MOI of disc D, about an axis passing through its centre and normal to its plane.
$\mathrm{I}_1=\frac{M R^2}{2}=\frac{2 \times(0.2)^2}{2}=0.04 \mathrm{~kg}-\mathrm{m}^2$
Initial angular velocity of disc $\mathrm{D}_1$, $\omega_1=50 \mathrm{rad} / \mathrm{s}$
MOI of disc $\mathrm{D}_2$ about an axis passing through its centre and normal to the plane.
$\begin{aligned}& 1_2=\frac{M R^2}{2}=\frac{4 \times(0.1)^2}{2} \\& =0.02 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}$
Initial angular velocity of disc $\mathrm{D}_2$, $\omega_2=200 \mathrm{rad} / \mathrm{s}$
Total angular momentum of the two discs, initially is
$L_i=I_1 \omega_1+I_2 \omega_2$
When both discs are brought in contact, axes of rotation coincide.
Consider both discs as a system. Hence, final angular momentum of the system is,
$\mathrm{L}_t=\left(\mathrm{K}_1+\mathrm{I}_2\right) \omega$
Here, $\omega$ is the final angular speed of the system.
According to the law of conservation of angular momentum, we get,
$\begin{aligned}& L_1=L_{+} \\& 1_1 \omega_1+l_2 \omega_2=\left(1_1+1_2\right) \omega \\& \rightarrow \omega=\frac{(0.04) \times(50)+(0.02) \times(200)}{(0.04+0.02)} \\& \rightarrow \omega=100 \mathrm{rad} / \mathrm{s}
\end{aligned}$