A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with angular velocity of `20 rad s^(-1)`. Its kinetic energy is

Options

A. 10 J
B. 100 J
C. 500 J
D. 250 J

Explanation

To find the kinetic energy of a rotating ring, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the ring, \( r = 0.5 \, \text{m} \)
- Mass of the ring, \( m = 10 \, \text{kg} \)
- Angular velocity, \( \omega = 20 \, \text{rad/s} \)

2. **Calculate the moment of inertia (I) of the ring about its diameter:**
The moment of inertia \( I \) for a ring rotating about its diameter is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 10 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ I = \frac{1}{2} \times 10 \times 0.25 \]
\[ I = \frac{10 \times 0.25}{2} = \frac{2.5}{2} = 1.25 \, \text{kg m}^2 \]

3. **Calculate the kinetic energy (KE):**
The kinetic energy of a rotating object is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
Substituting the values of \( I \) and \( \omega \):
\[ KE = \frac{1}{2} \times 1.25 \, \text{kg m}^2 \times (20 \, \text{rad/s})^2 \]
\[ KE = \frac{1.25}{2} \times 400 \]
\[ KE = 0.625 \times 400 = 250 \, \text{J} \]

4. **Final Result:**
The kinetic energy of the ring is \( 250 \, \text{J} \).

### Conclusion:
The correct answer is option 4: 250 joules.