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MHT CET 2026 (14 May, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from MHT CET 2026 (14 May, Shift 1)

Q3MHT CET 2026 · 14 MAY · SHIFT 1Medium

Angular Momentum & its Conservation

A particle of mass $m$ is projected with speed $v$ at $45^\circ$ with the horizontal. The magnitude of angular momentum of the particle about the point of projection when the particle is at its maximum height $h$ is $\left(\cos 45^\circ = \frac{1}{\sqrt{2}}\right)$

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Answer: (d)

At maximum height the velocity is horizontal: $v\cos 45^\circ = \dfrac{v}{\sqrt{2}}$ .
Its perpendicular distance from the point of projection is $h$ .
$L = m\dfrac{v}{\sqrt{2}}h = \dfrac{mvh}{\sqrt{2}}$ , so the answer is (d).

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