B

Step 1: Calculate the initial moment of inertia of the rotating disc (I1) using the formula for a circular disc:
\[ I_1 = \frac{1}{2}MR^2 \]
Step 2: The initial angular momentum (L1) of the first disc is given by:
\[ L_1 = I_1 \omega_1 = \frac{1}{2}MR^2 \omega_1 \]
Step 3: The moment of inertia of the second disc (stationary) is the same:
\[ I_2 = \frac{1}{2}MR^2 \]
Total moment of inertia after they stick together (I_total) is:
\[ I_{total} = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{2}MR^2 = MR^2 \]
Step 4: The final angular momentum (L_total) is equal to the initial angular momentum, so:
\[ L_{total} = (I_{total}) \omega_2 = MR^2 \omega_2 \]
Setting the angular momenta equal:
\[ \frac{1}{2}MR^2 \omega_1 = MR^2 \omega_2 \]
Step 5: Solve for \( \omega_2 \):
\[ \omega_2 = \frac{1}{2} \omega_1 \]
Step 6: Calculate the initial energy (E1):
\[ E_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} \left(\frac{1}{2}MR^2\right) \omega_1^2 = \frac{1}{4}MR^2 \omega_1^2 \]
Step 7: Calculate the final energy (E2):
\[ E_2 = \frac{1}{2} I_{total} \omega_2^2 = \frac{1}{2} (MR^2) \left(\frac{1}{2} \omega_1\right)^2 = \frac{1}{2} \cdot MR^2 \cdot \frac{1}{4} \omega_1^2 = \frac{1}{8}MR^2 \omega_1^2 \]
Step 8: Calculate energy lost:
\[ \Delta E = E_1 - E_2 = \frac{1}{4}MR^2 \omega_1^2 - \frac{1}{8}MR^2 \omega_1^2 = \frac{1}{8}MR^2 \omega_1^2 \]
Step 9: Calculate percentage of energy lost:
\[ p = \frac{\Delta E}{E_1} \times 100 = \frac{\frac{1}{8}MR^2 \omega_1^2}{\frac{1}{4}MR^2 \omega_1^2} \times 100 = \frac{1}{2} \times 100 = 50 \% \]
Therefore, the value of p is 50.