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QJEEM20Ph1afb: JEE Main 2020 Physics - Class 11 JEEM20Ph1afb Solution: Two different wires having lengths L 1 and L 2 , and respective …

Two different wires having lengths L1 and L2, and respective temperature coefficient of linear expansion α\alpha 1 and α\alpha 2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

Multiple Choice Options

  • A 2α1α22\sqrt {{\alpha _1}{\alpha _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}}}
  • C α1+α22{{{\alpha _1} + {\alpha _2}} \over 2}
  • D α1L1+α2L2L1+L2{{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 5th September Evening Slot Physics - Heat and Thermodynamics Question 262 English Explanation

L'1 = L1(1 + α\alpha 1 Δ\Delta T)

L'2 = L2(1 + α\alpha 2 Δ\Delta T)

L'eq = (L1 + L2) (1 + α\alpha avg Δ\Delta T)

∴\therefore (L1 + L2) (1 + α\alpha avg Δ\Delta T) = L1(1 + α\alpha 1 Δ\Delta T) + L2(1 + α\alpha 2 Δ\Delta T)

⇒\Rightarrow (L1 + L2) α\alpha avg = L1 α\alpha 1 + L2 α\alpha 2

⇒\Rightarrow α\alpha avg = α1L1+α2L2L1+L2{{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}

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