Thermal Properties of Matter

JEE Main level, numerical answer, medium

22 of 126

Question 22 of 126

A unit scale is to be prepared whose length does not change with temperature and remains 20cm20cm, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is 40cm40cm and length of iron will be _____ cmcm.
(αiron=1.2×10−5K−1\left({\alpha }_{iron}=1.2\times {10}^{-5}{K}^{-1}\right. and αbrass=1.8×10−5K−1)\left.{\alpha }_{brass}=1.8\times {10}^{-5}{K}^{-1}\right).

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Answer: 60

Linear expansion of solid with temperature is given by Lf=L0(1+α∆T){L}_{f}={L}_{0}\left(1+\alpha ∆T\right)
Here, lB(1+αBΔT)−li(1+αiΔT)=lB−li{l}_{B}\left(1+{\alpha }_{B}\Delta T\right)-{l}_{i}\left(1+{\alpha }_{i}\Delta T\right)={l}_{B}-{l}_{i}
⇒αBlB=liαi\Rightarrow {\alpha }_{B}{l}_{B}={l}_{i}{\alpha }_{i}
⇒1.8×10−5×40=li×1.2×10−5\Rightarrow 1.8\times {10}^{-5}\times 40={l}_{i}\times 1.2\times {10}^{-5}
⇒li=1.8×10−5×401.2×10−5=3×402=60\Rightarrow {l}_{i}=\dfrac{1.8\times {10}^{-5}\times 40}{1.2\times {10}^{-5}}=\dfrac{3\times 40}{2}=60
⇒li=60cm\Rightarrow {l}_{i}=60cm

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