Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Angular Momentum & its Conservation›Question 2

MHT CET 2026 (14 May, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from MHT CET 2026 (14 May, Shift 2)

Q2MHT CET 2026 · 14 MAY · SHIFT 2Medium

Angular Momentum & its Conservation

A particle of mass m is projected with velocity V making an angle of 45∘45^\circ with the horizontal. The magnitude of the angular momentum of the projectile about an axis of projection when the particle is at maximum height h is (gg = acceleration due to gravity) (sin⁡45∘=12) (cos⁡45∘=12)\left(\sin 45^\circ = \frac{1}{\sqrt{2}}\right)\ \left(\cos 45^\circ = \frac{1}{\sqrt{2}}\right)

Show answer and solution

Answer: (b)

  1. At maximum height the velocity is horizontal: Vcos⁡45∘=V2V\cos 45^\circ = \dfrac{V}{\sqrt{2}}; perpendicular distance from the point of projection is hh.
  2. h=V2sin⁡245∘2g=V24gh = \dfrac{V^2\sin^2 45^\circ}{2g} = \dfrac{V^2}{4g}
  3. L=mV2⋅V24g=mV342gL = m\dfrac{V}{\sqrt{2}}\cdot\dfrac{V^2}{4g} = \dfrac{mV^3}{4\sqrt{2}g}
  4. So the answer is (b).
Report a mistake

More questions on Angular Momentum & its Conservation