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JEE Main 202225 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2022 · 25 July · Morning Shift · Physics · Thermal Properties of Matter

A unit scale is to be prepared whose length does not change with...

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 40cm and length of iron will be _____cm.</p><p>(αiron=1.2×10−5K−1).</p><p>\left({\alpha}_{iron} = 1 . 2 \times {10}^{- 5}{K}^{- 1}\right) and {\alpha}_{brass} = 1 . 8 \times {10}^{- 5}{K}^{- 1}).

Correct answer

0

Step-by-step solution

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+αB\DeltaT)−li(1+αi\DeltaT)=lB−li{l}_{B}\left(1 + {\alpha}_{B}\DeltaT\right) - {l}_{i}\left(1 + {\alpha}_{i}\DeltaT\right) = {l}_{B} - {l}_{i}

⇒αBlB=liαi⇒ {\alpha}_{B}{l}_{B} = {l}_{i}{\alpha}_{i}

⇒1.8×10−5×40=li×1.2×10−5⇒ 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⇒ {l}_{i} = \dfrac{1 . 8 \times {10}^{- 5} \times 40}{1 . 2 \times {10}^{- 5}} = \dfrac{3 \times 40}{2} = 60

⇒li=60cm⇒ {l}_{i} = 60cm

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