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Question #42621

Difficulty: 8/10 | 🟢 Free

The Question

A physical pendulum consists of a massless rigid rod of length $L$ pivoted freely at one end $O$ . A thin straight rod (acting as the pendulum bob) of length $L$ is rigidly attached to the other end of the massless rod, such that the entire assembly forms a straight line of total length $2L$ . The linear mass density of the attached bob varies as $\lambda(x) = c x$ , where $x$ is the distance from the pivot $O$ and $c$ is a positive constant. If the pendulum undergoes small oscillations in a vertical plane, what is the time period of the oscillations?

Options

A.
$2\pi \sqrt{\frac{45L}{28g}}$
B.
$2\pi \sqrt{\frac{14L}{9g}}$
C.
$2\pi \sqrt{\frac{3L}{2g}}$
D.
$2\pi \sqrt{\frac{15L}{7g}}$

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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

Simple Pendulum Small Angle Approximation|Torque about non-fixed axes|Calculus (Integration)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze a physical pendulum with a non-uniform mass distribution, bypassing the standard simple pendulum approximation.

The primary trap is assuming the bob can be treated as a point mass located either at its geometric center or at its actual center of mass, which leads to incorrect time periods.

Students must recognize that for an extended, non-uniform body, the restorative torque depends on the exact center of mass, while the inertial resistance depends on the integrated moment of inertia about the pivot, requiring the rigorous application of $T = 2\pi\sqrt{I/Mgd}$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Oscillations - Class 11 Part 2
Concept	Simple Pendulum Small Angle Approximation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free