A boy stands over the centre of a horizontal platform which is rotating freely with a speed of `2 "revolutions"//s` about a vertical axis through the centre of the platform and straight up through the boy. He holds `2 kg` masses in each of his hands close to his body. The combined moment of inertia of the system is `1 kg xx metre^(2)`. The boy now stretches his arms so as to hold the masses far from his body. In this situation, the moment of inertia of the system increases to `2 kgxxmetre^(2)`, The kinetic energy of the system in the latter case as compared with that in the previous case will

Options

A. Remain unchanged
B. Decrease
C. Increase
D. Remain uncertain

Explanation

To solve the problem, we will analyze the situation using the principles of angular momentum and kinetic energy.

### Step 1: Understand the Initial Conditions
The boy is standing on a rotating platform with an initial moment of inertia \( I_i = 1 \, \text{kg} \cdot \text{m}^2 \) and an angular speed of \( \omega_i = 2 \, \text{revolutions/second} \).

### Step 2: Convert Angular Speed to Radians
Since we need to work with standard units, we convert revolutions per second to radians per second:
\[ \omega_i = 2 \, \text{revolutions/second} \times 2\pi \, \text{radians/revolution} = 4\pi \, \text{radians/second} \]

### Step 3: Calculate Initial Kinetic Energy
The kinetic energy \( K_i \) of the system can be calculated using the formula:
\[ K_i = \frac{1}{2} I_i \omega_i^2 \]
Substituting the values:
\[ K_i = \frac{1}{2} \times 1 \, \text{kg} \cdot \text{m}^2 \times (4\pi)^2 \]
Calculating \( (4\pi)^2 \):
\[ (4\pi)^2 = 16\pi^2 \]
So,
\[ K_i = \frac{1}{2} \times 1 \times 16\pi^2 = 8\pi^2 \, \text{Joules} \]

### Step 4: Understand the Final Conditions
When the boy stretches his arms, the moment of inertia increases to \( I_f = 2 \, \text{kg} \cdot \text{m}^2 \). Since there are no external torques acting on the system, the angular momentum is conserved:
\[ L_i = L_f \]
Where \( L = I \omega \). Therefore,
\[ I_i \omega_i = I_f \omega_f \]
Substituting the known values:
\[ 1 \cdot 4\pi = 2 \cdot \omega_f \]
Solving for \( \omega_f \):
\[ \omega_f = \frac{4\pi}{2} = 2\pi \, \text{radians/second} \]

### Step 5: Calculate Final Kinetic Energy
Now we can calculate the final kinetic energy \( K_f \):
\[ K_f = \frac{1}{2} I_f \omega_f^2 \]
Substituting the values:
\[ K_f = \frac{1}{2} \times 2 \, \text{kg} \cdot \text{m}^2 \times (2\pi)^2 \]
Calculating \( (2\pi)^2 \):
\[ (2\pi)^2 = 4\pi^2 \]
So,
\[ K_f = \frac{1}{2} \times 2 \times 4\pi^2 = 4\pi^2 \, \text{Joules} \]

### Step 6: Compare Kinetic Energies
Now we can compare the initial and final kinetic energies:
- Initial kinetic energy \( K_i = 8\pi^2 \, \text{J} \)
- Final kinetic energy \( K_f = 4\pi^2 \, \text{J} \)

### Conclusion
The kinetic energy of the system decreases when the boy stretches his arms. Therefore, the answer is that the kinetic energy in the latter case is less than in the previous case.