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Question #42390

Difficulty: 8/10 | 🟢 Free

The Question

A uniform rigid rod of mass $M$ and length $L$ is suspended vertically from a frictionless pivot at its upper end. A horizontal impulse $J$ is applied to the rod at its lower end, causing it to swing and reach a maximum angular displacement of $60^\circ$ from the vertical. If $g$ is the acceleration due to gravity, what is the magnitude of the impulsive reaction force exerted by the pivot on the rod during the strike?

Options

A.
$M \sqrt{\frac{g L}{24}}$
B.
$M \sqrt{\frac{3 g L}{8}}$
C.
$M \sqrt{\frac{g L}{6}}$
D.
$M \pi \sqrt{\frac{g L}{216}}$

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

Impulsive Torque and Angular Momentum|Physical Pendulum Inertia|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the application of impulsive torque and energy conservation in a physical pendulum, specifically targeting two classic JEE traps.

First, students often incorrectly apply linear momentum conservation ( $J = M v_{cm}$ ) without accounting for the impulsive reaction force at the hinge.

By requiring the calculation of this exact hinge impulse, the problem forces the use of angular impulse about the pivot to bypass the unknown force initially.

Second, the $60^\circ$ deflection acts as a trap for the small-angle approximation; students who incorrectly use $1 - \cos\theta \approx \theta^2/2$ for large angles will arrive at the distractor involving $\pi$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	System of Particles and Rotational Motion - Class 11 Part 1
Concept	Impulsive Torque and Angular Momentum
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free