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Question #42565

Difficulty: 8/10 | 🟢 Free

The Question

A conducting rod of length LL and mass mm is pivoted at one end and rotates with a constant angular velocity ω\omega in a horizontal plane. The other end slides on a frictionless, conducting circular ring. A magnetic field perpendicular to the plane of rotation exists in the region, varying with the radial distance rr from the pivot as B⃗=B0rLk^\vec{B} = B_0 \frac{r}{L} \hat{k}. A resistor of resistance RR connects the pivot and the ring. This resistor is placed inside a rigid, adiabatic container holding nn moles of an ideal diatomic gas. If an external agent maintains the constant angular velocity of the rod, what is the change in temperature ΔT\Delta T of the gas after a time interval tt? (Let RgR_g be the universal gas constant).

Options

  • A.

    B02ω2L4t10nRgR\frac{B_0^2 \omega^2 L^4 t}{10 n R_g R}

  • B.

    2B02ω2L4t27nRgR\frac{2 B_0^2 \omega^2 L^4 t}{27 n R_g R}

  • C.

    2B02ω2L4t45nRgR\frac{2 B_0^2 \omega^2 L^4 t}{45 n R_g R}

  • D.

    B02ω2L4t9nRgR\frac{B_0^2 \omega^2 L^4 t}{9 n R_g R}

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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

Work Done in Non-PV Diagrams|Motional EMF in Rotating Rods|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of rotational dynamics, electromagnetic induction, and thermodynamics.

The primary trap lies in the non-uniform magnetic field, which requires students to integrate both the motional EMF and the variable retarding magnetic torque from first principles rather than relying on standard formulas.

A common mistake is assuming a uniform magnetic field or incorrectly calculating the mechanical power as P=τωP = \tau \omega by using the average force instead of integrating the torque dτ=r(IBdr)d\tau = r(I B dr).

Additionally, students must correctly apply the First Law of Thermodynamics for a rigid container (isochoric process) using the specific heat capacity of a diatomic gas.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermodynamics - Class 11 Part 2
ConceptWork Done in Non-PV Diagrams
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree