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Question #42205

Difficulty: 8/10 | 🟢 Free

The Question

A uniform metal bar of length LL, cross-sectional area AA, Young's modulus YY, and coefficient of linear expansion α\alpha is placed on a smooth horizontal surface. One end of the bar is fixed to a rigid wall, and the other end is attached to a massless ideal spring of force constant kk. The other end of the spring is fixed to another rigid wall. Initially, the spring is in its relaxed state and the bar is at a uniform temperature T0T_0. The temperature of the bar is then slowly increased by ΔT\Delta T. Assuming YY and α\alpha are independent of temperature, what is the elastic potential energy stored in the metal bar due to thermal stress?

Options

  • A.

    k2YAL3α2ΔT22(kL+YA)2\frac{k^2 Y A L^3 \alpha^2 \Delta T^2}{2(k L + Y A)^2}

  • B.

    kY2A2L2α2ΔT22(kL+YA)2\frac{k Y^2 A^2 L^2 \alpha^2 \Delta T^2}{2(k L + Y A)^2}

  • C.

    YALα2ΔT22\frac{Y A L \alpha^2 \Delta T^2}{2}

  • D.

    k2Y2A2L3α2ΔT22(kL+YA)3\frac{k^2 Y^2 A^2 L^3 \alpha^2 \Delta T^2}{2(k L + Y A)^3}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Thermal Stress Calculation|First Law Application|Energy Stored in Inductor (Analogy)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the macroscopic work done against an external constraint and the internal elastic strain energy stored in a material.

A common trap is conflating the two, leading students to calculate the energy stored in the spring (Option B) instead of the bar.

Another mistake is ignoring the spring's yielding and assuming the bar is fully constrained, which leads to Option C.

The correct physical model treats the bar as an equivalent spring of constant kb=YA/Lk_b = YA/L in series with the external spring, recognizing that thermal strain does not store energy unless opposed by mechanical strain.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMechanical Properties of Solids - Class 11 Part 2
ConceptThermal Stress Calculation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree