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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA particle of mass \( m = 0.5\text{ kg} \) moves with a constant velocity \( \mathbf{v} = (4\hat{i} - 3\hat{j})\text{ m/s} \) along the straight line \( 3x + 4y = 12 \) (where \( x \) and \( y \) are in meters). The magnitude of the angular momentum of the particle about the origin is:
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Correct answer · Option B
\( 6.0\text{ J s} \)
Explanation
Perpendicular distance from origin to the line is \( r_\perp = \frac{|-12|}{\sqrt{3^2 + 4^2}} = 2.4\text{ m} \). Linear momentum \( p = m v = 0.5 \times 5 = 2.5\text{ kg m/s} \), so \( l = r_\perp p = 2.4 \times 2.5 = 6.0\text{ J s} \).
