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JEE Main 2024 (30 Jan, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2024 (30 Jan, Shift 1)

Q24JEE Main 2024 · 30 JAN · SHIFT 1Medium

Angular Momentum & its Conservation

A particle of mass $\mathrm{m}$ is projected with a velocity ' $\mathrm{u}$ ' 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 $\mathrm{h}$ is:

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Answer: (d)

At the top, velocity is horizontal $u\cos30^\circ$ at height $H=\frac{u^2\sin^2 30^\circ}{2g}=\frac{u^2}{8g}$ .
$L=m(u\cos30^\circ)H=m\cdot\frac{\sqrt3u}{2}\cdot\frac{u^2}{8g}$
$L=\frac{\sqrt3}{16}\frac{mu^3}{g}$
So the answer is (d).

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