A

Step 1: Identify the angular momentum of each cylinder. The angular momentum \(L\) of a cylinder is given by \(L = I \Omega\), where \(I\) is the moment of inertia and \(\Omega\) is the angular velocity. For a cylinder, \(I = \frac{1}{2}MR^2\).
Step 2: Calculate the initial angular momentum for both cylinders:
- For cylinder 1: \(L_1 = \frac{1}{2}M_1 R_1^2 \Omega_1\)
- For cylinder 2: \(L_2 = \frac{1}{2}M_2 R_2^2 \Omega_2\)
Step 3: When the cylinders touch, their final angular momentum will be equal to their initial angular momentum because angular momentum is conserved. Therefore, \(L_{final} = L_1 + L_2\).
Step 4: Let the final angular velocities of the cylinders after they touch be \(\Omega_f\). The final angular momentum for both cylinders combined is:
\(L_{final} = \frac{1}{2}M_1 R_1^2 \Omega_f + \frac{1}{2}M_2 R_2^2 \Omega_f\)
Step 5: Equate the initial and final angular momentum:
\(\frac{1}{2}M_1 R_1^2 \Omega_1 + \frac{1}{2}M_2 R_2^2 \Omega_2 = \left(\frac{1}{2}M_1 R_1^2 + \frac{1}{2}M_2 R_2^2\right) \Omega_f\
Step 6: Solving for \(\Omega_f\): \(\Omega_f = \frac{M_1 R_1^2 \Omega_1 + M_2 R_2^2 \Omega_2}{M_1 R_1^2 + M_2 R_2^2}\)
Therefore, A.