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JEE Main 2024 (29 Jan, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2024 (29 Jan, Shift 2)

Q26JEE Main 2024 · 29 JAN · SHIFT 2Medium

Angular Momentum & its Conservation

A body of mass $5\text{ kg}$ moving with a uniform speed $3\sqrt{2}\text{ ms}^{-1}$ in $X - Y$ plane along the line $y = x + 4$ . The angular momentum of the particle about the origin will be _________ $\text{kg m}^{2}\ \text{s}^{-1}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 60

$L=mv\,d$ , with $d$ = perpendicular distance of line $x-y+4=0$ from origin $=\frac{4}{\sqrt2}=2\sqrt2$ m.
$L=5\times3\sqrt2\times2\sqrt2$
$L=60$ kg m² s⁻¹

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