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Question #43520

Difficulty: 8/10 | 🟢 Free

The Question

A uniform solid sphere of mass MM and initial radius R0R_0 is isolated in deep space. The thermal internal energy of the material is given by Uth=McTU_{\text{th}} = M c T, where cc is a constant and TT is the absolute temperature. An internal heat source slowly raises the uniform temperature of the sphere by an amount ΔT\Delta T. Due to thermal expansion, the radius of the sphere increases to R=R0(1+αΔT)R = R_0(1 + \alpha \Delta T), where α\alpha is a constant. Assuming the sphere's mass density remains uniform throughout its volume during the expansion, what is the total heat QQ supplied to the sphere? (GG is the universal gravitational constant)

Options

  • A.

    McΔT+3GM2αΔT5R0(1+αΔT)M c \Delta T + \frac{3 G M^2 \alpha \Delta T}{5 R_0 (1 + \alpha \Delta T)}

  • B.

    McΔT+3GM2αΔT5R0M c \Delta T + \frac{3 G M^2 \alpha \Delta T}{5 R_0}

  • C.

    McΔT−3GM2αΔT5R0(1+αΔT)M c \Delta T - \frac{3 G M^2 \alpha \Delta T}{5 R_0 (1 + \alpha \Delta T)}

  • D.

    McΔT+3GM2αΔT5R0(1+αΔT)2M c \Delta T + \frac{3 G M^2 \alpha \Delta T}{5 R_0 (1 + \alpha \Delta T)^2}

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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

Gravitational Self-Energy|Thermal Expansion of Volume|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of the First Law of Thermodynamics with gravitational self-energy in a non-linear regime.

A common trap is for students to use a small-change linear approximation for the work done against gravity (ΔUg≈3GM25R0αΔT\Delta U_g \approx \frac{3GM^2}{5R_0} \alpha \Delta T) even when ΔT\Delta T is not specified as infinitesimally small, leading to Option B.

Another lethal mistake is incorrectly calculating the work done by taking the final gravitational force and multiplying it by the change in radius, leading to the (1+αΔT)2(1+\alpha\Delta T)^2 denominator in Option D.

Option C traps students who make a sign error in the self-energy formula.

The correct approach requires evaluating the exact difference between the initial and final gravitational self-energy states.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterGravitation - Class 11 Part 1
ConceptGravitational Self-Energy
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree