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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA particle of mass \( m \) moves along a line given by \( y = \sqrt{3}x + 4 \) in the \( xy \)-plane with a constant speed \( v_0 \). The magnitude of the angular momentum of this particle about the origin is:
Correct answer · Option 2
\( 2 m v_0 \)
Explanation
The perpendicular distance \( r_\perp \) from origin \( (0,0) \) to the line \( \sqrt{3}x - y + 4 = 0 \) is \( r_\perp = \frac{|0 - 0 + 4|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{4}{2} = 2\,\text{m} \). Angular momentum is \( l = p r_\perp = m v_0 (2) = 2 m v_0 \).
