A particle of mass 5 g is moving with a uniform speed of `3 sqrt(2) "cm s"^(-1)` in the XY-plane along the line `y = 2 sqrt(5)` cm. The magnitude of its angular momentum about the origin in `"g-cm"^(2) s^(-1)` is

Options

A. zero
B. 30
C. `30 sqrt(2)`
D. `30 sqrt(10)`

Explanation

To find the angular momentum of a particle moving in the XY-plane about the origin, we can follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 5 \, \text{g} \)
- Speed of the particle, \( v = 3\sqrt{2} \, \text{cm/s} \)
- The line of motion is given by \( y = 2\sqrt{5} \, \text{cm} \)

### Step 2: Determine the perpendicular distance from the origin to the line of motion
The line \( y = 2\sqrt{5} \) is horizontal, and the perpendicular distance from the origin (0, 0) to this line is simply the y-coordinate of the line:
- Perpendicular distance, \( d = 2\sqrt{5} \, \text{cm} \)

### Step 3: Use the formula for angular momentum
The formula for angular momentum \( L \) about the origin is given by:
\[ L = m \cdot v \cdot d \]
Substituting the values we have:
\[ L = 5 \, \text{g} \cdot (3\sqrt{2} \, \text{cm/s}) \cdot (2\sqrt{5} \, \text{cm}) \]

### Step 4: Calculate the angular momentum
Now, we can calculate \( L \):
\[ L = 5 \cdot 3\sqrt{2} \cdot 2\sqrt{5} \]
\[ L = 5 \cdot 6\sqrt{10} \]
\[ L = 30\sqrt{10} \, \text{g-cm}^2/\text{s} \]

### Step 5: Conclusion
The magnitude of the angular momentum about the origin is:
\[ L = 30\sqrt{10} \, \text{g-cm}^2/\text{s} \]

### Final Answer
The correct option is \( 30\sqrt{10} \, \text{g-cm}^2/\text{s} \).

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