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6.System of Particles and Rotational Motion
hard
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 $\mathrm{h}$ is :
A
$\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}$
B
$\frac{\sqrt{3}}{2} \frac{m u^2}{g}$
C
$\frac{m u^3}{\sqrt{2} g}$
D
zero
(JEE MAIN-2024)
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}$
Standard 11
Physics
