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Question #42749

Difficulty: 8/10 | 🟢 Free

The Question

A potentiometer wire of length LL, cross-sectional area AA, and thermal conductivity KK is connected to a constant current source I0I_0. The ends of the wire at x=0x=0 and x=Lx=L are clamped to massive blocks maintained at a constant temperature T0T_0. The wire is kept in a vacuum, so heat is only lost via conduction to the ends. The electrical resistivity of the wire varies with temperature TT as ρ(T)=ρ0[1+α(T−T0)]\rho(T) = \rho_0 [1 + \alpha (T - T_0)], where ρ0\rho_0 is the resistivity at T0T_0 and α\alpha is the temperature coefficient of resistivity. Assume the heat generated per unit volume is uniform and approximately equal to I02ρ0A2\frac{I_0^2 \rho_0}{A^2}. Two cells of EMFs E1E_1 and E2E_2 are balanced on this potentiometer wire at null points x1=L/3x_1 = L/3 and x2=2L/3x_2 = 2L/3 respectively, measured from the x=0x=0 end. Given the dimensionless parameter γ=αI02ρ0L22KA2\gamma = \frac{\alpha I_0^2 \rho_0 L^2}{2 K A^2}, what is the ratio E1E2\frac{E_1}{E_2}?

Options

  • A.

    54+7γ108+20γ\frac{54 + 7\gamma}{108 + 20\gamma}

  • B.

    12\frac{1}{2}

  • C.

    27+7γ54+10γ\frac{27 + 7\gamma}{54 + 10\gamma}

  • D.

    54+5γ108+14γ\frac{54 + 5\gamma}{108 + 14\gamma}

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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

Equipotential Surface Geometry|Potentiometer Null Point|Internal Energy Dependency

Why This Question Is Tricky

Common Mistakes

Students typically assume the potential gradient dV/dxdV/dx in a potentiometer wire is strictly constant, leading to the linear relation E∝xE \propto x.

This question introduces an electro-thermal feedback loop where internal Joule heating creates a parabolic temperature profile due to conductive heat loss to the clamped ends.

Because resistivity is temperature-dependent, the equipotential surfaces are non-linearly perturbed.

The trap is applying the standard geometric ratio E1/E2=x1/x2=1/2E_1/E_2 = x_1/x_2 = 1/2 without accounting for the spatial variation of resistivity caused by the internal energy dissipation, forcing the coupling of Kirchhoff's laws with Fourier's law of heat conduction.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptEquipotential Surface Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree