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Question #42107

Difficulty: 8/10 | 🟢 Free

The Question

A grounded spherical conducting liquid drop of radius RR and surface tension TT is situated at the center of a rigid, isolated spherical conducting shell of radius kRkR (k>1k > 1). The outer shell carries a total charge QQ. The space between them is vacuum. Assuming the effect of the grounding wire is negligible, what is the excess pressure ΔP\Delta P inside the liquid drop?

Options

  • A.

    2TR−Q232π2ϵ0k2R4\frac{2T}{R} - \frac{Q^2}{32\pi^2 \epsilon_0 k^2 R^4}

  • B.

    2TR−Q2(k−1)32π2ϵ0k2R4\frac{2T}{R} - \frac{Q^2 (k-1)}{32\pi^2 \epsilon_0 k^2 R^4}

  • C.

    2TR+Q232π2ϵ0k2R4\frac{2T}{R} + \frac{Q^2}{32\pi^2 \epsilon_0 k^2 R^4}

  • D.

    2TR−Q232π2ϵ0R4\frac{2T}{R} - \frac{Q^2}{32\pi^2 \epsilon_0 R^4}

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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

Molecular Forces of Cohesion|Energy Density Locality|Self-Energy of Charged Sphere|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of electrostatics and continuum mechanics, specifically focusing on the locality of energy density and the correct application of boundary conditions.

A lethal trap awaits students who use the global energy method: they might calculate the total electrostatic energy UE=Q2(k−1)8πϵ0k2RU_E = \frac{Q^2(k-1)}{8\pi\epsilon_0 k^2 R} and incorrectly differentiate it with respect to RR while treating kk as a constant ratio.

This yields the erroneous term Q2(k−1)32π2ϵ0k2R4\frac{Q^2(k-1)}{32\pi^2 \epsilon_0 k^2 R^4} (Option B).

Students must recognize that the outer shell is rigid, so the derivative ∂UE∂R\frac{\partial U_E}{\partial R} must be taken with the shell radius RshellR_{shell} held constant.

Using the local energy density PE=12ϵ0E2P_E = \frac{1}{2}\epsilon_0 E^2 elegantly bypasses this calculus trap.

Another conceptual trap is adding the electrostatic pressure instead of subtracting it, failing to realize that self-repulsion acts outward against the cohesive surface tension.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMechanical Properties of Fluids - Class 11 Part 2
ConceptMolecular Forces of Cohesion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree