JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2022
- Hard
- Numerical
A unit scale is to be prepared whose length does not change with temperature and remains \(20\,cm\), 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 \(40\,cm\) and length of iron will be\(...cm\) \(\left(\alpha_{\text {iron }}=1.2 \times 10^{-5} K ^{-1}\right.\) and \(\left.\alpha_{\text {brass }}=1.8 \times 10^{-5} K ^{-1}\right)\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
\(\ell_{ B }\left(1+\alpha_{ B } \Delta T \right)-\ell_{ i }\left(1+\alpha_{ i } \Delta T \right)=\ell_{ B }-\ell_{ i }\)
\(\alpha_{ B } \ell_{ B }=\ell_{ i } \alpha_{ i }\)
\(1.8 \times 10^{-5} \times 40=\ell_{ i } \times 1.2 \times 10^{-5}\)
\(\ell_{ i }=\frac{1.8 \times 10^{-5} \times 40}{1.2 \times 10^{-5}}=\frac{3 \times 40}{2}=60\) \(\ell_{ i }=60\,cm\)
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