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JEE Main 2022 (25 Jul, Shift 1) · Thermometry & Thermal Expansion

Thermometry & Thermal Expansion question from JEE Main 2022 (25 Jul, Shift 1)

Q45JEE Main 2022 · 25 JUL · SHIFT 1Medium

Thermometry & Thermal Expansion

A unit scale is to be prepared whose length does not change with temperature and remains 20 cm20\text{ 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 cm40\text{ cm}, then length of iron will be __________ cm⋅(α\text{cm} \cdot \left(\alpha\right.iron =1.2×10−5 K−1= 1.2 \times 10^{-5}\text{ K}^{-1} and α\alphabrass =1.8×10−5 K−1)= 1.8 \times 10^{-5}\text{ K}^{-1}\Big).

Numerical answer type. Work it out, then open the answer.

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Answer: 60

  1. Difference stays constant if both expand equally: Lbαb=LiαiL_b\alpha_b = L_i\alpha_i.
  2. Li=40×1.8×10−51.2×10−5=60L_i = \dfrac{40 \times 1.8\times10^{-5}}{1.2\times10^{-5}} = 60 cm
  3. Check: 60−40=2060 - 40 = 20 cm. Answer: 60 cm.
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