A horizonral circular plate is rotating about a vertical axis passing through its centre with an angular velocity `omega_(0)`. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is `K` intially, its final kinetic energy will be

Options

A. a, `2K`
B. b. `K//2`
C. c, `K`
D. d, `K//4`

Explanation

To solve the problem, we need to analyze the situation using the principles of rotational dynamics, specifically focusing on the conservation of angular momentum and the relationship between kinetic energy and moment of inertia.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- Let the initial moment of inertia of the system be \( I \).
- The initial angular velocity is \( \omega_0 \).
- The initial kinetic energy \( K \) is given by the formula:
\[ K = \frac{1}{2} I \omega_0^2 \]

2. **Final Conditions**:
- When the man stretches out his hands, the moment of inertia doubles, so the final moment of inertia \( I' \) is:
\[ I' = 2I \]

3. **Conservation of Angular Momentum**:
- Angular momentum \( L \) is conserved in the absence of external torques. Therefore:
\[ L = I \omega_0 = I' \omega' \]
- Substituting \( I' \) into the equation:
\[ I \omega_0 = 2I \omega' \]
- We can simplify this to find the final angular velocity \( \omega' \):
\[ \omega' = \frac{\omega_0}{2} \]

4. **Calculating Final Kinetic Energy**:
- The final kinetic energy \( K' \) can be calculated using the new moment of inertia and the new angular velocity:
\[ K' = \frac{1}{2} I' \omega'^2 \]
- Substituting \( I' = 2I \) and \( \omega' = \frac{\omega_0}{2} \):
\[ K' = \frac{1}{2} (2I) \left(\frac{\omega_0}{2}\right)^2 \]
- Simplifying this expression:
\[ K' = \frac{1}{2} (2I) \left(\frac{\omega_0^2}{4}\right) = \frac{I \omega_0^2}{4} \]
- Since \( K = \frac{1}{2} I \omega_0^2 \), we can express \( K' \) in terms of \( K \):
\[ K' = \frac{K}{2} \]

### Final Result:
The final kinetic energy \( K' \) of the system after the man stretches out his hands is:
\[ K' = \frac{K}{2} \]