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Question #42198

Difficulty: 8/10 | 🟢 Free

The Question

A uniform conducting rod of mass $m$ , length $L$ , and resistance $R$ is pivoted at one end and rotates in a horizontal plane with an initial angular velocity $\omega_0$ . A uniform vertical magnetic field $B$ exists in the region. The pivot exerts a constant frictional torque $\tau_f$ on the rod. The electrical power generated due to the motional EMF is continuously and entirely used as the work input for an ideal Carnot refrigerator operating between a cold reservoir at temperature $T_C$ and a hot reservoir at $T_H$ . Given that the system parameters satisfy the relation $B^2 L^4 \omega_0 = 8 R \tau_f$ , what is the total heat extracted from the cold reservoir before the rod comes to a complete stop?

Options

A.
$\frac{m L^2 \omega_0^2 T_C \ln 3}{12 (T_H - T_C)}$
B.
$\frac{m L^2 \omega_0^2 T_C}{6 (T_H - T_C)}$
C.
$\frac{m L^2 \omega_0^2 T_C \ln 2}{12 (T_H - T_C)}$
D.
$\frac{m L^2 \omega_0^2 T_C \ln 3}{6 (T_H - T_C)}$

Check Your Answer

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

Refrigerator Coefficient of Performance|Motional EMF in Rotating Rods|Work-Energy Theorem Validity

Why This Question Is Tricky

Common Mistakes

The core trap in this problem is the assumption that the initial kinetic energy of the rod is simply divided proportionally between friction and electrical work, or worse, that all kinetic energy converts to electrical work ( $W = \Delta K$ ).

Because the magnetic braking torque ( $\tau_m \propto \omega$ ) and electrical power ( $P \propto \omega^2$ ) are non-linear with respect to time, while the frictional torque is constant, the ratio of power dissipated by friction versus electrical resistance changes continuously as the rod slows down.

Students must set up the differential equation and integrate the non-linear power function over $\omega$ .

Failing to recognize this dynamic coupling leads to bypassing the integral and selecting the trap option B, which incorrectly assumes 100% energy conversion.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Refrigerator Coefficient of Performance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free