1. Identify the physical principle

When the second disc is placed gently and co-axially on the rotating disc, friction between the discs eventually makes them rotate together with a common angular velocity.

Since there is no external torque about the rotation axis, angular momentum is conserved.

2. Moment of inertia of each disc

For a thin circular disc about its central axis:

$I = \frac{1}{2}MR^2$

• For the first disc:

$I_1 = \frac{1}{2}MR^2$

• For the second disc of mass $\frac{M}{2}$:

$I_2 = \frac{1}{2}\left(\frac{M}{2}\right)R^2 = \frac{1}{4}MR^2$

3. Initial angular momentum

Initially only the first disc is rotating with angular velocity $\omega$.

So,

$L_i = I_1\omega = \frac{1}{2}MR^2\omega$

The second disc is initially at rest, so its angular momentum is zero.

4. Final angular momentum

After contact, both discs rotate together with angular velocity $\omega'$.

Total moment of inertia:

$I_f = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{4}MR^2 = \frac{3}{4}MR^2$

Thus final angular momentum is

$L_f = I_f\omega' = \frac{3}{4}MR^2\omega'$

5. Apply conservation of angular momentum

$L_i = L_f$

$\frac{1}{2}MR^2\omega = \frac{3}{4}MR^2\omega'$

Cancel $MR^2$:

$\frac{1}{2}\omega = \frac{3}{4}\omega'$

So,

$\omega' = \frac{\frac{1}{2}}{\frac{3}{4}}\omega = \frac{1}{2}\cdot \frac{4}{3}\omega = \frac{2}{3}\omega$

6. Match with options

$\omega' = \frac{2}{3}\omega$

So the correct option is D.