JEE Main 2020 — Physics Thermodynamics

2020mcqmedium

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

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Options

A
$\frac{{\alpha }_{1}{L}_{1}+{\alpha }_{2}{L}_{2}}{{L}_{1}+{L}_{2}}$
B
$2\sqrt{{\alpha }_{1}{\alpha }_{2}}$
C
$\frac{{\alpha }_{1}+{\alpha }_{2}}{2}$
D
$4\frac{{\alpha }_{1}{\alpha }_{2}}{{\alpha }_{1}+{\alpha }_{2}}\frac{{L}_{2}{L}_{1}}{{({L}_{2}+{L}_{1})}^{2}}$

Did you get this right?

Solution

The wire are joined end-to-end, so they are in series combination, then total length of wire is given by

$L={L}_{1}+{L}_{2}...(1)$

When the temperature is increase by $\Delta \theta$ , then its final length is given by using concept of linear expansion,

${L}^{'}=L(1+\alpha \Delta \theta )...(2)$

Using equation $(1)$ and $(2)$ for the series combination of wires, final length is given by

${L}^{'}={{L}_{1}}^{'}+{{L}_{2}}^{'}...(3)$

$\Rightarrow ({L}_{1}+{L}_{2})(1+{\alpha }_{\mathrm{eff}}\Delta \theta )={L}_{1}(1+{\alpha }_{1}\Delta \theta )+{L}_{2}(1+{\alpha }_{2}\Delta \theta )$

$\Rightarrow {\alpha }_{\mathrm{eff}}=\frac{{L}_{1}{\alpha }_{1}+{L}_{2}{\alpha }_{2}}{({L}_{1}+{L}_{2})}$

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Options

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

$2\sqrt{{\alpha }_{1}{\alpha }_{2}}$

$\frac{{\alpha }_{1}+{\alpha }_{2}}{2}$

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

Solution

The wire are joined end-to-end, so they are in series combination, then total length of wire is given by

$L={L}_{1}+{L}_{2}...(1)$

When the temperature is increase by $\Delta \theta$ , then its final length is given by using concept of linear expansion,

${L}^{'}=L(1+\alpha \Delta \theta )...(2)$

Using equation $(1)$ and $(2)$ for the series combination of wires, final length is given by

${L}^{'}={{L}_{1}}^{'}+{{L}_{2}}^{'}...(3)$

$\Rightarrow ({L}_{1}+{L}_{2})(1+{\alpha }_{\mathrm{eff}}\Delta \theta )={L}_{1}(1+{\alpha }_{1}\Delta \theta )+{L}_{2}(1+{\alpha }_{2}\Delta \theta )$

$\Rightarrow {\alpha }_{\mathrm{eff}}=\frac{{L}_{1}{\alpha }_{1}+{L}_{2}{\alpha }_{2}}{({L}_{1}+{L}_{2})}$

JEE Main 2020 — Physics Thermodynamics

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JEE Main 2020 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics