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Question #96105

Difficulty: 8/10 | 🟢 Free

The Question

A physical pendulum consists of a uniform rigid rod of mass $m$ and length $L$ pivoted freely at its upper end. A small container of negligible mass is attached to its lower end. Initially, the container holds a pure sample of radioactive isotope A. Isotope A decays into isotope B with decay constant $\lambda_A$ , and B decays into isotope C with decay constant $\lambda_B$ (where $\lambda_B \gg \lambda_A$ ). Isotope C is a stable gas that escapes the container isotropically and instantaneously. Let $m_c(t)$ be the total mass of the contents of the container at time $t$ , and $M(t) = m + m_c(t)$ be the total mass of the pendulum. The pendulum undergoes small oscillations in a vertical plane under gravity $g$ . Which of the following statements regarding the system is INCORRECT?

Options

A.
The correct equation of motion for small oscillations is $I(t)\ddot{\theta} + M(t)g y_{cm}(t)\theta = 0$ , where $I(t)$ and $y_{cm}(t)$ are the instantaneous moment of inertia and center of mass distance from the pivot.
B.
Applying $\tau_{ext} = \frac{d}{dt}(I\dot{\theta})$ yields a negative damping term $\dot{I}(t)\dot{\theta}$ , which correctly predicts an exponential growth in the oscillation amplitude.
C.
At secular equilibrium, the mass of isotope B in the container is approximately $m_B(t) = \left(\frac{\lambda_A M_B}{\lambda_B M_A}\right) m_A(t)$ , where $M_A$ and $M_B$ are their respective atomic masses.
D.
The angular frequency of the pendulum $\omega(t)$ gradually increases as the radioactive decay proceeds and gas C escapes.

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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

Successive Decay Equilibrium|Variable Mass Systems|Mass Defect vs. Binding Energy

Why This Question Is Tricky

Examiner Levers

Negative Thinking

Uses "NOT", "EXCEPT", or "INCORRECT" to flip the logic gate.

Difficulty Boost : +2

Extraneous Data

Provides unnecessary variables (Red Herrings) to test conceptual focus.

Difficulty Boost : +3

Common Mistakes

This question tests the student's ability to handle variable mass systems in rotational dynamics, combined with successive radioactive decay.

The primary trap lies in the misapplication of Newton's Second Law for rotation.

Students often memorize $\tau = \frac{dL}{dt} = \frac{d}{dt}(I\omega) = I\alpha + \dot{I}\omega$ , failing to realize this form is only valid for closed systems.

When mass escapes isotropically, it carries away its own angular momentum, exactly canceling the $\dot{I}\omega$ term.

Consequently, the equation of motion has no damping term, and the amplitude actually decreases due to the adiabatic increase in the natural frequency as the mass distribution shifts.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Nuclei - Class 12
Concept	Successive Decay Equilibrium
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free