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A particle of mass m = 5 is moving with a uniform speed v = √3√2 in the XOY plane along the line y = √3x + 4. The magnitude of the angular momentum of the particle about the origin is

1. (a) 60 units
2. (b) 40√2 units
3. (c) zero
4. (d) 7.5 units

Correct answer: (a) 60 units

Solution

The angular momentum is given by L = mvr sinθ, where r is the perpendicular distance from the origin to the line of motion. The line y = √3x + 4 has a perpendicular distance of 2 units from the origin. Substituting m = 5, v = √3√2, and r = 2, we get L = 5 × √3√2 × 2 = 60 units.

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