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Question #42727

Difficulty: 8/10 | 🟢 Free

The Question

A perfectly insulated cylindrical conductor of length $L$ (m), cross-sectional area $A$ (m $^2$ ), and density $D$ (kg/m $^3$ ) is initially solid at a temperature $T_0$ (K). A constant total direct current $I$ (A) is passed through it. The material has a melting point $T_m$ (K) (where $T_m > T_0$ ), specific heat capacity $c$ (J/kg $\cdot$ K) in the solid phase, and latent heat of fusion $L_f$ (J/kg). The electrical resistivities of the solid and liquid phases are $\rho_s$ ( $\Omega\cdot$ m) and $\rho_l$ ( $\Omega\cdot$ m) respectively, assumed independent of temperature. Assume infinite radial thermal conductivity (uniform temperature across any cross-section) and zero longitudinal heat conduction. As the conductor melts, the solid and liquid phases form parallel cylindrical channels extending the full length $L$ . What is the total time $\tau$ (s) required for the conductor to completely melt?

Options

A.
$\frac{D A^2}{I^2} \left[ \frac{c(T_m - T_0)}{\rho_s} + \frac{L_f (\rho_s + \rho_l)}{2 \rho_s \rho_l} \right]$
B.
$\frac{D A^2}{I^2 \rho_s} \left[ c(T_m - T_0) + L_f \right]$
C.
$\frac{D A^2}{I^2} \left[ \frac{c(T_m - T_0)}{\rho_s} + \frac{L_f}{\rho_l - \rho_s} \ln\left(\frac{\rho_l}{\rho_s}\right) \right]$
D.
$\frac{D A^2}{I^2} \left[ \frac{c(T_m - T_0)}{\rho_s} + \frac{L_f (\rho_s + \rho_l)}{\rho_s \rho_l} \right]$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Current Density Vector|Kirchhoff’s Junction Rule|Specific Heat Capacity of Phase Mixtures|Integration

Why This Question Is Tricky

Common Mistakes

Students often incorrectly assume that the macroscopic resistance of the wire remains constant at its initial solid-state value $R = \rho_s L / A$ throughout the entire melting process (Option B).

This ignores the electro-thermal coupling where the newly formed liquid phase provides a parallel path for the current, altering the equivalent resistance and the total Joule heating rate $I^2 R_{eq}$ .

Furthermore, students might misapply the geometry of the phase boundary, assuming a longitudinal series melting front (Option C) instead of a parallel configuration, which fundamentally changes the integration of the current density distribution.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Current Density Vector
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free