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JEE Main 2017 (9 Apr, Shift 1) · Thermometry & Thermal Expansion

Thermometry & Thermal Expansion question from JEE Main 2017 (9 Apr, Shift 1)

Q64JEE Main 2017 · 9 APR · SHIFT 1Medium

Thermometry & Thermal Expansion

A steel rail of length 5 m and area of cross section 40cm2 is prevented from expanding along its length while the temperature rises by 10oC. If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately:

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Answer: (b)

  1. Thermal force F=YAαΔTF = YA\alpha\Delta T.
  2. F=2×1011×40×10−4×1.2×10−5×10=9.6×104F = 2\times10^{11} \times 40\times10^{-4} \times 1.2\times10^{-5} \times 10 = 9.6\times10^4 N
  3. So F≈1×105F \approx 1\times10^5 N, answer (b).
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