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Question #42983

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The Question

A uniform conducting rod of mass $M$ and length $3l$ is pivoted at a point $P$ which is at a distance $l$ from one of its ends. The rod can rotate freely in a horizontal plane. A uniform vertical magnetic field $B$ exists throughout the region. The longer segment of the rod (of length $2l$ ) has its free end sliding on a fixed, perfectly conducting circular ring. A resistor of resistance $R$ connects the pivot $P$ and the circular ring. The shorter segment of the rod (of length $l$ ) is free and does not touch any ring.

Initially, the rod is rotating with an angular velocity $\omega_0$ and the resistor circuit is open. At $t=0$ , a small piece of putty of mass $m = M/3$ is dropped gently onto the free end of the shorter segment and sticks to it. Simultaneously, the resistor circuit is closed.

Assume the self-inductance of the circuit is negligible. Neglecting the resistance of the rod and friction at the pivot, find the total energy dissipated as Joule heating in the resistor $R$ until the rod comes to a complete stop.

Options

A.
$\frac{3}{8} M l^2 \omega_0^2$
B.
$\frac{1}{2} M l^2 \omega_0^2$
C.
$\frac{2}{3} M l^2 \omega_0^2$
D.
$\frac{3}{14} M l^2 \omega_0^2$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Motional EMF in Rotating Rods|Angular Momentum conservation|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of rotational collision dynamics with electromagnetic induction and the Work-Energy Theorem.

The primary trap is the 'Back-EMF Feedback Loop' where students often attempt to set up and integrate the complex differential equation for motional EMF ( $E = 2B\omega l^2$ ) and magnetic torque, risking algebraic errors.

The elegant 'JEE Advanced' bypass is recognizing that the total Joule heating exactly equals the kinetic energy of the system just after the inelastic collision.

A secondary lethal trap lies in the variable moment of inertia: students frequently miscalculate the initial moment of inertia of the off-center pivoted rod or fail to conserve angular momentum during the putty drop, incorrectly assuming the initial kinetic energy $\frac{1}{2}I_1\omega_0^2$ is entirely dissipated.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	Motional EMF in Rotating Rods
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free