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

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