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Question #42354

Difficulty: 8/10 | 🟢 Free

The Question

A rigid rod of length $L$ has a non-uniform linear mass density given by $\lambda = kx^2$ , where $x$ is the distance from its left end and $k$ is a positive constant. The rod is pivoted at its geometric center ( $x = L/2$ ) and is free to oscillate in a vertical plane under the acceleration due to gravity $g$ . What is the time period $T$ of small oscillations?

Options

A.
$2\pi \sqrt{\frac{2L}{5g}}$
B.
$2\pi \sqrt{\frac{12L}{5g}}$
C.
$2\pi \sqrt{\frac{3L}{20g}}$
D.
$2\pi \sqrt{\frac{7L}{15g}}$

Check Your Answer

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

Center of Mass (COM) Frame|Compound Pendulum Period|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to evaluate the physical pendulum period for a non-uniform mass distribution.

The primary trap lies in the misapplication of the Parallel Axis Theorem.

Students often incorrectly calculate the moment of inertia about the pivot by either using the moment of inertia about the end without shifting it to the COM first, or by forgetting to shift from the center of mass to the pivot.

Furthermore, it filters out rote memorization of standard uniform rod formulas by requiring integration to find both the shifted center of mass and the variable moment of inertia.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	System of Particles and Rotational Motion - Class 11 Part 1
Concept	Center of Mass (COM) Frame
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free