A circular disc is rotating about its natural axis with angular velocity of $ 10rad{s^{ - 1}} $ . A second disc of the same mass is joined to it coaxially. If the radius of the second disc is half of the first one, then they together rotate with an angular velocity of
$ \left( A \right)\,\,2.5rad{s^{ - 1}} $
$ \left( B \right)\,\,5rad{s^{ - 1}} $
$ \left( C \right)\,\,8rad{s^{ - 1}} $
$ \left( D \right)\,\,6rad{s^{ - 1}} $

Hint: Here in this question the concept of angular velocity will be used. From the question we can see that there is no external torque so from this we can say that the conservation of angular velocity will be valid. By using this information we can easily solve this question.

Complete step by step answer
So we have the angular velocity given as $ \omega = 10rad{s^{ - 1}} $
So the moment of inertia for the first disk will be equal to $ \dfrac{{Mr_1^2}}{2} $ and for the second it will be equal to $ \dfrac{{Mr_2^2}}{2} $ .
And since, $ {r_2} = \dfrac{{{r_1}}}{2} $ therefore, $ \dfrac{{Mr_2^2}}{2} = \dfrac{{Mr_1^2}}{8} $
Now from the hint we know that there is no external torque so from this we can say that the conservation of angular velocity will be valid. So,
 $ \Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2} $
Now on substituting the values, we will get the equation as
 $ \Rightarrow \dfrac{{Mr_2^2}}{2} \times 10 = \left( {\dfrac{{Mr_1^2}}{8} + \dfrac{{Mr_1^2}}{2}} \right) \times {\omega _2} $
Since, the numerator have the same quantity all the numerator will get cancel out each other, and on solving the above solution more, we will get the equation as
 $ \Rightarrow {\omega _2} = 8rad/s $
Therefore, the option $ \left( C \right) $ is correct.

Note:
Moment of inertia is the property of a body to resist angular acceleration due to external torque or we can say that moment of inertia is tendency to resist angular motion which is done by mass in case of linear motion.