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Question #42126

Difficulty: 8/10 | 🟢 Free

The Question

A faulty thermometer reads 5∘X5^\circ\text{X} at the melting point of ice and 95∘X95^\circ\text{X} at the boiling point of water at standard atmospheric pressure. A uniform metal rod is perfectly constrained between two rigid, immovable walls. The rod is heated, and the faulty thermometer records its initial and final temperatures as 14∘X14^\circ\text{X} and 86∘X86^\circ\text{X}, respectively. The rod has a constant coefficient of linear thermal expansion α=2×10−5 ∘C−1\alpha = 2 \times 10^{-5} \text{ } ^\circ\text{C}^{-1}. Its Young's modulus varies with the true Celsius temperature TT as Y(T)=Y0(1−βT)Y(T) = Y_0(1 - \beta T), where Y0=2×1011 N/m2Y_0 = 2 \times 10^{11} \text{ N/m}^2 and β=5×10−3 ∘C−1\beta = 5 \times 10^{-3} \text{ } ^\circ\text{C}^{-1}. Assuming the rod does not buckle, what is the magnitude of the thermal stress developed in the rod?

Options

  • A.

    240 MPa240 \text{ MPa}

  • B.

    320 MPa320 \text{ MPa}

  • C.

    288 MPa288 \text{ MPa}

  • D.

    216 MPa216 \text{ MPa}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Celsius-Fahrenheit Calibration Errors|Thermal Stress in Constrained Rods

Why This Question Is Tricky

Common Mistakes

This question tests the integration of linear thermometry calibration with temperature-dependent solid mechanics.

The primary trap is twofold: first, students often blindly plug the faulty thermometer's temperature difference (ΔX=72∘X\Delta X = 72^\circ\text{X}) directly into the stress formula without mapping it to the true Celsius scale.

Second, even if they find the true ΔT=80∘C\Delta T = 80^\circ\text{C}, they may incorrectly use the standard constant-modulus formula (σ=YαΔT\sigma = Y \alpha \Delta T) instead of recognizing that a temperature-dependent Young's Modulus requires integrating the differential stress (dσ=Y(T)αdTd\sigma = Y(T) \alpha dT) over the true temperature limits.

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Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptCelsius-Fahrenheit Calibration Errors
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree