + \alpha_2 L_2αeff​(L1​+L2​)=α1​L1​+α2​L2​

Therefore,

αeff=α1L1+α2L2L1+L2\boxed{\alpha_{\text{eff}} = \frac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}}αeff​=L1​+L2​α1​L1​+α2​L2​​​
  1. Option matching

This matches Option D:

α1L1+α2L2L1+L2\frac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}L1​+L2​α1​L1​+α2​L2​​

Hence, the correct answer is D.

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