A child is standing with folded hands at the center of a platform rotating about its central axis. The kinetic energy of the system is `K`. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is: a) K/4 b) K/2 c) 2K d) 4K

Options

A. K/4
B. K/2
C. 2 K
D. 4 K

Explanation

To solve the problem, we will use the principles of rotational motion, specifically the conservation of angular momentum and the relationship between kinetic energy, moment of inertia, and angular velocity.

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- The child is standing with folded hands at the center of a rotating platform.
- The initial kinetic energy of the system is given as \( K \).
- The moment of inertia of the system when the child has folded hands is denoted as \( I \).

2. **Write the Expression for Initial Kinetic Energy:**
- The kinetic energy \( K \) of a rotating system is given by the formula:
\[ K = \frac{1}{2} I \omega^2 \]
- Here, \( \omega \) is the angular velocity of the platform.

3. **Apply the Principle of Conservation of Angular Momentum:**
- According to the conservation of angular momentum, the initial angular momentum \( L_i \) must equal the final angular momentum \( L_f \):
\[ L_i = I \omega \]
- When the child stretches his arms, the moment of inertia doubles:
\[ I_f = 2I \]
- Let the new angular velocity be \( \omega_f \). Thus, we have:
\[ I \omega = 2I \omega_f \]
- From this, we can solve for \( \omega_f \):
\[ \omega_f = \frac{\omega}{2} \]

4. **Calculate the Final Kinetic Energy:**
- The final kinetic energy \( K' \) can be expressed as:
\[ K' = \frac{1}{2} I_f \omega_f^2 \]
- Substituting \( I_f = 2I \) and \( \omega_f = \frac{\omega}{2} \):
\[ K' = \frac{1}{2} (2I) \left(\frac{\omega}{2}\right)^2 \]
- Simplifying this:
\[ K' = \frac{1}{2} (2I) \left(\frac{\omega^2}{4}\right) = \frac{2I \omega^2}{8} = \frac{I \omega^2}{4} \]

5. **Relate Final Kinetic Energy to Initial Kinetic Energy:**
- We know from the initial kinetic energy that:
\[ K = \frac{1}{2} I \omega^2 \]
- Therefore, we can express \( K' \) in terms of \( K \):
\[ K' = \frac{I \omega^2}{4} = \frac{1}{2} \left(\frac{1}{2} I \omega^2\right) = \frac{1}{2} K \]

6. **Conclusion:**
- The final kinetic energy of the system after the child stretches his arms is:
\[ K' = \frac{K}{2} \]
- Thus, the correct answer is option (b) \( K/2 \).