Q89 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics

A particle of mass $m$ is 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:

  • Correct answer
    A
    $\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}$
  • B
    $\frac{\sqrt{3}}{2} \frac{\mathrm{mu}^2}{\mathrm{~g}}$
  • C
    $\frac{\mathrm{mu}^3}{\sqrt{2} \mathrm{~g}}$
  • D
    zero

Solution

Angular momentum, $L=m v H=m u \cos 30 H$ $=m u \cos 30^{\circ} \times \frac{u^2 \sin ^2 \theta}{2 g}\left[\because H=\frac{u^2 \sin ^2 \theta}{2 g}\right]$ $=\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g}$

Hence, the correct answer is option: 'A'.

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