Two different wires having lengths $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, 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:

(C) $\frac{\alpha_{1} L_{1}+a_{2} L_{2}}{L_{1}+L_{2}}$

(A) $\frac{\alpha_{1}+\alpha_{2}}{2}$

(B) $2 \sqrt{\alpha_{1} \alpha_{2}}$

(D) $4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{\mathrm{~L}_{1} \mathrm{~L}_{2}}{\left(\mathrm{~L}_{2}+\mathrm{L}_{1}\right)^{2}}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Heat & Thermodynamics

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:

(A) $\frac{α _{1} + α _{2}}{2}$
(B) $2 α_{1} α_{2}$
(C) $\frac{α _{1} L _{1} + a _{2} L _{2}}{L _{1} + L _{2}}$ ✓ Correct
(D) $4 \frac{α _{1} α _{2}}{α _{1} + α _{2}} \frac{L _{1} L _{2}}{( L _{2} + L _{1} ) ^{2}}$

Answer: (C)

Joined end to end, the two wires have combined length

L_{1} + L_{2}

at the starting temperature. For a rise

Δ T

each piece lengthens on its own:

L_{1}^{'} = L_{1} (1 + α_{1} Δ T)

and

L_{2}^{'} = L_{2} (1 + α_{2} Δ T)

, so the new total is

L_{1} + L_{2} + (L_{1} α_{1} + L_{2} α_{2}) Δ T

. Writing this same total as

(L_{1} + L_{2}) (1 + α_{a v g} Δ T)

and matching the coefficient of

Δ T

gives

(L_{1} + L_{2}) α_{a v g} = L_{1} α_{1} + L_{2} α_{2}

, so the effective coefficient is

α_{a v g} = \frac{L _{1} α _{1} + L _{2} α _{2}}{L _{1} + L _{2}}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option