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Question #42185

Difficulty: 8/10 | 🟢 Free

The Question

A horizontal cylindrical container of cross-sectional area AA has adiabatic walls and contains one mole of an ideal monatomic gas. A frictionless, adiabatic piston is pushed inwards at a constant speed v0v_0. Integrated into the piston is a conducting rod of length LL and electrical resistance ReR_e. The rod slides on two perfectly conducting, parallel rails that run along the inner length of the cylinder and are shorted at the closed end. A non-uniform magnetic field perpendicular to the plane of the rails exists in the region, given by B(x)=B0x−1/2B(x) = B_0 x^{-1/2}, where xx is the distance from the closed end. The entire electrical resistance ReR_e is exposed to the gas, so all Joule heating is transferred to the gas. If the initial volume and temperature of the gas are V0V_0 and T0T_0, and the gas is compressed to a volume V0/8V_0/8, find the final temperature of the gas. (Given: B02L2v0Re=RT0\frac{B_0^2 L^2 v_0}{R_e} = R T_0, where RR is the universal gas constant).

Options

  • A.

    4T04 T_0

  • B.

    5T05 T_0

  • C.

    7T07 T_0

  • D.

    194T0\frac{19}{4} T_0

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

First Law Application|Motional EMF for Arbitrary Shapes|Integration

Why This Question Is Tricky

Common Mistakes

This problem couples the First Law of Thermodynamics with electromagnetic induction.

The primary trap is ignoring the Joule heating entirely, leading to the standard adiabatic result of 4T_0.

A secondary, highly lethal trap tests the calculation of magnetic flux in a non-uniform field: students who incorrectly assume flux is B(x) multiplied by Area (instead of integrating B dA) will obtain an EMF that is off by a factor of 2, leading to the incorrect option 19/4 T_0.

The problem rigorously tests the differential formulation of the First Law where mechanical work and internal heat generation occur simultaneously.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermodynamics - Class 11 Part 2
ConceptFirst Law Application
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree