Solution:

The angular momentum

$L$

of a particle about the origin is given by:

$$L = m v r \sin\theta$$

where:

• 
  $m = 5$ 

  g (mass of the particle),

• 
  $v = 3\sqrt{2}$ 

  cm/s (speed of the particle),

• 
  $r = 2\sqrt{5}$ 

  cm (perpendicular distance from the origin),

• 
  $\theta = 90^\circ$ 

  (since the velocity is along a straight line parallel to the x-axis, the perpendicular distance is directly used).

Since

$\sin 90^\circ = 1$

, the equation simplifies to:

$$L = m v r$$

Substituting the given values:

$$L = (5) \times (3\sqrt{2}) \times (2\sqrt{5})$$

$$L = 5 \times 3 \times 2 \times \sqrt{10}$$

$$L = 30 \sqrt{10} \text{ g-cm²/s}$$

Thus, the magnitude of the angular momentum is:

$$\mathbf{30\sqrt{10} \text{ g-cm²/s}}$$