A disc of mass m and radius r is rotating with angular velocity, on a frictionless horizontal plane about a vertical axis passing through its centre. A ring of same mass and radius is gently placed on it in such a way that centres of both coincide. After some time it is found that both attain a common angular velocity

Options

A. If we consider disc and ring in a single system then its angular momentum is conserved.
B. `omega=(omega_(0))/(3)`
C. Loss in K.E. is double to that of final K.E. of the system
D. Loss in K.E. is 2/3 of initial K.E. of the system.

Explanation

To solve the problem, we need to analyze the situation step by step, applying the principles of conservation of angular momentum and kinetic energy.

### Step 1: Understand the System
We have a disc of mass \( m \) and radius \( r \) rotating with an initial angular velocity \( \omega_0 \). A ring of the same mass \( m \) and radius \( r \) is placed on top of the disc, and they eventually rotate together with a common angular velocity \( \omega \).

### Step 2: Moment of Inertia Calculation
1. **Moment of Inertia of the Disc**:
\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]

2. **Moment of Inertia of the Ring**:
\[ I_{\text{ring}} = m r^2 \]

3. **Total Moment of Inertia of the System** (when both are rotating together):
\[ I_{\text{total}} = I_{\text{disc}} + I_{\text{ring}} = \frac{1}{2} m r^2 + m r^2 = \frac{3}{2} m r^2 \]

### Step 3: Conservation of Angular Momentum
Since there is no external torque acting on the system, the angular momentum before and after the ring is placed on the disc must be conserved.

1. **Initial Angular Momentum**:
\[ L_{\text{initial}} = I_{\text{disc}} \cdot \omega_0 = \left(\frac{1}{2} m r^2\right) \cdot \omega_0 \]

2. **Final Angular Momentum** (after the ring is placed):
\[ L_{\text{final}} = I_{\text{total}} \cdot \omega = \left(\frac{3}{2} m r^2\right) \cdot \omega \]

3. **Setting Initial and Final Angular Momentum Equal**:
\[ \frac{1}{2} m r^2 \omega_0 = \frac{3}{2} m r^2 \omega \]

### Step 4: Solve for Common Angular Velocity
1. Cancel \( m r^2 \) from both sides:
\[ \frac{1}{2} \omega_0 = \frac{3}{2} \omega \]

2. Rearranging gives:
\[ \omega = \frac{\omega_0}{3} \]

### Step 5: Calculate Initial and Final Kinetic Energy
1. **Initial Kinetic Energy**:
\[ K_{\text{initial}} = \frac{1}{2} I_{\text{disc}} \omega_0^2 = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega_0^2 = \frac{1}{4} m r^2 \omega_0^2 \]

2. **Final Kinetic Energy**:
\[ K_{\text{final}} = \frac{1}{2} I_{\text{total}} \omega^2 = \frac{1}{2} \left(\frac{3}{2} m r^2\right) \left(\frac{\omega_0}{3}\right)^2 = \frac{1}{2} \cdot \frac{3}{2} m r^2 \cdot \frac{\omega_0^2}{9} = \frac{1}{12} m r^2 \omega_0^2 \]

### Step 6: Calculate Loss in Kinetic Energy
1. **Loss in Kinetic Energy**:
\[ \Delta K = K_{\text{initial}} - K_{\text{final}} = \frac{1}{4} m r^2 \omega_0^2 - \frac{1}{12} m r^2 \omega_0^2 \]

2. Finding a common denominator (12):
\[ \Delta K = \left(\frac{3}{12} - \frac{1}{12}\right) m r^2 \omega_0^2 = \frac{2}{12} m r^2 \omega_0^2 = \frac{1}{6} m r^2 \omega_0^2 \]

### Conclusion
The loss in kinetic energy is \( \frac{1}{6} m r^2 \omega_0^2 \).