question_answer

A particle of mass 'm' is projected with a velocity \[\upsilon \] 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 :     AIEEE  Solved  Paper (Held On 11 May  2011)

A)  zero

B) \[\frac{m{{\upsilon }^{3}}}{\sqrt{2}g}\]

C) \[\frac{\sqrt{3}}{16}\frac{m{{\upsilon }^{3}}}{g}\]

D) \[\frac{\sqrt{3}}{2}\frac{m{{\upsilon }^{2}}}{g}\]

Correct Answer: C

Solution :

                                \[{{L}_{0}}={{\Pr }_{\bot }}\]                 \[{{L}_{0}}=mv\cos \theta H\]                 \[=mg\frac{\sqrt{3}}{2}.\frac{{{v}^{2}}{{\sin }^{2}}{{30}^{o}}}{2g}=\frac{\sqrt{3}m{{v}^{3}}}{16g}.\]