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 follow these steps:

### Step 1: Understand the initial conditions
We have two discs:
- Disc 1 (rotating) with mass \( M \) and radius \( R \), rotating with angular speed \( \omega_1 \).
- Disc 2 (stationary) with mass \( M \) and radius \( \frac{R}{2} \).

### Step 2: Calculate the moment of inertia of both discs
The moment of inertia \( I \) of a disc about its central axis is given by:
\[ I = \frac{1}{2} M R^2 \]
For Disc 1:
\[ I_1 = \frac{1}{2} M R^2 \]
For Disc 2:
\[ I_2 = \frac{1}{2} M \left(\frac{R}{2}\right)^2 = \frac{1}{2} M \cdot \frac{R^2}{4} = \frac{1}{8} M R^2 \]

### Step 3: Calculate the initial angular momentum
The initial angular momentum \( L_i \) of the system is due to Disc 1 only, since Disc 2 is stationary:
\[ L_i = I_1 \omega_1 = \frac{1}{2} M R^2 \omega_1 \]

### Step 4: Calculate the final moment of inertia after both discs are combined
When both discs are rotating together, the total moment of inertia \( I_f \) is:
\[ I_f = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \left(\frac{4}{8} + \frac{1}{8}\right) M R^2 = \frac{5}{8} M R^2 \]

### Step 5: Apply conservation of angular momentum
Since there are no external torques acting on the system, the angular momentum is conserved:
\[ L_i = L_f \implies \frac{1}{2} M R^2 \omega_1 = I_f \omega_2 \]
Substituting \( I_f \):
\[ \frac{1}{2} M R^2 \omega_1 = \frac{5}{8} M R^2 \omega_2 \]
Cancelling \( M R^2 \) from both sides:
\[ \frac{1}{2} \omega_1 = \frac{5}{8} \omega_2 \]
Solving for \( \omega_2 \):
\[ \omega_2 = \frac{4}{5} \omega_1 \]

### Step 6: Calculate initial and final kinetic energy
The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega_1^2 = \frac{1}{4} M R^2 \omega_1^2 \]

The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} I_f \omega_2^2 = \frac{1}{2} \left(\frac{5}{8} M R^2\right) \left(\frac{4}{5} \omega_1\right)^2 \]
Calculating \( KE_f \):
\[ KE_f = \frac{1}{2} \cdot \frac{5}{8} M R^2 \cdot \frac{16}{25} \omega_1^2 = \frac{5}{8} M R^2 \cdot \frac{8}{25} \omega_1^2 = \frac{1}{5} M R^2 \omega_1^2 \]

### Step 7: Calculate the energy lost
The energy lost \( \Delta KE \) is:
\[ \Delta KE = KE_i - KE_f = \frac{1}{4} M R^2 \omega_1^2 - \frac{1}{5} M R^2 \omega_1^2 \]
Finding a common denominator (20):
\[ \Delta KE = \left(\frac{5}{20} - \frac{4}{20}\right) M R^2 \omega_1^2 = \frac{1}{20} M R^2 \omega_1^2 \]

### Step 8: Calculate the percentage of energy lost
The percentage of energy lost \( p \) is given by:
\[ p = \left(\frac{\Delta KE}{KE_i}\right) \times 100 = \left(\frac{\frac{1}{20} M R^2 \omega_1^2}{\frac{1}{4} M R^2 \omega_1^2}\right) \times 100 \]
Simplifying:
\[ p = \left(\frac{1/20}{1/4}\right) \times 100 = \left(\frac{1}{20} \cdot \frac{4}{1}\right) \times 100 = \frac{4}{20} \times 100 = 20\% \]

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