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6.System of Particles and Rotational Motion
medium
A particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt 2$ in the $XOY$ plane along the line $Y = X + 4$ . The magnitude of the angular momentum of the particle about the origin is .......
A
$0$
B
$60$
C
$7.5$
D
$40\sqrt 2$
(AIPMT-1991)
Solution

$\mathrm{d}=\mathrm{OA} \sin 45^{\circ}=4 \times \frac{1}{\sqrt{2}}$
$\mathrm{L}=\mathrm{mvd}$
$=5 \times 3 \sqrt{2} \times \frac{4}{\sqrt{2}}=60 \mathrm{unit}$
Standard 11
Physics