A circular disc of mass M and radius R is rotating about its axis with angular speed If about another stationary disc having radius `omega_1`.
`R/2` and same mass M is dropped co- axially on to the rotating disc. Gradually both discs attain constant angular speed `omega_2`. The energy lost in the process is p% of the initial energy . Value of p is ........

Answer

To solve the problem, we need to analyze the situation step by step, focusing on the conservation of angular momentum and the calculation of kinetic energy before and after the smaller disc is dropped onto the larger rotating disc.

### Step-by-Step Solution

1. **Identify the Moment of Inertia of the Discs:**
- For the larger disc (Disc A) of mass \( M \) and radius \( R \):
\[ I_A = \frac{1}{2} M R^2 \]
- For the smaller disc (Disc B) of mass \( M \) and radius \( \frac{R}{2} \):
\[ I_B = \frac{1}{2} M \left(\frac{R}{2}\right)^2 = \frac{1}{2} M \frac{R^2}{4} = \frac{1}{8} M R^2 \]

2. **Calculate the Initial Angular Momentum:**
- The initial angular momentum \( L_i \) of the system (only the larger disc is rotating):
\[ L_i = I_A \omega_1 = \left(\frac{1}{2} M R^2\right) \omega_1 \]

3. **Calculate the Final Angular Momentum:**
- After the smaller disc is dropped, both discs rotate together with a common angular speed \( \omega_2 \):
\[ I_{final} = I_A + I_B = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \frac{4}{8} M R^2 + \frac{1}{8} M R^2 = \frac{5}{8} M R^2 \]
- The final angular momentum \( L_f \):
\[ L_f = I_{final} \omega_2 = \left(\frac{5}{8} M R^2\right) \omega_2 \]

4. **Apply Conservation of Angular Momentum:**
- Setting initial angular momentum equal to final angular momentum:
\[ \frac{1}{2} M R^2 \omega_1 = \frac{5}{8} M R^2 \omega_2 \]
- Cancel \( M R^2 \) from both sides:
\[ \frac{1}{2} \omega_1 = \frac{5}{8} \omega_2 \]
- Rearranging gives:
\[ \omega_2 = \frac{4}{5} \omega_1 \]

5. **Calculate Initial and Final Kinetic Energy:**
- Initial kinetic energy \( K_i \):
\[ K_i = \frac{1}{2} I_A \omega_1^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega_1^2 = \frac{1}{4} M R^2 \omega_1^2 \]
- Final kinetic energy \( K_f \):
\[ K_f = \frac{1}{2} I_{final} \omega_2^2 = \frac{1}{2} \left(\frac{5}{8} M R^2\right) \left(\frac{4}{5} \omega_1\right)^2 \]
\[ K_f = \frac{1}{2} \left(\frac{5}{8} M R^2\right) \left(\frac{16}{25} \omega_1^2\right) = \frac{5}{8} M R^2 \cdot \frac{8}{25} \omega_1^2 = \frac{1}{5} M R^2 \omega_1^2 \]

6. **Calculate Energy Lost:**
- Energy lost \( \Delta K = K_i - K_f \):
\[ \Delta K = \frac{1}{4} M R^2 \omega_1^2 - \frac{1}{5} M R^2 \omega_1^2 = \left(\frac{5}{20} - \frac{4}{20}\right) M R^2 \omega_1^2 = \frac{1}{20} M R^2 \omega_1^2 \]

7. **Calculate Percentage Loss in Energy:**
- Percentage loss \( p \):
\[ p = \left(\frac{\Delta K}{K_i}\right) \times 100 = \left(\frac{\frac{1}{20} M R^2 \omega_1^2}{\frac{1}{4} M R^2 \omega_1^2}\right) \times 100 = \left(\frac{1/20}{1/4}\right) \times 100 = \left(\frac{1}{20} \times \frac{4}{1}\right) \times 100 = \frac{4}{20} \times 100 = 20 \]

### Final Answer
The value of \( p \) is \( 20 \).