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Question #42482

Difficulty: 8/10 | 🟢 Free

The Question

A uniform rigid rod of mass $M$ and length $L$ is pivoted smoothly at its upper end $O$ . A point charge $+Q$ of negligible mass is attached to the lower free end of the rod. Another point charge $+Q$ is fixed at a distance $L$ directly below the pivot $O$ . The system exists in a uniform downward gravitational field $g$ . The value of $Q$ is such that the rod is in stable equilibrium when it makes an angle $\theta_0 = \pi/3$ with the downward vertical. What is the time period of small oscillations of the rod about this equilibrium position?

Options

A.
$2\pi \sqrt{\frac{8L}{27g}}$
B.
$2\pi \sqrt{\frac{4L}{3g}}$
C.
$2\pi \sqrt{\frac{8L}{9g}}$
D.
$2\pi \sqrt{\frac{2L}{3g}}$

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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

Conservative Force Field|Physical Pendulum Inertia|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question targets the lethal misconception of assuming simple harmonic motion (SHM) relies solely on the linear restoring torque of gravity, ignoring the anharmonic corrections of a non-uniform conservative field.

Students often memorize the physical pendulum formula $T = 2\pi\sqrt{I/mgd}$ and blindly apply it, or they incorrectly linearize the electrostatic force by failing to compute the second derivative of the potential energy $U''(\theta_0)$ .

The trap exposes the necessity of deriving the effective torsional stiffness $k_{eff} = d^2U/d\theta^2$ at the equilibrium point, especially when the distance between interacting bodies is a non-linear function of the angular displacement (here, $r = 2L\sin(\theta/2)$ ).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Conservative Force Field
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free