A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\mathrm{\omega }$. Another disc of same thickness and radius but of mass 1/8 M is placed gently on the first disc co-axially. The angular velocity of the system is now

Mass M disc's moment of inertia around the axis that runs through its centre and is perpendicular to its plane.

${\mathrm{I}}_1=\frac{1}{2}{\mathrm{MR}}^2$

$\mathrm{angular} \mathrm{velocity} {\mathrm{\omega }}_1=\mathrm{\omega }$

Mass M/8 disc's moment of inertia around the axis that runs through it's centre and is perpendicular to its plane.

${\mathrm{I}}_2=\frac{1}{2}\frac{\mathrm{M}}{8}{\mathrm{R}}^2$

${\mathrm{I}}_2=\frac{1}{16}{\mathrm{MR}}^2$

Total moment of inertia,

$\mathrm{I}\text{'}={\mathrm{I}}_1+{\mathrm{I}}_2=\frac{9}{16}{\mathrm{MR}}^2$

Let $\omega \text{'}$ be the system's final angular velocity.

Using the conservation of angular momentum.
​${\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}$

${\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_1{\mathrm{\omega }}_2=\mathrm{I}\text{'}\mathrm{\omega }\text{'}$

${\mathrm{\omega }}_2=0$

$\frac{1}{2}{\mathrm{MR}}^2\mathrm{\omega }+0=\frac{9}{16}{\mathrm{MR}}^2\mathrm{\omega }\text{'}$

$\mathrm{\omega }\text{'}=\frac{8}{9}\mathrm{\omega }$

Hence the correct answer is $\frac{8}{9}\mathrm{\omega }.$