Two different wires having lengths L 1 and L 2, and respective temperature coefficient of linear expansion α1 and α2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

Question

Two different wires having lengths L 1 and L 2, and respective temperature coefficient of linear expansion α1 and α2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is : (A) 4 (α1 α2/α1+α2) ( L 2 L 1/( L 2+ L 1)2) (B) 2 √α1 α2 (C) (α1+α2/2) (D) (α1 L1+α2 L2/L1+L2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/05d58b4d8022bcc0d956ebce7fd60559-.png" /> where L1'=L1(1+α1 Δ T) L2'=L2(1+α2 Δ T) Le q'=(L1+L2)(1+α text avg Δ T) ⇒ (L1+L2)(1+α text avg Δ T)=L1+L2+L1 α1 Δ T+L2 a2 Δ T ⇒ (L1+L2) α text avg =L1 α1+L2 α2 ⇒ α text avg =(L1 α1+L2 α2/L1+L2)

Q. Two different wires having lengths $L_{1}$ and $L_{2}$ , 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 :

$4 \frac{α _{1} α _{2}}{α _{1} + α _{2}} \frac{L _{2} L _{1}}{( L _{2} + L _{1} ) ^{2}}$

$2 α_{1} α_{2}$

$\frac{α _{1} + α _{2}}{2}$

$\frac{α _{1} L _{1} + α _{2} L _{2}}{L _{1} + L _{2}}$

Q. Two different wires having lengths L1​ and L2​, and respective temperature coefficient of linear expansion α1​ and α2​, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

4α1​+α2​α1​α2​​(L2​+L1​)2L2​L1​​

2α1​α2​​

2α1​+α2​​

L1​+L2​α1​L1​+α2​L2​​

where L1′​=L1​(1+α1​ΔT) L2′​=L2​(1+α2​ΔT) Leq′​=(L1​+L2​)(1+αavg ​ΔT) ⇒(L1​+L2​)(1+αavg ​ΔT)=L1​+L2​+L1​α1​ΔT+L2​a2​ΔT ⇒(L1​+L2​)αavg ​=L1​α1​+L2​α2​ ⇒αavg ​=L1​+L2​L1​α1​+L2​α2​​

Q. Two different wires having lengths $L_{1}$ and $L_{2}$ , 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 :

$4 \frac{α _{1} α _{2}}{α _{1} + α _{2}} \frac{L _{2} L _{1}}{( L _{2} + L _{1} ) ^{2}}$

$2 α_{1} α_{2}$

$\frac{α _{1} + α _{2}}{2}$

$\frac{α _{1} L _{1} + α _{2} L _{2}}{L _{1} + L _{2}}$