Physics · Rotational Dynamics

JEE Main 2024 — 30 January, Shift 1 — Question 43

A particle of mass $m$ projected with a velocity ' $u$ ' making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :

Option A:
$\frac{\sqrt{3}}{16} \frac{m u^{3}}{g}$
Correct
Option B:
$\frac{\sqrt{3}}{2} \frac{m u^{2}}{g}$
Option C:
$\frac{m u^{3}}{\sqrt{2} g}$
Option D:
zero

Answer: A

Step-by-step solution

$\mathrm{L}=m u \cos \theta H$ $=m u \cos \theta \times \frac{u^{2} \sin ^{2} \theta}{2 g}$

$=\frac{m u^{3}}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^{2}=\frac{\sqrt{3} m u^{3}}{16 g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Angular Momentum and its Conservation

Consider a Disc of mass 5 kg , radius 2 m , rotating with angular velocity of 10 rad / s about an axis perpendicular to the plane of…

Answer: A

Step-by-step solution

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