Thermal Properties of Matter

JEE Main level, single correct, medium

32 of 126

Question 32 of 126

Two different wires having lengths L1{L}_{1} and L2{L}_{2} and respective temperature coefficient of linear expansion α1{\alpha }_{1} and α2,{\alpha }_{2}, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

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Answer: (A) α1L1+α2L2L1+L2\dfrac{{\alpha }_{1}{L}_{1}+{\alpha }_{2}{L}_{2}}{{L}_{1}+{L}_{2}}

The wire are joined end-to-end, so they are in series combination, then total length of wire is given by
L=L1+L2...(1)L={L}_{1}+{L}_{2}...(1)
When the temperature is increase by ∆θ∆\theta, then its final length is given by using concept of linear expansion,
L′=L(1+α∆θ)...(2){L}^{'}=L(1+\alpha ∆\theta )...(2)
Using equation (1)(1) and (2)(2) for the series combination of wires, final length is given by
L′=L1′+L2′...(3){L}^{'}={{L}_{1}}^{'}+{{L}_{2}}^{'}...(3)
⇒(L1+L2)(1+αeff∆θ)=L1(1+α1∆θ)+L2(1+α2∆θ)\Rightarrow \left({L}_{1}+{L}_{2}\right)(1+{\alpha }_{eff}∆\theta )={L}_{1}(1+{\alpha }_{1}∆\theta )+{L}_{2}(1+{\alpha }_{2}∆\theta )
⇒αeff=L1α1+L2α2(L1+L2)\Rightarrow {\alpha }_{eff}=\dfrac{{L}_{1}{\alpha }_{1}+{L}_{2}{\alpha }_{2}}{\left({L}_{1}+{L}_{2}\right)}

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